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

The answer is 'There does not exist any R such that f(x) is continous at x=0'. View full solution:

If a function f(x) = cases(x^2-2x), & x \ 0cases then

There does not exist any R such that f(x) is continous at x=0

f(x) is continous at x=0 for all R

f(x) is continous at x=0 for = 0

f(x) is continous at x=0 for = 14

CUET Mathematics 2026 21st May Shift 2

Calculus

Easy

core

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