RD Sharma Solutions Class 12 Mathematics Chapter 15 MCQ

Edited By Kuldeep Maurya | Updated on Jan 21, 2022 01:20 PM IST

The RD Sharma solutions are the most prescribed solution books by the CBSE schools to their students. Students would not face challenges in doing their mathematics sums when they possess this set of books. Especially when it comes to chapter 15 MCQs, students tend to confuse the options. The RD Sharma Class 12th MCQ will lend a helping hand in those circumstances.

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RD Sharma Class 12 Solutions Chapter 15 Tangents and Normals - Other Exercise
Tangents and Normals Excercise: MCQ
RD Sharma Chapter wise Solutions

Also Read - RD Sharma Solution for Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter 15 Tangents and Normals - Other Exercise

Chapter 15 - Tangents and Normals Ex 15.1

Chapter 15 - Tangents and Normals Ex 15.2

Chapter 15 - Tangents and Normals Ex 15.3

Chapter 15 -Tangents and Normals Ex-FBQ

Chapter 15 -Tangents and Normals Ex-VSA

Tangents and Normals Excercise: MCQ

Tangents and Normals Exercise Multiple Choice Questions Question 1

Answer:
$\left ( c \right )x+y=0$
Hint:
Use differentiation
Given:
$y=sin\: x$
Solution:
$y=\sin \: x$
$\frac{dy}{dx}=\cos \: x$
$\frac{dy}{dx}|\left ( _{0,0} \right )=\cos 0=1$
Equation of normal: slop of normal
$=\frac{-1}{\frac{dy}{dx}}=-1$
Equation of normal at $\left ( 0,0 \right )$
$y-0=-1\left ( x-0 \right )$
$y=-x$
$y+x=0$
$x+y=0$ Is the requried
Equation of normals to curve $y=\sin x$ at $\left ( 0,0 \right )$

Tangents and Normals Exercise Multiple Choice Questions Question 2

Answer:
$\left ( d \right )2x=\pi$
Hint:
Use differentiation
Given:
$y=x+\sin \: x\: \cos \: x$
Solution:
$\begin{aligned} &y=x+\sin x \cos x\\ &=x+\frac{\sin (2 x)}{2}\\ &y^{\prime}=1+\cos (2 x)\\ &\text { Substituting } x=\frac{\pi}{2}\\ &y^{\prime}\left(\frac{\pi}{2}\right)=1+\cos (\pi)\\ &=0 \end{aligned}$
Therefore, slope of tangent is 0. this means that the normal is a vertical line of the $x=c$
Since the point has x-coordinate $\frac{\pi }{2}$ , the equation of the line is
$x=\frac{\pi }{2}$
$2x=\pi$ is requried

Tangents and Normals Exercise Multiple Choice Questions Question 3

Answer:
$\left ( a \right )x-2y=2$
Hint:
Use slope of tangent $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
Given:
$y=x\left ( 2-x \right )$
Solution:

T he given curve is $y=2x-x^{2}$
i.e. $x^{2}-2x+y=0$
Now the equation of the tangent at the point $\left ( x_{1},x_{2} \right )=\left ( 2,0 \right )$
Or
$\begin{aligned} &x x_{1}-\left(x+x_{1}\right)+\frac{1}{2}\left(y+y_{1}\right)=0 \\ &2 x+(-1)(x+2)+\frac{1}{2}(y+0)=0 \\ &2 x-4+y=0 \\ &y=-2 x+4 \end{aligned}$
So, the slope of the tangent is $\left ( -2 \right )$
Therefore the slope of the normal= $\frac{1}{2}$
So, the equation of the normal at the point $\left ( 2,0 \right )$ is
$y-0=\frac{1}{2}\left ( x-2 \right )$
$2y=x-2$
I.e. $x-2y=2$ is required

Tangents and Normals Exercise Multiple Choice Questions Question 4

Answer:
$\left ( b \right )\left ( \frac{1}{4},\frac{1}{2} \right )$
Hint:
Use slope of tangent $m=\tan \theta$
Given:
$y^{2}=x$
Solution:
$\frac{dy}{dx}$ Slope of the tangent at that point of the curve
Given curve is $x=y^{2}$
Point $\left ( a^{2},a \right )$
$\begin{aligned} &\frac{d }{d x}\left(y^{2}-x\right)=0 \\ &\frac{d y}{d x}=\frac{1}{2 y}=\frac{1}{2 a}=\tan \left(45^{\circ}\right) \\ &\Rightarrow a=\frac{1}{2}, a^{2}=\frac{1}{4} \end{aligned}$
Point $\left ( \frac{1}{4},\frac{1}{2} \right )$ is requried.

