Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
2820 Upvotes
Verified Answer
The correct answer is:
$\alpha=\sin (1)$
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 $x=0$.
$\Rightarrow \lim _{x \rightarrow 0} f(x)=f(0) \Rightarrow \sin (1)=\alpha$
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 $x=0$.
$\Rightarrow \lim _{x \rightarrow 0} f(x)=f(0) \Rightarrow \sin (1)=\alpha$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.