Let f(x) ax^2+1,&x<1 +1 2 ,&x 1 . Then, f(x) is derivable at x = 1, if:

Let f(x) ax^2+1,&x<1 +1 2 ,&x 1 . Then, f(x) is derivable at x = …

MCQ

Let $\text{f(x)}\begin{cases}\text{ax}^2+1,&\text{x}<1\\\text{x}+\frac{1}{2},&\text{x}\leq1\end{cases}.$ Then, f(x) is derivable at x = 1, if:

A
a = 2
B
a = 1
C
a = 0
✓
$\text{a}=\frac{1}{2}$.

✓

Answer

Correct option: D.

$\text{a}=\frac{1}{2}$.

Given: $\text{f(x)}=\begin{cases}\text{ax}^2+1,&\text{x}<1\\\text{x}+\frac{1}{2},&\text{x}\leq1\end{cases}$
The function is derivable at x = 1, if left hand derivative and right hand derivative of the function are equal at x = 1.
(LHL at x = 1) $=\lim\limits_{\text{x}\rightarrow1^{-}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
(LHL at x = 1) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}(1)}{1-\text{h}-1}$
(LHL at x = 1) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}(1)}{-\text{h}}$
(LHL at x = 1) $=\lim\limits_{\text{h}\rightarrow0}\frac{\Big(1-\text{h}+\frac{1}{2}\Big)-\frac{3}{2}}{-\text{h}}=1$
(RHL at x = 1) $=\lim\limits_{\text{x}\rightarrow1^{+}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
(RHL at x = 1) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1+\text{h})-\text{f}(1)}{1+\text{h}-1}$
(RHL at x = 1) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1+\text{h})-\text{f}(1)}{\text{h}}$
(RHL at x = 1) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}(1+\text{h})^2+1-\frac{3}{2}}{\text{h}}$
(RHL at x = 1) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}(1-\text{h}^2+2\text{h})-\frac{1}{2}}{\text{h}}$
$\therefore\text{LHL}=\text{RHL}$
$\Rightarrow\text{a}-\frac{1}{2}=0$
$\Rightarrow\text{a}-\frac{1}{2}$