Monotonicity – JEE Main Previous Year Question with Solutions

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Mathematics

Monotonicity – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Math with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

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Mathematics

Maxima and Minima – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Math with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

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Mathematics

3D Geometry- JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Math with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

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Q. (A) Let $\mathrm{P}(3,2,6)$ be a point in space and $\mathrm{Q}$ be a point on the line $\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})+\mu(-3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+5 \hat{\mathrm{k}}) .$ Then the value of $\mu$ for which the vector $\overline{\mathrm{PQ}}$ is parallel to the plane $x-4 y+3 z=1$ is –

(A) $\frac{1}{4}$

(B) $-\frac{1}{4}$

(C) $\frac{1}{8}$

(D) $-\frac{1}{8}$

(B) A line with positive direction cosines passes through the point P (2, –1, 2) and makes equal angles with the coordinate axes. The line meets the plane 2x + y + z = 9 at point Q. The length of the line segment PQ equals –

(A) 1

(B) $\sqrt{2}$

(C) $\sqrt{3}$

(D) 2

(C) Let $(\mathrm{x}, \mathrm{y}, \mathrm{z})$ be points with integer coordinates satisfying the system of homogeneous equations $: 3 x-y-z=0 ;-3 x+z=0 ;-3 x+2 y+z=0 .$ Then the number of such points for which $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2} \leq 100$ is

[JEE 2009, 3+3+4]

Sol. ( (a) $A ;(b) C ;(c) 7$ )


Q. (A) Equation of the plane containing the straight line $\frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{3}=\frac{\mathrm{z}}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$

(A) x + 2y – 2z = 0

(B) 3x + 2y – 2z = 0

(C) x – 2y + z = 0

(D) 5x + 2y – 4z = 0

(B) If the distance of the point $\mathrm{P}(1,-2,1)$ from the plane $\mathrm{x}+2 \mathrm{y}-2 \mathrm{z}=\alpha$ where $\alpha>0,$ is $5,$ then the foot of the perpendicular from $P$ to the plane is-

(A) $\left(\frac{8}{3}, \frac{4}{3},-\frac{7}{3}\right)$

(B) $\left(\frac{4}{3},-\frac{4}{3}, \frac{1}{3}\right)$

(C) $\left(\frac{1}{3}, \frac{2}{3}, \frac{10}{3}\right)$

(D) $\left(\frac{2}{3},-\frac{1}{3}, \frac{5}{2}\right)$

(C) If the distance between the plane Ax – 2y + z = d and the plane containing the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ is $\sqrt{6},$ then $|d|$ is

(D) Match the statements in Column-I with the values in Column-II.

[JEE 2010, 3+5+3+(2+2+2+2)]

Sol. ((A) $\mathrm{C} ;(\mathrm{B}) \mathrm{A} ;(\mathrm{C}) 6 ;(\mathrm{D})(\mathrm{A}) \mathrm{t}(\mathrm{B}) \mathrm{p}, \mathrm{r}(\mathrm{C}) \mathrm{q}(\mathrm{D}) \mathrm{r}$ )

(a) Normal vector to the plane containing the

lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$ is

$\hat{n}=\left|\begin{array}{lll}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {3} & {4} & {2} \\ {4} & {2} & {3}\end{array}\right|=8 \hat{i}-\hat{j}-10 \hat{k}$

Let direction ratios of required plane be a, b, c.

Now 8a – b – 10c = 0

and $2 \mathrm{a}+3 \mathrm{b}+4 \mathrm{c}=0\left(\because \text { plane contains the line } \frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{3}=\frac{\mathrm{z}}{4}\right)$

$\Rightarrow \frac{a}{1}=\frac{b}{-2}=\frac{c}{1}$

$\cdot$ plane contains the line, which passes through origin, hence origin lies on a plane.

$\Rightarrow$ equation of required plane is $x-2 y+z=0$

(b) $\quad \because \quad\left|\frac{1-4-2-\alpha}{3}\right|=5$

$\Rightarrow \alpha=10,-20$

$\Rightarrow \alpha=10 \because \alpha>0$


Q. (A) The point P is the intersection of the straight line joining the points Q(2,3,5) and R(1,–1,4) with the plane 5x – 4y – z = 1. If S is the foot of the perpendicular drawn from the point T(2,1,4) to QR, then the length of the line segment PS is –

(A) $\frac{1}{\sqrt{2}}$

(B) $\sqrt{2}$

(C) 2

(D) $2 \sqrt{2}$

(B) The equation of a plane passing through the line of intersection of the planes x + 2y $+3 \mathrm{z}=2$ and $\mathrm{x}-\mathrm{y}+\mathrm{z}=3$ and at a distance $\frac{2}{\sqrt{3}}$ from the point $(3,1,-1)$ is

(A) $5 x-11 y+z=17$

(B) $\sqrt{2} x+y=3 \sqrt{2}-1$

(C) $x+y+z=\sqrt{3}$

(D) $x-\sqrt{2} y=1-\sqrt{2}$

(C) If the straight lines $\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar, then the plane(s) containing these two lines is(are)

(A) y + 2z = –1

(B) y + z = –1

(C) y – z = –1

(D) y – 2z = –1

[JEE 2012, 3+3+4]

Sol. ((a) $A ;(b) A ;(c) B, C$)

(a) Line QR :

$\frac{x-2}{1}=\frac{y-3}{4}=\frac{z-5}{1}=\lambda$

Any point on line QR :

$(\lambda+2,4 \lambda+3, \lambda+5)$

$\therefore$ Point of intersection with plane :

$5 \lambda+10-16 \lambda-12-\lambda-5=1$

$\Rightarrow \lambda=-\frac{2}{3}$

$\therefore \mathrm{P}\left(\frac{4}{3}, \frac{1}{3}, \frac{13}{3}\right)$


Q. Perpendiculars are drawn from points on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}$ to the plane $x$ $+\mathrm{y}+\mathrm{z}=3 .$ The feet of perpendiculars lie on the line

(A) $\frac{x}{5}=\frac{y-1}{8}=\frac{z-2}{-13}$

(B) $\frac{x}{2}=\frac{y-1}{3}=\frac{z-2}{-5}$

(C) $\frac{x}{4}=\frac{y-1}{3}=\frac{z-2}{-7}$

(D) $\frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}$

[JEE-Advanced 2013, 2]

Sol. (D)


Q. A line $\ell$ passing through the origin is perpendicular to the lines

$\ell_{1}:(3+\mathrm{t}) \hat{\mathrm{i}}+(-1+2 \mathrm{t}) \hat{\mathrm{j}}+(4+2 \mathrm{t}) \hat{\mathrm{k}},-\infty<\mathrm{t}<\infty$

$\ell_{2}:(3+2 s) \hat{\mathrm{i}}+(3+2 \mathrm{s}) \hat{\mathrm{j}}+(2+\mathrm{s}) \hat{\mathrm{k}},-\infty<\mathrm{s}<\infty$ Then, the coordinate(s) of the point(s) on $\ell_{2}$

at a distance of $\sqrt{17}$ from the point of intersection of $\ell$ and $\ell_{1}$ is (are) –

(A) $\left(\frac{7}{3}, \frac{7}{3}, \frac{5}{3}\right)$

(B) (–1,–1,0)

(C) (1,1,1)

(D) $\left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right)$

[JEE-Advanced 2013, 4, (–1)]

Sol. (B,D)


Q. Two lines $L_{1}: x=5, \frac{y}{3-\alpha}=\frac{z}{-2}$ and $L_{2}: x=\alpha, \frac{y}{-1}=\frac{z}{2-\alpha}$ are coplanar. Then $\alpha$ can take value(s)

(A) 1              (B) 2             (C) 3                  (D) 4

[JEE-Advanced 2013, 3, (–1)]

Sol. (A,D)

$\mathrm{L}_{1}: \frac{\mathrm{x}-5}{0}=\frac{\mathrm{y}}{3-\alpha}=\frac{\mathrm{z}}{-2}$

$\mathrm{L}_{2}: \frac{\mathrm{x}-\alpha}{0}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}}{2-\alpha}$

for lines to be coplanar

$\left|\begin{array}{ccc}{5-\alpha} & {0} & {0} \\ {0} & {3-\alpha} & {-2} \\ {0} & {-1} & {2-\alpha}\end{array}\right|=0$

$\Rightarrow \quad(5-\alpha)((3-\alpha)(2-\alpha)-2)=0$

$\Rightarrow \quad(5-\alpha)\left(\alpha^{2}-5 \alpha+4\right)=0$

$\Rightarrow \quad \alpha=1,4,5$


Q. Consider the lines $\mathrm{L}_{1}: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}+3}{1}, \mathrm{L}_{2}: \frac{\mathrm{x}-4}{1}=\frac{\mathrm{y}+3}{1}=\frac{\mathrm{z}+3}{2}$ and the planes $\mathrm{P}_{1}: 7 \mathrm{x}+\mathrm{y}+2 \mathrm{z}=3, \mathrm{P}_{2}: 3 \mathrm{x}+5 \mathrm{y}-6 \mathrm{z}=4 .$ Let $\mathrm{ax}+\mathrm{by}+\mathrm{cz}=\mathrm{d}$ be the

equation of the plane passing through the point of intersection of lines $L_{1}$ and $\mathrm{L}_{2}$ and perpendicular to planes $\mathrm{P}_{1}$ and $\mathrm{P}_{2} .$ Match List-I with List-II and select the correct answer using the code given below the lists.

[JEE-Advanced 2013, 3, (–1)]

Sol. (A)

For point of intersection of $L_{1}$ and $L_{2}$

$\left\{\begin{array}{l}{2 \lambda+1=\mu+4} \\ {-\lambda=\mu-3} \\ {\lambda-3=2 \mu-3}\end{array}\right.$

$\Rightarrow \mu=1$

$\Rightarrow$ point of intersction is $(5,-2,-1)$

Now, vector normal to the plane is $\overrightarrow{\mathrm{n}}_{1} \times \overrightarrow{\mathrm{n}}_{2}=\left|\begin{array}{ccc}{\hat{\mathrm{i}}} & {\hat{\mathrm{j}}} & {\hat{\mathrm{k}}} \\ {7} & {1} & {2} \\ {3} & {5} & {-6}\end{array}\right|$

$=-16(\hat{i}-3 \hat{j}-2 \hat{k})$

Let equation of required plane be $x-3 y-2 z=\alpha$

$\because$ it passes through $(5,-2,-1)$

$\therefore \alpha=13$

$\Rightarrow$ equation of plane is $x-3 y-2 z=13$


Q. From a point $\mathrm{P}(\lambda, \lambda, \lambda),$ perpendiculars $\mathrm{PQ}$ and $\mathrm{PR}$ are drawn respectively on the lines $\mathrm{y}=$ $\mathrm{x}, \mathrm{z}=1$ and $\mathrm{y}=-\mathrm{x}, \mathrm{z}=-1 .$ If $\mathrm{P}$ is such that $\angle \mathrm{QPR}$ is a right angle, then the possible value(s) of $\lambda$ is (are)

(A) $\sqrt{2}$              (B) 1            (C) –1              (D) $-\sqrt{2}$

[JEE(Advanced)-2014, 3]

Sol. (C)

Line $\mathrm{L}_{1}$ given by $\mathrm{y}=\mathrm{x} ; \mathrm{z}=1$ can be expressed as

$\mathrm{L}_{1}: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{1}=\frac{\mathrm{z}-1}{0}$

Similarly $\mathrm{L}_{2}(\mathrm{y}=-\mathrm{x} ; \mathrm{z}=-1)$ can be expressed as

$\mathrm{L}_{2}: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}+1}{0}$

Let any point $\mathrm{Q}(\alpha, \alpha, 1)$ on $\mathrm{L}_{1}$ and $\mathrm{R}(\beta,-\beta,-1)$ on $\mathrm{L}_{2}$

Given that $\mathrm{PQ}$ is perpendicular to $\mathrm{L}_{1}$

$\Rightarrow(\lambda-\alpha) .1+(\lambda-\alpha) \cdot 1+(\lambda-1) \cdot 0=0 \Rightarrow \lambda=\alpha$

$\therefore \mathrm{Q}(\lambda, \lambda, 1)$

Similarly PR is perpendicular to L $_{2}$ $(\lambda-\beta) \cdot 1+(\lambda+\beta)(-1)+(\lambda+1) \cdot 0=0 \Rightarrow \beta=0$

$\therefore \mathrm{R}(0,0,-1)$

Now as given

$\Rightarrow \overrightarrow{\mathrm{PR}} \cdot \overrightarrow{\mathrm{PQ}}=0$

$0 . \lambda+0 . \lambda+(\lambda-1)(\lambda+1)=0$

$\lambda \neq 1$ as $\mathrm{P} \& \mathrm{Q}$ are different points

$\Rightarrow \lambda=-1$


Q. In $\mathbb{D}^{3},$ consider the planes $P_{1}: y=0$ and $P_{2}: x+z=1 .$ Let $P_{3}$ be a plane, different from $\mathrm{P}_{1}$ and $\mathrm{P}_{2},$ which passes through the intersection of $\mathrm{P}_{1}$ and $\mathrm{P}_{2} .$ If the distance of the point $(0,1,0)$ from $P_{3}$ is 1 and the distance of a point $(\alpha, \beta, \gamma)$ from $P_{3}$ is $2,$ then which of the following relations is (are) true?

(A) $2 \alpha+\beta+2 \gamma+2=0$

(B) $2 \alpha-\beta+2 \gamma+4=0$

(C) $2 \alpha+\beta-2 \gamma-10=0$

(D) $2 \alpha-\beta+2 \gamma-8=0$

[JEE 2015, 4M, –2M]

Sol. (B,D)


Q. In $\square^{3},$ let $L$ be a straight line passing through the origin. Suppose that all the points on $L$ are at a constant distance from the two planes $P_{1}: x+2 y-z+1=0$ and $P_{2}: 2 x-y+$ $\mathrm{z}-1=0 .$ Let $\mathrm{M}$ be the locus of the feet of the perpendiculars drawn from the points on L to the plane $P_{1} .$ Which of the following points lie(s) on M?

[JEE 2015, 4M, –2M]

Sol. (A,B)

Straight line ‘L’ is parallel to line of intersection of plane $\mathrm{P}_{1} \&$ plane $\mathrm{P}_{2}$

$\therefore$ Equation of line $^{\prime} \mathbf{L}^{\prime}$ is

$\frac{x}{1}=\frac{y}{-3}=\frac{z}{-5}=\lambda$

$\frac{\alpha-\lambda}{1}=\frac{\beta+3 \lambda}{2}=\frac{\gamma+5 \lambda}{-1}=\mathrm{k}$

$\left.\begin{array}{l}{\alpha=\mathrm{k}+\lambda} \\ {\beta=2 \mathrm{k}-3 \lambda} \\ {\mathrm{y}=-\mathrm{k}-5 \lambda}\end{array}\right\}$ …(1)

satisfying in plane $\mathrm{P}_{1}$

$\mathrm{k}+\lambda+4 \mathrm{k}-6 \lambda+\mathrm{k}+5 \lambda+1=0$

$6 k=-1$

putting in ( 1) required locus is

$\mathrm{x}=-\frac{1}{6}+\lambda$

$y=-\frac{1}{3}-3 \lambda$

$z=\frac{1}{6}-5 \lambda$

Now check the options.


Q. Consider a pyramid OPQRS located in the first octant $(\mathrm{x} \geq 0, \mathrm{y} \geq 0, \mathrm{z} \geq 0)$ with $\mathrm{O}$ as origin, and OP and OR along the x-axis and the y-axis, respectively. The base OPQR of the pyramid is a square with $\mathrm{OP}=3 .$ The point $\mathrm{S}$ is directly above the mid-point $\mathrm{T}$ of diagonal OQ such that TS $=3 .$ Then-

(A) the acute angle between $\mathrm{OQ}$ and $\mathrm{OS}$ is $\frac{\mathrm{K}}{3}$

(B) the equaiton of the plane containing the triangle $\mathrm{OQS}$ is $\mathrm{x}-\mathrm{y}=0$

(C) the length of the perpendicular from $P$ to the plane containing the triangle OQS is $\frac{3}{\sqrt{2}}$

(D) the perpendicular distance from $\mathrm{O}$ to the straight line containing RS is $\sqrt{\frac{15}{2}}$

[JEE(Advanced) 2016]

Sol. (B,C,D)


Q. Let $P$ be the image of the point $(3,1,7)$ with respect to the plane $x-y+z=3 .$ Then the equation of the plane passing through $\mathrm{P}$ and containing the straight line $\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{1}$ is

(A) x + y – 3z = 0

(B) 3x + z = 0

(C) x – 4y + 7z = 0

(D) 2x – y = 0

[JEE(Advanced) 2016]

Sol. (C)

$\therefore x-4 y+7 z=0$


Q. The equation of the plane passing through the point (1,1,1) and perpendicular to the planes 2x + y – 2z = 5 and 3x – 6y – 2z = 7, is-

(A) 14x + 2y + 15z = 31

(B) 14x + 2y – 15z = 1

(C) –14x + 2y + 15z = 3

(D) 14x – 2y + 15z = 27

[JEE(Advanced) 2017]

Sol. (A)


Q. Let $P_{1}: 2 x+y-z=3$ and $P_{2}: x+2 y+z=2$ be two planes. Then, which of the following statement(s) is (are) TRUE ?

