Question Set D

cbse 12th Standard Maths Continuity and Differentiability Important Questions And Answers Study Material - QB365

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Questions:

Generic 3

10 x 3 = 30

If $y=\left(x+\sqrt{1+x^2}\right)^n$, then show that $\left(1+x^2\right) \frac{d^2 y}{d x^2}+x \frac{d y}{d x}=n^2 y$

If $y=e^{m \sin ^{-1} x}$, then show that $\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-m^2 y=0$

If $\sin y=x \cos (a+y)$, then show that $\frac{d y}{d x}=\frac{\cos ^2(a+y)}{\cos a}$.
Also, show that $\frac{d y}{d x}=\cos a$,when x=0.

If $(a+b x) e^{\frac{y}{x}}=x$, then prove that $x \frac{d^2 y}{d x^2}=\left(\frac{a}{a+b x}\right)^2$

If $e^y(x+1)=1$, then show that $\frac{d y}{d x}=-e^y$

If $y={ \left( { sin }^{ -1 }x \right) }^{ 2 }$, then prove that: $\left( 1-{ x }^{ 2 } \right) \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -x\frac { dy }{ dx } =2$

If $x=a\left( cos\quad t+log\quad tan\frac { t }{ 2 } \right) ,\quad y=a\quad sin\quad t$ , find $\frac { { d }^{ 2 }y }{ { dt }^{ 2 } } $and $\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } $

If $y={ X }^{ x }$prove that $\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -\frac { 1 }{ y } { \left( \frac { dy }{ dx } \right) }^{ 2 }-\frac { y }{ x } =0$

If $x=a\ cos\theta +b\ sin\theta ,\ y=a\ sin\theta -b\ cos\theta $, show that ${ y }^{ 2 }\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -x\frac { dy }{ dx } -y=0$

10.

If ${ x }^{ 16 }{ y }^{ 9 }={ \left( { x }^{ 2 }+y \right) }^{ 17 },$prove that $\frac { dy }{ dx } =\frac { 2y }{ x } $

Generic 2

4 x 2 = 8

11.

If $x=a \sec ^{3} \theta \text { and } y=a \tan ^{3} \theta, \text { then find } \frac{d y}{d x}$ $\theta=\frac{\pi}{3}$

12.

Find the points in the open interval (0, 3) where the gretest integer function f(x) = [x] is not differentiable

13.

Find the derivative of sin ($cos^{ 2 }\left( \sqrt { x } \right) $).

14.

If e^y (x+1) = 1, show that dy/dx = -e^y

Multiple Choice Question

16 x 1 = 16

15.

The derivative of $\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)$ with respect to $\sin ^{-1} x, \frac{1}{\sqrt{2}}<x<1$ is

(a)

(b)

$\frac{\pi}{2}-2$

(c)

$\frac{\pi}{2}$

(d)

-2

16.

Let $f(x)=\left|\begin{array}{cc}x^2 & \sin x \\ p & -1\end{array}\right|$, where p is a constant. Then, the value of p for which $f^{\prime}(0)=1$ is

(a)

(b)

(c)

(d)

-1

17.

The points, at which the function f given by $f(x)=\left\{\begin{array}{ll}\frac{x}{|x|}, & x<0 \\ -1, & x \geq 0\end{array}\right.$ is continuous, is/are

(a)

$x \in R$

(b)

x=0

(c)

$x \in R-\{0\}$

(d)

x = -1 and 1

18.

The value of k for which the function $f(x)=\left\{\begin{array}{cl}\frac{1-\cos 4 x}{8 x^2}, & \text { if } x \neq 0 \\ k, & \text { if } x=0\end{array}\right.$ is continuous at x = 0 is

(a)

(b)

-1

(c)

(d)

19.

The function f(x) = [x], where [x] denotes the greatest integer less than or equal to x, is continuous at

(a)

x = 1

(b)

x = 1.5

(c)

x = -2

(d)

x = 4

20.

