Question:

Let $f(x) = \frac{(1 - cos 4x)}{x^2}$, if x < 0 and a, if x = 0 and $\sqrt{\frac{x}{(\sqrt{(16+\sqrt{x})}-4)}}$, if x > 0}. If $f(x)$ is continuous at x = 0, determine the value of a.

Updated On: Jul 16, 2024

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Match List-I with List-II

List-I	List-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

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Match List-I with List-II

List-I	List-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
- CUET (UG) - 2025
- Mathematics
- Continuity and differentiability
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View More Questions

List-I	List-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

List-I	List-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