eSaral - JEE | NEET | Class 8-10 Exam Prep App

If

Question: If $T_{n}=\sin ^{n} x+\cos ^{n} x$, prove that (i) $\frac{T_{3}-T_{5}}{T_{1}}=\frac{T_{5}-T_{7}}{T_{3}}$ (ii) $2 T_{6}-3 T_{4}+1=0$ (iii) $6 T_{10}-15 T_{8}+10 T_{6}-1=0$ Solution: (i) LHS: $\frac{T_{3}-T_{5}}{T_{1}}=\frac{\left(\sin ^{3} x+\cos ^{3} x\right)-\left(\sin ^{5} x+\cos ^{5} x\right)}{\sin x+\cos x}$ $=\frac{\sin ^{3} x-\sin ^{5} x+\cos ^{3} x-\cos ^{5} x}{\sin x+\cos x}$ $=\frac{\sin ^{3} x\left(1-\sin ^{2} x\right)+\cos ^{3} x\left(1-\cos ^{2} x\right)}{\sin x+\cos x}$ $=...

Construct a △ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and ∠B = 45°.

Question: Construct a △ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and B = 45. Solution: Steps of Construction: 1. Construct a line segment BC of 4.5 cm. 2. At the point B, drawXBC = 45. 3. Keeping B as centre and radius 5.6cm draw an arc which intersects XB at D. 4. Join DC. 5. Draw the perpendicular bisector of DC which intersects DB at A. 6. Join AC. Hence △ABC is the required triangle...

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In the given figure, DE || BD. Determine AC and AE.

Question: In the given figure, DE || BD. Determine AC and AE. Solution: Given, $D E \| C B$. In $\triangle A B C$ and $\triangle A D E$ $\angle A D E=\angle C \quad$ (Corresponding angles) $\angle A=\angle A \quad$ (Common) $\triangle A B C \sim \triangle A D E \quad$ (A.A Similarity) $\frac{A E}{4}=\frac{12}{15}=\frac{14}{A C}$ $\frac{A E}{4}=\frac{12}{15}$ $A E \times 15=12 \times 4$ $4 E=\frac{12 \times 4}{15}$ $A E=\frac{4 \times 4}{5}$ $A E=\frac{16}{5}$ $\frac{A E}{4}=\frac{12}{15}=\frac{1...

Construct a △ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60°.

Question: Construct a △ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and B = 60. Solution: Steps of Construction: 1. Construct a line segment BC of 3.6 cm. 2. At the point B, drawXBC = 60. 3. Keeping B as center and radius 4.8cm draw an arc which intersects XB at D. 4. Join DC. 5. Draw the perpendicular bisector of DC which intersects DB at A. 6. Join AC. Hence △ABC is the required triangle....

Find the area between the curves

Question: Find the area between the curves $y=x$ and $y=x^{2}$ Solution: The required area is represented by the shaded area OBAO as The points of intersection of the curves, $y=x$ and $y=x^{2}$, is $\mathrm{A}(1,1)$. We draw AC perpendicular tox-axis. Area (OBAO) = Area (ΔOCA) Area (OCABO) (1) $=\int_{0}^{1} x d x-\int_{0}^{1} x^{2} d x$ $=\left[\frac{x^{2}}{2}\right]_{0}^{1}-\left[\frac{x^{3}}{3}\right]_{0}^{1}$ $=\frac{1}{2}-\frac{1}{3}$ $=\frac{1}{6}$ units...

Construct the angles of the following measurements:

Question: Construct the angles of the following measurements: (i) 30 (ii) 75 (iii) 105 (iv) 135 (v) 15 (vi) 22(1/2) Solution: (i) Steps of construction: 1. Draw a line segment AB. 2. Keeping A as the centre and any radius draw an arc which intersects AB at C. 3. Keeping C as center and the same radius draw an arc which intersects the previous arc at D. (ii) Steps of construction: 1. Draw a line segment AB 2. Keeping A as center and any radius draw an arc which intersects AB at C 3. Keeping C as ...