Tangents and Normals Exercise Multiple Choice Questions Question 6

Answer:
$\left ( b \right )\left ( 1,0 \right )$
Hint:
Use differentiation
Given:
$y=x^{2}-3x+2$
$y=x$
Solution:
Differentiate both the side w.r.t x we get
$\frac{dy}{dx}=1$
Let $\left ( x,y \right )$ be the required point
It is given that point lies on the curve
$\begin{aligned} &y=x_{1}^{2}-3 x_{1}+2 \\ &y=x^{2}-3 x+2 \end{aligned}$
Differentiating both the sides w.r.t x, we get
$\frac{dy}{dx}=2x-3$
Slope of the tangent= $\left ( \frac{dy}{dx} \right )\left ( _{x_{1},y_{1}} \right )=2x_{1}-3$
The tangent is perpendicular to the line
Slope of the tangent =-1/slope of the line= $\frac{-1}{1}=-1$
Now,
$\begin{aligned} &2 x_{1}-3=-1 \\ &2 x_{1}=2 \\ &x_{1}=1 \\ &y_{1}=x_{1}^{2}-3 x_{1}+2 \\ &y_{1}=(1)^{2}-3(1)+2 \\ &y_{1}=1-3+2 \\ &y_{1}= \\ &\left(x_{1}, y_{1}\right)=(1,0) \end{aligned}$

Tangents and Normals Exercise Multiple Choice Questions Question 7

Answer:
$\left ( b \right )\left ( \frac{1}{4}, \frac{1}{2}\right )$
Hint:
Use slop of the tangent $m=\tan \theta$
Given:
The curve $y^{2}=x$
Solution:
$\frac{dy}{dx}=$ Slope of the tangent at that point of the curve
Given curve is $x=y^{2}$
Keep $y=a$ then $x=y^{2}$
Point $\left ( a^{2},a \right )$
$\begin{aligned} &\frac{d y}{d x}\left(y^{2}-x\right)=0 \frac{d y}{d x}=\frac{1}{2 y}=\frac{1}{2 a}=\tan \left(45^{\circ}\right) \\ &\Rightarrow a=\frac{1}{2}, a^{2}=\frac{1}{4} \end{aligned}$
Point $\left ( \frac{1}{4},\frac{1}{2} \right )$

Tangents and Normals Exercise Multiple Choice Question Question 9

Answer:
$\left ( b \right )\tan^{-1}\frac{3}{4}$

Hint:

Use slop of the tangent $m=\tan \theta$
Given:
The curve $y^{2}=x$ and $x^{2}=y^{2}$ at $\left ( 1,1 \right )$
Solution:

Slope of the tangent to curve
$y=x^{2}$ At $\left ( 1,1 \right )$ is $m_{1}=2$
Slope of the tangent to the curve
$x=y^{2}$ At $\left ( 1,1 \right )$ is $m_{2}=\frac{1}{2}$
Angle between tangents to the curve is $\tan \theta =\frac{3}{4}$
Or $\theta =tan^{-1}\frac{3}{4}$

Tangents and Normals Exercise Multiple Choice Question Question 10

Answer:
$\text { (c) } x+3 y \pm 8=0$
Hint:
Use differentiation
Given:
$3x^{2}-y^{2}=8$ Is parallel to $x+3y=8$
Solution:
Given the equation of the line $3x^{2}-y^{2}=8$ now differentiating both sides with respect to x, we get $\frac{d\left ( 3x^{2}-y^{2} \right )}{dx}=\frac{d\left ( 8 \right )}{dx}$

Now applying the sum rule of differentiation an differentiation of constant =0,so we get

$\begin{aligned} &\frac{d\left(3 x^{2}\right)}{d x}-\frac{d\left(y^{2}\right)}{d x}=0 \\ &3 \frac{d\left(x^{2}\right)}{d x}-\frac{d\left(y^{2}\right)}{d x}=0 \\ &3 \times 2\left(x^{2-1}\right) \times \frac{d(x)}{d x}-2\left(y^{2-1}\right) \times \frac{d(y)}{d x}=0 \\ &6 x-2 y \times \frac{d y}{d x}=0 \\ &6 x=2 y \times \frac{d y}{d x} \\ &\frac{d y}{d x}=\frac{6 x}{2 y}=\frac{3 x}{y} \end{aligned}$

So, this is the slope of the given curve. We know the slope of the normal to the curve is

