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

Construct an angle of 67.5° by using the ruler and compasses.

Question: Construct an angle of 67.5 by using the ruler and compasses. Solution: Steps of Construction:1. Draw a line TS.2. ConstructPOS = 90.3.Draw OQ,bisector ofPOT. Thus,QOS=135is obtained.4. Draw the bisector ofQOS.5.ROS thus obtained is equal to 67.5....

A sinusoidal voltage of peak value 250{~V} is applied to a series LCR circuit,

Question: A sinusoidal voltage of peak value $250 \mathrm{~V}$ is applied to a series LCR circuit, in which $\mathrm{R}=8 \Omega, \mathrm{L}=24 \mathrm{mH}$ and $\mathrm{C}=60 \mu \mathrm{F}$. The value of power dissipated at resonant condition is ' $x$ ' $k W$. The value of $x$ to the nearest integer is Solution: At resonance power ( $\mathrm{P}$ ) $P=\frac{\left(V_{r m s}\right)^{2}}{R}$ $P=\frac{(250 / \sqrt{2})^{2}}{8}=3906.25 W$ $\approx 4 \mathrm{~kW}$...

In each of the following cases, given reasons to show that the construction of ∆ABC is not possible:

Question: In each of the following cases, given reasons to show that the construction of ∆ABCis not possible:(i)AB= 6 cm, A= 40 and (BC+AC) = 5.8 cm.(ii)AB= 7 cm, A= 50 and (BCAC) = 8 cm.(iii)BC= 5 cm, B= 80, C= 50 and A= 60.(iv)AB= 4 cm,BC= 3 cm andAC= 7 cm. Solution: (i)AB= 6 cm, A= 40 and (BC+AC) = 5.8 cm.Here BC + AC is not greater than AB so, this triangle is not possible.(ii)AB= 7 cm, A= 50 and (BCAC) = 8 cm.HereBCACis not less than AB so this triangle is not possible.(iii)BC= 5 cm, B= 80,...

Construct a ∆ABC whose perimeter is 11.6 cm and the base angles are 45° and 60°

Question: Construct a ∆ABCwhose perimeter is 11.6 cm and the base angles are 45 and 60 Solution: Steps of construction:1. Draw a line segmentPQ = 11.6 cm. 2. Construct an angle of $45^{\circ}$ and bisect it to get $\angle Q P X$. 3. Construct an angle of $60^{\circ}$ and bisect it to get $\angle P Q Y$. 4. The rayXPandYQintersect atA.5. Draw the right bisectors of APandAQ, cuttingPQatBandC, respectively.6. JoinABandAC. Thus, $\triangle A B C$ is the required triangle....

Let $f:

Question: Let $f:[-1,1] \rightarrow R$ be defined as $f(x)=a x^{2}+b x+c$ for all $\mathrm{x} \in[-1,1]$, where $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{R}$ such that $\mathrm{f}(-1)=2, \mathrm{f}^{\prime}(-1)=1$ and for $x \in(-1,1)$ the maximum value of $f^{\prime \prime}(x)$ is $\frac{1}{2}$. If $f(x) \leq \alpha$ $x \in[-1,1]$, then the least value of $\alpha$ is equal to Solution: $\mathrm{f}:[-1,1] \rightarrow \mathrm{R}$ $f(x)=a x^{2}+b x+c$ $f(-1)=a-b+c=2$ $f^{\prime}(-1)=-2 a+b=1...

Construct a triangle whose perimeter is 10.4 cm and the base angles are 45° and 120°.

Question: Construct a triangle whose perimeter is 10.4 cm and the base angles are 45 and 120. Solution: Steps of construction:1. Draw a line segmentPQ = 10.4 cm. 2. Construct an angle of $45^{\circ}$ and bisect it to get $\angle Q P X$. 3. Construct an angle of $120^{\circ}$ and bisect it to get $\angle P Q Y$. 4. The rayXPandYQintersect atA.5. Draw the right bisectors of APandAQ, cuttingPQatBandC, respectively.6. JoinABandAC. Thus, $\triangle A B C$ is the required triangle....