(A) The line of intersection of $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$ has direction ratios $1,2,-1$

(B) The line $\frac{3 x-4}{9}=\frac{1-3 y}{9}=\frac{z}{3}$ is perpendicular to the line of intersection of $P_{1}$ and $P_{2}$

(C) The acute angle between $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$ is $60^{\circ}$

(D) If $P_{3}$ is the plane passing through the point $(4,2,-2)$ and perpendicular to the line of intersection of $\mathrm{P}_{1}$ and $\mathrm{P}_{2},$ then the distance of the point $(2,1,1)$ from the plane $\mathrm{P}_{2}$ is $\frac{2}{\sqrt{3}}$

[JEE(Advanced) 2018]

Sol. (C,D)

D.C. of line of intersection $(a, b, c)$

$\begin{aligned} \Rightarrow \quad & 2 \mathrm{a}+\mathrm{b}-\mathrm{c}=0 \\ & \mathrm{a}+2 \mathrm{b}+\mathrm{c}=0 \end{aligned}$

$\frac{a}{1+2}=\frac{b}{-1-2}=\frac{c}{4-1}$

$\therefore \quad \mathrm{D} . \mathrm{C} .$ is $(1,-1,1)$

B) $\frac{3 x-4}{9}=\frac{1-3 y}{9}=\frac{z}{3}$

$\Rightarrow \quad \frac{x-4 / 3}{3}=\frac{y-1 / 3}{-3}=\frac{z}{3}$

$\Rightarrow \quad$ lines are parallel.

(C) Acute angle between $\mathrm{P}_{1}$ and $\mathrm{P}_{2}=\cos ^{-1}\left(\frac{2 \times 1+1 \times 2-1 \times 1}{\sqrt{6} \sqrt{6}}\right)$ \[ =\cos ^{-1}\left(\frac{3}{6}\right)=\cos ^{-1}\left(\frac{1}{2}\right)=60^{\circ} \]

(D) Plane is given by $(x-4)-(y-2)+(z+2)=0$ \[ \Rightarrow \quad x-y+z=0 \]

Distance of $(2,1,1)$ from plane $=\frac{2-1+1}{\sqrt{3}}=\frac{2}{\sqrt{3}}$


Q. Consider the cube in the first octant with sides $\mathrm{OP}, \mathrm{OQ}$ and $\mathrm{OR}$ of length $1,$ along the x-axis,

y-axis and z-axis, respectively, where $\mathrm{O}(0,0,0)$ is the origin. Let $\mathrm{S}\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)$ be the centre

of the cube and $T$ be the vertex of the cube opposite to the origin $O$ such that $S$ lies on the diagonal OT.If $\overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{SP}}, \overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{SQ}}, \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{SR}}$ and $\overrightarrow{\mathrm{t}}=\overrightarrow{\mathrm{ST}},$ then the value of $|(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}) \times(\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{t}})|$ is

[JEE(Advanced) 2018]

Sol. 8


Q. Let P be a point in the first octant, whose image Q in the plane x + y = 3 (that is, the line segment PQ is perpendicular to the plane x + y = 3 and the mid-point of PQ lies in the plane x + y = 3) lies on the z-axis. Let the distance of P from the x-axis be 5. If R is the image of P in the xy-plane, then the length of PR is

Sol. 8


Definite integration – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Math with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral also provides complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

Q. Let ƒ be a non-negative function defined on the interval $[0,1] .$ If $\int_{0}^{x} \sqrt{1-\left(f^{\prime}(t)\right)^{2}} d t=\int_{0}^{x} f(t) d t$ $0 \leq \mathrm{x} \leq 1,$ and $f(0)=0,$ then $-$

(A) $f\left(\frac{1}{2}\right)<\frac{1}{2}$ and $f\left(\frac{1}{3}\right)>\frac{1}{3}$

(B) $f\left(\frac{1}{2}\right)>\frac{1}{2}$ and $f\left(\frac{1}{3}\right)>\frac{1}{3}$

(C) $f\left(\frac{1}{2}\right)<\frac{1}{2}$ and $f\left(\frac{1}{3}\right)<\frac{1}{3}$

(D) $f\left(\frac{1}{2}\right)>\frac{1}{2}$ and $f\left(\frac{1}{3}\right)<\frac{1}{3}$

[JEE 2009, 3]

Sol. (C)

$\int_{0}^{x} \sqrt{1-\left(f^{\prime}(t)\right)^{2}} \mathrm{d} t=\int_{0}^{x} f(t) d t, 0 \leq x \leq 1$

differentiating both the sides & squreing

$\Rightarrow 1-\left(f^{\prime}(\mathrm{x})\right)^{2}=f^{2}(\mathrm{x})$

$\Rightarrow \frac{f^{\prime}(x)}{\sqrt{1-f^{2}(x)}}=1$

$\Rightarrow \sin ^{-1} f(\mathrm{x})=\mathrm{x}+\mathrm{c}$

$f(0)=0$

$\Rightarrow f(\mathrm{x})=\sin \mathrm{x}$

$\Rightarrow \because \sin \mathrm{x} \leq \mathrm{x}$ for $\mathrm{x} \in[0,1]$

$\Rightarrow f\left(\frac{1}{2}\right)<\frac{1}{2}$ and $f\left(\frac{1}{3}\right)<\frac{1}{3}$


Q. If $\mathrm{I}_{\mathrm{n}}=\int_{-\pi}^{\pi} \frac{\sin \mathrm{nx}}{\left(1+\pi^{\mathrm{x}}\right) \sin \mathrm{x}} \mathrm{d} \mathrm{x}, \mathrm{n}=0,1,2, \ldots, \mathrm{then}-$

(A) $\mathrm{I}_{\mathrm{n}}=\mathrm{I}_{\mathrm{n}+2}$

(B) $\sum_{\mathrm{m}=1}^{10} \mathrm{I}_{2 \mathrm{m}+1}=10 \pi$

(C) $\sum_{\mathrm{m}=1}^{10} \mathrm{I}_{2 \mathrm{m}}=0$

(D) $\mathrm{I}_{\mathrm{n}}=\mathrm{I}_{\mathrm{n}+1}$

[JEE 2009, 4]

Sol. (A,B,D)

$\mathrm{I}_{\mathrm{n}}=\int_{-\pi}^{\pi} \frac{\sin \mathrm{nx}}{\left(1+\pi^{\mathrm{x}}\right) \sin \mathrm{x}} \mathrm{dx}$

$\mathrm{I}_{\mathrm{n}}=\int_{-\pi}^{\pi} \frac{\pi^{\mathrm{x}} \sin \mathrm{nx}}{\left(1+\pi^{\mathrm{x}}\right) \sin \mathrm{x}} \mathrm{dx}$

$2 \mathrm{I}_{\mathrm{n}}=\int_{-\pi}^{\pi} \frac{\sin \mathrm{nx}}{\sin \mathrm{x}} \mathrm{dx}$ ..(i)

$2 \mathrm{I}_{\mathrm{n}+2}=\int_{-\pi}^{\pi} \frac{\sin (\mathrm{n}+2) \mathrm{x}}{\sin \mathrm{x}} \mathrm{dx} \quad \ldots(\mathrm{i})$

(ii) – (i)

$\Rightarrow 2\left(\ln _{+2}-\mathrm{I}_{\mathrm{n}}\right)=\int_{-\pi}^{\pi} \cos (\mathrm{n}+1) \mathrm{x}=0$

$\Rightarrow \quad \mathrm{I}_{\mathrm{n}+2}=\mathrm{I}_{\mathrm{n}}$

$\sum_{m=1}^{10} \mathrm{I}_{2 \mathrm{m}}=10 \sum_{\mathrm{m}=1}^{10} \mathrm{I}_{2}=\frac{10}{2} \int_{-\pi}^{\pi} \frac{\sin 2 \mathrm{x}}{\sin \mathrm{x}} \mathrm{d} \mathrm{x}=0$

Put n = 1 in equation (i)

$2 \mathrm{I}_{1}=\int_{-\pi}^{\pi} \frac{\sin \mathrm{x} \mathrm{d} \mathrm{x}}{\sin \mathrm{x}}=2 \pi$

$\mathrm{I}_{1}=\pi$

$\sum_{m=1}^{10} I_{2 m+1}=10 \pi$


Q. Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be a continuous function which satisfies $\mathrm{f}(\mathrm{x})=\int_{0}^{\mathrm{x}} \mathrm{f}(\mathrm{t}) \mathrm{d} \mathrm{t}$ Then the value of f(ln 5) is……..

[JEE 2009, 4]

Sol. 0

$\mathrm{f}(\mathrm{x})=\int_{0}^{\mathrm{x}} \mathrm{f}(\mathrm{t}) \mathrm{dt}$

$\mathrm{f}^{\mathrm{l}}(\mathrm{x})=\mathrm{f}(\mathrm{x})$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}$

$\Rightarrow \int \frac{d y}{y}=\int d x$

$\Rightarrow \ln y=x+c$

$\Rightarrow y=e^{x+c}$

$\Rightarrow y=0$

$\left(\begin{array}{c}{\text { at } x=0, y=0} \\ {c \rightarrow-\infty}\end{array}\right)$

$f(x)=0$

$f(\ell n 5)=0$


Q. The value of $\lim _{x \rightarrow 0} \frac{1}{x^{3}} \int_{0}^{x} \frac{t \ell n(1+t)}{t^{4}+4} d t$ is

(A) 0

(B) $\frac{1}{12}$

(C) $\frac{1}{24}$

(D) $\frac{1}{64}$

[JEE 2010, 3 (–1)]

Sol. (B)

Applying L-Hospital rule,


Q. The value(s) of $\int_{0}^{1} \frac{\mathrm{x}^{4}(1-\mathrm{x})^{4}}{1+\mathrm{x}^{2}} \mathrm{dx}$ is (are)

(A) $\frac{22}{7}-\pi$

(B) $\frac{2}{105}$

(C) 0

(D) $\frac{71}{15}-\frac{3 \pi}{2}$

[JEE 2010, 3]

Sol. (A)


Q. Let $f$ be a real-valued function defined on the interval $(-1,1)$ such that

$e^{-x} f(x)=2+\int_{0}^{x} \sqrt{t^{4}+1} d t,$ for all $x \in(-1,1),$ and let $f^{-1}$ be the inverse function of $f$ Then $\left(f^{-1}\right)^{\prime}(2)$ is equal to-

(A) 1

(B) $\frac{1}{3}$

(C) $\frac{1}{2}$

(D) $\frac{1}{\mathrm{e}}$

[JEE2010, 5 (–2)]

Sol. (B)

from $(2), f^{-1}(2)=\frac{1}{3}$


Q. For any real number x, let [x] denote the largest integer less than or equal to x. Let f be a real valued function defined on the interval [–10, 10] by $\mathrm{f}(\mathrm{x})=\left\{\begin{aligned} \mathrm{x}-[\mathrm{x}] & \text { if }[\mathrm{x}] \text { is odd } \\ 1+[\mathrm{x}]-\mathrm{x} & \text { if }[\mathrm{x}] \text { is even } \end{aligned}\right.$ Then the value of $\frac{\pi^{2}}{10} \int_{-10}^{10} \mathrm{f}(\mathrm{x}) \cos \pi \mathrm{x} \mathrm{d} \mathrm{x}$ is

[JEE 2010, 3]

Sol. 4


Q. The value of $\int_{\sqrt{\mathrm{in} 2}}^{\sqrt{\mathrm{n} 3}} \frac{\mathrm{x} \sin \mathrm{x}^{2}}{\sin \mathrm{x}^{2}+\sin \left(\mathrm{ln} 6-\mathrm{x}^{2}\right)} \mathrm{dx}$ is

(A) $\frac{1}{4} \ln \frac{3}{2}$

(B) $\frac{1}{2} \ln \frac{3}{2}$

(C) $\ln \frac{3}{2}$

(D) $\frac{1}{6} \ln \frac{3}{2}$

[JEE 2011, 3 (–1)]

Sol. (A)


Q. Let $S$ be the area of the region enclosed by $y=e^{-x^{2}}, y=0, x=0,$ and $x=1 .$ Then

(A) $\mathrm{S} \geq \frac{1}{\mathrm{e}}$

(B) $\mathrm{S} \geq 1-\frac{1}{\mathrm{e}}$

(C) $S \leq \frac{1}{4}\left(1+\frac{1}{\sqrt{\mathrm{e}}}\right)$

(D) $S \leq \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{\mathrm{e}}}\left(1-\frac{1}{\sqrt{2}}\right)$

[JEE 2012, 4M]

Sol. (A,B,D)


Q. The value of the integral $\int_{-\pi / 2}^{\pi / 2}\left(\mathrm{x}^{2}+\ln \frac{\pi+\mathrm{x}}{\pi-\mathrm{x}}\right) \cos \mathrm{xd} \mathrm{x}$ is

(A) 0

(B) $\frac{\pi^{2}}{2}-4$

(C) $\frac{\pi^{2}}{2}+4$

(D) $\frac{\pi^{2}}{2}$c

[JEE 2012, 3M, –1M]

Sol. (B)


Q. For a $\in \mathrm{R}$ (the set of all real numbers), a\neq-1.

$\lim _{n \rightarrow \infty} \frac{\left(1^{a}+2^{a}+\ldots \ldots+n^{a}\right)}{(n+1)^{a-1}[(n a+1)+(n a+2)+\ldots \ldots+(n a+n)]}=\frac{1}{60}$ Then $a=$

(A) 5 (B) 7 (C) $\frac{-15}{2}$ (D) $\frac{-17}{2}$

[JEE(Advanced) 2013, 3, (–1)]

Sol. (B)


Q. Let $f:[\mathrm{a}, \mathrm{b}] \rightarrow[1, \infty)$ be a continuous function and let $\mathrm{g}: \square \rightarrow \square$ be defined as

Then

(A) g(x) is continuous but not differentiable at a

(B) g(x) is differentiable on 

(C) g(x) is continuous but not differentiable at b

(D) g(x) is continuous and differentiable at either a or b but not both.

[JEE(Advanced)-2014, 3]

Sol. (A,C)


Q. The value of $\int_{0}^{1} 4 x^{3}\left\{\frac{d^{2}}{d x^{2}}\left(1-x^{2}\right)^{5}\right\} d x$ is

[JEE(Advanced)-2014, 3]

Sol. 2


Q. The following integral $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(2 \csc x)^{17} d x$ is equal to

(A) $\int_{0}^{\log (1+\sqrt{2})} 2\left(e^{\mathfrak{u}}+e^{-\mathfrak{u}}\right)^{16} \mathrm{d} \mathfrak{u}$

(B) $\int_{0}^{\log (1+\sqrt{2})}\left(\mathrm{e}^{\mathrm{u}}+\mathrm{e}^{-\mathrm{u}}\right)^{17} \mathrm{du}$

(C) $\int_{0}^{\log (1+\sqrt{2})}\left(e^{\mathfrak{u}}-e^{-\mathfrak{u}}\right)^{17} \mathrm{d} \mathfrak{u}$

(D) $\int_{0}^{\log (1+\sqrt{2})} 2\left(e^{\mathfrak{u}}-e^{-\mathfrak{u}}\right)^{16} d \mathfrak{u}$

[JEE(Advanced)-2014, 3(–1)]

Sol. (A)


Q. Let $f:[0,2] \rightarrow \square$ be a function which is continuous on $[0,2]$ and is differentiable on $(0,2)$ with $f(0)=1 .$ Let $F(x)=\int_{0}^{x^{2}} f(\sqrt{t}) d t$ for $x \in[0,2] .$ If $F^{\prime}(x)=f^{\prime}(x)$ for all $x \in(0,2)$ then $F(2)$ equals $-$

(A) $\mathrm{e}^{2}-1$

(B) $\mathrm{e}^{4}-1$

(C) e – 1

(D) e $^{4}$

[JEE(Advanced)-2014, 3(–1)]

Sol. (B)


Given that for each a $\in(0,1)$, $\lim _{\mathrm{h} \rightarrow 0^{+}} \int_{\mathrm{h}}^{1-\mathrm{h}} \mathrm{t}^{-\mathrm{a}}(1-\mathrm{t})^{\mathrm{a}-1}$ $\mathrm{dt}$ exists. Let this limit be g(a). In addition,

it is given that the function g(a) is differentiable on (0,1).

