MCQ 11 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A)$ If $\text{y}=\log_7(\text{x}^2+7\text{x}+4),$ then $\frac{\text{dy}}{\text{dx}}=\frac{(2\text{x}+7)}{(\text{x}^2+7\text{x}+4),}$
Reason $(R) \log_\text{b}=\frac{\log_\text{e}}{\log_\text{e}\text{b}}$
- 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.
- ✓
$A$ is false but $R$ is true.
AnswerCorrect option: D. $A$ is false but $R$ is true.
View full question & answer→MCQ 21 Mark
Directions: In the following questions, the Assertions (A) and Reason(s) (R) have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion (A) If $x^2 + 2xy + y^3 = 42$, Then $\frac{\text{dy}}{\text{dx}}=\frac{2(\text{x+y})}{(2\text{x+3}\text{y}^2)}$
Reason(R) $\frac{\text{dy}^\text{n}}{\text{dx}}=\text{ny}^{(\text{n-1})}$
- 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
- ✓
View full question & answer→MCQ 31 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A) \frac{\text{dx}^{\sin\text{x}}}{\text{dx}}=\text{x}^{\sin\text{x}}[(\cos)\log\text{x}+\frac{\sin\text{x}}{\text{x}}]$
Reason $(R)$ if $y = x^{f(x)}$ then $\frac{\text{dy}}{\text{dx}}=\text{x}^\text{f(x)}[\text{f '(x)}\log\text{x}+\frac{\text{f(x)}}{\text{x}}]$
- ✓
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
AnswerCorrect option: A. Both $A$ and $R$ are true and $R$ is the correct explanation of $A$
View full question & answer→MCQ 41 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A):$ Acontinuous funection is always differentiable.
Reason $(R):$ Adifferentiable function is always continuous.
- 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.
- ✓
$A$ is false but $R$ is true.
AnswerCorrect option: D. $A$ is false but $R$ is true.
View full question & answer→MCQ 51 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A)$ if $\text{y}=\sin^{-1}\frac{2\text{x}}{1+\text{x}^2}$ then $\frac{\text{dy}}{\text{dx}}=\frac{2}{1+\text{x}^2}$
Reason $(R) \sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$
- ✓
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.
AnswerCorrect option: A. Both $A$ and $R$ are true and $R$ is the correct explanation of $A.$
View full question & answer→MCQ 61 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A) f(x) = x - 1 + x - 2$ is continuous but not differentiable at $x = 1, 2.$
Reason $(R)$ Every differentiable function is continuous
- A
Both $A$ and $R$ are true and $R$ is the correct explanation of $A.$
- ✓
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.
AnswerCorrect option: B. Both $A$ and $R$ are true but $R$ is not the correct explanation of $A.$
View full question & answer→MCQ 71 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion: $\text{u}=\text{f}(\cot\text{x})\&\text{f}(1)=\sqrt2$ and $\text{g}(\sqrt{2})=2$ then $\Big(\frac{\text{du}}{\text{dv}}\Big)_{\text{x}=\frac{\text{x}}{4}}=1.$
Reason: If $u = f(x), v = g(x)$ then derivative of $\text{f w.r.t.}$ to $g$ is $\frac{\text{du}}{\text{dv}}=\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}.$
- ✓
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
- B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.
- C
Assertion is correct but Reason is incorrect.
- D
Both Assertion and Reason are incorrect.
AnswerCorrect option: A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
View full question & answer→MCQ 81 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A)$ The value of the constant $'k\ ’$ so that $\text{f(x)}=\begin{cases}\text{kx}^2,\text{if x}\leq2\\3,\text{if x}>2\end{cases}$ is continuous at $x = 2$ is $\text{k}=\frac{4}{3}$
Reason $(R)$ A function $f(x)$ is continuous at a point $x= a$ of its domain if $\lim\limits_{\text{x}\rightarrow 0}\text{f(x)}=\text{f(x)}$
- 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.
- ✓
$A$ is false but $R$ is true.
AnswerCorrect option: D. $A$ is false but $R$ is true.