Derivative of $e^{\sin ^2 x}$ with respect to cos x is

(a)

sin x$e^{\sin ^2 x}$

(b)

cos x$e^{\sin ^2 x}$

(c)

- 2 cos x$e^{\sin ^2 x}$

(d)

-2 sin² x cos x$e^{\sin ^2 x}$

21.

If Xe^y = 1, then the value of $\frac{dy}{dx}$ at x = 1 is

(a)

-1

(b)

(c)

-e

(d)

- $\frac{1}{e}$

22.

If $y=(\cos x)^{(\cos x)^{(\cos x) \ldots \infty}}$,then $\frac{d y}{d x}$ is equal to

(a)

$\frac{y \tan x}{y \log \cos x-1}$

(b)

$\frac{y^{2} \tan x}{y \log \cos x-1}$

(c)

$\frac{y \tan x}{1+y \log \cos x}$

(d)

None ofthese

23.

If $y^{x}=e^{y-x}, \text { then } \frac{d y}{d x}$ is equal to

(a)

$\frac{1+\log y}{y \log y}$

(b)

$\frac{(1+\log y)^{2}}{y \log y}$

(c)

$\frac{1+\log y}{(\log y)^{2}}$

(d)

$\frac{(1+\log y)^{2}}{\log y}$

24.

Derivate of $\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$ $0<x<\frac{\pi}{2}$ is

(a)

$\frac{1}{2}$

(b)

(c)

(d)

None of these

25.

If $f(x)=\left\{\begin{array}{cl} \frac{\sqrt{1+k x}-\sqrt{1-k x}}{x}, & \text { for }-1 \leq x<0 \\ 2 x^{2}+3 x-2, & \text { for } 0 \leq x \leq 1 \end{array}\right.$ is continuous at x = 0, then k is equal to

(a)

-4

(b)

-3

(c)

-2

(d)

-1

26.

If $f(x)=\left\{\begin{array}{ll} \lambda\left(x^{2}-2 x\right), & \text { if } x \leq 0 \\ 4 x+1, & \text { if } x>0 \end{array}\right.$ then which one of the following is correct.

(a)

f(x) is continuous at x = 0 for any value of λ

(b)

f(x) is discontinuous at x = 0 for any value of λ

(c)

f(x) is discontinuous at x = 1for any value of

(d)

None of the above

27.

The function f(x) = $\begin{cases} \frac { { e }^{ 1/x }-1 }{ { e }^{ 1/x }+1 } ,x\neq 0 \\ 0\quad \quad x=0 \end{cases}$

(a)

is continuous at x = 0

(b)

Continuous everywhere

(c)

Not continuous at x = 0 but can be made continuous

(d)

Not continuous at x = 0

28.

Discuss the continuity of function f(x) = |x - 1| + |x + 1|

(a)

discontinuous at x = 1

(b)

discontinuous at x = ±1

(c)

continuous everywhere

(d)

discontinuous at x = -1

29.

A function f is said to be continuous for x ∈ R, if

(a)

it is continuous at x = 0

(b)

differentiable at x = 0

(c)

continuous at two points

(d)

differentiable for x ∈ R

30.

$\lim _{ x\rightarrow 0 }{ \frac { \sqrt { \frac { 1 }{ 2 } (1-cosx) } }{ x } } $ is equal to

(a)

(b)

-1

(c)

(d)

none of this

Assertion and reason

2 x 1 = 2

31.

Assertion: If a function f is discontinuous at c, then c is called a point of discontinuity.
Reason: A function is continuous at x = c, if the function is defined at x = c and the value of the function at x = c equals the limit of the function at x = c.
(a) Assertion is correct, Reason is correct; Reason is a correct explanation for assertion.
(b) Assertion is correct, Reason is correct; Reason is not a correct explanation for Assertion
(c) Assertion is correct, Reason is incorrect
(d) Assertion is incorrect, Reason is correct.

32.