Q. The value of $\mathrm{g}\left(\frac{1}{2}\right)$ is –

(A) $\pi$

(B) $2 \pi$

(C) $\frac{\pi}{2}$

(D) $\frac{\pi}{4}$

[JEE(Advanced)-2014, 3(–1)]

Sol. (A)


Q. The value of $\mathrm{g}^{\prime}\left(\frac{1}{2}\right)$ is-

(A) $\frac{\pi}{2}$

(B) $\pi$

(C) $-\frac{\pi}{2}$

(D) 0

[JEE(Advanced)-2014, 3(–1)]

Sol. (D)


Q.

[JEE(Advanced)-2014, 3(–1)]

Sol. (C)


Q. Let $f: \square \rightarrow \square$ be a function defined by $f(x)$ $=\left\{\begin{array}{ccc}{[\mathrm{x}]} & {,} & {\mathrm{x} \leq 2} \\ {0} & {,} & {\mathrm{x}>2}\end{array}\right.$ where [x] is the greatest integer less than or equal to x. If $\mathrm{I}=\int_{-1}^{2} \frac{\mathrm{x} f\left(\mathrm{x}^{2}\right)}{2+f(\mathrm{x}+1)} \mathrm{dx}$ , then the value of (4I – 1) is

[JEE 2015, 4M, –0M]

Sol. (A)


Q. If $\alpha=\int_{0}^{1}\left(\mathrm{e}^{9 \mathrm{x}+3 \tan ^{-1} \mathrm{x}}\right)\left(\frac{12+9 \mathrm{x}^{2}}{1+\mathrm{x}^{2}}\right) \mathrm{d} \mathrm{x}$ where $\tan ^{-1} \mathrm{x}$ takes only principal values, then the value of $\left(\log _{\mathrm{e}}|1+\alpha|-\frac{3 \pi}{4}\right)$ is

[JEE 2015, 4M, –0M]

Sol. 0


Q. Let $f: \mathbb{U} \rightarrow \square$ be a continuous odd function, which vanishes exactly at one point and $f(1)=\frac{1}{2}$ Suppose that $\mathrm{F}(\mathrm{x})=\int_{-1}^{\mathrm{x}} f(\mathrm{t}) \mathrm{dt}$ for all $\mathrm{x} \in[-1,2]$ and $\mathrm{G}(\mathrm{x})$ $=\int_{-1}^{x} \mathfrak{t}|f(f(\mathfrak{t}))| d \mathfrak{t}$ for all $x \in[-1,2]$. If $\lim _{x \rightarrow 1} \frac{F(x)}{G(x)}=\frac{1}{14},$ then the value of $f\left(\frac{1}{2}\right)$ is

[JEE 2015, 4M, –0M]

Sol. 9


Q. The option(s) with the values of a and L that satisfy the following equation is(are)

(A) $a=2, L=\frac{e^{4 \pi}-1}{e^{\pi}-1}$

(B) $a=2, L=\frac{e^{4 \pi}+1}{e^{\pi}+1}$c

(C) $a=4, L=\frac{e^{4 \pi}-1}{e^{\pi}-1}$

(D) $a=4, L=\frac{e^{4 \pi}+1}{e^{\pi}+1}$

[JEE 2015, 4M, –0M]

Sol. 7


Q. Let $f(\mathrm{x})=7 \tan ^{8} \mathrm{x}+7 \tan ^{6} \mathrm{x}-3 \tan ^{4} \mathrm{x}-3 \tan ^{2} \mathrm{x}$ for all $\mathrm{x} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) .$ Then the correct expression(s)is(are)

(A) $\int_{0}^{\pi / 4} \mathrm{x} f(\mathrm{x}) \mathrm{d} \mathrm{x}=\frac{1}{12}$

(B) $\int_{0}^{\pi / 4} f(\mathrm{x}) \mathrm{d} \mathrm{x}=0$

(C) $\int_{0}^{\pi / 4} \mathrm{x} f(\mathrm{x}) \mathrm{d} \mathrm{x}=\frac{1}{6}$

(D) $\int_{0}^{\pi / 4} f(\mathrm{x}) \mathrm{d} \mathrm{x}=1$

[JEE 2015, 4M, –0M]

Sol. (A,C)


Q. Let $f^{\prime}(x)=\frac{192 x^{3}}{2+\sin ^{4} \pi x}$ for all $\mathrm{x} \in \square$ with $f$ $\left(\frac{1}{2}\right)$ $=0 .$ If $\mathrm{m} \leq \int_{1 / 2}^{1} f(\mathrm{x}) \mathrm{d} \mathrm{x} \leq \mathrm{M}$ then the possible values of m and M are

(A) m = 13, M = 24

(B) $\quad \mathrm{m}=\frac{1}{4}, \mathrm{M}=\frac{1}{2}$

(C) m = –11, M = 0

(D) m = 1, M = 12

[JEE 2015, 4M, –0M]

Sol. (A,B)


Let $\mathrm{F}: \mathbb{U} \rightarrow \square$ be a thrice differentiable function. Suppose that $\mathrm{F}(1)=0, \mathrm{F}(3)=-4 \mathrm{F}^{\prime}(\mathrm{x})<$ 0 for all $\mathrm{x} \in(1 / 2,3) .$ Let $f(\mathrm{x})=\mathrm{xF}(\mathrm{x})$ for all $\mathrm{x} \in \mathbb{D}$.

Q. The correct statement(s) is(are)

(A) $f^{\prime}(1)<0$

(B) $f(2)<0$

(C) $f^{\prime}(\mathrm{x}) \neq 0$ for any $\mathrm{x} \in(1,3)$

(D) $f^{\prime}(x)=0$ for some $x \in(1,3)$

[JEE 2015, 4M, –0M]

Sol. (D)


Q. If $\int_{1}^{3} \mathrm{x}^{2} \mathrm{F}^{\prime}(\mathrm{x}) \mathrm{d} \mathrm{x}=-12$ and $\int_{1}^{3} \mathrm{x}^{3} \mathrm{F}^{\prime \prime}(\mathrm{x}) \mathrm{d} \mathrm{x}=40,$ then the correct expression(s) is (are)

(A) 9ƒ'(3) + ƒ'(1) – 32 = 0

(B) $\int_{1}^{3} f(\mathrm{x}) \mathrm{d} \mathrm{x}=12$

(C) 9ƒ'(3) – ƒ'(1) + 32 = 0

(D) $\left.\int_{1}^{3} f(x) d x=-12\right]$

[JEE 2015, 4M, –0M]

Sol. (A,B,C)


Q. The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^{2} \cos x}{1+e^{x}} d x$ is equal to

(A) $\frac{\pi^{2}}{4}-2$

(B) $\frac{\pi^{2}}{4}+2$

(C) $\pi^{2}-\mathrm{e}^{\frac{\pi}{2}}$

(D) $\pi^{2}+\mathrm{e}^{\frac{\pi}{2}}$

[JEE(Advanced)2016]

Sol. (C,D)


Q. Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be a differentiable function such that $\mathrm{f}(0)=0, \mathrm{f}\left(\frac{\pi}{2}\right)=3$ and $\mathrm{f}^{\prime}(0)=1$ If $\mathrm{g}(\mathrm{x})=\int_{\mathrm{x}}^{\frac{\pi}{2}}\left[\mathrm{f}^{\prime}(\mathrm{t}) \csc \mathrm{t}-\cot t \csc \mathrm{t} \mathrm{f}(\mathrm{t})\right] \mathrm{d} \mathrm{t}$ for $\mathrm{x} \in\left(0, \frac{\pi}{2}\right],$ then $\lim _{\mathrm{x} \rightarrow 0} \mathrm{g}(\mathrm{x})=$

[JEE(Advanced)-2017]

Sol. 2


Q. If $\mathrm{I}=\sum_{\mathrm{k}=1}^{98} \int_{\mathrm{k}}^{\mathrm{k}+1} \frac{\mathrm{k}+1}{\mathrm{x}(\mathrm{x}+1)} \mathrm{d} \mathrm{x},$ then

(A) $\mathrm{I}<\frac{49}{50}$

(B) $\mathrm{I}<\log _{\mathrm{e}} 99$

(C) $\mathrm{I}>\frac{49}{50}$

(D) $\mathrm{I}>\log _{\mathrm{e}} 99$

[JEE(Advanced)-2017]

Sol. (B,C)


Q. If $\mathrm{g}(\mathrm{x})=\int_{\sin \mathrm{x}}^{\sin (2 \mathrm{x})} \sin ^{-1}(\mathrm{t}) \mathrm{dt},$ then

(A) $\mathrm{g}^{\prime}\left(\frac{\pi}{2}\right)=-2 \pi$

(B) $\mathrm{g}^{\prime}\left(-\frac{\pi}{2}\right)=2 \pi$

(C) $\mathrm{g}^{\prime}\left(\frac{\pi}{2}\right)=2 \pi$

(D) $\mathrm{g}^{\prime}\left(-\frac{\pi}{2}\right)=-2 \pi$

[JEE(Advanced)-2017]

Sol. (Bonus)


Q. The value of the integral $\int_{0}^{\frac{1}{2}} \frac{1+\sqrt{3}}{\left((x+1)^{2}(1-x)^{6}\right)^{\frac{1}{4}}} d x$ is

[JEE(Advanced)-2018]

Sol. 2


Maxima-Minima – JEE Main Previous Year Question with Solutions

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Mathematics

Tangent & Normal – JEE Main Previous Year Question with Solutions

JEE Main Previous Year Question of Math with Solutions are available at eSaral. Practicing JEE Main Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

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Download eSaral app for free study material and video tutorials.

Mathematics

Quadratic Equation – JEE Main Previous Year Question with Solutions

JEE Main Previous Year Question of Math with Solutions are available at eSaral. Practicing JEE Main Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

Simulator

Previous Years AIEEE/JEE Mains Questions

Q. If the roots of the equation $b x^{2}+c x+a=0$ be imaginary, then for all real values of x, the expression $3 b^{2} x^{2}+6 b c x+2 c^{2}$ is :-

(1) Greater than –4ab (2) Less than –4ab (3) Greater than 4ab (4) Less than 4ab

[AIEEE-2010]

Sol. (1)

since the roots of $\mathrm{bx}^{2}+\mathrm{cx}+\mathrm{a}=0$ are imaginary

$\because c^{2}-4 a b<0 \Rightarrow c^{2}<4 a b$

for exp. $3 b^{2} x^{2}+6 b c x+2 c^{2}$

Min. value $=\frac{-\mathrm{D}}{4 \mathrm{a}}=\frac{\left(36 \mathrm{b}^{2} \mathrm{c}^{2}-24 \mathrm{b}^{2} \mathrm{c}^{2}\right)}{12 \mathrm{b}^{2}}=\frac{-12 \mathrm{b}^{2} \mathrm{c}^{2}}{12 \mathrm{b}^{2}}=-\mathrm{c}^{2}$

$\therefore-\mathrm{c}^{2}>-4 \mathrm{ab}$

So exp. is greater than (– 4ab)


Q. If $\alpha$ and $\beta$ are the roots of the equation $x^{2}-x+1=0,$ then $\alpha^{2009}+\beta^{2009}=$

(1) –2 (2) –1 (3) 1 (4) 2

[AIEEE-2010]

Sol. (3)

Roots of equation $\mathrm{x}^{2}-\mathrm{x}+1=0 \mathrm{ar}$

$\alpha=-\omega, \quad \beta=-\omega^{2}$

$\alpha^{2009}+\beta^{2009}=(-\omega)^{2009}+\left(-\omega^{2}\right)^{2009}$c

$=-\left(\omega^{2}+\omega\right)=1$


Q. Let for $a \neq a_{1} \neq 0, f(x)=a x^{2}+b x+c, g(x)=a_{1} x^{2}+b_{1} x+c_{1}$ and $p(x)=f(x)-g(x)$

If $\mathrm{p}(\mathrm{x})=0$ only for $\mathrm{x}=-1$ and $\mathrm{p}(-2)=2,$ then the value of $\mathrm{p}(2)$ is:

(1) 18 (2) 3 (3) 9 (4) 6

[AIEEE-2011]

Sol. (1)

$\mathrm{P}(\mathrm{x})=\mathrm{k}(\mathrm{x}+1)^{2}$

$\mathrm{P}(-2)=2=\mathrm{k}(-1)^{2}$

$\Rightarrow \mathrm{k}=2$

$\therefore \mathrm{P}(\mathrm{x})=2(\mathrm{x}+1)^{2}$

$\Rightarrow \mathrm{P}(2)=18$

Aliter :

$\mathrm{P}(\mathrm{x})=\left(\mathrm{a}-\mathrm{a}_{1}\right) \mathrm{x}^{2}+\left(\mathrm{b}-\mathrm{b}_{1}\right) \mathrm{x}+\left(\mathrm{c}-\mathrm{c}_{1}\right)=0$

only x = –1

P(x) = 0

So roots are equal.

it means D = 0

$\left(b-b_{1}\right)^{2}=4\left(a-a_{1}\right)\left(c-c_{1}\right) \Rightarrow q^{2}=4 p r$

Let

$\mathrm{a}-\mathrm{a}_{1}=\mathrm{p}$

$\mathrm{b}-\mathrm{b}_{1}=\mathrm{q}$

$\mathrm{c}-\mathrm{c}_{1}=\mathrm{r}$

P(–1) = 0

p – q + r = 0 …… (1)

4p – 2q + r = 2 ….. (2)

4p + 2q + r = ?

4p + 2q + r = ?

$(p+r)^{2}-4 p r=0$

$(p-r)^{2}=0$                       

from eq. (1) q = 2r

So from eq. (2) 4r – 4r + r = 2

r = 2

So 4p + 2q + r = 4r + 4r + r = 9r = 18


Q. Sachin and Rahul attempted to solve a quadratic equation. Sachin made a mistake in writing down the constant term and ended up in roots (4, 3). Rahul made a mistake in writing down coefficient of x to get roots (3, 2). The correct roots of equation are:

(1) –4, –3 (2) 6, 1 (3) 4, 3 (4) –6, –1

[AIEEE-2011]

Sol. (2)


Q. The equation $\mathrm{e}^{\mathrm{sinx}}-\mathrm{e}^{-\mathrm{sinx}}-4=0$ has :

(1) exactly four real roots.

(2) infinite number of real roots.

(3) no real roots.

(4) exactly one real root.