$=-\frac{1}{\frac{dy}{dx}}$

$=-\frac{1}{\frac{3x}{y}}=\left ( -\frac{y}{3x} \right )$ (1)

Now the given equation of the line $x+3y=8$

$3y=8-x$

Differentiating w.r.t x we get

$\begin{aligned} &\frac{d(3 y)}{d x}=\frac{d(8-x)}{d x} \\ &3 \frac{d y}{d x}=-1 \\ &\frac{d y}{d x}=-\frac{1}{3} \end{aligned}$
So, the slope of the line is $-\frac{1}{3}$
Now, as the normal to the curve is parallel to this line, hence the slope of the line should be equal to the slope of the normal to the given curve,
$\begin{aligned} &\therefore\left(-\frac{y}{3 x}\right)=-\frac{1}{3} \\ &3 y=3 x \\ &y=x \end{aligned}$

On substituting this value of the given equation of the curve, we get

$\begin{aligned} &3 x^{2}-y^{2}=8 \\ &3 x^{2}-(x)^{2}=8 \\ &2 x^{2}=8 \\ &x^{2}=4 \\ &x=\pm 2 \end{aligned}$

When x=2 the equation of the curve becomes,

$3x^{2}-y^{2}=8$

$3\left ( 2 \right )^{2}-y^{2}=8$

$3\left ( 4 \right )-y^{2}=8$

$12-8=y^{2}$

$y^{2}=4$

$y=\pm 2$

When x=-2, the equation of the curve becomes,

$\begin{aligned} &3 x^{2}-y^{2}=8 \\ &3(-2)^{2}-y^{2}=8 \\ &3(4)-y^{2}=8 \\ &12-8=y^{2} \\ &y^{2}=4 \\ &y=\pm 2 \end{aligned}$
So, the points at which normal is parallel to the given line are $\left ( \pm 2,\pm 2 \right )$
And required equation of the normal to the curve at $\left ( \pm 2,\pm 2 \right )$ is
$\begin{aligned} &y-(\pm 2)=-\frac{1}{3}[x-(\pm 2)] \\ &3(y-(\pm 2))=-(x-(\pm 2)) \\ &3 y-(\pm 6)=-x+(\pm 2) \\ &x+3 y-(\pm 6)-(\pm 2)=0 \\ &x+3 y+(\pm 8)=0 \end{aligned}$
Hence the equation of normal to the curve
$3x^{2}-y^{2}=8$ Which is parallel to the line $x+3y=8$ is $x+3y+\left ( \pm 8 \right )=0$

Tangents and Normals Exercise Multiple Choice Questions Question 15

Answer:
$\left ( c \right )6$
Hint:
Use slope of the tangent $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
Given:
$ay+x^{2}=7$ And $x^{3}=y$
Solution:
Slope of the tangent at (1,1) to $ay+x^{2}=7$ is $m_{1}=-\frac{2}{a}$
Slope of the tangent at (1,1) to $x^{3}=y$ is $m_{2}=3$
Curve cuts through orthogonally $m_{1}\times m_{2}=3$
$\begin{aligned} &\therefore m_{1} \times m_{2}=-\frac{2}{a} \cdot 3=-1 \\ &\therefore a=6 \end{aligned}$

Tangents and Normals Exercise Multiple Choice Question Question 16

Answer:
$\left ( b \right )b=-1,c=1$
Hint:
Use $y=mx+c$
Given:
$y=x$ Touches the curve $y=x^{2}+bx+c$ at the point (1,1)
Solution:
$y=x$
Clearly , $m=1$
$y=x^{2}+bx+c$ (1)

Differentiate w.r.t, $x_{1}$ we get

$\frac{dy}{dx}=2x+b$
Putting the point $\left ( 1,1 \right )$ and m=1 in above curve
$1=2\left ( 1 \right )+b$
$b=-1$
Putting the point $\left ( 1,1 \right )$ in (1)
$\begin{aligned} &1=(1)^{2}+b(1)+c \\ &1=1-1+c \quad \text { [Since } b=-1] \\ &c=1 \\ &\therefore b=-1, c=1 \end{aligned}$

Tangents and Normals Exercise Multiple Choice Questions Question 17

Answer:
(b) Is the correct option
Hint:
Substituting $x=1$ in $x=3t^{2}-1$
Given:
$x=3t^{2}+1,y=t^{3}-1$
Solution:
$x=3t^{2}+1,$ (1)

