If the function \(f(x) = \begin{cases}\displaystyle\frac{\sin^{2}(\lambda x)}{x^{2}}, & x \ne 0\\ 4, & x = 0\end{cases}\) is continuous at \(x = 0\), then all the possible values of \(\lambda\) are given by
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\lambda = \pm 2\)
For continuity at 0, the limit of f at 0 must equal f(0) = 4. (If \(\lambda = 0\), f(x) = 0 for x ≠ 0, which fails, so \(\lambda \ne 0\).)
Using \(\lim_{t \to 0}\frac{\sin t}{t} = 1\) with \(t = \lambda x\):
\( \displaystyle \frac{\sin^{2}(\lambda x)}{x^{2}} = \lambda^{2}\left(\frac{\sin \lambda x}{\lambda x}\right)^{2} \)
\( \displaystyle \lim_{x \to 0} f(x) = \lambda^{2} \)
So
\(\lambda^{2} = 4\)
\(\lambda = 2\) or \(\lambda = -2\)
Both values work, because the limit depends only on \(\lambda^{2}\).
Why the other options are incorrect
Option (1) \(\lambda = 2\) only — Incorrect. \(\lambda = -2\) also gives \(\lambda^{2} = 4\), so it is left out wrongly.
Option (2) \(\lambda = -2\) only — Incorrect. \(\lambda = 2\) also gives \(\lambda^{2} = 4\), so it is left out wrongly.
Option (4) \(\lambda = \pm 4\) — Incorrect. Then the limit is \(\lambda^{2} = 16\), not 4.