Find the area under the given curves and given lines:

Question: Find the area under the given curves and given lines: (i) $y=x^{2}, x=1, x=2$ and $x$-axis (ii) $y=x^{4}, x=1, x=5$ and $x$-axis Solution: i. The required area is represented by the shaded area ADCBA as Area $\mathrm{ADCBA}=\int_{1}^{2} y d x$ $=\int_{1}^{2} x^{2} d x$ $=\left[\frac{x^{3}}{3}\right]_{1}^{2}$ $=\frac{8}{3}-\frac{1}{3}$ $=\frac{7}{3}$ units ii. The required area is represented by the shaded area ADCBA as\ Area $\mathrm{ADCBA}=\int_{1}^{5} x^{4} d x$ $=\left[\frac{x^{5}}{...

Prove the:

Question: Prove the: $\left|\sqrt{\frac{1-\sin x}{1+\sin x}}\right|+\left|\sqrt{\frac{1+\sin x}{1-\sin x}}\right|=-\frac{2}{\cos x}$, where $\frac{\pi}{2}x\pi$ Solution: $\mathrm{LHS}=\left|\sqrt{\frac{1-\sin x}{1+\sin x}}\right|+\left|\sqrt{\frac{1+\sin x}{1-\sin x}}\right|$ $=\left|\sqrt{\frac{(1-\sin x)(1-\sin x)}{(1+\sin x)(1-\sin x)}}\right|+\left|\sqrt{\frac{(1+\sin x)(1+\sin x)}{(1-\sin x)(1+\sin x)}}\right|$ $=\left|\sqrt{\frac{(1-\sin x)(1-\sin x)}{(1+\sin x)(1-\sin x)}}\right|+\left|\s...

If a=sec

Question: If $a=\sec [x-\tan x$ and $b=\operatorname{cosec} x+\cot x$, then shown that $a b+a-b+1=0$. Solution: $a=\sec x-\tan x \quad$ And,$b=\operatorname{cosec} x+\cot x$ $=\frac{1-\sin x}{\cos x} \quad$ And, $b=\frac{1+\cos x}{\sin x}$ Now, we have : $a b+a-b+1$ $\left(\frac{1-\sin x}{\cos x}\right)\left(\frac{1+\cos x}{\sin x}\right)+\frac{1-\sin x}{\cos x}-\left(\frac{1+\cos x}{\sin x}\right)+1$ $=\frac{1-\sin x+\cos x-\sin x \cos x+\sin x-\sin ^{2} x-\cos x-\cos ^{2} x+\sin x \cos x}{\sin...

Area lying between the curve

Question: Area lying between the curve $y^{2}=4 x$ and $y=2 x$ is A. $\frac{2}{3}$ B. $\frac{1}{3}$ C. $\frac{1}{4}$ D. $\frac{3}{4}$ Solution: The area lying between the curve, $y^{2}=4 x$ and $y=2 x$, is represented by the shaded area $\mathrm{OBAO}$ as The points of intersection of these curves are O (0, 0) and A (1, 2). We draw AC perpendicular tox-axis such that the coordinates of C are (1, 0). Area OBAO = Area (OCABO) Area (ΔOCA) $=\int_{0}^{1} 2 \sqrt{x} d x-\int_{0}^{1} 2 x d x$ $=2\left...

If sin x

Question: If $\sin x+\cos x=m$, then prove that $\sin ^{6} x+\cos ^{6} x=\frac{4-3\left(m^{2}-1\right)^{2}}{4}$, where $m^{2} \leq 2$ Solution: $\sin x+\cos x=m \quad($ Given $)$ To prove : $\sin ^{6} x+\cos ^{6} x=\frac{4-3\left(m^{2}-1\right)^{2}}{4}$, where $m^{2} \leq 2$ Proof : LHS : $\sin ^{6} x+\cos ^{6} x$ $=\left(\sin ^{2} x\right)^{3}+\left(\cos ^{2} x\right)^{3}$ $=\left(\sin ^{2} x+\cos ^{2} x\right)^{3}-3 \sin ^{2} x \cos ^{2} x\left(\sin ^{2} x+\cos ^{2} x\right)$ $=1-3 \sin ^{2} x...