Differentiate w.r.t t

$\begin{aligned} &\frac{d x}{d t}=6 t\\ &\text { And, } y=t^{3}-1\\ &\text { Differentiating w.r.t t }\\ &\frac{d y}{d x}=3 t^{2} \end{aligned}$

$\begin{aligned} &\text { Substituting } x=1 \text { in (1) }\\ &1=3 t^{2}+1\\ &3 t^{2}=0\\ &t=0\\ &\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{3 t^{2}}{6 t}=\frac{t}{2}\\ &\left[\frac{d y}{d x}\right]_{t=0}=0\\ &\therefore \text { slope }=0 \end{aligned}$

Tangents and Normals Exercise Multiple Choice Question Question 18

Answer:
(c) Is the correct option
Hint:
Use $slope \left ( 1 \right )\times slope\left ( 2 \right )=-1$
Given:
$y=2e^{x},y=be^{-x}$
Solution:
$y=ae^{x}$ (1)
Differentiating w.r.t x, we get
$\frac{dy}{dx}=ae^{x}$
And $y=be^{-x}$ (2)
Since equation (1) and equation (2) intersect orthogonally
$\begin{aligned} &\text { slope }(1) \times \operatorname{slope}(2)=-1 \\ &a e^{x} \times b e^{-x}=1 \\ &a b=1 \end{aligned}$

Tangents and Normals Exercise Multiple Choice Questions Question 19

Answer:
(c) Is the correct option
Hint:
Equation of normal $y-y_{1}=m\left ( x-x_{1} \right )$
Given:
$x=a \cos ^{3} \theta, y=a \sin ^{3} \theta$
Solution:
$x=a \cos ^{3} \theta$ (1)
Differentiating w.r.t $\theta$
$\begin{aligned} &\frac{d x}{d \theta}=a \cdot 3 \cos ^{2} \theta(-\sin \theta) \\ &=-3 a \cos ^{2} \theta \sin \theta \\ &\left.\frac{d x}{d \theta}\right|_{\theta-\frac{\pi}{4}}=-3 a \times \frac{1}{2} \times \frac{1}{\sqrt{2}}=-\frac{3 a}{2 \sqrt{2}} \\ &y=a \sin ^{3} \theta \end{aligned}$
Differentiating w.r.t $\theta$
$\begin{aligned} &\frac{d y}{d \theta}=a \times 3 \sin ^{2} \theta \cos \theta \\ &\left.\frac{d y}{d \theta}\right|_{\theta-\frac{\pi}{4}}=3 a \times \frac{1}{2} \times \frac{1}{\sqrt{2}}=\frac{3 a}{2 \sqrt{2}} \end{aligned}$
Slope of normal of the given curve $M(N)=-\frac{d x}{d y}=-\left(\frac{\frac{d x}{d \theta}}{\frac{d y}{d \theta}}\right)=-\left(\frac{\frac{3 \alpha}{2 \sqrt{2}}}{\frac{-3 \alpha}{2 \sqrt{2}}}\right)=1$
Equation of normal at $\theta =\frac{\pi }{4}$
$\begin{aligned} &y-\frac{a}{(\sqrt{2})^{3}}=1\left(x-\left(\frac{a}{\sqrt{2}}\right)^{3}\right) \\ &y=x \end{aligned}$

Tangents and Normals Exercise 15 point 1 Question 20

Answer:
(a) $\frac{1}{2}$ Is the correct option
Hint:
$slope\left ( 1 \right )\times slope\left ( 2 \right )=-1$
Given:
$y=2e^{x}$ And $y=ae^{-x}$
Solution:
$y=2e^{x}$ (1)
Differentiating w.r.t x, we get
$\frac{dy}{dx}=2e^{x}$
Again $y=ae^{x}$ (2)
Differentiating w.r.t x we get
$\frac{dy}{dx}=ae^{x}$
Since (1) and (2) intersect orthogonally
$\begin{aligned} &\text { slope }(1) \times \operatorname{slope}(2)=-1 \\ &a e^{x} \times-a e^{-x}=-1 \\ &-2 a=-1 \\ &a=\frac{1}{2} \end{aligned}$

