Online Test — Continuity and Differentiability
10 Questions • 20 min • Chapter MCQ
20:00
Question 1 of 10
easy
$f(x)=|x|$ is:
Differentiable everywhere
Continuous but not differentiable at $x=0$
Neither continuous nor differentiable at $x=0$
Differentiable but not continuous at $x=0$
Explanation: $|x|$ is continuous everywhere. At $x=0$, LHD $=-1 \ne$ RHD $=+1$. Not differentiable.
Question 2 of 10
easy
$\frac{d}{dx}(\ln\sin x)=$
$\cot x$
$\tan x$
$\frac{1}{\sin x}$
$-\cot x$
Explanation: Chain rule: $\frac{1}{\sin x}\cdot\cos x = \cot x$.
Question 3 of 10
medium
If $y=e^{\sin^{-1}x}$, then $\frac{dy}{dx}=$
$\frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}}$
$e^{\sin^{-1}x}\sqrt{1-x^2}$
$\frac{1}{\sqrt{1-x^2}}$
$e^{\cos x}$
Explanation: $dy/dx = e^{\sin^{-1}x} \cdot \frac{1}{\sqrt{1-x^2}}$ by chain rule.
Question 4 of 10
hard
If $x=t^2$ and $y=t^3$, then $\frac{d^2y}{dx^2}=$
$\frac{3}{4t}$
$\frac{3t}{2}$
$\frac{3}{2}$
$\frac{1}{2t}$
Explanation: $dy/dx=(3t^2)/(2t)=3t/2$. $\frac{d}{dt}(dy/dx)=3/2$. $d^2y/dx^2=(3/2)/(2t)=3/(4t)$.
Question 5 of 10
medium
$\frac{d}{dx}\sec^{-1}x=$
$\frac{1}{\sqrt{x^2-1}}$
$\frac{1}{|x|\sqrt{x^2-1}}$
$\frac{-1}{|x|\sqrt{x^2-1}}$
$\frac{1}{x\sqrt{x^2+1}}$
Explanation: Standard result: $\frac{d}{dx}\sec^{-1}x = \frac{1}{|x|\sqrt{x^2-1}}$.
Question 6 of 10
medium
The function $f(x) = \begin{cases}kx^2 & x\le2 \\ 4x-4 & x>2\end{cases}$ is continuous at $x=2$ when $k=$
1
2
$1/2$
3
Explanation: $\lim_{x\to2^+}(4x-4)=4$. At $x=2$: $kx^2=4k$. Continuous requires $4k=4$, so $k=1$.
Question 7 of 10
medium
Rolle's theorem is NOT applicable for $f(x)=|x|$ on $[-1,1]$ because:
$f(-1)\ne f(1)$
$f$ is not differentiable at $x=0$
$f$ is not continuous on $[-1,1]$
The function is not bounded
Explanation: $f(-1)=f(1)=1$ and $f$ is continuous. But $f$ is NOT differentiable at $x=0\in(-1,1)$. Rolle's theorem requires differentiability on the open interval.
Question 8 of 10
hard
If $y = (\sin x)^{\cos x}$, then $\frac{dy}{dx}=$
$(\sin x)^{\cos x}(\cos x \cdot \cot x - \sin x \cdot \ln\sin x)$
$(\sin x)^{\cos x}\cos x\cot x$
$\cos x(\sin x)^{\cos x-1}$
$(\sin x)^{\cos x}\ln(\sin x)$
Explanation: $\ln y=\cos x\ln\sin x$. Differentiate: $(1/y)y'=-\sin x\ln\sin x+\cos x\cot x$. So $y'=(\sin x)^{\cos x}(\cos x\cot x-\sin x\ln\sin x)$.
Question 9 of 10
hard
If $x^y = e^{x-y}$, then $\frac{dy}{dx}=$
$\frac{\ln x}{(1+\ln x)^2}$
$\frac{1+\ln x}{\ln x}$
$\frac{\ln x}{1+\ln x}$
$\frac{1}{1+\ln x}$
Explanation: Taking log: $y\ln x=x-y$. $y(1+\ln x)=x$. $y=x/(1+\ln x)$. $dy/dx=\frac{(1+\ln x)-x(1/x)}{(1+\ln x)^2}=\frac{\ln x}{(1+\ln x)^2}$.
Question 10 of 10
hard
The value of $c$ in Lagrange's MVT for $f(x)=\sqrt{x}$ on $[1,4]$ is:
$9/4$
$5/2$
$3/2$
$2$
Explanation: $f'(c)=(f(4)-f(1))/(4-1)=(2-1)/3=1/3$. $f'(x)=1/(2\sqrt{x})=1/3 \Rightarrow \sqrt{x}=3/2 \Rightarrow x=9/4$.