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  • Let   <img alt="\mathrm{f(x)=\left\{\begin{array}{c}\frac{1}{|x|} \text { for }|x| \geq 1 \\ a x^2+b \text { for }|x|<1\end{array}\right.}" src="https://entrancecorner.oncodecogs.com/g

Let   \mathrm{f(x)=\left\{\begin{array}{c}\frac{1}{|x|} \text { for }|x| \geq 1 \\ a x^2+b \text { for }|x|<1\end{array}\right.}. If f(x) is continuous and differentiable at any point, then

Option: 1

\mathrm{a=\frac{1}{2}, b=-\frac{3}{2}}


Option: 2

\mathrm{a=-\frac{1}{2}, b=\frac{3}{2}}


Option: 3

\mathrm{a=1, b=-1}


Option: 4

none of these


Answers (1)

best_answer

The given function is clearly continuous at all points except possible at \mathrm{x= \pm 1} . For f(x) to be continuous at x=1, we must have

\mathrm{ \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} f(x)=f(1) \\ }

\mathrm{ \Rightarrow \quad \lim _{x \rightarrow 1} a x^2+b=\lim _{x \rightarrow 1} \frac{1}{|x|} \Rightarrow a+b=1 }

Clearly, f(x) is differentiable for all x, excetp possibly at \mathrm{x= \pm 1 }  As f(x) is an evern function, so we need to check its differentiabilify at x=1 only.
For f(x) to be differentiable at x=1, we must have
\mathrm{ \lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1} }

\mathrm{ \Rightarrow \lim _{x \rightarrow 1} \frac{a x^2+b-1}{x-1}=\lim _{x \rightarrow 1} \frac{\frac{1}{|x|}-1}{x-1} \\ }

\mathrm{ \Rightarrow \quad \lim _{x \rightarrow 1} \frac{a x^2-a}{x-1}=\lim _{x \rightarrow 1} \frac{\frac{1}{x}-1}{x-1} \quad\{\because a+b=1 \therefore b-1=-a \mid \\ }

\mathrm{ \Rightarrow \lim _{x \rightarrow 1} a(x+1)=\lim _{x \rightarrow 1} \frac{-1}{x} \Rightarrow 2 a=-1 \Rightarrow a=-1 / 2 }

\mathrm{ Putting ~d=-1 / 2~ in (i), we~ get ~b=3 / 2. }

Posted by

avinash.dongre

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