Consider the function f(x) = [sin x], x $\in$[0, $\pi$]
Assertion: f(x) is not continuous at x = $\frac{\pi}{2}$
Reason: $lim_{x\rightarrow \frac{\pi }{2}}$f(x) does not exist
(a) Assertion is correct, Reason is correct; Reason is a correct explanation for assertion.
(b) Assertion is correct, Reason is correct; Reason is not a correct explanation for Assertion
(c) Assertion is correct, Reason is incorrect
(d) Assertion is incorrect, Reason is correct.

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Answers Key:

Generic 3

10 x 3 = 30

Given, $y=\left(x+\sqrt{1+x^2}\right)^n$
On differentiating both sides w.r.t. x, we get
$\frac{d y}{d x}=n\left(x+\sqrt{1+x^2}\right)^{n-1}\left(1+\frac{2 x}{2 \sqrt{1+x^2}}\right)$
[by using chain rule of derivative]
$\Rightarrow \frac{d y}{d x}=n\left(x+\sqrt{1+x^2}\right)^{n-1}\left(\frac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}}\right) $
$\Rightarrow \frac{d y}{d x}=\frac{n\left(x+\sqrt{1+x^2}\right)^n}{\sqrt{1+x^2}} $
$\Rightarrow \frac{d y}{d x}=\frac{n y}{\sqrt{1+x^2}}$
$\Rightarrow \sqrt{1+x^2} \frac{d y}{d x}=n y$
Again, differentiating both sides w.r.t. $, we get
$ \sqrt{1+x^2} \frac{d^2 y}{d x^2}+\frac{2 x}{2 \sqrt{1+x^2}} \cdot \frac{d y}{d x}=n \frac{d y}{d x}$
$\Rightarrow \left(1+x^2\right) \frac{d^2 y}{d x^2}+x \frac{d y}{d x}=n \cdot \sqrt{1+x^2} \frac{d y}{d x} $
$\text { [multiplying both sides by } \sqrt{1+x^2} \text { ] }$
$\Rightarrow \left(1+x^2\right) \frac{d^2 y}{d x^2}+x \frac{d y}{d x}=n \sqrt{1+x^2} \cdot \frac{n y}{\sqrt{1+x^2}}$
[from Eq. (ii)]
$\therefore\left(1+x^2\right) \frac{d^2 y}{d x^2}+x \frac{d y}{d x}=n^2 y$
Hence proved.

Given, $y=e^{m \sin ^{-1} x}$
On differentiating both sides of Eq.(i) w.r.t. x, we get $\frac{d y}{d x}=e^{m \sin ^{-1} x} \frac{d}{d x}\left(m \sin ^{-1} x\right)$
[by using chain rule of derivative]
$=e^{m \sin ^{-1} x} \cdot m \frac{1}{\sqrt{1-x^2}}$
$\Rightarrow \sqrt{1-x^2} \frac{d y}{d x}=m y$
Now, on squaring both sides, we get
$\left(1-x^2\right)\left(\frac{d y}{d x}\right)^2=m^2 y^2$
On differentiating both sides of Eq. (ii) w.r.t. x, we get
$\left(1-x^2\right) 2\left(\frac{d y}{d x}\right) \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2(-2 x)=2 m^2 y\left(\frac{d y}{d x}\right)$
$\Rightarrow \quad\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}=m^2 y$
$ {\left[\text { dividing both sides by } 2\left(\frac{d y}{d x}\right)\right]}$
$ \therefore \quad\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-m^2 y=0$
Hence proved.

Given $\sin y=x \cos (a+y)$
$\Rightarrow x = \frac{\sin y }{\cos (a+y)}$
On differentiating both sides w.r.t. y, we get
$\frac{d x}{d y} =\frac{\cos (a+y) \frac{d}{d y}(\sin y)-\sin y \frac{d}{d y} \cos (a+y)}{\cos ^2(a+y)} $
$\frac{d x}{d y} =\frac{\cos (a+y) \cos y+\sin y \sin (a+y)}{\cos ^2(a+y)} =\frac{\cos (a+y-y)}{\cos ^2(a+y)}$
$\Rightarrow \frac{d x}{d y} =\frac{[\because \cos A \cos B+\sin A \sin B=\cos (A-B)]}{\cos ^2(a+y)} \Rightarrow \frac{d y}{d x}=\frac{\cos ^2(a+y)}{\cos a}$
Put x = 0 in Eq. (i), we get y=0
Now, $\frac{d y}{d x}=\frac{\cos ^2(a+0)}{\cos a}=\frac{\cos ^2 a}{\cos a}=\cos a$
Hence proved.

