Assertion (A) & Reason (B) MCQ

MCQ 11 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 $\frac{d y}{d x}=\frac{d y}{d v} \times \frac{d v}{d x}$.

✓
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
C
(A) is true but (R) is false.
D
(A) is false but (R) is true.

Answer

Correct option: A.

Both (A) and (R) are true and (R) is the correct explanation of (A).

(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 $u=\frac{2}{3} x^3+5 \Rightarrow \frac{d u}{d x}=\frac{2}{3} \cdot 3 x^2=2 x^2$
$
\begin{array}{l}
\therefore \quad \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
\end{array}
$

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MCQ 21 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): $\frac{d^2 y}{d x^2}=\left(\frac{d y}{d t}\right)^2 \times\left(\frac{d t}{d x}\right)^2$

A
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
C
(A) is true but (R) is false.
D
(A) is false but (R) is true.

Answer

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

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MCQ 31 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}
$

A
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
✓
(A) is true but (R) is false.
D
(A) is false but (R) is true.

Answer

Correct option: C.

(A) is true but (R) is false.

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MCQ 41 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) : $\frac{d}{d x}(x y)=0 \Rightarrow \frac{d y}{d x}=\frac{-y}{x}$

A
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
C
(A) is true but (R) is false.
D
(A) is false but (R) is true.

Answer

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

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MCQ 51 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 $\frac{d}{d x}\left(\tan ^{-1} \frac{1}{x}\right)=1+x^2$.

A
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
✓
(A) is true but (R) is false.
D
(A) is false but (R) is true.

Answer

Correct option: C.

(A) is true but (R) is false.

(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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MCQ 61 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 $\frac{d u}{d v}=\frac{d u / d x}{d v / d x}$.

A
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
C
(A) is true but (R) is false.
D
(A) is false but (R) is true.

Answer

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

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MCQ 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 ' $a$ '.

✓
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
C
(A) is true but (R) is false.
D
(A) is false but (R) is true.

Answer

Correct option: A.

Both (A) and (R) are true and (R) is the correct explanation of (A).

(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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MCQ 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.

A
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
✓
(A) is true but (R) is false.
D
(A) is false but (R) is true.

Answer

Correct option: C.

(A) is true but (R) is false.

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

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