[AIEEE-2012]

Sol. (3)


Q. If the equations $\mathrm{x}^{2}+2 \mathrm{x}+3=0$ and $\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}=0, \mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{R}$, have a common root, then a : b : c is :

(1) 1 : 2 : 3 (2) 3 : 2 : 1 (3) 1 : 3 : 2 (4) 3 : 1 : 2

[JEE-MAIN-2013]

Sol. (1)

$\mathrm{x}^{2}+2 \mathrm{x}+3=0$

$\mathrm{D}<0$

$\therefore \quad \mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}=0 \mathrm{has}$

both roots common

$\therefore \quad a: b: c=1: 2: 3$


Q. Let  and $\beta$ be the roots of equation $\mathrm{x}^{2}-6 \mathrm{x}-2=0 .$ If $\mathrm{a}_{\mathrm{n}}=\alpha^{\mathrm{n}}-\beta^{\mathrm{n}},$ for $\mathrm{n} \geq$ 1, then the value of $\frac{\mathrm{a}_{10}-2 \mathrm{a}_{8}}{2 \mathrm{a}_{9}}$ is equal to :

(1) 3 (2) – 3 (3) 6 (4) – 6

[JEE-MAIN-2015]

Sol. (1)


Q. The sum of all real values of x satisfying the equation $\left(x^{2}-5 x+5\right)^{x^{2}+4 x-60}=1$ is :

(1) 5 (2) 3 (3) –4 (4) 6

[JEE-MAIN-2016]

Sol. (2)

$\mathrm{x}^{2}-5 \mathrm{x}+5=1 \Rightarrow \mathrm{x}=1,4$

$x^{2}-5 x+5=-1 \Rightarrow x=2,3$

but 3 is rejected

$\mathrm{x}^{2}+4 \mathrm{x}-60=0 \Rightarrow \mathrm{x}=-10,6$

Sum = 3


Q. If $\alpha, \beta \in \mathrm{C}$ are the distinct roots of the equation $\mathrm{x}^{2}-\mathrm{x}+1=0,$ then $\alpha^{101}+\beta^{107}$ is equal to-

If $\alpha, \beta \in \mathrm{C}$ are the distinct roots of the equation $\mathrm{x}^{2}-\mathrm{x}+1=0,$ then $\alpha^{101}+\beta^{107}$ is equal to-

[JEE-MAIN-2018]

Sol. (2)


Definite Integration – JEE Main Previous Year Question with Solutions

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Q. $\int_{0}^{\pi}[\cot x] d x,$ where $[.]$ denotes the greatest integer function, is equal to –

(1) –1 $         

(2)-\frac{\pi}{2}$           

(3) $\frac{\pi}{2}$           

(4) 1

[AIEEE-2009] 

Sol. (2)


Q. Let p(x) be a function defined on R such that p'(x) = p'(1 – x), for all x $\in$0, 1], p(0) = 1 and p(1) = 41. Then $\int_{0}^{1}$ p(x) dx equals :-

(1) $\sqrt{41}$

(2) 21

(3) 41

(4) 42

[AIEEE-2010]

Sol. (2)


Q. The value of $\int_{0}^{1} \frac{8 \log (1+x)}{1+x^{2}} d x$ is :-

(1) $\frac{\pi}{2} \log 2$

(2) $\log 2$

(3) $\pi \log 2$

(4) $\frac{\pi}{8} \log 2$

[AIEEE-2011]

Sol. (3)


Q. Let [.] denote the greatest integet function then the value of $\int_{0}^{1.5} \mathrm{x}\left[\mathrm{x}^{2}\right] \mathrm{dx}$ is :-

( 1)$\frac{5}{4}$ (2) 0 (3) $\frac{3}{2}$ (4) $\frac{3}{4}$

[AIEEE-2011]

Sol. (4)


Q. If $\mathrm{g}(\mathrm{x})=\int_{0}^{\mathrm{x}} \cos 4 \mathrm{t} \mathrm{dt},$ then $\mathrm{g}(\mathrm{x}+\pi)$ equals :

(1) $\mathrm{g}(\mathrm{X}) \cdot \mathrm{g}(\pi)$

(2) $\frac{\mathrm{g}(\mathrm{x})}{\mathrm{g}(\pi)}$

(3) $\mathrm{g}(\mathrm{x})+\mathrm{g}(\pi)$

(4) $\mathrm{g}(\mathrm{x})-\mathrm{g}(\pi)$

[AIEEE-2012]

Sol. (3,4)


Q. Statement-I : The value of the integral $\int_{\pi / 6}^{\pi / 3} \frac{\mathrm{dx}}{1+\sqrt{\tan \mathrm{x}}}$ is equal to $\frac{\pi}{6}$

Statement-II : $\int_{a}^{b} f(x) d x-\int_{a}^{b} f(a+b-x) d x$

(1) Statement-I is true, Statement-II is true; Statement-II is a correct explanation for Statement- I.

(2) Statement-I is true, Statement-II is true; Statement-II is not a correct explanation for Statement-I.

(3) Statement-I is true, Statement-II is false.

(4) Statement-I is false, Statement-II is true.

[JEE-MAIN-2013]

Sol. (4)


Q. The integral $\int_{0}^{\pi} \sqrt{1+4 \sin ^{2} \frac{x}{2}-4 \sin \frac{x}{2}} d x$ equals :

(1) $\pi-4$

(2) $\frac{2 \pi}{3}-4-4 \sqrt{3}$

(3) $4 \sqrt{3}-4$

(4) $4 \sqrt{3}-4-\frac{\pi}{3}$

[JEE-MAIN-2014]

Sol. (4)


Q. The integral $\int_{2}^{4} \frac{\log x^{2}}{\log x^{2}+\log \left(36-12 x+x^{2}\right)} d x$ is equal to :

(1) 1 (2) 6 (3) 2 (4) 4

[JEE-MAIN-2015]

Sol. (1)


Q. The integral $\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{\mathrm{dx}}{1+\cos x}$ is equal to :-

(1) –1 (2) –2 (3) 2 (4) 4

[JEE-MAIN-2017]

Sol. (3)


Q. The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^{2} x}{1+2^{x}} d x$ is :

( 1)$\frac{\pi}{2}$

(2) $4 \pi$

(3) $\frac{\pi}{4}$

(4) $\frac{\pi}{8}$

[JEE-MAIN-2018]

Sol. (3)


 

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Mathematics

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Download eSaral app for free study material and video tutorials.

Q. Let the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lie in the plane $x+3 y-\alpha z+\beta=0 .$ Then $(\alpha, \beta)$ equals

(1) (5, – 15)             (2) (–5, 5)             (3) (6, –17)               (4) (–6, 7)

[AIEEE-2009]

Sol. (4)

Line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2} \ldots .(1)$

Plane $\mathrm{x}+3 \mathrm{y}-\alpha \mathrm{z}+\beta=0 \quad \ldots .(2)$

Point $(2,1,-2)$ put in ( 2)

$2+3+2 \alpha+\beta=0 \Rightarrow 2 \alpha+\beta=-5$

$\mathrm{Now} \mathrm{a}_{1} \mathrm{a}_{2}+\mathrm{b}_{1} \mathrm{b}_{2}+\mathrm{c}_{1} \mathrm{c}_{2}=0$

$3-15-2 \alpha=0 \Rightarrow-12-2 \alpha=0 \Rightarrow \alpha=-6$

$-12+\beta=-5 \Rightarrow \beta=7$

$\therefore \alpha=-6, \beta=7$


Q. The projections of a vector on the three coordinate axis are 6, –3, 2 respectively. The direction cosines of the vector are :-

(1) $\frac{6}{7}, \frac{-3}{7}, \frac{2}{7}$

(2) $\frac{-6}{7}, \frac{-3}{7}, \frac{2}{7}$

(3) 6, –3, 2

(4) $\frac{6}{5}, \frac{-3}{5}, \frac{2}{5}$

[AIEEE-2009]

Sol. (1)


Q. Statement–1 : The point A(3,1,6) is the mirror image of the point B(1, 3, 4) in the plane

x – y + z = 5.

Statement–2 : The plane x – y + z = 5 bisects the line segment joining A(3, 1, 6) and B(1, 3, 4).

(1) Statement–1 is true, Statement–2 is true; Statement–2 is a correct explanation for Statement– 1.

(2) Statement–1 is true, Statement–2 is true ; Statement–2 is not a correct explanation for statement– 1.

(3) Statement–1 is true,n Statement–2 is false.

(4) Statement–1 is false, Statement–2 is true.

[AIEEE-2010]

Sol. (2)

Mirror image of $\mathrm{B}(1,3,4)$ in plane $\mathrm{x}-\mathrm{y}+\mathrm{z}=5$

$\frac{x-1}{1}=\frac{y-3}{-1}=\frac{z-4}{1}=-2 \frac{(1-3+4-5)}{1+1+1}=2$

$\Rightarrow x=3, y=1, z=6$

$\therefore$ mirror image of $\mathrm{B}(1,3,4)$

is $\mathrm{A}(3,1,6)$

statement-1 is correct

statement-2 is true but it is not the correct explanation.

because it is not neccessary that to bisect the line joining (1,3,4) and (3,1,6) plane is always perpendicular to line AB


Q. If the angle between the line $x=\frac{y-1}{2}=\frac{z-3}{\lambda}$ and the plane $x+2 y+3 z=4$ is $\cos ^{-1}(\sqrt{\frac{5}{14}})$, then $\lambda$ equals:-

( 1)$\frac{2}{5}$

( 2)$\frac{5}{3}$

( 3)$\frac{2}{3}$

( 4)$\frac{3}{2}$

[AIEEE-2011]

Sol. (3)

$\frac{x}{1}=\frac{y-1}{2}=\frac{z-3}{\lambda}$ equation of line

equation of plane $x+2 y+3 z=4$

$\sin \theta=\frac{1+4+3 \lambda}{\sqrt{14} \sqrt{1+4}+\lambda^{2}}$

$\Rightarrow \lambda=\frac{2}{3}$


Q. Statement-1: The point $\mathrm{A}(1,0,7)$ is the mirror image of the point $\mathrm{B}(1,6,3)$ in the line:

\[ \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} \]

Statement-2: The line : $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ bisects the line segment joining $A(1,0,7)$ and B(1, 6, 3).

(1) Statement-1 is true, Statement-2 is false.

(2) Statement-1 is false, Statement-2 is true

(3) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

(4) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

[AIEEE-2011]

Sol. (4)

$1(1-1)+2(0-6)+3(7-3)$

$=0-12+12=0$

mid point $\mathrm{AB}(1,3,5)$

lies on $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$


Q. The distance of the point (1, –5, 9) from the plane x – y + z = 5 measured along a straight line x = y = z is :

(1) $3 \sqrt{5}$

(2) $10 \sqrt{3}$

(3) $5 \sqrt{3}$

(4) $3 \sqrt{10}$

[AIEEE-2011]

Sol. (2)

Distance $\mathrm{PM}=\sqrt{100+100+100}=10 \sqrt{3}$


Q. An equation of a plane parallel to the plane x – 2y + 2z – 5 = 0 and at a unit distance from the origin is :

(1) x – 2y + 2z + 5 = 0

(2) x – 2y + 2z – 3 = 0

(3) x – 2y + 2z + 1 = 0

(4) x – 2y + 2z – 1 = 0

[AIEEE-2012]

Sol. (2)

Equation of plane parallel to

$x-2 y+2 z-5=0$

is $x-2 y+2 z=k$

Distance of plane from origen $=1$

$\left|\frac{k}{\sqrt{9}}\right|=1$

$\mathrm{k}=\pm 3$

Equation of required plane is

$x-2 y+2 z \pm 3=0$


Q. If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect, then $k$ is equal to :

(1) 0

(2) – 1

( 3)$\frac{2}{9}$

( 4)$\frac{9}{2}$

[AIEEE-2012]

Sol. (4)

$\left|\begin{array}{ccc}{3-1} & {\mathrm{K}+1} & {0-1} \\ {2} & {3} & {4} \\ {1} & {2} & {1}\end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc}{2} & {\mathrm{K}+1} & {-1} \\ {2} & {3} & {4} \\ {1} & {2} & {1}\end{array}\right|=0$

$\Rightarrow 2 \mathrm{K}-9=0 \Rightarrow \mathrm{K}=\frac{9}{2}$


Q. Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is :-

( 1)$\frac{3}{2}$

( 2)$\frac{5}{2}$

(3) $\frac{7}{2}$

( 4)$\frac{9}{2}$

[JEE-MAIN 2013]

Sol. (3)

$4 x+2 y+4 z+5=0$

$4 x+2 y+4 z-16=0$

$\Rightarrow \quad d=\left|\frac{21}{\sqrt{36}}\right|=\frac{7}{2}$


Q. If the lines $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}$ and $\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}$ are coplanar, then $k$ can have :

(1) any value

(2) exactly one value

(3) exactly two values

(4) exactly three values.

[JEE-MAIN 2013]

Sol. (3)

$\Rightarrow(\bar{a}-\bar{b}) \cdot(\bar{c} \times \bar{d})=0$

$\Rightarrow\left|\begin{array}{ccc}{1} & {-1} & {-1} \\ {1} & {1} & {-k} \\ {k} & {2} & {1}\end{array}\right|=0$

$\Rightarrow \quad(1+2 k)+\left(1+k^{2}\right)-(2-k)=0$

$\Rightarrow \quad \mathrm{k}^{2}+3 \mathrm{k}=0<\begin{array}{c}{0} \\ {-3}\end{array}$


Q. A vector $\overrightarrow{\mathrm{n}}$ is inclined to x-axis at $45^{\circ},$ to $\mathrm{y}$ -axis at $60^{\circ}$ and at an acute angle to z-axis. If $\overrightarrow{\mathrm{n}}$ is a normal to a plane passing through the point $(\sqrt{2},-1,1),$ then the equation of the plane is :

(1) $\sqrt{2} x-y-z=2$

(2) $\sqrt{2} x+y+z=2$

(3) $3 \sqrt{2} x-4 y-3 z=7$

(4) $4 \sqrt{2} x+7 y+z=2$

[JEE-MAIN Online 2013]

Sol. (2)

Let $\alpha, \beta, \gamma$ v be direction cosine of vector $\vec{n}$

$\alpha^{2}+\beta^{2}+\gamma^{2}=1$

$\frac{1}{2}+\frac{1}{4}+\gamma^{2}=1$

$\gamma=\frac{1}{2} \quad\left\{-\frac{1}{2} \text { rejected }\right\}$

$\overline{\mathrm{n}}=\frac{\hat{\mathrm{i}}}{\sqrt{2}}+\frac{\hat{\mathrm{j}}}{2}+\frac{\hat{\mathrm{k}}}{2}=\sqrt{2 \hat{\mathrm{i}}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$

$\overline{\mathrm{a}}=\sqrt{2} \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$

$d=\bar{a} \cdot \bar{n}=2-1+1=2$

$\overline{\mathrm{r}} \cdot(\sqrt{2} \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=2$

$\sqrt{2} x+y+z=2$


Q. The acute angle between two lines such that the direction cosines $\ell, \mathrm{m}, \mathrm{n}$ of each of them satisfy the equations $\ell+\mathrm{m}+\mathrm{n}=0$ and $\ell^{2}+\mathrm{m}^{2}-\mathrm{n}^{2}=0$ is :-

(1) $30^{\circ}$ (2) $45^{\circ}$ (3) $60^{\circ}$ (4) $15^{\circ}$

[JEE-MAIN Online 2013]

Sol. (3)

$\begin{array}{rl}{1+\mathrm{m}+\mathrm{n}} & {=0 ; \ell^{2}+\mathrm{m}^{2}-\mathrm{n}^{2}=0} \\ {\mathrm{a} \ell=-(\mathrm{m}+\mathrm{n})} & {} \\ {(\mathrm{m}+\mathrm{n})^{2}+\mathrm{m}^{2}-\mathrm{n}^{2}} & {=0} \\ {(\mathrm{m}+\mathrm{n})[\mathrm{m}+\mathrm{n}+\mathrm{m}-\mathrm{n}]} & {=0} \\ {\mathrm{m}+\mathrm{n}=0} & {\mathrm{m}=0} \\ {\mathrm{m}=-\mathrm{n}}\end{array}$

Dr’s of line $1 \Rightarrow 0,-\mathrm{n}, \mathrm{n}$ or $0,-1,1$

Dr’s of line $2 \Rightarrow-\mathrm{n}, 0, \mathrm{n}$ or $-1,0,1$

$\cos \theta=\frac{1}{\sqrt{2} \sqrt{2}}$

$\theta=60^{\circ}$


Q. Let Q be the foot of perpendicular from the origin to the plane 4x – 3y + z + 13 = 0 and R be a point (–1, 1, –6) on the plane. Then length QR is :

(1) $3 \sqrt{\frac{7}{2}}$

(2) $\sqrt{14}$

(3) $\sqrt{\frac{19}{2}}$

(4) $\frac{3}{\sqrt{2}}$

[JEE-MAIN Online 2013]

Sol. (1)


Q. If the projections of a line segment on thex, y and z-axes in 3-dimensional space are 2, 3 and 6 respectively, then the length ofthe line segment is :

(1) 7             (2) 9           (3) 12               (4) 6

[JEE-MAIN Online 2013]

Sol. (1)

i.e. $[\mathrm{OM}=2, \mathrm{ON}=3, \mathrm{OQ}=6]$

$\therefore \mathrm{P}(2,3,6)$

$\begin{aligned} \therefore \mathrm{OP} &=\sqrt{2^{2}+3^{2}+6^{2}} \\ &=\sqrt{49}=7 \end{aligned}$


Q. If two lines $L_{1}$ and $L_{2}$ in space, are defined by $\mathrm{L}_{1}=\{\mathrm{x}=\sqrt{\lambda} \mathrm{y}+(\sqrt{\lambda}-1)$

$\mathrm{z}=(\sqrt{\lambda}-1) \mathrm{y}+\sqrt{\lambda}\}$ and

$\mathrm{L}_{2}=\{\mathrm{x}=\sqrt{\mu} \mathrm{y}+(1-\sqrt{\mu})$

$\mathrm{z}=(1-\sqrt{\mu}) \mathrm{y}+\sqrt{\mu}\}$

then $\mathrm{L}_{1}$ is perpendicular to $\mathrm{L}_{2},$ for all non-negative reals $\lambda$ and $\mu,$ such that :

(1) $\lambda=\mu$

(2) $\lambda \neq \mu$

(3) $\sqrt{\lambda}+\sqrt{\mu}=1$

(4) $\lambda+\mu=0$

[JEE-MAIN Online 2013]

Sol. (1,4)

$\mathrm{L}_{1}=\{\mathrm{x}=\sqrt{\lambda} \mathrm{y}+(\sqrt{\lambda}-1), \mathrm{z}=(\sqrt{\lambda}-1) \mathrm{y}+\sqrt{\lambda}\}$

