Q76 / 160 · WBJEE 2015 · Mathematics

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
    $\alpha=\sin (-1)$
  • Correct answer
    C
    $\alpha=\sin (1)$
  • D
    $\alpha=1$

Solution

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 continuous at $x=0$. $\Rightarrow \lim _{x \rightarrow 0} f(x)=f(0) \Rightarrow \sin (1)=\alpha$

Hence, the correct answer is option: 'C'.

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