Reaction of Grignard reagent,

Question: Reaction of Grignard reagent, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{MgBr}$ with $\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}$ followed by hydrolysis gives compound "A" which reacts instantly with Lucas reagent to give compound $\mathrm{B}, \mathrm{C}_{10} \mathrm{H}_{13} \mathrm{Cl}$.The Compound $\mathrm{B}$ is : Correct Option: , 3 Solution: (3)...

Construct a ∆PQR whose perimeter is 12 cm and the lengths of whose sides are in the ratio 3 : 2 : 4.

Question: Construct a∆PQRwhose perimeter is 12 cm and the lengths of whose sides are in the ratio 3 : 2 : 4. Solution: 1. Draw a line segmentXY= 12 cm.2. In the downward direction, construct an acute angle withXYatX.3. FromX, set off (3 + 2 + 4) = 9 arcs of equal distances alongXZ.4. Mark pointsL, MandNsuch that XL= 3 units,LM= 2 units andMN= 4 units.5. JoinNY.6. Through L and M, drawLQ∥∥NYandMR∥∥NYcuttingXYatQandR,respectively.7. WithQas the centre and radiusQX, draw an arc.8. WithRas the centr...

Consider the function

Question: Consider the function $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ defined by $f(x)=\left\{\begin{array}{c}\left(2-\sin \left(\frac{1}{x}\right)\right)|x|, x \neq 0 \\ 0, x=0\end{array}\right.$. Then $\mathrm{f}$ is (1) monotonic on $(-\infty, 0) \cup(0, \infty)$ (2) not monotonic on $(-\infty, 0)$ and $(0, \infty)$(3) monotonic on $(0, \infty)$ only(4) monotonic on $(-\infty, 0)$ onlyCorrect Option: , 2 Solution: $f(x)=\left\{\begin{array}{cc}-x\left(2-\sin \left(\frac{1}{x}\right)...

Construct a ∆ABC in which base AB = 5 cm, ∠A = 30° and AC – BC = 2.5 cm.

Question: Construct a ∆ABCin which baseAB= 5 cm, A= 30 andACBC= 2.5 cm. Justify your construction. Solution: Steps of construction:1. Draw a line segmentAB = 5 cm. 2. Construct $\angle B A X=30^{\circ}$. 3. Set offAD= 2.5 cm.4. JoinDB.5. Draw the right bisector ofDB, meetingDBproduced atC.6. JoinCB. Thus, $\triangle A B C$ is the required triangle. Justification:Point C lies on the perpendicular bisector of DB.⇒ CD = BCNowAD = AC DC = AC BC...

Construct a ∆ABC in which BC = 6 cm, ∠B = 30° and AB – AC = 3.5 cm.

Question: Construct a ∆ABCin whichBC= 6 cm, B= 30 andABAC= 3.5 cm. Justify your construction. Solution: Steps of construction:1. Draw a line segmentBC= 6 cm. 2. Construct $\angle C B X=30^{\circ}$ 3. Set offBD= 3.5 cm.4. JoinDC.5. Draw the right bisector ofDC, meetingBDproduced atA.6. JoinAC. Thus, $\triangle A B C$ is the required triangle. Justification:Point A lies on the perpendicular bisector of DC.⇒ AD = ACNowBD = AB AD = AB AC...

An RC circuit as shown in the figure is driven by a AC source generating a square wave.

Question: An $\mathrm{RC}$ circuit as shown in the figure is driven by a $\mathrm{AC}$ source generating a square wave. The output wave pattern monitored by CRO would look close to : Correct Option: 3, Solution: For $\mathrm{t}_{1}-\mathrm{t}_{2}$ Charging graph $\mathrm{t}_{2}-\mathrm{t}_{3}$ Discharging graph...

The product "A" in the above reaction is:

Question: The product "A" in the above reaction is:Correct Option: , 2 Solution:...

Construct a ∆ABC in which AB = 5.8 cm, ∠B = 60° and BC + CA = 8.4 cm.

