WBJEE · Maths · Continuity and Differentiability
Let \([x]\) denotes the greatest integer less than or equal to \(x\). Then, the value of \(\alpha\) for which the function \(f(x)=\left\{\begin{array}{c}\frac{\sin \left[-x^{2}\right]}{\left[-x^{2}\right]}, x \neq 0 \\ \alpha, \quad x=0\end{array}\right.\) is continuous at \(x=0,\) is
- A \(\alpha=0\)
- B \(\alpha=\sin (-1)\)
- C \(\alpha=\sin (1)\)
- D \(\alpha=1\)
Answer & Solution
Correct Answer
(C) \(\alpha=\sin (1)\)
Step-by-step Solution
Detailed explanation
We have, \(f(x)=\left\{\begin{array}{ll}\frac{\sin \left(-x^{2}\right)}{[-x^{2}]} & , x \neq 0 \\ \alpha, & x=0\end{array}\right.\) Now, \(\lim _{x \rightarrow 0} \frac{\sin \left(-x^{2}\right)}{\left.\mid-x^{2}\right]}=\frac{\sin (-1)}{(-1)}=\sin (1)\) Since, \(f(x)\) is…
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