$\therefore \mathrm{L}_{1}: \frac{\mathrm{x}-(\sqrt{\lambda}-1)}{\sqrt{\lambda}}=\frac{\mathrm{y}-0}{1}=\frac{\mathrm{Z}-\sqrt{\lambda}}{(\sqrt{\lambda}-1)} \quad \& \mathrm{L}_{2}:\{\mathrm{x}=\sqrt{\mu} \mathrm{y}+(1-\sqrt{\mu}), \mathrm{z}=(1-\sqrt{\mu}) \mathrm{y}+\sqrt{\mu}\}$

$\therefore \mathrm{L}_{2}: \frac{\mathrm{x}-(1-\sqrt{\mu})}{\sqrt{\mu}}=\frac{\mathrm{y}-0}{1}=\frac{\mathrm{z}-\sqrt{\mu}}{(1-\sqrt{\mu})}$

$\therefore \mathrm{L}_{1} \perp^{r} \mathrm{L}_{2} \Rightarrow(\sqrt{\lambda})(\sqrt{\mu})+1+(\sqrt{\lambda}-1)(1-\sqrt{\mu})=0$

$\Rightarrow \sqrt{\lambda} \sqrt{\mu}+1-\sqrt{\lambda} \sqrt{\mu}+\sqrt{\lambda}+\sqrt{\mu}-1=0$

$\Rightarrow \sqrt{\lambda}+\sqrt{\mu}=0 \Rightarrow \lambda=\mu=0$


Q. The equation of a plane through the line of intersection of the planes x + 2y = 3, y – 2z+ 1 = 0, and perpendicular to the first plane is :

(1) 2x – y + 7z = 11

(2) 2x – y + 10 z = 11

(3) 2x – y – 9z = 10

(4) 2x – y – 10z = 9

[JEE-MAIN Online 2013]

Sol. (2)

$(x+2 y-3)+\lambda(y-2 z+1)=0$

$(1)(1)+(2+\lambda) 2+(-2 \lambda) \times 0=0$

$1+4+2 \lambda=0$

$\lambda=-\frac{5}{2}$

$(x+2 y-3)-\frac{5}{2}(y-2 z+1)=0$

$2 \mathrm{x}-\mathrm{y}+10 \mathrm{z}-11=0$


Q. Let ABC be a triangle with vertices at points $\mathrm{A}(2,3,5), \mathrm{B}(-1,3,2)$ and $\mathrm{C}(\lambda, 5, \mu)$ in three dimensional space. If the median through A is equally inclined with the axes, then $(\lambda, \mu$.) is equal to :

(1) (10, 7)                  (2) (7.5)               (3) (7, 10)                    (4) (5,7)

[JEE-MAIN Online 2013]

Sol. (3)


Q. The angle between the lines whose direction cosines satisfy the equations $\ell+\mathrm{m}+\mathrm{n}=0$ and $\ell^{2}=\mathrm{m}^{2}+\mathrm{n}^{2}$ is

( 1)$\frac{\pi}{3}$

( 2)$\frac{\pi}{4}$

( 3)$\frac{\pi}{6}$

(4) $\frac{\pi}{2}$

[JEE-MAIN 2014]

Sol. (1)


Q. The image of the line $\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5}$ in the plane $2 x-y+z+3=0$ is the line:

(1) $\frac{x+3}{3}=\frac{y-5}{1}=\frac{z-2}{-5}$

(2) $\frac{x+3}{-3}=\frac{y-5}{-1}=\frac{z+2}{5}$

(3) $\frac{x-3}{3}=\frac{y+5}{1}=\frac{z-2}{-5}$

(4) $\frac{x-3}{-3}=\frac{y+5}{-1}=\frac{z-2}{5}$

[JEE-MAIN 2014]

Sol. (1)

$\mathrm{L}: \frac{\mathrm{x}-1}{3}=\frac{\mathrm{y}-3}{1}=\frac{\mathrm{z}-4}{-5}$

$P: 2 x-y+z+3=0$

It can be observed given line is parallel to given plane.

Image of $(1,3,4)$ in given plane can be calculated as

$\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=\frac{-2(2 \times 1-3+4+3)}{6}=-2$

$\Rightarrow x=-3 ; y=5 ; z=2$


Q. The equation of the plane containing the line 2x – 5y + z = 3 ; x + y + 4z = 5, and parallel to the plane, x + 3y + 6z = 1, is :

(1)x + 3y + 6z = 7

(2) 2x + 6y + 12z = – 13

(3) 2x + 6y + 12z = 13

(4) x + 3y + 6z = – 7

[JEE(Main)-2015]

Sol. (1)

Let equation of plane parallel to $\mathrm{x}+3 \mathrm{y}+6 \mathrm{z}=1$ is $\mathrm{x}+3 \mathrm{y}+6 \mathrm{z}=\lambda$ point on line of intersection is $(4,1,0),$ it also lie on plane $x+3 y+6 z=\lambda$ so required plane is $x+3 y+6 z=7$


Q. The distance of the point (1, 0, 2) from the point of intersection of the line $\frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}+1}{4}=\frac{\mathrm{z}-2}{12}$ and the plane $\mathrm{x}-\mathrm{y}+\mathrm{z}=16,$ is :

(1) $3 \sqrt{21}$

(2) 13

(3) $2 \sqrt{14}$

(4) 8

[JEE(Main)-2015]

Sol. (2)


Q. The distance of the point (1, –5, 9) from the plane x – y + z = 5 measured along the line x = y = z is :

(1) $\frac{20}{3}$

(2) $3 \sqrt{10}$

(3) $10 \sqrt{3}$

( 4)$\frac{10}{\sqrt{3}}$

[JEE(Main)-2016]

Sol. (3)

Equation of line parallel to $\mathrm{x}=\mathrm{y}=\mathrm{z}$ through $(1,-5,9)$ is $\frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}+5}{1}=\frac{\mathrm{z}-9}{1}=\lambda$

If $\mathrm{P}(\lambda+1, \lambda-5, \lambda+9)$ be point of intesection of line and plane.

$\Rightarrow$ Required distance $=10 \sqrt{3}$


Q. If the line, $\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}$ lies in the plane, $1 x+m y-z=9,$ then $1^{2}+m^{2}$ is equal to :-

(1) 2               (2) 26              (3) 18                  (4) 5

[JEE(Main)-2016]

Sol. (1)

Given line

\[ \frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3} \]

and Given plane is $\ell \mathrm{x}+\mathrm{my}-\mathrm{z}=9$

Now, it is given that line lies on plane $\therefore 2 \ell-\mathrm{m}-3=0 \Rightarrow 2 \ell-\mathrm{m}=3$

Also, $(3,-2,-4)$ lies on plane

\[ 3 \ell-2 m=5 \]

Solving ( 1) and $(2),$ we get

\[ \ell=1, \mathrm{m}=-1 \]

$\therefore \ell^{2}+\mathrm{m}^{2}=2$


Q. If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to line, $\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{4}=\frac{\mathrm{z}}{5}$ is $\mathrm{Q},$ then $\mathrm{PQ}$ is equal to :-

(1) $6 \sqrt{5}$

(2) $3 \sqrt{5}$

(3) $2 \sqrt{42}$

(4) $\sqrt{42}$

[JEE(Main)-2017]

Sol. (3)

Line $\mathrm{PQ} ; \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}+2}{4}=\frac{\mathrm{z}-3}{5}$

Let $\mathrm{F}(\lambda+1,4 \lambda-2,5 \lambda+3)$

                         

F lies on the plane

$2(\lambda+1)+3(4 \lambda-2)-4(5 \lambda+3)+22=0$

$\Rightarrow-6 \lambda+6=0 \Rightarrow \lambda=1$

$\mathrm{F}(2,2,8)$

$\mathrm{PQ}=2 \quad \mathrm{PF}=2 \sqrt{42}$


Q. The distantce of the point (1, 3, –7) from the plane passing through the point (1, 1, –1), havingnormal perpendicular to both the lines $\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3}$ and $\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1},$ is :

(1) $\frac{10}{\sqrt{74}}$

(2) $\frac{20}{\sqrt{74}}$

(3) $\frac{10}{\sqrt{83}}$

(4) $\frac{5}{\sqrt{83}}$

[JEE(Main)-2017]

Sol. (1)

Normal vector

$\left|\begin{array}{ccc}{\hat{\mathrm{i}}} & {\hat{\mathrm{j}}} & {\hat{\mathrm{k}}} \\ {1} & {-2} & {3} \\ {2} & {-1} & {-1}\end{array}\right|=5 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$

So plane is 5(x – 1) + 7(y + 1) + 3(z + 1) = 0

$\Rightarrow 5 x+7 y+3 z+5=0$

Distance $\frac{5+21-21+5}{\sqrt{25+49+9}}=\frac{10}{\sqrt{83}}$


Q. The length of the projection of the line segment joining the points (5, –1, 4) and (4, –1, 3) on the plane, x + y + z = 7 is :

( 1)$\frac{2}{3}$

(2) $\frac{1}{3}$

(3) $\sqrt{\frac{2}{3}}$

(4) $\frac{2}{\sqrt{3}}$

[JEE(Main)-2018]

Sol. (3)


Q. If $\mathrm{L}_{1}$ is the line of intersection of the planes $2 \mathrm{x}-2 \mathrm{y}+3 \mathrm{z}-2=0, \mathrm{x}-\mathrm{y}+\mathrm{z}+1=0$ and $\mathrm{L}_{2}$ is the line of intersection of the planes $\mathrm{x}+2 \mathrm{y}-\mathrm{z}-3=0,3 \mathrm{x}-\mathrm{y}+2 \mathrm{z}-1=0,$ then the distance of the origin from the plane, containing the lines $L_{1}$ and $L_{2}$ is $z$

(1) $\frac{1}{3 \sqrt{2}}$

(2) $\frac{1}{2 \sqrt{2}}$

(3) $\frac{1}{\sqrt{2}}$

( 4)$\frac{1}{4 \sqrt{2}}$

[JEE(Main)-2018]

Sol. (1)

Plane passes through line of intersectuion of first two planes is $(2 x-2 y+3 z-2)+\lambda(x-y+z+1)=0$

$\mathrm{x}(\lambda+2)-\mathrm{y}(2+\lambda)+\mathrm{z}(\lambda+3)+(\lambda-2)=0 \ldots \ldots(1)$

is having infinite number of solution with x + 2y – z – 3 = 0 and 3x – y + 2z – 1 = 0 then

$\left|\begin{array}{ccc}{(\lambda+2)} & {-(\lambda+2)} & {(\lambda+3)} \\ {1} & {2} & {-1} \\ {3} & {-1} & {2}\end{array}\right|=0$

Solving $\lambda=5$ $7 x-7 y+8 z+3=0$

perpendicular distance from $(0,0,0)$

is $\frac{3}{\sqrt{162}}=\frac{1}{3 \sqrt{2}}$


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This set of question papers serves as a very important source to revise the important topics and gain an understanding into the pattern of questions asked in previous years. Practicing these papers will positively help students to gain confidence over their learning.
The benefits of solving IIT-JEE previous years papers is that aspirants get to know the type of questions asked in the JEE exam. JEE aspirants can evaluate their preparation after finishing the entire syllabus, topics and chapters. They can get the experience of giving real exam while solving a past year JEE question paper to become confident for the upcoming JEE exam.

While preparing for the IIT-JEE exam, aspirants should be aware about the question paper structure and the format of questions to be asked in this exam. This will help to make an effective preparation strategy for the exam. JEE Advanced Previous Year Question Papers are the best resources to prepare for exam. This will help an individual to understand the exam pattern of JEE. This will also enhance your level of preparation. JEE aspirants must solve multiple sample papers and analyse their performances in order to recognize their strengths and weaknesses.

 

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Click Here to Download Topic-wise JEE Main Previous Year Questions with Solutions

We have tried our best to provide you last 10 years question with solutions.
This set of question papers serves as a very important source to revise the important topics and gain an understanding into the pattern of questions asked in previous years. Practicing these papers will positively help students to gain confidence over their learning.
The benefits of solving IIT-JEE previous years papers is that aspirants get to know the type of questions asked in the JEE exam. JEE aspirants can evaluate their preparation after finishing the entire syllabus, topics and chapters. They can get the experience of giving real exam while solving a past year JEE question paper to become confident for the upcoming JEE exam.

Matrices – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Math with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral also provides complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

Q. Let A be a 2 × 2 matrix

Statement- $1: \operatorname{adj}(\operatorname{adj} A)=A$

Statement-2: $|$ adj $A|=| A |$

(1) Statement–1 is true, Statement–2 is false.

(2) Statement–1 is false, Statement–2 is true.

(3) Statement–1 is true, Statement–2 is true;Statement–2 is a correct explanation for Statement–1.

(4) Statement–1 is true, Statement–2 is true; Statement–2 is not a correct explanation for statement–1.

[AIEEE- 2009]

Sol. (4)

Let $A=\left(\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right)$

$\operatorname{adj}(\mathrm{A})=\left(\begin{array}{cc}{\mathrm{d}} & {-\mathrm{b}} \\ {-\mathrm{c}} & {\mathrm{a}}\end{array}\right)$

$\operatorname{adj}(\operatorname{adj} A)=\left(\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right)=A$ statement is right

Statement 2 We know $\mathrm{A}$ adj $(\mathrm{A})=|\mathrm{A}| \mathrm{I}_{\mathrm{n}}$ taking determinant $|\mathrm{A} \cdot \operatorname{adj}(\mathrm{A})|=\| \mathrm{A}\left|\mathrm{I}_{\mathrm{n}}\right|$

$\Rightarrow|\operatorname{adj}(\mathrm{A})|=|\mathrm{A}|^{\mathrm{n}-1}$

Here $\mathrm{n}=2$ (order)

so $|\operatorname{adj} \mathrm{A}||\mathrm{A}|^{2-1}=|\mathrm{A}|$

so statement 2 is also true and 2 is not explanation for statement 1


Q. The number of 3× 3 non-singular matrices, with four entries as 1 and all other entries as 0, is :-

(1) Less than 4            (2) 5             (3) 6              (4) At least 7

[AIEEE-2010]

Sol. (4)


Q. Let A be a $2 \times 2$ matrix with non-zero entries and let $\mathrm{A}^{2}=\mathrm{I},$ where I is $2 \times 2$ identity matrix. Define $\operatorname{Tr}(\mathrm{A})=$ sum of diagonal elements of $\mathrm{A}$ and $|\mathrm{A}|=$ determinant of matrix $\mathrm{A}$. Statement- $1: \operatorname{Tr}(\mathrm{A})=0$

Statement-2: $|\mathrm{A}|=1$

(1) Statement–1 is true, Statement–2 is true; Statement–2 is a correct explanation for

Statement–1.

(2) Statement–1 is true, Statement–2 is true; Statement–2 is not a correct explanation for

statement–1.

(3) Statement–1 is true, Statement–2 is false.

(4) Statement–1 is false, Statement–2 is true.

[AIEEE-2010]

Sol. (3)

Statement 1:

Let $\mathrm{A}=\left(\begin{array}{ll}{\mathrm{a}} & {\mathrm{b}} \\ {\mathrm{c}} & {\mathrm{d}}\end{array}\right) \mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d},$ are non zero

$A^{2}=\left(\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right)\left(\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right)=\left(\begin{array}{ll}{1} & {0} \\ {0} & {1}\end{array}\right)$

$\Rightarrow a^{2}+b c=1$

$\Rightarrow \mathrm{ab}+\mathrm{bd}=0 \Rightarrow \mathrm{b}(\mathrm{a}+\mathrm{d})=0$

so $\mathrm{b} \neq 0,(\mathrm{a}+\mathrm{d})=0$

$a+d=0 \Rightarrow \operatorname{tr}(A)=0$

Statement 2:

\[ |A|=a d-b c=-a^{2}-b c=-\left(a^{2}+b c\right)=-1 \]

So Statement 2 is false.


Q. Let A and B be two symmetric matrices of order 3.

Statement-1 : A(BA) and (AB)A are symmetric matrices.

Statement-2 : AB is symmetric matrix if matrix multiplication of A with B is commutative.

(1) Statement-1 is true, Statement-2 is false.

