If a function f(x) = cases(x^2-2x), & x ≤ 0\\4x+1, & x>0cases then Option 1: There does not exist any R such that f(x) is continous at x=0 Option 2: f(x) is continous at x=0 for all R Option 3: f(x) is continous at x=0 for = 0 Option 4: f(x) is continous at x=0 for = 1/4

Answer: There does not exist any R such that f(x) is continous at x=0. Solution: The left-hand limit at 0 is (0-0) = 0 and f(0)=0, for every . The right-hand limit is 4(0)+1 = 1. Since 0 1 regardless of, no value of makes f continuous at x=0.

Calculus

Continuity & Differentiability

Easy

core

If a function $f(x) = \begin{cases}\lambda(x^2-2x), & x \le 0\\4x+1, & x>0\end{cases}$ then

There does not exist any $\lambda \in \mathbb{R}$ such that $f(x)$ is continous at $x=0$

$f(x)$ is continous at $x=0$ for all $\lambda \in \mathbb{R}$

$f(x)$ is continous at $x=0$ for $\lambda = 0$

$f(x)$ is continous at $x=0$ for $\lambda = \dfrac{1}{4}$

✅ Correct Option: 1

The left-hand limit at $0$ is $\lambda(0-0) = 0$ and $f(0)=0$ , for every $\lambda$ .

The right-hand limit is $4(0)+1 = 1$ .

Since $0 \ne 1$ regardless of $\lambda$ , no value of $\lambda$ makes $f$ continuous at $x=0$ .

2025: 13 May Shift 1

If $f(x) = x^3\log_e x$ , Then $f''(e^2)$ is equal to

2025: 16 May Shift 1

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

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:

2026: 15 May Shift 1

Match the LIST-I with LIST-II

LIST-I (Function in parametric form)	LIST-II ( $\dfrac{dy}{dx}=$ )
A. $x=t, y=t$	I. 1
B. $x=t^2, y=2t$	II. $\dfrac{1}{t}$
C. $x=3t, y=t^2$	III. $\dfrac{2t}{3}$
D. $x=2t, y=3t^2$	IV. $3t$

Choose the correct answer from the options given below:

2025: 22 May Shift 2

If the function

$f(x) = \begin{cases} ax + 2, & x \leq 1 \\ x^2 + 3x + b, & x > 1 \end{cases}$

is differentiable at $x = 1$ , then the value of $(2a + b)$ is

2026: 19 May Shift 1

If $y = \cot^{-1}\left(\frac{1-x}{1+x}\right)$ then $\frac{dy}{dx}$ equals

If a function f(x)={λ(x2−2x),x≤04x+1,x>0f(x) = \begin{cases}\lambda(x^2-2x), & x \le 0\\4x+1, & x>0\end{cases}f(x)={λ(x2−2x),4x+1,​x≤0x>0​ then

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If a function $f(x) = \begin{cases}\lambda(x^2-2x), & x \le 0\\4x+1, & x>0\end{cases}$ then