In ∆ABC, points P and Q are on CA and CB, respectively such that CA = 16 cm,

Question: In $\triangle \mathrm{ABC}$, points $\mathrm{P}$ and $\mathrm{Q}$ are on $\mathrm{CA}$ and $\mathrm{CB}$, respectively such that $\mathrm{CA}=16 \mathrm{~cm}, \mathrm{CP}=10 \mathrm{~cm}, \mathrm{CB}=30 \mathrm{~cm}$ and $\mathrm{CQ}=25 \mathrm{~cm}$. Is $\mathrm{PQ} \| \mathrm{AB}$ ? Solution: Given: $A C=16 \mathrm{~cm}, C P=10 \mathrm{~cm}, C B=30 \mathrm{~cm}$ and $C Q=25 \mathrm{~cm}$, we get We will check whether $\frac{C P}{A C}=\frac{C Q}{B C}$ or not to conclude whether $P Q \...

Construct the following angles at the initial point of a given ray and justify the construction:

Question: Construct the following angles at the initial point of a given ray and justify the construction: (i) 45 (ii) 90 Solution: (i) Steps of construction: 1. Draw a line segment AB and produce BA to C. 2. Keeping A as the center and any radius draw an arc which intersects AC at D and AB at E. 3. Keeping D and E as center and radius more than half of DE draw arcs which intersect each other at F. 4. Join FA which intersects the arc in (2) at G. 5. Keeping G and E as center and radius more than...

Smaller area enclosed by the circle

Question: Smaller area enclosed by the circle $x^{2}+y^{2}=4$ and the line $x+y=2$ is A. $2(\pi-2)$ B. $\pi-2$ C. $2 \pi-1$ D. $2(\pi+2)$ Solution: The smaller area enclosed by the circle, $x^{2}+y^{2}=4$, and the line, $x+y=2$, is represented by the shaded area ACBA as It can be observed that, Area ACBA = Area OACBO Area (ΔOAB) $=\int_{0}^{2} \sqrt{4-x^{2}} d x-\int_{0}^{2}(2-x) d x$ $=\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{0}^{2}-\left[2 x-\frac{x^{2}}{2}\r...

What values of x will make DE || AB in the given figure?

Question: What values ofxwill make DE || AB in the given figure? Solution: If $D E \| A B$, then $\frac{C E}{C B}=\frac{C D}{A C}$ $\frac{x}{x+3 x+4}=\frac{x+3}{4 x+22}$ $\frac{x}{4 x+4}=\frac{x+3}{4 x+22}$ $x \times(4 x+22)=x+3 \times(4 x+4)$ $4 x^{2}+22 x=4 x^{2}+12 x+4 x+12$ $22 x=16 x+12$ $22 x-16 x=12$ $6 x=12$ $x=\frac{12}{6}$ $x=2$ Hence, the value of $x$ is 2 ....

Using a protractor, draw an angle of measure 72°.

Question: Using a protractor, draw an angle of measure 72. With this angle as given draw angles of measure 36 and 54. Solution: Steps of construction: 1. Draw anABCof720with the help of a protractor. 2. Keeping B as center and any radius draw an arc which intersects AB at D and BC at E. 3. Keeping D and E as center and radius more than half of DE draw two arcs which intersect each other at F. 4. Join FB which intersects the arc in (2) at G. 5. Keeping D and G as center and radius more than half ...

Using integration find the area of the triangular region whose sides have the equations

Question: Using integration find the area of the triangular region whose sides have the equations $y=2 x+1, y=3 x+1$ and $x=4$. Solution: The equations of sides of the triangle are $y=2 x+1, y=3 x+1$, and $x=4$. On solving these equations, we obtain the vertices of triangle as $A(0,1), B(4,13)$, and $C(4,9)$. It can be observed that, Area (ΔACB) = Area (OLBAO) Area (OLCAO) $=\int_{0}^{4}(3 x+1) d x-\int_{0}^{4}(2 x+1) d x$ $=\left[\frac{3 x^{2}}{2}+x\right]_{0}^{4}-\left[\frac{2 x^{2}}{2}+x\righ...

If cot x

Question: If $\cot x(1+\sin x)=4 m$ and $\cot x(1-\sin x)=4 n$, prove that $\left(m^{2}+n^{2}\right)^{2}=m n$ Solution: Given : $4 m=\cot x(1+\sin x)$ and $4 n=\cot x(1-\sin x)$ Multiplying both the equations: $\Rightarrow 16 m n=\cot ^{2} x\left(1-\sin ^{2} x\right)$ $\Rightarrow 16 m n=\cot ^{2} x \cdot \cos ^{2} x$ $\Rightarrow m n=\frac{\cos ^{4} x}{16 \sin ^{2} x}$ ....(1) Squaring the given equation : $16 m^{2}=\cot ^{2} x(1+\sin x)^{2}$ and $16 n^{2}=\cot ^{2} x(1-\sin x)^{2}$ $\Rightarro...