(2) Statement-1 is false, Statement-2 is true

(3) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

(4) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

[AIEEE-2011]

Sol. (4)

$A^{T}=A$

$\mathrm{B}^{\mathrm{T}}=\mathrm{B}$

Statement- 1 $(\mathrm{A}(\mathrm{BA}))^{\mathrm{T}}=(\mathrm{BA})^{\mathrm{T}} \mathrm{A}^{\mathrm{T}}$ \[ =\mathrm{A}^{\mathrm{T}} \mathrm{B}^{\mathrm{T}} \mathrm{A}^{\mathrm{T}}=\mathrm{A}(\mathrm{BA}) \rightarrow \text { symmetric } \]

$((\mathrm{AB}) \mathrm{A}))^{\mathrm{T}}=\mathrm{A}^{\mathrm{T}} \mathrm{B}^{\mathrm{T}} \mathrm{A}^{\mathrm{T}}=(\mathrm{AB}) \mathrm{A} \rightarrow$ symmetric

Statement – 1 is true

Statement- 2:

$(\mathrm{AB})^{\mathrm{T}}=\mathrm{B}^{\mathrm{T}} \mathrm{A}^{\mathrm{T}}=\mathrm{B} \mathrm{A}$

if $\mathrm{AB}=\mathrm{B} \mathrm{A}$ then

$(\mathrm{AB})^{\mathrm{T}}=\mathrm{B} \mathrm{A}=\mathrm{AB}$

Statement- 2 is true

but Not a correct expalnation.


Q. Statement-1 : Determinant of a skew-symmetric matrix of order 3 is zero.

Statement-1 : For any matrix A, det(AT) = det(A) and det(–A) = –det(A).

Where det(B) denotes the determinant of matrix B. Then :

(1) Statement-1 is true and statement-2 is false

(2) Both statements are true

(3) Both statements are false

(4) Statement-1 is false and statement-2 is true.

[AIEEE-2011]

Sol. (1)

Statement- 1: The value of determinant of skew symmetric matrix of odd order is always zero. So Statement-I. is true. Statement-II : This Statement is not always true depends on the order of matrix. $|-A|=-|A|$ if order is odd, so Statement–II is wrong. Statement-I is true and Statement-II is false.


Q. Let $\mathrm{A}=\left(\begin{array}{lll}{1} & {0} & {0} \\ {2} & {1} & {0} \\ {3} & {2} & {1}\end{array}\right) .$ If $\mathrm{u}_{1}$ and $\mathrm{u}_{2}$ are column matrices such that $\mathrm{Au}_{1}=\left(\begin{array}{l}{1} \\ {0} \\ {0}\end{array}\right)$ and $\mathrm{Au}_{2}=\left(\begin{array}{l}{0} \\ {1} \\ {0}\end{array}\right),$ then $\mathrm{u}_{1}+\mathrm{u}_{2}$ is equal to :

[AIEEE-2012]

Sol. (1)


Q. If $P=\left[\begin{array}{lll}{1} & {\alpha} & {3} \\ {1} & {3} & {3} \\ {2} & {4} & {4}\end{array}\right]$ is the adjoint of a $3 \times 3$ matrix $A$ and $|A|=4,$ then $\alpha$ is equal to

(1) 4                  (2) 11                      (3) 5                     (4) 0

[JEE(Main) – 2013]

Sol. (2)

P = adj (A)

taking determinant


Q. If $\mathrm{A}$ is an $3 \times 3 \times 3$ non-singular matrix such that $\mathrm{AA}^{\prime}=\mathrm{A}^{\prime} \mathrm{A}$ and $\mathrm{B}=\mathrm{A}^{-1} \mathrm{A}^{\prime},$ the BB’ equals :

(1) I + B (2) I (3) $\mathrm{B}^{-1}$ (4) $\left(B^{-1}\right)^{\prime}$

[JEE(Main) – 2014]

Sol. (2)


Q. If $A=\left[\begin{array}{ccc}{1} & {2} & {2} \\ {2} & {1} & {-2} \\ {a} & {2} & {b}\end{array}\right]$ is a matrix satisfying the equation $A A^{T}=9$, where I is $3 \times 3$ identity matrix, then the ordered pair $(\mathrm{a}, \mathrm{b})$ is equal to :

(1) (2, 1)               (2) (–2, –1)                (3) (2, –1)                (4) (–2, 1)

[JEE(Main)-2015]

Sol. (2)


Q. If $\mathrm{A}=\left[\begin{array}{cc}{5 \mathrm{a}} & {-\mathrm{b}} \\ {3} & {2}\end{array}\right]$ and $\mathrm{A}$ adj $\mathrm{A}=\mathrm{A} \mathrm{A}^{\mathrm{T}},$ then $5 \mathrm{a}+\mathrm{b}$ is equal to :

(1) 13                (2) –1                (3) 5                       (4) 4

[JEE(Main)-2016]

Sol. (3)


Q. If $A=\left[\begin{array}{cc}{2} & {-3} \\ {-4} & {1}\end{array}\right],$ then adj $\left(3 A^{2}+12 A\right)$ is equal to :-

[JEE(Main)-2017]

Sol. (3)


Q. Let $\mathrm{A}=\left[\begin{array}{lll}{1} & {0} & {0} \\ {1} & {1} & {0} \\ {1} & {1} & {1}\end{array}\right]$ and $\mathrm{B}=\mathrm{A}^{20} .$ Then the sum of the elements of the first column of $\mathrm{B}$ is :

(1) 211               (2) 251               (3) 231                 (4) 210

[JEE(Main)-2018]

Sol. (3)


Q. Let A be a matrix such that A. $\left[\begin{array}{ll}{1} & {2} \\ {0} & {3}\end{array}\right]$ is a scalar matrix and $|3 \mathrm{A}|=108 .$ Then $\mathrm{A}^{2}$ equals :

[JEE(Main)-2018]

Sol. (1)


Q. Suppose $\mathrm{A}$ is any $3 \times 3$ non-singular matrix and $(\mathrm{A}-3 \mathrm{I})(\mathrm{A}-5 \mathrm{I})=0,$ where $\mathrm{I}=\mathrm{I}_{3}$ and $\mathrm{O}$ $=\mathrm{O}_{3} .$ If $\alpha \mathrm{A}+\beta \mathrm{A}^{-1}=4 \mathrm{I},$ then $\alpha+\beta$ is equal to :

(1) 13                 (2) 7                  (3) 12                    (4) 8

[JEE(Main)-2018]

Sol. (4)


Logarithm – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Math with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral also provides complete chapter-wise notes of Class 11th and 12th both for all subjects.

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Logarithm

Maxima & Minima – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Math with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral also provides complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

 

 

Limit – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Math with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral also provides complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

Q. Let $\mathrm{L}=\lim _{x \rightarrow 0} \frac{\mathrm{a}-\sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}}-\frac{\mathrm{x}^{2}}{4}}{\mathrm{x}^{4}}, \mathrm{a}>0 .$ If $\mathrm{L}$ is finite, then $:-$

(A) a = 2

(B) a = 1

(C) $\mathrm{L}=\frac{1}{64}$

(D) $\mathrm{L}=\frac{1}{32}$

[JEE 2009, 4]

Sol. (A,C)

$\mathrm{a}-\mathrm{a}\left(1-\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}\right)^{\frac{1}{2}}-\frac{\mathrm{x}^{2}}{4} \quad \mathrm{a}-\mathrm{a}\left(1-\frac{\mathrm{x}^{2}}{2 \mathrm{a}^{2}}-\frac{1}{8} \frac{\mathrm{x}^{4}}{\mathrm{a}^{4}}\right)-\frac{\mathrm{x}^{2}}{4}$

$\mathrm{a}=2,\left(\mathrm{coefficient} \text { of } \mathrm{x}^{2}=0\right)$

$\therefore \mathrm{L}=\frac{1}{64}$


Q. If $\lim _{x \rightarrow 0}\left[1+x \ell n\left(1+b^{2}\right)\right]^{\frac{1}{x}}=2 b \sin ^{2} \theta, b>0$ and $\theta \in(-\pi, \pi],$ then the value of $\theta$ is-

[JEE 2011, 3M, –1M]

Sol. (D)


Q. If $\lim _{x \rightarrow \infty}\left(\frac{x^{2}+x+1}{x+1}-a x-b\right)=4,$ then $-$

(A) a = 1, b = 4 (B) a = 1, b = –4] (C) a = 2, b = –3 (D) a = 2, b = 3

[JEE 2012, 3M, –1M]

Sol. (B)


Q. Let $\alpha(\mathrm{a})$ and $\beta(\mathrm{a})$ be the roots of the equation $(\sqrt[3]{1+a}-1) x^{2}+(\sqrt{1+a}-1) x+(\sqrt[6]{1+a}-1)=0$ where $a>-1 .$ Then $\lim _{a \rightarrow 0^{+}} \alpha(a)$ and $\lim _{a \rightarrow 0^{+}} \beta(a)$ are

(A) $-\frac{5}{2}$ and 1

(B) $-\frac{1}{2}$ and $-1$

(C) $-\frac{7}{2}$ and 2

(D) $-\frac{9}{2}$ and 3

[JEE 2012, 3M, –1M]

Sol. (B)


Q. The largest value of the non-negative integer a for which $\lim _{x \rightarrow 1}\left\{\frac{-a x+\sin (x-1)+a}{x+\sin (x-1)-1}\right\}^{\frac{1-x}{1-\sqrt{x}}}=\frac{1}{4}$ is

[JEE(Advanced)-2014, 3]

Sol. 0


Q. Let $\alpha, \beta \in \mathrm{R}$ be such that $\lim _{x \rightarrow 0} \frac{x^{2} \sin (\beta x)}{\alpha x-\sin x}=1 .$ Then $6(\alpha+\beta)$ equals

[JEE(Advanced)-2016]

Sol. 7


Q. Let $\mathrm{f}(\mathrm{x})=\frac{1-\mathrm{x}(1+|1-\mathrm{x}|)}{|1-\mathrm{x}|} \cos \left(\frac{1}{1-\mathrm{x}}\right)$ for $\mathrm{x} \neq 1 .$ Then

[JEE(Advanced)-2017]

Sol. (A,C)


Q. For any positive integer n, define $f_{\mathrm{n}}:(0, \infty) \rightarrow \square$ as $f_{\mathrm{n}}(\mathrm{x})=\sum_{\mathrm{j}=1}^{\mathrm{n}} \tan ^{-1}\left(\frac{1}{1+(\mathrm{x}+\mathrm{j})(\mathrm{x}+\mathrm{j}-1)}\right)$ for all $\mathrm{x} \in(0, \infty)$

(Here, the inverse trigonometric function $\tan ^{-1} \mathrm{x}$ assume values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) .$ )

Then, which of the following statement(s) is (are) TRUE?

(A) $\sum_{j=1}^{5} \tan ^{2}\left(f_{j}(0)\right)=55$

(B) $\sum_{j=1}^{10}\left(1+f_{j}^{\prime}(0)\right) \sec ^{2}\left(f_{j}(0)\right)=10$

(C) For any fixed positive integer $n, \lim _{x \rightarrow \infty} \tan \left(f_{\mathrm{n}}(x)\right)=\frac{1}{n}$

(D) For any fixed positive integer $n, \operatorname{limsec}_{x \rightarrow \infty} \operatorname{ec}^{2}\left(f_{\mathrm{n}}(x)\right)=1$

[JEE(Advanced)-2018]

Sol. (D)


Q. For each positive integer $n,$ let $y_{n}=\frac{1}{n}(n+1)(n+2) \ldots(n+n)^{1 / n}$ For $x \in \square,$ let $[x]$ be the greatest integer less than or equal to $x$. If $\lim _{n \rightarrow \infty} y_{n}=L,$ then the value of $[L]$ is

[JEE(Advanced)-2018]

Sol. 1


Vector – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Math with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral also provides complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

Q. (A) If $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ are unit vectors such that $(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d})=1$ and $\vec{a} \cdot \vec{c}=\frac{1}{2},$ then $-$

(A) $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ are non-coplanar

(B) $\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}, \overrightarrow{\mathrm{d}}$ are non-coplanar

(C) $\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{d}}$ arenon-parallel

(D) $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{d}}$ are parallel and $\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ are parallel

(B) Match the statements / expressions given in Column I with the values given in Column II.

[JEE 2009, 3+8]

Sol. ((A) C (B) (A) Q,S              (B) P,R,S,T             (C) T             (D) R )

(A) Let angle between $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ be $\theta_{1}, \overrightarrow{\mathrm{c}}$ and $\overrightarrow{\mathrm{d}}$ be $\theta_{2}$ and $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}}$ be $\theta$

since $(\vec{a} \times \vec{b}) .(\vec{c} \times \vec{d})=1$

$\sin \theta_{1} \sin \theta_{2} \cos \theta=1$

$\theta_{1}=90, \theta_{2}=90, \theta=0$

$\overrightarrow{\mathrm{a}} \perp \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}} \perp \overrightarrow{\mathrm{d}},(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \|(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}})$

so $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\mathrm{k}(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}})$

and $\quad(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{c}}=\mathrm{k}(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}}) \cdot \overrightarrow{\mathrm{c}}$

Case I $(\vec{a} \times \vec{b}) \cdot \vec{d}=k(\vec{c} \times \vec{d}) . \vec{d} \quad \because[\vec{a} \vec{b} \vec{c}]=0$

Case II $\overrightarrow{\mathrm{b}} \cdot(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})=\mathrm{k} \overrightarrow{\mathrm{b}} \cdot(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}}) \therefore[\overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}}]=0$

(A) and (B) are in correct

let $\overrightarrow{\mathrm{b}} \| \overrightarrow{\mathrm{d}} \Rightarrow \overrightarrow{\mathrm{b}}=\pm \overrightarrow{\mathrm{d}}$

As $(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \cdot(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}})=1$

$\Rightarrow(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{b})=\pm 1$

$[\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}} \overrightarrow{\mathrm{b}}]=\pm 1$

$[\overrightarrow{\mathrm{c}} \quad \overrightarrow{\mathrm{b}} \quad \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}]=\pm 1$

$\overrightarrow{\mathrm{c}} \cdot[\overrightarrow{\mathrm{b}} \times(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})]=\pm 1$

$\overrightarrow{\mathrm{c}} \cdot[\overrightarrow{\mathrm{a}}-(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}) \overrightarrow{\mathrm{b}}]=\pm 1$

$\overrightarrow{\mathrm{c} .} \overrightarrow{\mathrm{a}}=\pm 1 \quad \because \quad \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0$

Which is contradiction so option (c) is correct let option (d) is correct

$\overrightarrow{\mathrm{d}}=\pm \overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{c}}=\pm \overrightarrow{\mathrm{b}}$

$(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d})=1$

$(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{a})=\pm 1$

so (d) incorrect

(B) $(\mathrm{A}) 2 \sin ^{2} \theta+\sin ^{2} 2 \theta=2$

$\sin ^{2} 2 \theta=2 \cos ^{2} \theta$

$4 \sin ^{2} \theta \cos ^{2} \theta=2 \cos ^{2} \theta$

$\cos ^{2} \theta=0$ or $\sin ^{2} \theta=\frac{1}{2}$

$\theta=\frac{\pi}{2}$ or $\sin 0=\pm \frac{1}{\sqrt{2}}$

$\theta=\frac{\pi}{2}, \pm \frac{\pi}{4}$

(B) $\mathrm{f}(\mathrm{x})=\left[\frac{6 \mathrm{x}}{\pi}\right] \cos \left[\frac{3 \mathrm{x}}{\pi}\right]$

possible point of discontinuty of $\left[\frac{6 \mathrm{x}}{\pi}\right]$

$\frac{6 \mathrm{x}}{\pi}=\mathrm{n}, \mathrm{n} \in \mathrm{I}$

$x=\frac{n \pi}{6}$

$\Rightarrow \mathrm{x}=\frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}, \pi$

Lt $f(x)=1 \cos 0=1$

$\mathrm{x} \rightarrow \frac{\pi^{+}}{6}$

$\mathrm{Lt}-\mathrm{f}(\mathrm{x})=0 \cos 0=0$

$\mathrm{X} \rightarrow \frac{\pi^{-}}{6}$

disconti at $x=\frac{\pi}{6}$

similarly $x=\frac{\pi}{3}, \frac{\pi}{2}, \pi$

(C) Here $\mathrm{v}=\left|\begin{array}{ccc}{1} & {1} & {0} \\ {1} & {2} & {0} \\ {1} & {1} & {\pi}\end{array}\right|=\pi \mathrm{cm}^{3}$

(d) Given $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\sqrt{3} \overrightarrow{\mathrm{c}}=0$

$\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=-\sqrt{3} \overrightarrow{\mathrm{c}}$

$|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^{2}=|\sqrt{3} \overrightarrow{\mathrm{c}}|^{2}$

$a^{2}+b^{2}+2 \vec{a} \cdot \vec{b}=3 c^{2}$

$2+\cos \theta=3$

$\cos \theta=\frac{1}{2}$

$\theta=\frac{\pi}{3}$


Q. (A) Two adjacent sides of a parallelogram ABCD are given by $\overrightarrow{\mathrm{AB}}=2 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+11 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{AD}}=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} .$ The side $\mathrm{AD}$ is rotated by an acute angle $\alpha$ in the plane of the parallelogram so that AD becomes AD’. If AD’ makes a right angle with the side AB, then the cosine of the angle $\alpha$ is given by –

