Let the function $f(x)=\left(x^2+1\right)\left|x^2-a x+2\right|+\cos |x|$ be not differentiable at the two points $x=\alpha=2$ and $x=\beta$. Then the distance of the point $(\alpha, \beta)$ from the line $12 x+5 y+10=0$ is equal to :

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Aicharya · Accepted Answer

$f(x)=\left(x^2+1ight)\left|x^2-a x+2ight|+\cos |x|$ Notice that $\cos (-x)=\cos x=\cos |x|$ which means $\cos |x|$ is differentiable everywhere in $x \in R$ $\begin{aligned} & \Rightarrow f(x) 	ext { can be non differentiable where }\left|x^2-a x+2ight| \ & =0 \end{aligned}$ $\Rightarrow x^2-a x+2=0$ $\Rightarrow 4-2 a+2=0 \Rightarrow a=3$ $\Rightarrow \quad\left(x^2-3 x+2ight)=0 \quad \Rightarrow \quad x=1,2$ $\beta=1$ distance of $(\alpha, \beta)$ from line $\begin{aligned} & 12 x+5 y+10=0 \ & \Rightarrow \frac{|2(12)+5(1)+10|}{13}=\frac{39}{13}=3 \end{aligned}$

Let the function $f(x)=\left(x^2+1\right)\left|x^2-a x+2\right|+\cos |x|$ be not differentiable at the two…

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