Tangents and Normals Exercise Multiple Choice Question , Question 21

Answer : (b) is the correct option
Hint : $\tan \theta=\left|\frac{m_{1}-1}{1-m_{1}}\right|$
Given : $y=6 x-x^{2}$
Solution :
$y=6 x-x^{2}$ (1)
Differentiating (1) w.r.t x we get
$\frac{d y}{d x}=6-2 x=m_{1}$
And $x+y=2$ (2)
Since
$\begin{aligned} &\tan \theta=\left|\frac{m_{1}+1}{1-m_{1}}\right| \\ &\tan \frac{\pi}{4}=\left|\frac{m_{1}+1}{1-m_{1}}\right| \\ &1=\frac{m_{1}+1}{1-m_{1}} \end{aligned}$
$m_{1}+1=\pm\left(1-m_{1}\right)$
Taking
$\begin{aligned} &m_{1}+1=1-m_{1} \\ &2 m_{1}=0 \\ &m_{1}=0 \\ &6-2 x=0 \\ &x=3 \end{aligned}$
Putting the values of x in (1) we get $y=9$ .
$\therefore$ Point is $(3,9)$

Tangents and Normals Exercise Multiple Choice Question , Question 22

Answer : (c) is the correct option.
Hint :
$\tan \theta=\left|\frac{m_{1}-m_{2}}{1-m_{1} m_{2}}\right|$
Given : $y^{2}=4 a x \text { And } x^{2}=4 a y$
Solution :
$\begin{aligned} &y^{2}=4 a x\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; (1)\\ &x^{2}=4 a y\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; (2) \end{aligned}$
Differentiating (1) w.r.t x we get
$\begin{aligned} &2 y \frac{d y}{d x}=4 a \\ &\left|\frac{d y}{d x}\right|_{(0,0)}=\frac{4 a}{2 x}=0=m_{1} \end{aligned}$
And
$\begin{aligned} &2 x=4 a \frac{d y}{d x} \\ &\left|\frac{d y}{d x}\right|_{(0,0)}=\left|\frac{4 a}{2 x}\right|=0=m_{2} \end{aligned}$
$\begin{aligned} &\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|= \\ &=\left|\frac{0-0}{1+0 \cdot 0}\right|=\infty \\ &\theta=\frac{\pi}{2} \end{aligned}$

Tangents and Normals Exercise Multiple Choice Question , Question 23

Answer : (c) is the correct option
Hint :
$\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
Given : $y=2 \sin ^{2} x, y=\cos 2 x$
Solution :
$y=2 \sin ^{2} x$ (1)
Differentiating (1) w.r.t x, we get
$\begin{aligned} &\frac{d y}{d x}=4 \sin x \cos x \\ &{\left[\frac{d y}{d x}\right]_{x=\frac{\pi}{6}}=4 \sin \frac{\pi}{6} \cos \frac{\pi}{6}=4 \times \frac{1}{2} \times \frac{\sqrt{3}}{2}=\sqrt{3}=m_{1}} \end{aligned}$
And $y=\cos\; 2x$ (2)
Differentiating (2) w.r.t y, we get
$\begin{aligned} &\frac{d y}{d x}=-2 \sin 2 x \\ &{\left[\frac{d y}{d x}\right]_{x=\frac{\pi}{6}}=-2 \sin \left(2 \times \frac{\pi}{6}\right)=-2 \times \sin \frac{\pi}{3}=-2 \times \frac{\sqrt{3}}{2}=-\sqrt{3}} \end{aligned}$
$\begin{aligned} &\tan \theta=\left|\frac{m_{1}-m_{2}}{1-3}\right| \\ &=\left|\frac{\sqrt{3}+\sqrt{3}}{-2}\right|=\left|\frac{2 \sqrt{3}}{-2}\right|=+\sqrt{3} \end{aligned}$
$\begin{aligned} &\tan \theta=+\sqrt{3} \\ &\theta=\frac{\pi}{3} \end{aligned}$

Tangents and Normals Exercise Multiple Choice Question , Question 24

Answer : (c) is the correct option
Hint : Obtain the slope
Given : $y=2 x^{7}+3 x+5$
Solution :
$y=2 x^{7}+3 x+5$ (1)
Differentiating (1) w.r.t x
$\frac{d y}{d x}=14 x^{6}+3$
Here the slope cannot be zero. Hence it make an acute angle with x-axis