(A) $\frac{8}{9}$

(B) $\frac{\sqrt{17}}{9}$

(C) $\frac{1}{9}$

(D) $\frac{4 \sqrt{5}}{9}$

(B) If $\vec{a}$ and $\vec{b}$ are vectors in space given by $\vec{a}=\frac{\hat{i}-2 \hat{j}}{\sqrt{5}}$ and $\vec{b}=\frac{2 \hat{i}+\hat{j}+3 \hat{k}}{\sqrt{14}},$ then the value of $(2 \vec{a}+\vec{b}) \cdot[(\vec{a} \times \vec{b}) \times(\vec{a}-2 \vec{b})]$ is

[JEE 2010, 5+3]

Sol. ( (A) B (B) 5 )


Q. (A) Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}-\hat{k}$ be three vectors. A vector $\vec{v}$ in the plane of $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}},$ whose projection on $\overrightarrow{\mathrm{c}}$ is $\frac{1}{\sqrt{3}},$ is given by

(A) $\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$

(B) $-3 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-\hat{\mathrm{k}}$

(C) $3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$

(D) $\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$

(B) The vector(s) which is/are coplanar with vectors $\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ and $\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}},$ and perpendicular to the vector $\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ is/are

(A) $\hat{\mathrm{j}}-\hat{\mathrm{k}}$

(B) $-\hat{\hat{\mathrm{i}}}+\hat{\hat{\mathrm{j}}}$

(C) $\hat{\mathrm{i}}-\hat{\mathrm{j}}$

$(D)-\hat{j}+\hat{k}$

(C) Let $\vec{a}=-\hat{i}-\hat{k}, \vec{b}=-\hat{i}+\hat{j}$ and $\vec{c}=\hat{i}+2 \hat{j}+3 \hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{r} .} \overrightarrow{\mathrm{a}}=0,$ then the value of $\overrightarrow{\mathrm{r}} . \overrightarrow{\mathrm{b}}$ is

[JEE 2011, 3+4+4]

Sol. ( (A) C (B) A,D (C) 9 )

$\overrightarrow{\mathrm{v}}=\mathrm{x} \overrightarrow{\mathrm{a}}+\mathrm{y} \overrightarrow{\mathrm{b}}$

$=\hat{\mathbf{i}}(\mathrm{x}+\mathrm{y})+\hat{\mathrm{j}}(\mathrm{x}-\mathrm{y})+\hat{\mathrm{k}}(\mathrm{x}+\mathrm{y}) \ldots .(\mathrm{i})$

Given $, \overrightarrow{\mathrm{v}} . \hat{\mathrm{c}}=\frac{1}{\sqrt{3}}$

$\Rightarrow \frac{x+y-x+y-x-y}{\sqrt{3}}=\frac{1}{\sqrt{3}}$

$\mathrm{y}-\mathrm{x}=1$

$\Rightarrow \mathrm{x}-\mathrm{y}=-1$

using (ii) in (i) we get $\overrightarrow{\mathrm{v}}=(\mathrm{x}+\mathrm{y}) \hat{\mathrm{i}}-\hat{\mathrm{j}}+(\mathrm{x}+\mathrm{y}) \hat{\mathrm{k}}$

(b) $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$

$\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$

$\overrightarrow{\mathrm{c}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$

$\overrightarrow{\mathrm{v}}=\lambda((\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}})=\lambda((\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}) \overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}) \overrightarrow{\mathrm{a}}$

$\overrightarrow{\mathrm{v}}=\lambda[4(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})-4(\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}})]$

$\overrightarrow{\mathrm{v}}=4 \lambda(\hat{j}-\hat{\mathrm{k}})$

(c) $\overrightarrow{\mathrm{a}}=-\hat{\mathrm{i}}-\hat{\mathrm{k}}$

$\overrightarrow{\mathrm{b}}=-\hat{\mathrm{i}}+\hat{\mathrm{j}}$

$\overrightarrow{\mathrm{c}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}}$

Taking cross product by $\overrightarrow{\mathrm{a}}$

$(\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{a}}=(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{a}}$

$\Rightarrow(\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}) \overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}) \overrightarrow{\mathrm{r}}=(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}) \overrightarrow{\mathrm{b}})-(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}) \overrightarrow{\mathrm{c}}$

$\Rightarrow 0-\overrightarrow{\mathrm{r}}=(-1-3)(-\hat{\mathrm{i}}+\hat{\mathrm{j}})-(1)(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$

$\overrightarrow{\mathrm{r}}=-3 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{b}}=3+6=9$


Q. (A) If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors satisfying $|\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9,$ then

\[|2 \overrightarrow{\mathrm{a}}+5 \overrightarrow{\mathrm{b}}+5 \overrightarrow{\mathrm{c}}| \text { is }\]

(B) If $\vec{a}$ and $\vec{b}$ are vectors such that $|\vec{a}+\vec{b}|=\sqrt{29}$ and $\overrightarrow{\mathrm{a}} \times(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \times \overrightarrow{\mathrm{b}},$ then a possible value of $(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \cdot(-7 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$ is

(A) 0 (B) 3 (C) 4 (D) 8

[JEE 2012, 4+3]

Sol. ( (A) 3 (B) C)

$|\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9$

$\Rightarrow 6-2 \Sigma \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=9$

$\Rightarrow \Sigma \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=-\frac{3}{2}$

$|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}|^{2} \geq 0$

$\Sigma \overrightarrow{\mathrm{a}}^{2}+2 \Sigma \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}} \geq 0$

$\sum \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}} \geq-\frac{3}{2}$

for equality $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}|=0$

$\Rightarrow \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=0$

$5 \overrightarrow{\mathrm{b}}+5 \overrightarrow{\mathrm{c}}=-5 \overrightarrow{\mathrm{a}}$

$2 \vec{a}+5 \vec{b}+5 \vec{c}=-3 \vec{a}$

$|2 \vec{a}+5 \vec{b}+5 \vec{c}|=3|\vec{a}|=3$

(b) $(\vec{a}+\vec{b}) \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=0$

$\Rightarrow \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})$

$|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=\sqrt{29} \Rightarrow|\lambda|=1$

$\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})$

$(\vec{a}+\vec{b}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})=-14+6+12=4$


Q. Let $\overrightarrow{\mathrm{PR}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$ and $\overline{\mathrm{SQ}}=\hat{\mathrm{i}}-3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}$ determine diagonals of a parallelogram PQRS and $\overrightarrow{\mathrm{PT}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ be another vector. Then the volume of the parallelepiped determined by the vectors $\overrightarrow{\mathrm{PT}}, \overrightarrow{\mathrm{PQ}}$ and $\overrightarrow{\mathrm{PS}}$ is

(A) 5          (B) 20          (C) 10               (D) 30

[JEE-Advanced 2013, 2M]

Sol. (C)

$\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{PR}} \& \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{QS}}$

$\overrightarrow{\mathrm{a}}=\frac{\overrightarrow{\mathrm{PR}}+\overrightarrow{\mathrm{QS}}}{2} \& \overrightarrow{\mathrm{b}}=\frac{\overrightarrow{\mathrm{PR}}-\overrightarrow{\mathrm{QS}}}{2}$

$\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-3 \hat{\mathrm{k}} \& \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$

Volume $=\left|\begin{array}{ccc}{2} & {-1} & {-3} \\ {1} & {2} & {1} \\ {1} & {2} & {3}\end{array}\right|$

$2(4)+(3-1)-3(2-2)$

$8+2=10$


Q. Consider the set of eight vectors $\mathrm{V}=\{\mathrm{a} \hat{\mathrm{i}}+\mathrm{b} \hat{\mathrm{j}}+\mathrm{c} \hat{\mathrm{k}}: \mathrm{a}, \mathrm{b}, \mathrm{c} \in\{-1,1\}\} .$ Three non- coplanar vectors can be chosen from $\mathrm{V}$ in $2^{\mathrm{p}}$ ways. Then $\mathrm{p}$ is

[JEE-Advanced 2013, 4, (–1)]

Sol. 5

The 8 vectors will represent $\mathrm{O}$ is at the centre of cube $\overrightarrow{\mathrm{OA}}, \overrightarrow{\mathrm{OB}} \ldots \ldots \overrightarrow{\mathrm{OD}}, \overrightarrow{\mathrm{OP}}, \ldots \ldots . \overrightarrow{\mathrm{OS}}$

ABCDPQRS any three out of these 8 will be coplanar when two of them are

There are 4 pairs of collinear vectors $\overrightarrow{\mathrm{OA}} \& \overrightarrow{\mathrm{OR}}, \overrightarrow{\mathrm{OB}} \& \overrightarrow{\mathrm{OS}}, \overrightarrow{\mathrm{OC}} \& \overrightarrow{\mathrm{OP}}, \overrightarrow{\mathrm{OD}} \& \overrightarrow{\mathrm{OQ}}$

(it will generate $4 \times 6=24$ set of coplanar vectors) rest of the combination of 3 vectors will form three edges of a tetrahedron so they will be not coplanar. So number of non-coplanar vectors $^{8} \mathrm{C}_{3}-4.6=32$


Q. Match List-I with List-II and select the correct answer using the code given

below the lists.

[JEE-Advanced 2013, 3, (–1)]

Sol. (C)

(P) Given $[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]=2$

$[2(\vec{a} \times \vec{b}) 3(\vec{b} \times \vec{c})(\vec{c} \times \vec{a})]=6[\vec{a} \times \vec{b} \quad \vec{b} \times \vec{c} \vec{c} \times \vec{a}]$

$=6[\vec{a} \vec{b} \vec{c}]^{2}=24$

(Q) Given $[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]=5$

$[3(\vec{a}+\vec{b})(\vec{b}+\vec{c}) 2(\vec{c}+\vec{a})]=12[\vec{a} \vec{b} \vec{c}]=60$

(R) Given $\frac{1}{2}|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=20$

$\Rightarrow|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=40$

$\left|\frac{1}{2}(2 \vec{a}+3 \vec{b}) \times(\vec{a}-\vec{b})\right|=\frac{1}{2}|0+3 \vec{b} \times \vec{a}-2 \vec{a} \times \vec{b}|$

$=\frac{1}{2}|-5 \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=\frac{5}{2}|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=\frac{5}{2} \cdot 40=100$

(S) Given $|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=30$

$|(\vec{a}+\vec{b}) \times \vec{a}|=|0+\vec{b} \times \vec{a}|=30$


Q. Let $\overrightarrow{\mathrm{x}}, \overrightarrow{\mathrm{y}}$ and $\overrightarrow{\mathrm{z}}$ be three vectors each of magnitude $\sqrt{2}$ and the angle between each pair of them is $\frac{\pi}{3} .$ If $\overrightarrow{\mathrm{a}}$ is a nonzero vector perpendicular to $\overrightarrow{\mathrm{x}}$ and $\overrightarrow{\mathrm{y}} \times \overrightarrow{\mathrm{z}}$ and $\overrightarrow{\mathrm{b}}$ is nonzero vector perpendicular to $\overrightarrow{\mathrm{y}}$ and $\overrightarrow{\mathrm{z}} \times \overrightarrow{\mathrm{x}},$ then

(A) $\overrightarrow{\mathrm{b}}=(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{z}})(\overrightarrow{\mathrm{z}}-\overrightarrow{\mathrm{x}})$

(B) $\overrightarrow{\mathrm{a}}=(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{y}})(\overrightarrow{\mathrm{y}}-\overrightarrow{\mathrm{z}})$

(C) $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=-(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{y}})(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{z}})$

$(D) \quad \vec{a}=(\vec{a} \cdot \vec{y})(\vec{z}-\vec{y})$

[JEE(Advanced)-2014, 3]

Sol. (A,B,C)

Given that $|\overrightarrow{\mathrm{x}}|=|\overrightarrow{\mathrm{y}}|=|\overrightarrow{\mathrm{z}}|=\sqrt{2}$

and angle between each pair is $\frac{\pi}{3}$

$\therefore \overrightarrow{\mathrm{x}} \cdot \overrightarrow{\mathrm{y}}=\overrightarrow{\mathrm{y} .} \cdot \overrightarrow{\mathrm{z}}=\overrightarrow{\mathrm{z}} \cdot \overrightarrow{\mathrm{x}}=\sqrt{2} \cdot \sqrt{2} \cdot \frac{1}{2}=1$

Now $\vec{a}$ is $\perp$ to $\vec{x} \&(\vec{y} \times \vec{z})$

Let $\overrightarrow{\mathrm{a}}=\lambda(\overrightarrow{\mathrm{x}} \times(\overrightarrow{\mathrm{y}} \times \overrightarrow{\mathrm{z}}))$

$=\lambda((\overrightarrow{\mathrm{x}} \cdot \overrightarrow{\mathrm{z}}) \overrightarrow{\mathrm{y}}-(\overrightarrow{\mathrm{x}} \cdot \overrightarrow{\mathrm{y}}) \overrightarrow{\mathrm{z}})=\lambda(\overrightarrow{\mathrm{y}}-\overrightarrow{\mathrm{z}})$

$\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{y}}=\lambda(\overrightarrow{\mathrm{y}} \cdot \overrightarrow{\mathrm{y}}-\overrightarrow{\mathrm{y}} \cdot \overrightarrow{\mathrm{z}})=\lambda(2-1)=\lambda$

$\Rightarrow \overrightarrow{\mathrm{a}}=(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{y}})(\overrightarrow{\mathrm{y}}-\overrightarrow{\mathrm{z}})$

Now let $\overrightarrow{\mathrm{b}}=\mu(\overrightarrow{\mathrm{y}} \times(\overrightarrow{\mathrm{z}} \times \overrightarrow{\mathrm{x}}))=\mu(\overrightarrow{\mathrm{z}}-\overrightarrow{\mathrm{x}})$

$\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{z}}=\mu(2-1)=\mu$

$\Rightarrow \overrightarrow{\mathrm{b}}=(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{z}})(\overrightarrow{\mathrm{z}}-\overrightarrow{\mathrm{x}})$

Now $\vec{a} \cdot \vec{b}=(\vec{a} \cdot \vec{y})(\vec{y}-\vec{z}) \cdot(\vec{b} \cdot \vec{z})(\vec{z}-\vec{x})$

$=(\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z})(\vec{y} \cdot \vec{z}-\vec{y} \cdot \vec{x}-\vec{z} \cdot \vec{z}+\vec{z} \cdot \vec{x})$

$=(\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z})(1-1-2+1)$

$=-(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{y}})(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{z}})$


Q. Let $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}},$ and $\overrightarrow{\mathrm{c}}$ be three non-coplanar unit vectors such the angle between every pair of them is $\frac{\pi}{3} \cdot$ If $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\mathrm{p} \overrightarrow{\mathrm{a}}+\mathrm{qb}+\mathrm{r} \overrightarrow{\mathrm{c}},$ where $\mathrm{p}, \mathrm{q}$ and $\mathrm{r}$ are scalars, then the value of

$\frac{\mathrm{p}^{2}+2 \mathrm{q}^{2}+\mathrm{r}^{2}}{\mathrm{q}^{2}}$ is

[JEE(Advanced)-2014, 3]

Sol. 4


Q. Let $\Delta \mathrm{PQR}$ be a triangle. Let $\overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{QR}}, \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{RP}}$ and $\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{PQ}} .$ If $|\overrightarrow{\mathrm{a}}|=4 \sqrt{3}$ and $\overrightarrow{\mathrm{b} . \overrightarrow{\mathrm{c}}}=24$ then which of the following is (are) true?

