Question

Let $f\left(x\right)=\{\begin{array}{cc}\frac{1}{|x|}:& |x|\ge 1\\ a{x}^2+b:& |x|<1\end{array}$ be continuous and differentiable every where. Then a and b are:

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

Answer 1

The correct option is A $\frac{-1}{2},\frac{3}{2}$

Let $f\left(x\right)=\{\begin{array}{cc}-\frac{1}{x}:& x\le -1\\ a{x}^2+b:& -1<x<1\\ \frac{1}{x}:& x\ge 1\end{array}$
f(x) is continuous $\Rightarrow a+b=1$.
$f^{\prime }\left(x\right)=\{\begin{array}{cc}\frac{1}{x^2}:& x\le -1\\ 2ax:& -1<x<1\\ \frac{-1}{x^2}:& x\ge 1\end{array}$
f(x) is differentiable at $x=-1,1\Rightarrow 1=-2a\Rightarrow a=\frac{-1}{2}$.
$\therefore a=\frac{-1}{2},b=\frac{3}{2}$