Tangents and Normals Exercise Multiple Choice Question , Question 25

Answer : (a) is the correct option
Hint : Slope of the mormal is 1 or -1
Given : $9y^{2}=x^{3}$
Solution :
$9y^{2}=x^{3}$ (1)
Differentiating (1) w.r.t x, we get
$\begin{aligned} &18 y \frac{d y}{d x}=3 x^{2} \\ &\frac{d y}{d x}=\frac{3 x^{2}}{18 y}=\frac{x^{2}}{6 y} \end{aligned}$
$\begin{aligned} &\text { Slope of normal }=-\frac{d x}{d y}=-\frac{6 y}{x^{2}}=1 \text { or }-1 \\ &\text { Case } 1, \mathrm{~m}=1 \end{aligned}$
$\begin{aligned} &-\frac{6 y}{x^{2}}=1 \\ &-6 y=x^{2} \\ &-6 y_{1}=x_{1}^{2} \\ &y_{1}=-\frac{x_{1}^{2}}{6} \end{aligned}$
$\begin{aligned} &P\left(x, \frac{x^{2}}{6}\right) \\ &\text { Putting } P\left(x, \frac{x^{2}}{6}\right) \ln (1) \end{aligned}$
$\begin{aligned} &9\left(\frac{x^{2}}{6}\right)^{2}=x^{3} \\ &\Rightarrow 9 \times \frac{x^{4}}{36}=x^{3} \\ &\Rightarrow x^{3}(x-4)=0 \end{aligned}$
$\Rightarrow x= 0,4$
Putting the values of x in (1) we get $y=-\frac{8}{3}$
Similarly in case (2) if $y=\pm\frac{8}{3}$
$\therefore$ The point $\left(4, \pm \frac{8}{3}\right)$

Tangents and Normals Exercise Multiple Choice Question , Question 26

Answer : (b) is the correct option
Hint : $\text { Putting } \mathrm{x}=2 \text { and } \mathrm{y}=1 \text { in }(1) \text { and }(2)$
Solution :
$x = t^{2}+3t-8$ (1)
$y = 2t^{2}-2t-5$ (2)
$\begin{aligned} &\frac{d x}{d t}=2 t+3 \\ &\frac{d y}{d t}=4 t-2 \\ &\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{4 t-2}{2 t+3} \end{aligned}$ (3)
Putting $x=2$ in (1)
$\begin{aligned} &2=t^{2}+3 t-8 \\ &t^{2}+3 t-10=0 \\ &t^{2}+5 t-2 t-10=0 \end{aligned}$
$\begin{aligned} &t(t+5)-2(t+5)=0 \\ &(t+5)(t-2)=0 \\ &t=-5,2 \end{aligned}$
Putting $y = -1$ in (2)
$\begin{aligned} &-1=2 t^{2}-2 t-5 \\ &2 t^{2}-2 t-4=0 \\ &2 t^{2}-t-2=0 \\ &t^{2}-2 t+t-2=0 \end{aligned}$
$\begin{aligned} &t(t-2)+1(t-2)=0 \\ &(t-2)(t+1)=0 \\ &t=2,-1 \end{aligned}$
Putiing $t =2$ in (3)
$\frac{d y}{d t}=\frac{4(2)-2}{2(2)+3}=\frac{8-2}{4+3}=\frac{6}{7}$
The slope of tangent is $\frac{6}{7}$

Tangents and Normals Exercise Multiple Choice Question , Question 27

Answer : (a) is the correct option
Hint :
Use $c=\frac{a}{m}$
Given : The line $y= mx+1$ is the tangent of the curve $y^{2}= 4x$
Solution :
$y^{2}=4 x=4 \times 1 \times x$ (1)
From equation (1) we equate with $y^{2}=4ax$ so that we get the value of $c=1$ , the y intercept of the parabola
$c=\frac{a}{m},$ Where m is slope
From (1) , the value of a is 1
$\\1 =\frac{1}{m} \\ m=1$
The slope of the curve is 1
Hence the slope of the tangent is 1

Tangents and Normals Exercise Multiple Choice Question , Question 28

Answer : (b) is the correct option
Hint :
Equation of normal $\frac{y-y_{1}}{x-x_{1}}=m(N)$
Given : $2 y+x^{2}=3$
Solution :
$2 y+x^{2}=3$ (1)
Differentiating equation (1) w.r.t y
$\begin{aligned} &2+2 x \frac{d x}{d y}=0 \\ &\frac{d x}{d y}=-\frac{1}{x} \end{aligned}$
$\begin{aligned} &-\frac{d x}{d y}=\frac{1}{x} \\ &{\left[-\frac{d y}{d x}\right]_{(1,1)}=\frac{1}{1}=1=m(N)} \end{aligned}$
Equation of the normal at the point $(1,1)$ is $\frac{y-1}{x-1}=m(N)$
$\begin{aligned} &\frac{y-1}{x-1}=1 \\ &y-1=x-1 \\ &x-y=0 \end{aligned}$