(A) $\frac{|\overrightarrow{\mathrm{c}}|^{2}}{2}-|\overrightarrow{\mathrm{a}}|=12$

(B) $\frac{|\overrightarrow{\mathrm{c}}|^{2}}{2}+|\overrightarrow{\mathrm{a}}|=30$

(C) $|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}|=48 \sqrt{3}$

(D) $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=-72$

[JEE 2015, 4M, –2M]

Sol. (A,C,D)


Q. Suppose that $\overrightarrow{\mathrm{p}}, \overrightarrow{\mathrm{q}}$ and $\overrightarrow{\mathrm{r}}$ are three non-coplanar vectors in $\mathrm{U}^{3} .$ Let the components of a vector $\overrightarrow{\mathrm{s}}$ along $\overrightarrow{\mathrm{p}}, \overrightarrow{\mathrm{q}}$ and $\overrightarrow{\mathrm{r}}$ be $4,3$ and $5,$ respectively. If the components of this vector $\overrightarrow{\mathrm{s}}$ along $(-\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}),(\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}})$ and $(-\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}})$ are $\mathrm{x}, \mathrm{y}$ and $\mathrm{z},$ respectively, then the value of $2 \mathrm{x}+\mathrm{y}+\mathrm{z}$ is

[JEE 2015, 4M, –0M]

Sol. (Bonus)

Although the language of the question is not appropriate (in complete information) and it must be declare as bonus but as per the theme of problem it must be as follows

$\overrightarrow{\mathrm{s}}=4 \overrightarrow{\mathrm{p}}+3 \overrightarrow{\mathrm{q}}+5 \overrightarrow{\mathrm{r}}$

$\overrightarrow{\mathrm{s}}=\mathrm{x}(-\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}})+\mathrm{y}(\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}})+\mathrm{z}(-\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}})$

$\overrightarrow{\mathrm{s}}=\overrightarrow{\mathrm{p}}(-\mathrm{x}+\mathrm{y}-\mathrm{z})+\overrightarrow{\mathrm{q}}(\mathrm{x}-\mathrm{y}-\mathrm{z})+\overrightarrow{\mathrm{r}}(\mathrm{x}+\mathrm{y}+\mathrm{z})$

$-x+y-z=4$

$x-y-z=3$

$x+y+z=5$

$\Rightarrow \mathrm{x}=4, \mathrm{y}=\frac{9}{2}, \mathrm{z}=-\frac{7}{2}$

$\Rightarrow 2 \mathrm{x}+\mathrm{y}+\mathrm{z}=8-\frac{7}{2}+\frac{9}{2}=9$


Q. Let $\hat{\mathbf{u}}=\mathbf{u}_{1} \hat{\mathbf{i}}+\mathbf{u}_{2} \hat{\mathbf{j}}+\mathbf{u}_{3} \hat{\mathbf{k}}$ be a unit vector in $\mathbb{U}^{2}$ and $\hat{\mathbf{w}}=\frac{1}{\sqrt{6}}(\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) .$ Given that there exists a vector $\overrightarrow{\mathrm{v}}$ in $\square^{3}$ such that $|\hat{\mathrm{u}} \times \overrightarrow{\mathrm{v}}|=1$ and $\hat{\mathrm{w}} .(\hat{\mathrm{u}} \times \overrightarrow{\mathrm{v}})=1 .$ Which of the following statement(s) is(are) correct?

(A) There is exactly one choice for such $\overrightarrow{\mathrm{v}}$

(B) There are infinitely many choice for such $\overrightarrow{\mathrm{v}}$

(C) If \hat{u lies in the xy-plane then } $\left|\mathrm{u}_{1}\right|=\left|\mathrm{u}_{2}\right|$

(D) If ú lies in the xz-plane then $2\left|u_{1}\right|=\left|u_{3}\right|$

[JEE(Advanced) 2016]

Sol. (B,C)


Q. Let O be the origin and let PQR be an arbitrary triangle. The point $\mathrm{S}$ is such that $\overrightarrow{\mathrm{OP}} \cdot \overrightarrow{\mathrm{OP}}+\overrightarrow{\mathrm{OR}} \cdot \overrightarrow{\mathrm{OP}}=\overrightarrow{\mathrm{OR}} \cdot \overrightarrow{\mathrm{OP}}+\overrightarrow{\mathrm{OQ}} \cdot \overrightarrow{\mathrm{OS}}=\overrightarrow{\mathrm{OQ}} \cdot \overrightarrow{\mathrm{OR}}+\overrightarrow{\mathrm{OP}} \cdot \overrightarrow{\mathrm{OS}}$ Then the triangle PQR has S as its

(A) incentre

(B) orthocenter

(C) circumcentre

(D) centroid

[JEE(Advanced) 2017]

Sol. (B)

Let position vector of $\mathrm{P}(\overrightarrow{\mathrm{p}}), \mathrm{Q}(\overrightarrow{\mathrm{q}}), \mathrm{R}(\overrightarrow{\mathrm{r}}) \& \mathrm{S}(\overrightarrow{\mathrm{r}})$ with respect to $\mathrm{O}(\overrightarrow{\mathrm{o}})$

Now, $\overrightarrow{\mathrm{OP}} \cdot \overrightarrow{\mathrm{OQ}}+\overrightarrow{\mathrm{OR}} \cdot \overrightarrow{\mathrm{OS}}=\overrightarrow{\mathrm{OR}} \cdot \overrightarrow{\mathrm{OP}}+\overrightarrow{\mathrm{OQ}} \cdot \overrightarrow{\mathrm{OS}}$

$\Rightarrow \overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{s}}=\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}} \cdot \overrightarrow{\mathrm{s}}$

$\Rightarrow(\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{s}}) \cdot(\overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{r}})=0$ …..(1)

Also, $\overrightarrow{\mathrm{OR}} \cdot \overrightarrow{\mathrm{OP}}+\overrightarrow{\mathrm{OQ}} \cdot \overrightarrow{\mathrm{OS}}=\overrightarrow{\mathrm{OQ}} \cdot \overrightarrow{\mathrm{OR}}+\overrightarrow{\mathrm{OP}} \cdot \overrightarrow{\mathrm{OS}}$

$\Rightarrow \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}} \cdot \overrightarrow{\mathrm{s}}=\overrightarrow{\mathrm{q}} \cdot \overrightarrow{\mathrm{r}}+\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{s}}$

\[ \Rightarrow(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{s}}) \cdot(\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{q}})=0 \] ……(2)


Let O be the origin, and $\overrightarrow{\mathrm{OX}}, \overrightarrow{\mathrm{OY}}, \overrightarrow{\mathrm{OZ}}$ be three unit vectors in the directions of the sides $\overrightarrow{\mathrm{QR}}, \overrightarrow{\mathrm{RP}}, \overrightarrow{\mathrm{PQ}},$ respectively, of a triangle $\mathrm{PQR}$

Q. $|\overrightarrow{\mathrm{OX}} \times \overrightarrow{\mathrm{OY}}|=$

(A) sin(Q + R)

(B) sin(P + R)

(C) sin 2R

(D) sin(P + Q)

[JEE(Advanced) 2017]

Sol. (D)

$\overrightarrow{\mathrm{OX}}=\frac{\overrightarrow{\mathrm{QR}}}{\mathrm{QR}}$

$\overrightarrow{\mathrm{OY}}=\frac{\overrightarrow{\mathrm{RP}}}{\mathrm{RP}}$

$|\overrightarrow{\mathrm{OX}} \times \overrightarrow{\mathrm{OY}}|=\sin \mathrm{R}=\sin (\mathrm{P}+\mathrm{Q})$


Q. If the triangle PQR varies, then the minimum value of cos(P + Q) + cos(Q + R) + cos(R + P) is

[JEE(Advanced) 2017]

Sol. (B)

$-(\cos P+\cos Q+\cos R) \geq-\frac{3}{2}$ as we know $\cos P+\cos Q+$ cosR will take

its maximum value when $\mathrm{P}=\mathrm{Q}=\mathrm{R}=\frac{\pi}{3}$


Q. Let $\vec{a}$ and $\vec{b}$ be two unit vectors such that $\vec{a} . \vec{b}=0 .$ For some $\mathbf{x}, y \in \square,$ let $\vec{c}=x \vec{a}+y \vec{b}+(\vec{a} \times \vec{b})$ If $|\overrightarrow{\mathrm{c}}|=2$ and the vector $\overrightarrow{\mathrm{c}}$ is inclined at the same angle $\alpha$ to both $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}},$ then the value of $8 \cos ^{2} \alpha$ is

[JEE(Advanced) 2018]

Sol. 3

$\overrightarrow{\mathrm{c}}=\mathrm{x} \overrightarrow{\mathrm{a}}+\mathrm{y} \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$

$\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=\mathrm{x}$ and $\mathrm{x}=2 \cos \alpha$

$\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{b}}=\mathrm{y}$ and $\mathrm{y}=2 \cos \alpha$

Also, $|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=1$

$\begin{aligned} \therefore \quad & \overrightarrow{\mathrm{c}}=2 \cos (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} \\ \overrightarrow{\mathrm{c}}^{2} &=4 \cos ^{2} \alpha(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})^{2}+(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})^{2}+2 \cos \alpha(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \cdot(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \\ & 4=8 \cos ^{2} \alpha+1 \\ & 8 \cos ^{2} \alpha=3 \end{aligned}$


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Q. The number of values of $\theta$ in the interval $\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$ such that $\theta \neq \frac{n \pi}{5}$ for $n=0, \pm 1, \pm 2$ and $\tan \theta=\cot 5 \theta$ as well as $\sin 2 \theta=\cos 4 \theta,$ is

[JEE 2010, 3]

Sol. 3


Q. The positive integer value of n > 3 satisfying the equation

$\frac{1}{\sin \left(\frac{\pi}{\mathrm{n}}\right)}=\frac{1}{\sin \left(\frac{2 \pi}{\mathrm{n}}\right)}+\frac{1}{\sin \left(\frac{3 \pi}{\mathrm{n}}\right)}$ is

[JEE 2011, 4]

Sol. 7

$\frac{1}{\sin \frac{\pi}{\mathrm{n}}}=\frac{1}{\sin \frac{2 \pi}{\mathrm{n}}}+\frac{1}{\sin \frac{3 \pi}{\mathrm{n}}}$

$\Rightarrow \frac{1}{\sin \frac{\pi}{\mathrm{n}}}-\frac{1}{\sin \frac{3 \pi}{\mathrm{n}}}=\frac{1}{\sin \frac{2 \pi}{\mathrm{n}}}$

$\Rightarrow \frac{\sin \frac{3 \pi}{n}-\sin \frac{\pi}{n}}{\sin \frac{\pi}{n} \sin \frac{3 \pi}{n}}=\frac{1}{\sin \frac{2 \pi}{n}}$

$\Rightarrow \frac{2 \cos \frac{2 \pi}{\mathrm{n}} \sin \frac{\pi}{\mathrm{n}}}{\sin \frac{\pi}{\mathrm{n}} \sin \frac{3 \pi}{\mathrm{n}}}=\frac{1}{\sin \frac{2 \pi}{\mathrm{n}}}$

$\Rightarrow 2 \cos \frac{2 \pi}{\mathrm{n}} \sin \frac{2 \pi}{\mathrm{n}}=\sin \frac{3 \pi}{\mathrm{n}}$

$\Rightarrow \sin \frac{4 \pi}{\mathrm{n}}=\sin \frac{3 \pi}{\mathrm{n}} \Rightarrow \frac{4 \pi}{\mathrm{n}}=\mathrm{K} \pi+(-1)^{\mathrm{K}} \frac{3 \pi}{\mathrm{n}}$

If $\mathrm{K}=2 \mathrm{m} \quad \Rightarrow \quad \frac{\pi}{\mathrm{n}}=2 \mathrm{m} \pi$

$\Rightarrow \quad n=\frac{1}{2 m} \quad \Rightarrow n=\frac{1}{2}, \frac{1}{4}, \frac{1}{6} \ldots \ldots$

If $\mathrm{K}=2 \mathrm{m}+1 \Rightarrow \frac{7 \pi}{\mathrm{n}}=(2 \mathrm{m}+1) \pi$

$\Rightarrow \mathrm{n}=\frac{7}{2 \mathrm{m}+1} \quad \Rightarrow \quad \mathrm{n}=7, \frac{7}{3}, \frac{7}{5} \ldots \ldots$

Possible value of n is 7


Q. Let $\theta, \varphi \in[0,2 \pi]$ be such that $2 \cos \theta(1-\sin \varphi)=\sin ^{2} \theta\left(\tan \frac{\theta}{2}+\cot \frac{\theta}{2}\right) \cos \varphi-1, \tan (2 \pi-\theta)>0$

and $-1<\sin \theta<-\frac{\sqrt{3}}{2} .$ Then $\varphi$ cannot satisfy-

(A) $0<\varphi<\frac{\pi}{2}$

(B) $\frac{\pi}{2}<\varphi<\frac{4 \pi}{3}$

(C) $\frac{4 \pi}{3}<\varphi<\frac{3 \pi}{2}$

(D) $\frac{3 \pi}{2}<\varphi<2 \pi$

[JEE 2012, 4M]

Sol. (A,C,D)


Q. For $\mathrm{x} \in(0, \pi),$ the equation $\sin \mathrm{x}+2 \sin 2 \mathrm{x}-\sin 3 \mathrm{x}=3 \mathrm{has}$

(A) infinitely many solutions

(B) three solutions

(C) one solution

(D) no solution

[JEE(Advanced)-2014, 3(–1)]

Sol. (D)


Q. The number of distinct solutions of equation $\frac{5}{4} \cos ^{2} 2 x+\cos ^{4} x+\sin ^{4} x+\cos ^{6} x+\sin ^{6} x=2$ in the interval $[0,2 \pi]$ is

[JEE 2015, 4M, –0M]

Sol. 8


Q. Let $S=\left\{x \in(-\pi, \pi): x \neq 0, \pm \frac{\pi}{2}\right\} .$ The sum of all distinct solution of the equation $\sqrt{3} \sec x+\csc x+2(\tan x-\cot x)=0$ in the set $S$ is equal to $-$

(A) $-\frac{7 \pi}{9}$

(B) $-\frac{2 \pi}{9}$

(C) 0

(D) $\frac{5 \pi}{9}$

[JEE(Advanced)-2016]

Sol. (C)

$\sqrt{3} \sin x+\cos x=2 \cos 2 x$

$\Rightarrow \cos 2 x=\cos \left(x-\frac{\pi}{3}\right)$

$\Rightarrow 2 x=2 n \pi \pm\left(x-\frac{\pi}{3}\right)$

$\quad \quad x=(6 n-1) \frac{\pi}{3}$ or $(6 n+1) \frac{\pi}{9}$

$\Rightarrow x=-\frac{\pi}{3}, \frac{\pi}{9}, \frac{7 \pi}{9}$ and $-\frac{5 \pi}{9}$ in $(-\pi, \pi)$

$\Rightarrow \operatorname{sum}=0$


Q. Let $a, b, c$ be three non-zero real numbers such that the equation $\sqrt{3} a \cos x+2 b \sin x=c, \quad x$ $\in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ has two distinct real roots $\alpha$ and $\beta$ with $\alpha+\beta=\frac{\pi}{3} .$ Then the value of $\frac{b}{a}$ is $-$

$=$

[JEE(Advanced)-2018]

Sol. 0.5


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Q. Let $\mathrm{P}, \mathrm{Q}, \mathrm{R}$ and $\mathrm{S}$ be the points on the plane with position vectors $-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}, 4 \hat{\mathrm{i}}, 3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}$ and $-3 \hat{\mathrm{j}}$ and $-3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}$ respectively. The quadrilateral PQRS must be a

(A) parallelogram, which is neither a rhombus nor a rectangle

(B) square

(C) rectangle, but not a square

(D) rhombus, but not a square

[JEE 2010, 3]

Sol. (A)

$\Rightarrow$ PQRS is a parallelogram but neither a rhombus nor a rectangle.


Q. A straight line L through the point $(3,-2)$ is inclined at an angle $60^{\circ}$ to the line $\sqrt{3} x+y=1$. If $L$ also intersect the x-axis, then the equation of $L$ is

(A) $y+\sqrt{3} x+2-3 \sqrt{3}=0$

(B) $\mathrm{y}-\sqrt{3} \mathrm{x}+2+3 \sqrt{3}=0$

(C) $\sqrt{3} y-x+3+2 \sqrt{3}=0$

(D) $\sqrt{3} y+x-3+2 \sqrt{3}=0$

[JEE 2011, 3 (–1)]

Sol. (B)


Q. For a > b > c > 0, the distance between (1, 1) and the point of intersection of the lines $a x+b y+c=0$ and $b x+a y+c=0$ is less than $2 \sqrt{2} .$ Then

(A) a + b – c > 0

(B) a – b + c < 0

(C) a – b + c > 0

(D) a + b – c < 0

[JEE-Advanced 2013, 2]

Sol. (A or C or A,C)

Point of intersection of both lines is $\left(-\frac{c}{(a+b)},-\frac{c}{(a+b)}\right)$

Distance between $\left(-\frac{c}{(a+b)},-\frac{c}{(a+b)}\right) \&(1,1)$ is

Distance $=\sqrt{\frac{(a+b+c)^{2}}{(a+b)^{2}} \times 2}<2 \sqrt{2}$

$a+b+c<2(a+b)$

$a+b-c>0$

According to given condition option (C) also correct.


Q. For a point $P$ in the plane, let $d_{1}(P)$ and $d_{2}(P)$ be the distances of the point $P$ from the lines $x-y=0$ and $x+y=0$ respectively. The area of the region $R$ consisting of all points $P$ lying in the first quadrant of the plane and satisfying $2 \leq d_{1}(P)+d_{2}(P) \leq 4,$ is

[JEE(Advanced)-2014, 3]

Sol. 6