If a function \( f(x)=\begin{cases}\lambda(x^{2}-2x), & x\le 0\\4x+1, & x\gt 0\end{cases} \) then
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → There does not exist any \( \lambda\in\mathrm{R} \) such that \( f(x) \) is continuous at \( x=0 \)
f is continuous at x = 0 only if the left-hand limit, the right-hand limit and f(0) are all equal.
\( \displaystyle f(0)=\lambda(0-0)=0 \)
\( \displaystyle \lim_{x\to 0^{-}}f(x)=\lim_{x\to 0^{-}}\lambda(x^{2}-2x)=0 \)
\( \displaystyle \lim_{x\to 0^{+}}f(x)=\lim_{x\to 0^{+}}(4x+1)=1 \)
The left-hand limit is 0 for every value of λ, but the right-hand limit is 1. They can never be equal, so there is no λ for which f is continuous at x = 0.
NCERT (2026-27 edition):
Continuity
