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STD 12 - 1. relation and function — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 1. relation and function4 Marks
MCQ
Let $f: R \rightarrow R$ be a function defined by $f(x)=\left\{\begin{array}{l}\frac{\sin \left(x^2\right)}{x} \text { if } x \neq 0 \\ 0 \text { if } x=0\end{array}\right\}$ Then, at $x=0, f$ is
  • A
    not continuous
  • B
    continuous but not differentiable
  • C
    differentiable and the derivative is not continuous
  • ✓
    differentiable and the derivative is continuous
✓

Answer

Correct option: D.
differentiable and the derivative is continuous
Given function
$f(x)=\left[\begin{array}{cc} \frac{\sin \left(x^2\right)}{x} & , x \neq 0 \\ 0 & , \text { if } x=0\end{array}\right.$
then $\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\sin x^2}{x}=\lim _{x \rightarrow 0}$
$x \frac{\sin x^2}{x^2}=0=f(0)$
Hence, $f(x)$ is continuous at $x=0$
Now, for differentiability
$\text{RHD} ($ at $x=0)$
$=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{\sin h^2}{h^2}=1$
and $\text{LHD} ($at $x=0)$
$=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)-\lim _{h \rightarrow 0} \sin h^2}{-h}=1$
So, $f(x)$ is differentiable at $x=0$
$\therefore \quad f^{\prime}(x)=\left[\begin{array}{cc}2 \cos \left(x^2\right)-\frac{\sin x^2}{x^2} & , \text { if } x \neq 0 \\ 1 & , \text { if } x=0\end{array}\right]$
$\because \lim _{x \rightarrow 0} f^{\prime}(x)=\lim _{x \rightarrow 0}\left[2 \cos \left(x^2\right)-\frac{\sin x^2}{x^2}\right]$ $=2-1=1$
$\therefore \lim _{x \rightarrow 0} f^{\prime}(x)=f^{\prime}(0)$ So,f $f(x)$ is differentiable and the

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