Given, $(a+b x) e^{\frac{y}{x}}=x \Rightarrow e^{\frac{y}{x}}=\frac{x}{a+b x}$
Taking $\log _e$ both sides, we get
$\log _e e^{\frac{y}{x}}=\log _e\left(\frac{x}{a+b x}\right)$
$ \Rightarrow \quad \frac{y}{x} \log _e e=\log _e\left(\frac{x}{a+b x}\right)$
$\Rightarrow \quad \frac{y}{x}=\log _e \frac{x}{a+b x}$
${\left[\because \log _e e=1\right]}$
$\Rightarrow \frac{y}{x}=\log _e x-\log _e(a+b x)$
Differentiating w.r.t. x, we get
$\frac{x \frac{d y}{d x}-y}{x^2} =\frac{1}{x}-\frac{b}{a+b x}$
$\Rightarrow \quad \frac{x d y}{d x}-y =x^2\left(\frac{1}{x}-\frac{b}{a+b x}\right)=\frac{a x}{a+b x}$
Differentiating again w.r.t x, we get
$\Rightarrow \frac{x d^2 y}{d x^2}+\frac{d y}{d x}-\frac{d y}{d x}=\frac{(a+b x) a-a x(b)}{(a+b x)^2} $
$ \Rightarrow \frac{x d^2 y}{d x^2}=\frac{a^2}{(a+b x)^2}=\left(\frac{a}{a+b x}\right)^2$

Given, $e^y(x+1)=1$
On differentiating both sides w.r.t. x, we get
$e^y+(x+1) e^y \frac{d y}{d x}=0 $
$\Rightarrow e^y+1 \cdot \frac{d y}{d x}=0$[using Eq. (i)]
$\Rightarrow \quad e^y+\frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-e^y \text { Hence proved. }$

We have: $y={ \left( { sin }^{ -1 }x \right) }^{ 2 }$
${ y }_{ 1 }=2\left( { sin }^{ -1 }x \right) .\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } $
$\sqrt { 1-{ x }^{ 2 } } { y }_{ 1 }=2\left( { sin }^{ -1 }x \right) $
Squaring $\left( 1-{ x }^{ 2 } \right) { { y }_{ 1 } }^{ 2 }=4{ \left( { sin }^{ -1 }x \right) }^{ 2 }$
$ \left( 1-{ x }^{ 2 } \right) { { y }_{ 1 } }^{ 2 }=4y$
Diff. w.r.t.x, $\left( 1-{ x }^{ 2 } \right) 2{ y }_{ 1 }{ y }_{ 2 }+(-2x){ { y }_{ 1 } }^{ 2 }=4{ y }_{ 1 }$
Hence, $\left( 1-{ x }^{ 2 } \right) \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -x\frac { dy }{ dx } =2$