View full question & answer→MCQ 91 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A):\ \text{f(x)}=\sin\text{x}$ is continuous $x = 0.$
Reason $(R):\ \sin\text{x}$ is differentiable at $x = 0.$
- 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.
AnswerCorrect option: C. $A$ is true but $R$ is false.
View full question & answer→MCQ 101 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A):\ \text{f(x)}=\sin\text{x}$ is continuous for all $\text{x }\in\text{ R}$
Reason $(R):\ \sin\text{x}$ and $\text{x}$ are continuous at on $R.$
- ✓
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.
AnswerCorrect option: A. Both $A$ and $R$ are true and $R$ is the correct explanation of $A.$
View full question & answer→MCQ 111 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A) f(x) = [x]$ greatest integer function is not differentiable at $x = 2$
Reason $(R)$ The greatest integer function is not continuous at any integer
- ✓
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.
AnswerCorrect option: A. Both $A$ and $R$ are true and $R$ is the correct explanation of $A.$
View full question & answer→MCQ 121 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A)$ The function $\text{f(x)}=\begin{cases}12\text{x} -13 , \text{if x}\leq3\\2\text{x}^2+5,\text{if x}>3\end{cases}$ is differentiable at $x = 3.$
Reason $(R)$ The function $f(x)$ is differentiable at $x = c$ of its domain if Left hand derivative of $f$ at $c =$ Right hand derivative of $f$ at $c.$
- A
Both $A$ and $R$ are true and $R$ is the correct explanation of $A.$
- ✓
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.
AnswerCorrect option: B. Both $A$ and $R$ are true but $R$ is not the correct explanation of $A.$
View full question & answer→MCQ 131 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A): \text{f(x)}=\tan^2\text{x}$ is continuous at $\text{x}=\frac{\pi}{2}$
Reason ($R): ?^2$ is continuous at $\text{x}=\frac{\pi}{2}$
- 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.
- ✓
$A$ is false but $R$ is true.
AnswerCorrect option: D. $A$ is false but $R$ is true.
View full question & answer→MCQ 141 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion$(A)$ if $\text{y}=\tan^{-1}\Big(\frac{\cos\text{x}+\sin\text{x}}{\sin\text{x}-\cos\text{x}}\Big) ,\frac{-\pi}{4}<\text{x}<\frac{\pi}{4},$then$\frac{\text{dy}}{\text{dx}}=-1$
Reason $(R) \frac{\cos\text{x}+\sin\text{x}}{\sin\text{x}-\cos\text{x}}=\tan\Big(\text{x}+\frac{\pi}{4}\Big)$
- 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.
- ✓
$A$ is false but $R$ is true.
AnswerCorrect option: D. $A$ is false but $R$ is true.
View full question & answer→MCQ 151 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A)$ If $\text{f(x)}=\cos,\text{then}\text{ f '}\Big(\frac{\pi}{4}\Big)=\frac{-1}{\sqrt{2}}$ and $\text{ f '}\Big(\frac{3\pi}{4}\Big)=\frac{1}{\sqrt{2}}$
Reason $(R)\ \text{f(x)}=\cos=\begin{cases}\cos\text{x },0\leq\text{x} \leq\frac{\pi}{2}\\-\cos\text{x },\text{if }\frac{\pi}{2}<\text{x}\leq\pi\end{cases}$
- ✓
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.
AnswerCorrect option: A. Both $A$ and $R$ are true and $R$ is the correct explanation of $A.$
View full question & answer→MCQ 161 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion: $\text{f}(\text{x})=\begin{cases}\text{x}^2\sin\big(\frac{1}{\text{x}}\big), &\text{x}=0\\0, &\text{x}=0\end{cases}$ is continuous at $x = 0.$
Reason: Both $\text{h}(\text{x})=\text{x}^2,\text{g}(\text{x})=\begin{cases}\text{x}^2\sin\big(\frac{1}{\text{x}}\big), &\text{x}=0\\0, &\text{x}=0\end{cases}$ are continuous at $x = 0.$
- A
Both $A$ and $R$ are true and $R$ is the correct explanation of $A.$
- B
Both $A$ and $R$ are true and $R$ is not the correct explanation of $A.$
- ✓
$A$ is true but $R$ is false.
