CUET Mathematics: CalculusContinuity & Differentiability. Free, no login required.

If $f(x) = \begin{cases} \frac{\tan(\frac{\pi}{4} - x)}{\cot 2x} & , x ≠ \frac{\pi}{4} \\ 2K + 1 & , x = \frac{\pi}{4} \end{cases}$ is continuous at $x = \frac{\pi}{4}$, then the value of K is equal to

If $e^x + e^y = e^{x+y}$, then $\frac{dy}{dx}$ =

If $y^{1/m} + y^{-1/m} = 2x$, then the value of $(x^2 - 1)\frac{d^2y}{dx^2} + x\frac{dy}{dx}$ is:

The value of k for which the function $f(x) = \begin{cases} \frac{1-\cos 8x}{16x^2}, & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases}$ is continuous at $x = 0$ is:

If $x = e^{\cos 2t}$, $y = e^{\sin 2t}$, then $\frac{dy}{dx}$ equals to

If $y = \sqrt{x + \sqrt{x + \sqrt{x + ...\ ...\ ...}}}$, then

If $y = (log x)^{(log x)}$, $x > 1$ then $\frac{dy}{dx}$ is equal to

If $g(x) = \begin{cases} \frac{αx}{|x|}, & \text{if } x < 0 \\ 5, & \text{if } x ≥ 0 \end{cases}$ is continuous at x = 0, then the value of α is

If $y = e^{acos^{-1}x}, -1 < x < 1$, then $(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx}$ is equal to

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If $y = \left(\log\left(x + \sqrt{x^2+a^2}\right)\right)^2$ and $x \neq \frac{1-a^2}{2}$, then $(x^2+a^2)\frac{d^2y}{dx^2} + x\frac{dy}{dx}$ is equal to:

Match List-I with List-II

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For $x > y > 0$, if $x^5 y^6 = (x + y)^{11}$, then $\frac{d^2y}{dx^2}$ is

If $x = a\cos\alpha + b\sin\alpha$ and $y = a\sin\alpha - b\cos\alpha$, then $\left(x\frac{dy}{dx} - y^2\frac{d^2y}{dx^2}\right)$ is equal to:

The value of k for which the function, defined by, $f(x) = \begin{cases} \frac{3x + 4 \tan x}{x} & : x \neq 0 \\ k & : x = 0 \end{cases}$ is continuous at $x = 0$, is

Which of the following functions $f(x)$ are differentiable at $x = 0$?

(A) $|x|$

(B) $|x - 1|$

(C) $|\sin x|$

(D) $|\cos x|$

(E) $x^2$

Choose the correct answer from the options given below:

If $e^y(x + 1) = 1$, then

If $x = 4t$ and $y = \frac{4}{t}$, then $\frac{d^2y}{dx^2}$ is

If the function defined by $f(x) = \begin{cases} \ kx^2 + 1, & \text{if } x \le 1 \\ 2 , & \text{if } x > 1 \end{cases}$ is continuous at $x = 1$, then k is equal to

The function $f: \mathbb{R} \to \mathbb{R}$ defined by

$f(x) = \begin{cases} x^2, & x \ge 1 \\ x, & x < 1 \end{cases}$

is

If $y = \frac{1}{1+x^{b-a}+x^{c-a}} + \frac{1}{1+x^{c-b}+x^{a-b}} + \frac{1}{1+x^{a-c}+x^{b-c}}$ then $\frac{d^2y}{dx^2}$ is

Match List-I with List-II

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Match List-I with List-II. Here [x] denotes the greatest integer function

$\begin{array}{|l|l|} \hline \rule{0pt}{2.8ex}\text{List-I} & \text{List-II} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(A) } f(x) = [x] & \text{(I) is continuous everywhere but not differentiable at } x=-1 \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(B) } f(x) = |x-1| & \text{(II) is continuous everywhere except at all integral values} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(C) } f(x) = e^{|x|} & \text{(III) is continuous everywhere but not differentiable at } x=1 \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(D) } f(x) = |x+1| & \text{(IV) is continuous everywhere but not differentiable at } x=0 \\[1.2ex] \hline \end{array}$

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If $y = x\sin y$, then $\frac{dy}{dx}$ is:

If $x^2 - y^2 = t - \frac{1}{t}$, and $x^4 + y^4 = t^2 + \frac{1}{t^2}$, then which of the following is correct?

If $y = 3e^{2x} + 2e^{3x}$, then $\frac{d^2y}{dx^2} + 6y$ is equal to

Match List-I with List-II

$\begin{array}{|l|l|} \hline \rule{0pt}{2.8ex}\text{List-I} & \text{List-II} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(A) } f(x) = |x| & \text{(I) Not differentiable at } x=-2 \text{ only} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(B) } f(x) = |x+2| & \text{(II) Not differentiable at } x=0 \text{ only} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(C) } f(x) = |x^2-4| & \text{(III) Not differentiable at } x=2 \text{ only} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(D) } f(x) = |x-2| & \text{(IV) Not differentiable at } x=2,-2 \text{ only} \\[1.2ex] \hline \end{array}$

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Let $y = \sin(\cos x^2)$, then the value of $\frac{dy}{dx}$ at $x = \frac{\sqrt{\pi}}{2}$ is equal to

If the function $f(x) = \begin{cases} \frac{k\cos x}{\pi - 2x} & : x \neq \frac{\pi}{2} \\ 3 & : x = \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$, then $k$ is equal to

If $e^y = \log x$, then which of the following is true?