$x=a\left( cos\quad t+log\quad tan\frac { t }{ 2 } \right) ,$
$y=a\quad sin\ t$
$\frac { dx }{ dt } =a\left( -sin\quad t+\frac { 1 }{ tan\frac { t }{ 2 } } { sec }^{ 2 }\frac { t }{ 2 } \frac { 1 }{ 2 } \right) $
$=a\left( -sin\quad t+\frac { cos\frac { t }{ 2 } }{ sin\frac { t }{ 2 } } \frac { 1 }{ { cos }^{ 2 }\frac { t }{ 2 } } \frac { 1 }{ 2 } \right)$
$=a\left( -sin\quad t+\frac { 1 }{ { 2sin\frac { t }{ 2 } cos }\frac { t }{ 2 } } \right) $
$=a\left( -sin\quad t+\frac { 1 }{ sin\quad t } \right) $
$=a\left( \frac { 1-{ sin }^{ 2 }t }{ sin\quad t } \right) =a\frac { { cos }^{ 2 }t }{ sin\quad t } $
$\frac { dy }{ dt } =a\quad cos\ t$
$\frac { { d }^{ 2 }y }{ { dt }^{ 2 } } =-a\quad sin\ t$
$\frac { dy }{ dx } =\frac { dy/dt }{ dx/dt } =\frac { a\quad cos\quad t }{ \frac { { acos }^{ 2 }t }{ sin\quad t } } =\frac { sin\quad t }{ cos\quad t } =tan\ t$
$\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ sec }^{ 2 }t\frac { dt }{ dx } $
$={ sec }^{ 2 }t\frac { 1 }{ \frac { { acos }^{ 2 }t }{ sin\quad t } } =\frac { 1 }{ a } { sec }^{ 4 }t\ sin\ t$

We have: $y={ X }^{ x }$
Taking logs $logy\quad y=x\quad log\quad x$
$loy\quad y=x\quad log\quad x$
$\frac { 1 }{ y } \frac { dy }{ dx } =x.\frac { 1 }{ x } +log\quad x.1$
$\frac { dy }{ dx } =y(1+log\quad x)...(1)$
$\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =y.\frac { d }{ dx } (1+log\quad x)+\frac { dy }{ dx } (1+log\quad x)$
$\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =y.\left( 0+\frac { 1 }{ x } \right) +\frac { dy }{ dx } .\frac { \frac { dy }{ dx } }{ y } $
$\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =\frac { y }{ x } +\frac { 1 }{ y } { \left( \frac { dy }{ dx } \right) }^{ 2 }$
$\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -\frac { 1 }{ y } { \left( \frac { dy }{ dx } \right) }^{ 2 }-\frac { y }{ x } =0$

We have: $x=a\ cos\theta +b\ sin\theta ,\ y=a\ sin\theta -b\ cos\theta $
$\frac { dx }{ d\theta } =-a\quad sin\theta +b\quad cos\theta $
$\frac { dy }{ d\theta } =a\quad cos\theta +b\quad sin\theta $
$\frac { dy }{ dx } =\frac { dy/d\theta }{ dx/d\theta } =\frac { a\quad cos\theta +b\quad sin\theta }{ -a\quad sin\theta +b\quad cos\theta } $
$\frac { dy }{ dx } =\frac { x }{ -y } $
$\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =\frac { y(1)-x\frac { dy }{ dx } }{ { y }^{ 2 } } $
${ y }^{ 2 }\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =-y+x\frac { dy }{ dx } $
${ y }^{ 2 }\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -x\frac { dy }{ dx } +y=0$

10.

We have: ${ x }^{ 16 }{ y }^{ 9 }={ \left( { x }^{ 2 }+y \right) }^{ 17 },$
Taking logs., $log({ x }^{ 16 }{ y }^{ 9 })=log{ \left( { x }^{ 2 }+y \right) }^{ 17 }$
$16logx+9logy=17log\left( { x }^{ 2 }+y \right) $
$ \frac { 16 }{ x } +\frac { 9 }{ y } .\frac { dy }{ dx } =17\frac { 1 }{ { x }^{ 2 }+y } \left[ 2x+\frac { dy }{ dx } \right] $
$=\left[ \frac { 9 }{ y } -\frac { 17 }{ { x }^{ 2 }+y } \right] \frac { dy }{ dx } =\frac { 34x }{ { x }^{ 2 }+y } -\frac { 16 }{ x } $
$=\frac { { 9x }^{ 2 }+9y-17y }{ y\left( { x }^{ 2 }+y \right) } \frac { dy }{ dx } =\frac { { 34x }^{ 2 }-{ 16x }^{ 2 }-16y }{ x\left( { x }^{ 2 }+y \right) }$
$=\frac { 1 }{ y } \left( { 9x }^{ 2 }-8y \right) \frac { dy }{ dx } =\frac { 1 }{ x } \quad \left( { 18x }^{ 2 }-16y \right) $
$\frac { 1 }{ y } \quad \frac { dy }{ dx } =\frac { 2 }{ x } $
$\frac { dy }{ dx } =\quad \frac { 2y }{ x } $

Generic 2

4 x 2 = 8

11.

