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If $f(x)=\left\{\begin{aligned} e^x+a x, & x \lt 0 \\ b(x-1)^2, & x \geq 0\end{aligned}\right.$ is differentiable at $x=0$, then $(a, b)$ is
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The correct answer is:
$(-3,1)$
Given $f(x)$ is differentiable at $x=0$. Hence, $f(x)$ will be continuous at $x=0$.
$\begin{aligned} & \lim _{x \rightarrow 0^{-}}\left(e^x+a x\right)=\lim _{x \rightarrow 0^{+}} b(x-1)^2 \\ \Rightarrow & e^0+a \times 0=b(0-1)^2 \Rightarrow b=1...(i)\end{aligned}$
But $f(x)$ is differentiable at $x=0$, then
$\begin{aligned} & L f^{\prime}(x)=R f^{\prime}(x) \Rightarrow \frac{d}{d x}\left(e^x+a x\right)=\frac{d}{d x} b(x-1)^2 \\ & \Rightarrow e^x+a=2 b(x-1) \\ & \text { At } x=0, e^0+a=-2 b \Rightarrow a+1=-2 b \Rightarrow a=-3 \\ & \Rightarrow(a, b)=(-3,1) .\end{aligned}$
$\begin{aligned} & \lim _{x \rightarrow 0^{-}}\left(e^x+a x\right)=\lim _{x \rightarrow 0^{+}} b(x-1)^2 \\ \Rightarrow & e^0+a \times 0=b(0-1)^2 \Rightarrow b=1...(i)\end{aligned}$
But $f(x)$ is differentiable at $x=0$, then
$\begin{aligned} & L f^{\prime}(x)=R f^{\prime}(x) \Rightarrow \frac{d}{d x}\left(e^x+a x\right)=\frac{d}{d x} b(x-1)^2 \\ & \Rightarrow e^x+a=2 b(x-1) \\ & \text { At } x=0, e^0+a=-2 b \Rightarrow a+1=-2 b \Rightarrow a=-3 \\ & \Rightarrow(a, b)=(-3,1) .\end{aligned}$
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