Let $\mathrm{f}(\mathrm{x})=$ $\left\{\begin{array}{l}a x^2+1, x 1 \ x+a, x \leq 1\end{array}\right.$. Then $f(x)$ is derivable at $x=1$, if

Question

Let $\mathrm{f}(\mathrm{x})=$ $\left\{\begin{array}{l}a x^2+1, x > 1 \ x+a, x \leq 1\end{array}\right.$. Then $f(x)$ is derivable at $x=1$, if, $a=2$, $a=1$, $a=0$, $a=1 / 2$

MARKS App · Accepted Answer

$\begin{aligned} & \operatorname{Lf}^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h} \ & =\lim _{h \rightarrow 0} \frac{(1-h+a)-(1+a)}{-h}=\lim _{h \rightarrow 0} \frac{-h}{-h}=1 \ & \operatorname{Rf}^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \ & =\lim _{h \rightarrow 0} \frac{\left[a(1+h)^2+1\right]-(1+a)}{h} \ & =\lim _{h \rightarrow 0}(a h+2 a)=2 a \end{aligned}$ Since f' (1) exists, $\therefore L f^{\prime}(1)=R f^{\prime}(1) \Rightarrow a=1 / 2$

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