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

12.

{1,2}

13.

$\frac { -cos\left( { cos }^{ 2 }\sqrt { x } \right) sin\sqrt { x } cos\sqrt { x } }{ \sqrt { x } } $

14.

On differentiating e^y (x+1) = 1
e^y+(x+1)e^ydy/dx = 0
$\Rightarrow { e }^{ y }+\frac { dy }{ dx } =0$
$\Rightarrow \frac { dy }{ dx } ={ -e }^{ y }$

Multiple Choice Question

16 x 1 = 16

15.

(a)
2

16.

(d)
-1

17.

(a)
$x \in R$

18.

19.

(b)
x = 1.5

20.

21.

(a)
-1

22.

Given that,$y=(\cos x)^{(\cos x)^{(\cos x)} \cdots^{\infty}}$
$\Rightarrow \quad y=(\cos x)^{y}$
Taking log on both sides, we get
log y = y log(cos x)
Now, differentiating w.r.t. x, we get
$\frac{1}{y} \cdot \frac{d y}{d x}=y \cdot \frac{1}{\cos x}(-\sin x)+\log (\cos x) \cdot \frac{d y}{d x}$
$\Rightarrow \frac{d y}{d x}=y\left\{-y \tan x+\log \cos x \cdot \frac{d y}{d x}\right\} $
$\Rightarrow(1-y \log \cos x) \frac{d y}{d x}=-y^{2} \tan x $
$\Rightarrow \frac{d y}{d x}=\frac{y^{2} \tan x}{(y \log \cos x-1)} $

23.

We have $y^{x}=e^{y-x}$
Taking log both sides, we get
$ x \log y =y-x $
$ \Rightarrow x(\log y+1) =y $
$\Rightarrow \frac{y}{1+\log y} =x $
$\left(\frac{(1+\log y)-\left(\frac{1}{y}\right) y}{(1+\log y)^{2}}\right) \frac{d y}{d x}=1$
$ \frac{\log y}{(1+\log y)^{2}} \frac{d y}{d x}=1 $
$\Rightarrow \frac{d y}{d x}=\frac{(1+\log y)^{2}}{\log y} $

24.

$ \sqrt{1+\sin x}=\cos \frac{x}{2}+\sin \frac{x}{2} \text { and }\\ \sqrt{1-\sin x}=\cos \frac{x}{2}-\sin \frac{x}{2} $

25.

$\begin{array}{l} \text { LHL }=\lim _{x \rightarrow 0^{-}} \frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} \\ =\lim _{x \rightarrow 0^{-}} \frac{2 k x}{x(\sqrt{1+k x}+\sqrt{1-k x})}=k \end{array}$
$\begin{aligned} \mathrm{RHL} &=\lim _{x \rightarrow 0^{+}}\left(2 x^{2}+3 x-2\right)=-2 \\ f(0) &=-2 \end{aligned}$
$\therefore$ It is given that f(x) is continuous at x = 0.
$\therefore$ LHL= RHL = f(0) ⇒ k = - 2

26.

$ \lim _{x \rightarrow 0^{-}} f(x)=0 \text { and } \lim _{x \rightarrow 0^{+}} f(x)=1$

27.

(d)
Not continuous at x = 0

28.

29.

As differentiable functions is continuous also

30.

As $\lim _{ x\rightarrow 0 }{ \frac { \sqrt { \frac { 1 }{ 2 } (1-cosx) } }{ x } } $
$=\lim _{ x\rightarrow 0 }{ \frac { |sin \ x| }{ x } } $
and LHL ≠ RHL at x = 0

Assertion and reason

2 x 1 = 2

31.

(b) Assertion is correct, Reason is correct; Reason is not a correct explanation for Assertion

32.