Question: Construct a ∆ABCin which AB = 5.8 cm, B= 60 andBC+CA= 8.4 cm. Justify your construction. Solution: Steps of construction:1. Draw a line segmentAB = 5.8 cm. 2. Construct $\angle A B X=60^{\circ}$. 3. Set offBP= 8.4 cm.4. JoinPA.5. Draw the right bisector ofPA, meetingBPatC.6. JoinAC. Thus, $\triangle A B C$ is the required triangle. Justification: In $\triangle \mathrm{APC}$, CAP = CPA (By construction)⇒ CP = AC (Sides opposite to equal angles are equal)NowBC = PB PC = PB AC⇒ BC + AC = ...

Which of the following statements is correct for the function

Question: Which of the following statements is correct for the function $g(\alpha)$ for $\alpha \in R$ such that $g(\alpha)=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin ^{\alpha} x}{\cos ^{a} x+\sin ^{\alpha} x} d x$(1) $g(\alpha)$ is a strictly increasing function (2) $\mathrm{g}(\alpha)$ has an inflection point at $\alpha=-\frac{1}{2}$(3) $\mathrm{g}(\alpha)$ is a strictly decreasing function(4) $\mathrm{g}(\alpha)$ is an even functionCorrect Option: 4, Solution: $g(\alpha)=\int_{\frac{\pi}...

Construct a ∆ABC in which BC = 4.5 cm, ∠B = 45° and AB + AC = 8 cm.

Question: Construct a ∆ABCin whichBC= 4.5 cm, B= 45 andAB+AC= 8 cm. Justify your construction. Solution: Steps of construction:1. Draw a line segmentBC= 4.5 cm. 2. Construct $\angle C B X=45^{\circ}$. 3. Set offBP= 8 cm.4. JoinPC.5. Draw the right bisector ofPC, meetingBPatA.6. JoinAC. Thus, $\triangle A B C$ is the required triangle. Justification: In $\triangle \mathrm{APC}$, $\angle \mathrm{ACP}=\angle \mathrm{APC}$ (By construction) ⇒ AC = AP (Sides opposite to equal angles are equal)NowAB =...

Construct a right-angled triangle whose hypotenuse measures 5 cm

Question: Construct a right-angled triangle whose hypotenuse measures 5 cm and the length of one of whose sides containing the right angle measures 4.5 cm. Solution: Steps of construction:1. Draw a line segmentBC= 5 cm.2. Find the midpointOofBC.3. WithOas the centre and radiusOB, draw a semicircle onBC.4. WithBas the centre and radius equal to 4.5 cm, draw an arc cutting the semicircle atA.4. JoinABandAC.Thus, ABCis the required triangle....

Let $mathrm{f}$ be a real valued function, defined on

Question: Let $\mathrm{f}$ be a real valued function, defined on $\mathrm{R}-\{-1,1\}$ and given by $f(x)=3 \log _{e}\left|\frac{x-1}{x+1}\right|-\frac{2}{x-1}$ Then in which of the following intervals, function $\mathrm{f}(\mathrm{x})$ is increasing?(1) $(-\infty,-1) \cup\left(\left[\frac{1}{2}, \infty\right)-\{1\}\right)$(2) $(-\infty, \infty)-\{-1,1\}$(3) $\left(-1, \frac{1}{2}\right]$(4) $\left(-\infty, \frac{1}{2}\right]-\{-1\}$Correct Option: 1, Solution: $f(x)=3 \ln (x-1)-3 \ln (x+1)-\fra...

Construct an equilateral triangle each of whose altitudes measures 5.4 cm.

Question: Construct an equilateral triangle each of whose altitudes measures 5.4 cm. Measure each of its sides. Solution: Steps of construction:1. Draw a lineXY.2. Mark any pointP.3. FromPdrawPQXY. 4. From $P$, set off $P A=5.4 \mathrm{~cm}$, cutting $P Q$ at $A$. 5. Construct $\angle P A B=30^{\circ}$ and $\angle P A C=30^{\circ}$, meeting $X Y$ at $B$ and $C$, respectively. Thus, $\triangle A B C$ is the required triangle. Measure of each side is $6 \mathrm{~cm}$....

Construct an equilateral triangle each of whose sides measures 5 cm.

