The function f(x)= cc x^2 & for x<1 2-x & for x 1 . is - Vidyadip

MCQ

The function $f(x)=\left\{\begin{array}{cc}x^2 & \text { for } x<1 \\ 2-x & \text { for } x \geq 1\end{array}\right.$ is

A
not differentiable at $x=1$
B
differentiable at $x=1$
C
not continuous at $x=1$
D
neither continuous nor differentiable at $x=1$

✓

Answer

At $x=1 \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1} x^2=1$
And $\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} 2-x=1$
Also, $f(1)=2-1=1 \because \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(x)$
$\therefore f(x)$ is continuous at $x=1$
Now, L.H.D. $=\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1} \frac{x^2-1}{x-1}=\lim _{x \rightarrow 1}(x+1)=2$
R.H.D. $=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1} \frac{(2-x)-1}{x-1}=-1$
$\because \quad$ L.H.D. $\neq$ R.H.D. $\therefore f(x)$ is not differentiable at $x=1$.

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