For which value of \(\lambda\) is the function, \(f(x) = \begin{cases} \lambda(x^2 - 2x) & \text{if } x \le 0 \\ 4x + 1 & \text{if } x \gt 0 \end{cases}\) continuous at \(x = 0\)?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → for no value of \(\lambda\)
For continuity at x = 0, the left-hand limit, the right-hand limit and f(0) must all be equal.
\(f(0) = \lambda(0 - 0) = 0\)
\( \displaystyle \lim_{x \to 0^-} \lambda(x^2 - 2x) = \lambda \times 0 = 0 \)
\( \displaystyle \lim_{x \to 0^+} (4x + 1) = 1 \)
The left-hand limit and f(0) are 0 whatever \(\lambda\) is, because \(\lambda\) multiplies \(x^2 - 2x\), which is 0 at x = 0. The right-hand limit is 1. Since \(0 \ne 1\), no value of \(\lambda\) makes f continuous at x = 0.
NCERT (2026-27 edition):
Continuity