- D
$R$ is true but $A$ is false.
AnswerCorrect option: C. $A$ is true but $R$ is false.
Assertion: $\text{f}(0)=\lim\limits_{\text{x}\rightarrow0}\text{x}^2\sin\big(\frac{1}{\text{x}}\big)=0^2\times($finite value$)=0$
$\therefore$ It is continuous at $x = 0$
Reason: $h(x) = x^2$ is continuous but $g(x)$ is not continuous
$\lim\limits_{\text{x}\rightarrow0}\sin\big(\frac{1}{\text{x}}\big) =$ not defined $($value oscillates$)$
$\lim\limits_{\text{x}\rightarrow0}\sin\big(\frac{1}{\text{x}}\big)=0$
$\therefore$ not continuous.
View full question & answer→MCQ 171 Mark
Directions: In the following questions, the Assertions (A) and Reason(s) (R) have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion (A) If $x = at^2$ and $y = 2$ at where ‘t’ is the parameter and ‘a’ is a constant, then $\frac{\text{d}^2\text{y}}{\text{dy}^2}= \frac{-1}{\text{t}^2}.$
Reason(R ) $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}^2\text{y}}{\text{dt}^2}\div\frac{\text{d}^2\text{x}}{\text{dt}^2}$
- 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
- ✓
View full question & answer→MCQ 181 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion$(A): f(x) = [x]$ is not differentiableat $x = 2.$
Reason$(R): f(x) = [x]$ is not continuous at $x = 2.$
- ✓
Both $A$ and $R$ are true and $R$ is the correct explanation of $A$
- B
Both $A$ and $R$ are true but $R$ is $\text{NOT}$ the correct explanation of $A$
- C
$A$ is true but $R$ is false.
- D
$A$ is false but $R$ is true.
AnswerCorrect option: A. Both $A$ and $R$ are true and $R$ is the correct explanation of $A$
View full question & answer→MCQ 191 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A):$ If $\text{f(x)}=\text{x}+\begin{vmatrix}\text{x}+2&\text{ab}\\\text{ab}&\text{x}+\text{b}^2\end{vmatrix}$ then $\text{f'(x)}=2\text{x}+\text{a}^2+\text{b}^2$
Reason$(R):$ If $\triangle=\begin{vmatrix}\text{f(x)}&\text{g(x)} \\ \text{u(x)}&\text{g(x)} \end{vmatrix},$ Then $\frac{\text{d}\triangle}{\text{dx}}=\begin{vmatrix}\text{f'(x)}&\text{g'(x)} \\ \text{u(x)}&\text{g(x)} \end{vmatrix}+\begin{vmatrix}\text{f(x)}&\text{g(x)} \\ \text{u'(x)}&\text{g'(x)} \end{vmatrix}$
- ✓
Both $A$ and $R$ are true and $R$ is the correct explanation of $A$
- B
Both $A$ and $R$ are true but $R$ is $\text{NOT}$ the correct explanation of $A.$
- C
$A$ is true but $R$ is false
- D
$A$ is false but $R$ is true
AnswerCorrect option: A. Both $A$ and $R$ are true and $R$ is the correct explanation of $A$
View full question & answer→MCQ 201 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A):$ The derivative of $\log\sin\text{x}\text{ w.r.t}\sqrt{\cos\text{x}}$ is $2\sqrt{\cos\text{x}} \cos\text{x } \text{cosec x}$
Reason$(R):$ The derivative of $\text{u w.r.t. v}$ is $\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}$
- 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 $\text{NOT}$ the correct explanation of $A.$
- C
$A$ is true but $R$ is false
- D
$A$ is false but $R$ is true
View full question & answer→MCQ 211 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 $(A).$
- ✓
$(A)$ is true but $(R)$ is false.
- D
$(A)$ is false but $(R)$ is true.
AnswerCorrect option: C. $(A)$ is true but $(R)$ is false.