In each of the figures [(i)-(iv)] given below,

Question: In each of the figures [(i)-(iv)] given below, a line segment is drawn parallel to one side of the triangle and the lengths of certain line-segment are marked. Find the value ofxin each of the following : Solution: In each of the figure, we have to find the value ofx By cross multiplication on both sides, we get $1 \times(d+x)=d \times(1+c)$ $d+x=d+d c$ $x=d+d c-d$ Hence the value of $x$ is $d c$. By cross multiplication on both sides, we get $a \times(b+x)=b \times(a+1)$ $a b+a x=a b+...

Using integration finds the area of the region bounded by the triangle whose vertices are

Question: Using integration finds the area of the region bounded by the triangle whose vertices are $(-1,0),(1,3)$ and $(3,2)$. Solution: BL and CM are drawn perpendicular tox-axis. It can be observed in the following figure that, Area (ΔACB) = Area (ALBA) + Area (BLMCB) Area (AMCA) (1) Equation of line segment AB is $y-0=\frac{3-0}{1+1}(x+1)$ $y=\frac{3}{2}(x+1)$ $\therefore$ Area $($ ALBA $)=\int_{-1}^{1} \frac{3}{2}(x+1) d x=\frac{3}{2}\left[\frac{x^{2}}{2}+x\right]_{-1}^{1}=\frac{3}{2}\left[...

Using rulers and compasses only, draw an angle of measure 135°.

Question: Using rulers and compasses only, draw an angle of measure 135. Solution: Steps of construction: 1. Draw a line segment AB and produce BA to C. 2. Keeping A as the center and any radius draw an arc which intersects AC at D and AB at E. 3. Keeping D and E as center and radius more than half of DE draw arcs which intersect each other at F. 4. Join FA which intersects the arc in (2) at G. 5. Keeping G and D as center and radius more than half of GD draw arcs which intersect each other at H...

Using rulers and compasses only, draw a right angle.

Question: Using rulers and compasses only, draw a right angle. Solution: Steps of construction: 1. Draw a line segment AB. 2. Keeping A as the center and any radius draw an arc which intersects AB at C. 3. Keeping C as center and the same radius draw an arc which intersects the previous arc at D. 4. Keeping D as the center and same radius draw an arc which intersects arc in (2) at E. 5. Keeping E and D as center and radius more than half of ED draw arcs which intersect each other at F. 6. Join F...

Find the area of the region bounded by the curves

Question: Find the area of the region bounded by the curves $y=x^{2}+2, y=x, x=0$ and $x=3$ Solution: The area bounded by the curves, $y=x^{2}+2, y=x, x=0$, and $x=3$, is represented by the shaded area OCBAO as Then, Area OCBAO = Area ODBAO Area ODCO $=\int_{0}^{3}\left(x^{2}+2\right) d x-\int_{0}^{3} x d x$ $=\left[\frac{x^{3}}{3}+2 x\right]_{0}^{3}-\left[\frac{x^{2}}{2}\right]_{0}^{3}$ $=[9+6]-\left[\frac{9}{2}\right]$ $=15-\frac{9}{2}$ $=\frac{21}{2}$ units...

If cosec

Question: If $\operatorname{cosec} x-\sin x=a^{3}$, sec $x-\cos x=b^{3}$, then prove that $a^{2} b^{2}\left(a^{2}+b^{2}\right)=1$ Solution: $\operatorname{cosec} x-\sin x=a^{3}$ $\therefore \frac{1}{\sin x}-\sin =a^{3}$ $\Rightarrow \frac{1-\sin ^{2} x}{\sin x}=a^{3}$ $\Rightarrow \frac{\cos ^{2} x}{\sin x}=a^{3}$ $\Rightarrow a=\left(\frac{\cos ^{2} x}{\sin x}\right)^{\frac{1}{3}} \quad \ldots(\mathrm{i})$ Also, sec $x-\cos x=b^{3}$ $\Rightarrow \frac{1}{\cos x}-\cos =b^{3}$ $\Rightarrow \frac{...

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