Q58 / 150 · MHT CET 2024 (11 May Shift 1) · Mathematics
If $\mathrm{f}(x)=\left\{\begin{array}{cc}\frac{\mathrm{a}}{2}(x-|x|), & \text { for } x \lt 0 \\ 0, & \text { for } x=0 \\ b x^2 \sin \left(\frac{1}{x}\right), & \text { for } x\gt0\end{array}\right.$ is continuous at $x=0$, then
- Correct answerAa is any real value and b is any real value
- B$a$ is only rational value and $b$ is any real value
- Ca is only irrational value and b is any real value
- Da is only rational value and b is only rational value
Solution
$\begin{array}{ll}
& \lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\mathrm{f}(0)=\lim _{x \rightarrow 0^{+}} \mathrm{f}(x) \\
\therefore \quad & \lim _{x \rightarrow 0} \frac{\mathrm{a}}{2}(x-|x|)=0 \\
\therefore \quad & \lim _{x \rightarrow 0} \frac{\mathrm{a}}{2}[x-(-x)]=0 \quad \ldots[\because x \lt 0 \Rightarrow|x|=-x] \\
\therefore \quad & \lim _{x \rightarrow 0} x=0
\end{array}$
Which is true for any real value of a. $\lim _{x \rightarrow 0^{+}} f(x)=f(0)$ $\therefore \quad \operatorname{limb}_{x \rightarrow 0} \mathrm{~b} x^2 \sin \left(\frac{1}{x}\right)=0$ Note that $x \neq 0$ $\therefore \quad-1 \leq \sin \left(\frac{1}{x}\right) \leq 1$ $\therefore \quad$ for any real value of b, above limit will be 0.
Which is true for any real value of a. $\lim _{x \rightarrow 0^{+}} f(x)=f(0)$ $\therefore \quad \operatorname{limb}_{x \rightarrow 0} \mathrm{~b} x^2 \sin \left(\frac{1}{x}\right)=0$ Note that $x \neq 0$ $\therefore \quad-1 \leq \sin \left(\frac{1}{x}\right) \leq 1$ $\therefore \quad$ for any real value of b, above limit will be 0.
Hence, the correct answer is option: 'A'.