Solve this following

Question:

If $f(x)=\left\{\begin{array}{ll}\frac{1}{|x|} & ;|x| \geq 1 \\ a x^{2}+b & ;|x|<1\end{array}\right.$ is differentiable at

every point of the domain, then the values of a and b are respectively :

  1. $\frac{1}{2}, \frac{1}{2}$

  2. $\frac{1}{2},-\frac{3}{2}$

  3. $\frac{5}{2},-\frac{3}{2}$

  4. $-\frac{1}{2}, \frac{3}{2}$


Correct Option: , 4

Solution:

$f(x)=\left\{\begin{array}{cc}\frac{1}{|x|}, & |x| \geq 1 \\ a x^{2}+b, & |x|<1\end{array}\right.$

at $\mathrm{x}=1$ function must be continuous

So, $1=a+b$           $\ldots(1)$

differentiability at $x=1$

$\left(-\frac{1}{x^{2}}\right)_{x=1}=(2 a x)_{x=1}$

$\Rightarrow-1=2 \mathrm{a} \Rightarrow \mathrm{a}=-\frac{1}{2}$

(1) $\Rightarrow \mathrm{b}=1+\frac{1}{2}=\frac{3}{2}$

 

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