Calculus

Continuity & Differentiability

Easy

common

If $y = (1+x)(1+x^2)(1+x^4)$ , then $\frac{dy}{dx}$ at $x = -1$ is

$\frac{1}{4}$

$\frac{1}{2}$

✅ Correct Option: 1

By the product rule, every term except the one in which $(1+x)$ is differentiated still contains the factor $(1+x)$ , which vanishes at $x=-1$ .

So $\frac{dy}{dx}\big|_{-1} = 1\cdot(1+1)(1+1) = 4$ .

2023: 30 May Shift 3

The derivative of $\sec(\tan \sqrt{x})$ with respect to x is :

2024: 16 May Shift 1

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

2025: 3 June Shift 1

Match List-I with List-II

List-I	List-II
Function	Points of discontinuity
(A) $f(x) = \frac{x^2 + 1}{x}$	(I) $x = 4$
(B) $f(x) = \frac{\vert x - 1 \vert}{x - 1}$	(II) $x = 2$
(C) $f(x) = \begin{cases} x - 1, & x < 2 \\ x + 1, & x \ge 2 \end{cases}$	(III) $x = 0$
(D) $f(x) = \frac{1 - x}{(x - 4)}$	(IV) $x = 1$

Choose the correct answer from the options given below:

2025: 13 May Shift 1

The function $f(x) = [x]$ , where $[x]$ denotes the greatest integer function, is continuous at $x =$

(A) 2.9

(B) 5

(D) 6.5

Choose the correct answer from the options given below:

2026: 29 May Shift 2

The function $f(x)=\begin{cases}\frac{e^{3x}-e^{-5x}}{x}, & x\neq 0\\ \lambda, & x=0\end{cases}$ is continuous at x=0 for the value of ' $\lambda$ ' as:

2026: 29 May Shift 1

If $e^y(1+x)=1$ , then $\dfrac{d^2y}{dx^2}$ is:

If y=(1+x)(1+x2)(1+x4)y = (1+x)(1+x^2)(1+x^4)y=(1+x)(1+x2)(1+x4), then dydx\frac{dy}{dx}dxdy​ at x=−1x = -1x=−1 is

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If $y = (1+x)(1+x^2)(1+x^4)$ , then $\frac{dy}{dx}$ at $x = -1$ is