Tangents and Normals Exercise Multiple Choice Question , Question 29

Answer : (a) is the correct option
Hint :
Slope $\frac{y-y_{1}}{x-x_{1}}$
Given : $x^{2}=4 y$
Solution :
$x^{2}=4 y$ (1)
Differentiating (1) w.r.t x
$\begin{aligned} &4 \frac{d y}{d x}=2 x \\ &\frac{d y}{d x}=\frac{x}{2} \\ &-\frac{d x}{d y}=-\frac{2}{x_{1}} \end{aligned}$
Slope of normal $=-\frac{2}{x_{1}}$
Since the normal passes through $(1,2)$
$\begin{aligned} &\text { Slope }=\frac{y_{1}-2}{x_{1}-1} \\ &-\frac{2}{x_{1}}=\frac{\frac{x^{2}}{4}-2}{x_{1}-2} \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\text { From }(1)] \end{aligned}$
$-\frac{2}{x_{1}}=\frac{x_{1}^{2}-8}{4\left(x_{1}-2\right)} \Rightarrow x_{1}^{3}-8 x_{1}=-8 x_{1}-8 \Rightarrow x_{1}=2$
Putting the value of $x_{1}= 2$ in (1) , was $y=1$
Point of contact $(2,1)$
Equation of normal
$\begin{aligned} &y-1=-\frac{2}{3}(x-2) \\ &y-1=-(x-2) \\ &x+y=3 \end{aligned}$

Tangents and Normals Exercise Multiple Choice Question , Question 30

Answer : (a) is the correct option
Hint : Equation of the normal, $y-y_{1}=N\left(x-x_{1}\right)$
Given : $3 y=6 x-5 x^{3}$
Solution :
Let $\left(x_{1}, y_{1}\right)$ be the point on the curve $3 y=6 x-5 x^{3}$ at which normal passes through origin
$3 y=6 x-5 x^{3}$ ....(1)
Differentiate (1) w.r.t x, we get
$\begin{aligned} &3 \frac{d y}{d x}=6-15 x^{2} \\ &\frac{d y}{d x}=2-5 x^{2} \end{aligned}$
Slope of the tangent at P is $2-5x_{1}^{2}$
Slope of normal at P is $\frac{1}{5x{_{1}}^{2}-2}$
Equation of normal at P is
$y-y_{1}=\frac{1}{5 x_{1}^{2}-2}\left(x-x_{1}\right)$
Since it passes through the origin
$y=\frac{x}{5 x_{1}^{2}-2}$ .....(2)
On solving eqn 1 and 2
We get $\mathrm{x}=1 \text { and } \mathrm{y}=\frac{1}{3}$
As we know, abscissa is the x-coordinate
So, option (a) satisfies above equation i.e., 1

Tangents and Normals Exercise Multiple Choice Question, Question 31

Answer : (b) is the correct option
Hint : Multiply the slope of both curve
Given : $x^{3}-3 x y^{2}+2=0 \text { And } 3 x^{2} y-y^{3}=2$
Solution :
Given $x^{3}-3 x y^{2}+2=0$ (1)
Differentiating (1) we get
$\begin{aligned} &3 x^{2}-3 y^{2}-3 x(2 y) \frac{d y}{d x}=0 \\ &3\left(x^{2}-y^{2}\right)=6 x y m_{1} \\ &m_{1}=\frac{x^{2}-y^{2}}{2 x y} \end{aligned}$
And $3 x^{2} y-y^{3}-2=0$ (2)
Differentiating (2), we get
$\begin{aligned} &6 x y-3 x^{2} \frac{d y}{d x}-3 y^{2} \frac{d y}{d x}=0 \\ &6 x y=6 m_{2}\left(y^{2}-x^{2}\right) \\ &m_{2}=\frac{2 x y}{y^{2}-x^{2}} \end{aligned}$
$m_{1} \times m_{2}=\frac{x^{2}-y}{2 x y} \times \frac{2 x y}{y^{2}-x^{2}}=-1$
Equation (1) and (2) at right angle

Tangents and Normals Exercise Multiple Choice Question, Question 32

Answer : (d) is the correct option
Hint :
$\text { Put } t=\frac{\pi}{4} \text { in } \frac{d y}{d x}$
Given : $x=e^{t} \cos t, y=e^{t} \sin t$
Solution :
$x=e^{t} \cos t$ (1)
Differentiating (1) w.r.t t, we get
$\begin{aligned} &\frac{d x}{d y}=e^{t}(\cos t-\sin t) \\ &y=e^{t} \sin t \end{aligned}$ (2)
Differentiating (2) w.r.t t, we get
$\begin{aligned} &\frac{d y}{d t}=e^{t}(\cos t+\sin t) \\ &\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\cos t+\sin t}{\cos t-\sin t} \end{aligned}$
$\begin{aligned} &{\left[\frac{d y}{d x}\right]_{(0,0)}=\infty} \\ &\tan ^{-1}(\infty)=\frac{\pi}{2} \end{aligned}$

