List of top Mathematics Questions on Continuity and differentiability asked in CUET (UG)

Let \( y=\sin(\cos(x^2)) \). Find \( \frac{dy}{dx} \) at \( x=\frac{\sqrt{\pi}}{2} \).
Let \( y=\sin(\cos(x^2)) \). Find \( \frac{dy}{dx} \) at \( x=\frac{\sqrt{\pi}}{2} \).
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

Match List-I with List-II

List-IList-II
(A) \( f(x) = |x| \)(I) Not differentiable at \( x = -2 \) only
(B) \( f(x) = |x + 2| \)(II) Not differentiable at \( x = 0 \) only
(C) \( f(x) = |x^2 - 4| \)(III) Not differentiable at \( x = 2 \) only
(D) \( f(x) = |x - 2| \)(IV) Not differentiable at \( x = 2, -2 \) only


Choose the correct answer from the options given below:

Match List-I with List-II

List-IList-II
(A) \( f(x) = |x| \)(I) Not differentiable at \( x = -2 \) only
(B) \( f(x) = |x + 2| \)(II) Not differentiable at \( x = 0 \) only
(C) \( f(x) = |x^2 - 4| \)(III) Not differentiable at \( x = 2 \) only
(D) \( f(x) = |x - 2| \)(IV) Not differentiable at \( x = 2, -2 \) only


Choose the correct answer from the options given below:

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
Let [x] denote the greatest integer function. Then match List-I with List-II:
[x] denote the greatest integer function
\(\text{ If } f(x), \text{ defined by } f(x) = \begin{cases}  kx + 1 & \text{if } x \leq \pi \\  \cos x & \text{if } x > \pi  \end{cases} \text{ is continuous at } x = \pi, \text{ then the value of } k \text{ is:}\)
If f(x)= \(\frac{\sqrt{4} + x - 2}{x}, If \ x \neq 0 \\ k \ If \ x \neq 0\) ,is continuous at x = 0, then the value of k is:
If f(x) =\(\begin{cases}\frac{x^2-9}{x-3}, x≠3 \\ 5, x=3 \end {cases}\) then f(x):
\(f(x) = \begin{cases} 3x-8 & \text{if } x \leq 5 \\ 2k & \text{if } x > 5 \end{cases}\)
If \(f(x)=\begin{cases} \frac{1-\cos4x}{x^2} & x\ne0 \\ k & x=0 \end{cases}\) is continuous at x = 0, then the value of k is :
Let\[f(x)=\begin{cases} 2x-1, x<1\\ 1, x=1 \\ x^2,x>1 \end{cases}\]then at x = 1
The function f(x) = \(|x - 1|\) is
The points of discontinuity of the function \(f\) defined by \(f(x) = \begin{cases} x+2 & x≤1 \\ x-2 &1<x<2\\ 0& x≥2\end{cases}\) are:
for which value of \(\lambda\) is the function ,\(f(x) = \begin{cases} \lambda(x^2-2x) & \text{if } x \leq 0 \\ 4x+1& \text{if } x > 0 \end{cases}\) continuous at \(x=0 ?\)