Question

If $\left\{\begin{array}{l}\frac{1}{\left|x\right|},x\ge 1\\ a{x}^2+b,x<1\end{array}\right.$is differentiable at every point of the domain, then the values of $a$ and $b$ are respectively:

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

Answer 1

The correct option is B

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

Finding the values of $a$ and $b$:

Let the given,

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

Given, $f\left(x\right)$ is continuous at $x=1$

$\Rightarrow$ $1=a+b........\left(i\right)$

Also, $f\left(x\right)$ is differentiable at $x=1$

$\Rightarrow$$-1=2a........\left(ii\right)$

From equation $\left(i\right)\&\left(ii\right)$.

$a=-\frac{1}{2}\&b=\frac{3}{2}$

Hence, correct answer is option $\left(B\right)$.