Assertion (A) & Reason (B) MCQ

Question 11 Mark

Assertion $(A)$ : If $x=a t^2$ and $y=2 a t$, then $\left[\frac{d^2 y}{d x^2}\right]_{t=2}=\frac{-1}{16 a}$
Reason $(R) : \frac{d^2 y}{d x^2}=\left(\frac{d y}{d t}\right)^2 \times\left(\frac{d t}{d x}\right)^2$

Answer

$(c)$ : We have $, x=a t^2, y=2 a t$
$\Rightarrow \frac{d x}{d t}=2 a t $ and $\frac{d y}{d t} =2 a $
$\Rightarrow \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{1}{t}$
$\therefore \frac{d^2 y}{d x^2}=\frac{-1}{t^2} \times \frac{d t}{d x}=\frac{-1}{t^2} \times \frac{1}{2 a t}=\frac{-1}{2 a t^3}$
${\left[\frac{d^2 y}{d x^2}\right]_{t=2}=\frac{-1}{2 \times a \times(2)^3}=\frac{-1}{16 a}}$

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Question 21 Mark

Assertion (A) : If $y=\log _{10} x+\log _e x$, then $\frac{d y}{d x}=\frac{\log _{10} e}{x}+\frac{1}{x}$.
Reason (R): $\frac{d}{d x}\left(\log _{10} x\right)=\frac{\log x}{\log 10}$ and
$
\frac{d}{d x}\left(\log _e x\right)=\frac{\log x}{\log e}
$

Answer

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Question 31 Mark

Assertion $(A)$: If $e^{x y}+\log (x y)+\cos (x y)+5=0$, then $\frac{d y}{d x}=-\frac{y}{x}$.
Reason $(R) : \frac{d}{d x}(x y)=0 \Rightarrow \frac{d y}{d x}=\frac{-y}{x}$

Answer

$\text { (a) : } e^{x y}+\log (x y)+\cos (x y)+5=0$
$\therefore e^{x y} \frac{d}{d x}(x y)+\frac{1}{(x y)} \frac{d}{d x}(x y)-\sin (x y) \frac{d}{d x}(x y)=0$
$\Rightarrow \frac{d}{d x}(x y)\left\{e^{x y}+\frac{1}{x y}-\sin (x y)\right\}=0$
$\because e^{x y}+\frac{1}{x y}-\sin (x y) \neq 0 $
$\therefore \frac{d}{d x}(x y)=0$
$\Rightarrow x \frac{d y}{d x}+y \cdot 1=0 $
$\Rightarrow \frac{d y}{d x}=-\frac{y}{x}$

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Question 41 Mark

Assertion (A) : If $y=\cot ^{-1}\left(\frac{1+x \sqrt{x}}{\sqrt{x}-x}\right)$, then $\frac{d y}{d x}=\frac{-1}{1+x^2}+\frac{1}{2 \sqrt{x}(1+x)}$
Reason (R) : $\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^2}$ and $\frac{d}{d x}\left(\tan ^{-1} \frac{1}{x}\right)=1+x^2$.

Answer

(c) : We have, $y=\cot ^{-1}\left(\frac{1+x \sqrt{x}}{\sqrt{x}-x}\right)$
Let $x=\cot \phi$ and $\sqrt{x}=\cot \theta$
Then $y=\cot ^{-1}\left(\frac{1+\cot \theta \cot \phi}{\cot \theta-\cot \phi}\right)=\phi-\theta=\tan ^{-1} \frac{1}{x}-\tan ^{-1} \frac{1}{\sqrt{x}}$
$
\therefore \quad \frac{d y}{d x}=\frac{-1}{1+x^2}+\frac{1}{1+x} \times \frac{1}{2 \sqrt{x}}
$

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Question 51 Mark

Assertion $(A) :$ If $u=f(\sin x), v=g(\cos x)$ and $f^{\prime}\left(\frac{1}{\sqrt{2}}\right)=2, g^{\prime}\left(\frac{1}{\sqrt{2}}\right)=4$, then $\left(\frac{d u}{d v}\right)_{x=\pi / 4}=\frac{1}{\sqrt{2}}$.
Reason $(R):$ If $u=f(x), v=g(x)$, then the derivative of $f$ with respect to $g$ is $\frac{d u}{d v}=\frac{d u / d x}{d v / d x}$.

