Question

If $\left\{\begin{array}{ll}\frac{1}{\left|x\right|}& x\ge 1\\ a{x}^2+b& \left|x\right|<1\end{array}\right.$

is differentiable at every point of the domain, then the values of $a$ and $b$ are respectively:

• A. $\left(\frac{5}{2},-\frac{3}{2}\right)$
• B. $\left(-\frac{1}{2},\frac{3}{2}\right)$
• C. $\left(\frac{1}{2},\frac{1}{2}\right)$
• D. $\left(\frac{1}{2},-\frac{3}{2}\right)$

Answer 1

The correct option is B

$\left(-\frac{1}{2},\frac{3}{2}\right)$

Explanation for the correct option:

Step 1. Evaluating the given conditions:

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

At $x=1$ function must be continuous

So, $1=a+b$ …(1)

Differentiability at $x=1$

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

$\Rightarrow$ $-1=2a$

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

Step 2. Put the value of $a$ in equation (1), we get

$\begin{array}{rcl}b& =& 1+\frac{1}{2}\\ & =& \frac{3}{2}\end{array}$

$\mathbf{\therefore }\mathbf{(}\mathit{a}\mathbf{,}\mathbf{}\mathit{b}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\left(-\frac{1}{2},\frac{3}{2}\right)$

Hence, Option ‘B’ is Correct.