Question: Construct an equilateral triangle each of whose sides measures 5 cm. Solution: Steps of construction:1. Draw a line segmentAB= 5 cm.2. WithAas the centre and a radius equal toAB, draw an arc.3. WithBas the centre and the same radius as above, draw another arc cutting the previously drawn arc atC.4. JoinACandBC. Thus, $\triangle A B C$ is the required triangle....

Construct a ΔABC in which BC = 4.8 cm,

Question: Construct a ΔABC in which BC = 4.8 cm, B = 45 and C = 75. Measure A. Solution: Steps of construction: 1. Draw a line segmentBC= 4.8 cm. 2. Construct $\angle C B X=45^{\circ}$. 3. Construct $\angle B C Y=75^{\circ}$. 4. The ray $B X$ and $C Y$ intersect at $A$. Thus, $\triangle A B C$ is the required triangle. When we measure $\angle A$, we get $\angle A=60^{\circ}$....

the locus of the mirror image of a point on the parabola

Question: Let $\mathrm{C}$ be the locus of the mirror image of a point on the parabola $\mathrm{y}^{2}=4 \mathrm{x}$ with respect to the line $\mathrm{y}=\mathrm{x}$. Then the equation of tangent to $\mathrm{C}$ at $\mathrm{P}(2,1)$ is :(1) $x-y=1$(2) $2 x+y=5$(3) $x+3 y=5$(4) $x+2 y=4$Correct Option: 1, Solution: Given $y^{2}=4 x$ Mirror image on $y=x \Rightarrow C: x^{2}=4 y$ $2 \mathrm{x}=4 \cdot \frac{\mathrm{dy}}{\mathrm{dx}} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}}{2}$...

Show that each one of the following progressions is a G.P. Also,

Question: Show that each one of the following progressions is a G.P. Also, find the common ratio in each case: (i) $4,-2,1,-1 / 2, \ldots$ (ii) $-2 / 3,-6,-54, \ldots$ (iii) $a, \frac{3 a^{2}}{4}, \frac{9 a^{3}}{16}, \ldots$ (iv) $1 / 2,1 / 3,2 / 9,4 / 27, \ldots$ Solution: (i) We have, $a_{1}=4, a_{2}=-2, a_{3}=1, a_{4}=-\frac{1}{2}$ Now, $\frac{a_{2}}{a_{1}}=\frac{-2}{4}=\frac{-1}{2}, \frac{a_{3}}{a_{2}}=\frac{1}{-2}, \frac{a_{4}}{a_{3}}=\frac{-\frac{1}{2}}{1}=\frac{-1}{2}$ $\therefore \quad \...

Construct a ∆ABC in which BC = 5 cm, AB = 3.8 cm and AC = 2.6 cm

Question: Construct a∆ABCin whichBC= 5 cm,AB= 3.8 cm andAC= 2.6 cm. Bisect the largest angle of this triangle. Solution: Steps of construction:1. Draw a line segmentBC= 5 cm.2. WithBas the centre and a radius equal to 3.8 cm, draw an arc.3. WithCas the centre and a radius equal to 2.6 cm, draw an arc cutting the previously drawn arc atA.4. JoinABandAC.Thus,ABCis the required triangle.Take the largest angle and draw the angle bisector....

which the function

Question: The range of a $\in \mathbb{R}$ for which the function $f(x)=(4 a-3)\left(x+\log _{e} 5\right)+2(a-7) \cot \left(\frac{x}{2}\right) \sin ^{2}\left(\frac{x}{2}\right)$ $\mathrm{x}+2 \mathrm{n} \pi, \mathrm{n} \in \mathbb{N}$, has critical points, is (1) $(-3,1)$(2) $\left[-\frac{4}{3}, 2\right]$(3) $[1, \infty)$(4) $(-\infty,-1]$Correct Option: , 2 Solution: $f(x)=(4 a-3)\left(x+\log _{e} 5\right)+(a-7) \sin x$ $f(x)=(4 a-3)(1)+(a-7) \cos x=0$ $\Rightarrow \quad \cos x=\frac{3-4 a}{a-7}...

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