Answer

$\text { (d): Given, } u=f(\sin x) \Rightarrow \frac{d u}{d x}=f^{\prime}(\sin x) \cdot \cos x$
$\text { and } v=g(\cos x) \Rightarrow \frac{d v}{d x}=-g^{\prime}(\cos x) \cdot \sin x$
$\therefore \frac{d u}{d v}=\frac{(d u / d x)}{(d v / d x)}=\frac{f^{\prime}(\sin x)}{g^{\prime}(\cos x)} \cdot\left(\frac{-\cos x}{\sin x}\right)$
$\therefore\left(\frac{d u}{d v}\right)_{x=\pi / 4}=\frac{f^{\prime}\left(\frac{1}{\sqrt{2}}\right)}{g^{\prime}\left(\frac{1}{\sqrt{2}}\right)} \cdot(-1)=\frac{2}{4} \cdot(-1)=-\frac{1}{2}$

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Question 61 Mark

Assertion $(A)$ : If $y=\frac{1}{4} u^4$ and $u=\frac{2}{3} x^3+5$ then $\frac{d y}{d x}=\frac{2}{27} x^2\left(2 x^3+15\right)^3$.
Reason $(R)$ : If $y$ is a function of $v$ and $v$ is a function of $x$, then $\frac{d y}{d x}=\frac{d y}{d v} \times \frac{d v}{d x}$.

Answer

$(a)$ : We have, $y=\frac{1}{4} u^4 $
$\Rightarrow \frac{d y}{d u}=\frac{1}{4} \cdot 4 u^3=u^3$ and $u=\frac{2}{3} x^3+5 $
$\Rightarrow \frac{d u}{d x}=\frac{2}{3} \cdot 3 x^2=2 x^2$
$\therefore \frac{d y}{d x}=\frac{d y}{d u} \cdot \frac{d u}{d x}$
$=u^3 \cdot 2 x^2=\left(\frac{2}{3} x^3+5\right)^3\left(2 x^2\right)$
$=\frac{2}{27} x^2\left(2 x^3+15\right)^3$

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Question 71 Mark

Consider the function $f(x)=\left\{\begin{array}{cc}x^2, & x \geq 1 \\ x+1, & x<1\end{array}\right.$
Assertion (A) : $f$ is not derivable at $x=1$ as $\lim _{x \rightarrow 1^{-}} f(x) \neq \lim _{x \rightarrow 1^{+}} f(x)$.
Reason (R) : If a function $f$ is derivable at a point ' $a$ ', then it is continuous at ' $a$ '.

Answer

(a) : Reason is a standard result.
Also $\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}(x+1)=2$
and $\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} x^2=1$
$
\Rightarrow \quad \lim _{x \rightarrow 1^{-}} f(x) \neq \lim _{x \rightarrow 1^{+}} f(x)
$
$\Rightarrow f$ is not continuous at $x=1$
$\Rightarrow f$ is not derivable at $x=1$ (From Reason)

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Question 81 Mark

Consider the function $f(x)=[\sin x], x \in[0, \pi]$.
Assertion $(A) : f(x)$ is not continuous at $x=\frac{\pi}{2}$.
Reason $(R) : \lim _{x \rightarrow \frac{\pi}{2}} f(x)$ does not exist.

Answer

$(c) :$ We know that for all
$x \in\left[0, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, \pi\right], 0<\sin x<1$
$\Rightarrow \quad[\sin x]=0$
$\therefore \lim _{x \rightarrow \frac{\pi}{2}}[\sin x]=0$
Thus, we see that the Reason is not true.
Also, $f\left(\frac{\pi}{2}\right)=\left[\sin \frac{\pi}{2}\right]=1$
$\Rightarrow \lim _{x \rightarrow \frac{\pi}{2}} f(x) \neq f\left(\frac{\pi}{2}\right)$
$\therefore f$ is not continuous at $x=\frac{\pi}{2}$

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MATHS Assertion (A) & Reason (B) MCQ

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