Tangents and Normals Exercise Multiple Choice Question, Question 33

Answer : (b) is the correct option
Hint : Equation of tangent $y-y_{1}=m\left(x-x_{1}\right)$
Given : $y=e^{2 x}$
Solution :
$y=e^{2 x}$ (1)
Differentiating (1) w.r.t x, we get
$\begin{aligned} &\frac{d y}{d x}=2 e^{2 x} \\ &\left.\frac{d y}{d x}\right|_{x=0}=2 e^{0}=2 \end{aligned}$
Equation of the tangent at the point (0, 1) and having the slope 2 is
$\begin{aligned} &y-1=2(x-0) \\ &y-1=2 x \\ &y=2 x+1 \end{aligned}$
Since it (2) meets in x-axis
$\therefore y=0$
Putting $y=0$ in (2)
$0=2 x+1 \Rightarrow x=-\frac{1}{2}$
$\therefore$ The tangent meets at the point $P\left(\frac{-1}{2}, 0\right)$

Tangents and Normals Exercise Multiple Choice Question, Question 34

Answer : (a) is the correct option
Hint :
Equation of tangent $y-y_{1}=m\left(x-x_{1}\right)$
Given :
$y\left(1+x^{2}\right)=2-x$
Solution :
$y\left(1+x^{2}\right)=2-x$ (1)
Since it crosses x -axis
$\begin{aligned} &2-x=0 \\ &x=2 \end{aligned}$
The point of contact $(2,0)$
Differentiating (1) w.r.t x
$\begin{aligned} &y(2 x)+\left(1+x^{2}\right) \frac{d y}{d x}=-1 \\ &\text { At }(2,0) \\ &0(2 \times 2)+\left(1+2^{2}\right) \frac{d y}{d x}=-1 \\ &\frac{d y}{d x}=-\frac{1}{5} \end{aligned}$
Equation of tangent at $(1,2)$
$\begin{aligned} &y-0=-\frac{1}{5}(x-2) \\ &5 y=-x+2 \\ &x+5 y=2 \end{aligned}$

Tangents and Normals Exercise Multiple Choice Question, Question 35

Answer : (d) is the correct option
Hint : $\text { Parallel to } x \text {-axis } \Rightarrow y=0$
Given : $y=x^{3}-12 x+18$
Solution :
$y=x^{3}-12 x+18$ (1)
Differentiating (1) w.r.t x, we get
$\frac{d y}{d x}=3 x^{2}-12$
Since it's parallel to x-axis
$\begin{aligned} &\frac{d y}{d x}=0 \\ &3 x^{2}-12=0 \\ &x^{2}=4 \\ &x=\pm 2 \end{aligned}$
When $x=2$
$\begin{aligned} &y=(2)^{3}-12(2)+18 \\ &=8-24+18 \\ &=2 \end{aligned}$
When $x = -2$
$\begin{aligned} &y=(-2)^{3}-12(-2)+18 \\ &=-8+24+18 \\ &=34 \end{aligned}$
The points are $(2,2)$ and $(-2,34)$

Tangents and Normals Exercise Multiple Choice Question, Question 36

Answer : (a) is the correct option
Hint : $\text { Put }(0,0) \text { in } \frac{d y}{d x}$
Given :
$y=x^{\frac{1}{5}}$
Solution :
$y=x^{\frac{1}{5}}$ (1)
Differentiating (1) w.r.t x, we get
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{5} x^{\frac{1}{5}-1} \\ &=\frac{1}{5} x^{-\frac{4}{5}} \\ &\left(\frac{d y}{d x}\right)_{(0,0)}=\infty \end{aligned}$
$\therefore y=x^{\frac{1}{5}}$ Has vertical tangent

Chapter 15, Tangents and Normals, is where many students lose marks due to a lack of clarity. This chapter consists of three exercises, ex 15.1 to ex 15.3. RD Sharma Class 12 Solutions Tangents and Normals MCQ contains questions from concepts like an equation to the normal curve, the point of the curve when the tangent is perpendicular, the angle between curves, the slope of the tangent, and the angle of intersection. There are around 36 MCQ questions present in this chapter. The RD Sharma Class 12 Chapter 15 MCQ will lend a helping hand.

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RD Sharma Chapter wise Solutions

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