We have, $x=a t^2, y=2 a t$
$\Rightarrow \frac{d x}{d t}=2 a t$ and $\frac{d y}{d t}=2 a$
$\Rightarrow \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{1}{t}$
$\therefore \frac{d^2 y}{d x^2}=\frac{-1}{t^2} \times \frac{d t}{d x}=\frac{-1}{t^2} \times \frac{1}{2 a t}=\frac{-1}{2 a t^3}$
${\left[\frac{d^2 y}{d x^2}\right]_{t=2}=\frac{-1}{2 \times a \times(2)^3}=\frac{-1}{16 a}}$
View full question & answer→MCQ 221 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.
- ✓
$(A)$ is false but $(R)$ is true.
AnswerCorrect option: D. $(A)$ is false but $(R)$ is true.
Given, $u=f(\sin x)$
$\Rightarrow \frac{d u}{d x}=f^{\prime}(\sin x) \cdot \cos x$
and $v=g(\cos x)$
$\Rightarrow \frac{d v}{d x}=-g^{\prime}(\cos x) \cdot \sin x$
$\therefore \frac{d u}{d v}=\frac{(d u / d x)}{(d v / d x)}=\frac{f^{\prime}(\sin x)}{g^{\prime}(\cos x)} \cdot\left(\frac{-\cos x}{\sin x}\right)$
$\therefore\left(\frac{d u}{d v}\right)_{x=\pi / 4}=\frac{f^{\prime}\left(\frac{1}{\sqrt{2}}\right)}{g^{\prime}\left(\frac{1}{\sqrt{2}}\right)} \cdot(-1)=\frac{2}{4} \cdot(-1)=-\frac{1}{2}$
View full question & answer→MCQ 231 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.
AnswerCorrect option: C. (A) is true but (R) is false.
(c) : We have, $y=\log _{10} x+\log _e x$
$
\therefore \quad \frac{d y}{d x}=\frac{1}{x} \log _{10} e+\frac{1}{x}
$
View full question & answer→MCQ 241 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}$
- ✓
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.
AnswerCorrect option: A. Both $(A)$ and $(R)$ are true and $(R) $ is the correct explanation of $(A)$.
$e^{x y}+\log (x y)+\cos (x y)+5=0$
$\therefore e^{x y} \frac{d}{d x}(x y)+\frac{1}{(x y)} \frac{d}{d x}(x y)-\sin (x y) \frac{d}{d x}(x y)=0$
$\Rightarrow \frac{d}{d x}(x y)\left\{e^{x y}+\frac{1}{x y}-\sin (x y)\right\}=0$
$\because e^{x y}+\frac{1}{x y}-\sin (x y) \neq 0$
$ \therefore \frac{d}{d x}(x y)=0$
$\Rightarrow x \frac{d y}{d x}+y \cdot 1=0 $
$\Rightarrow \frac{d y}{d x}=-\frac{y}{x}a$
View full question & answer→MCQ 251 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.
AnswerCorrect option: A. Both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A).$
We have, $y=\frac{1}{4} u^4$
$\Rightarrow \frac{d y}{d u}=\frac{1}{4} \cdot 4 u^3=u^3$ and $u=\frac{2}{3} x^3+5$
$\Rightarrow \frac{d u}{d x}=\frac{2}{3} \cdot 3 x^2=2 x^2$
$\therefore \frac{d y}{d x}=\frac{d y}{d u} \cdot \frac{d u}{d x}$
$=u^3 \cdot 2 x^2$
$=\left(\frac{2}{3} x^3+5\right)^3\left(2 x^2\right)$
$=\frac{2}{27} x^2\left(2 x^3+15\right)^3$
View full question & answer→MCQ 261 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.
AnswerCorrect 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}}
$
View full question & answer→MCQ 271 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.
AnswerCorrect 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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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.
AnswerCorrect option: C. $(A)$ is true but $(R)$ is false.
We know that for all
$x \in\left[0, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, \pi\right], 0<\sin x<1$
$\Rightarrow[\sin x]=0$
$\therefore \lim _{x \rightarrow \frac{\pi}{2}}[\sin x]=0$
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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