Question

If $f(x)=a{x}^2-b,|x|<1=\frac{1}{|x|},|x|\ge 1$ is differentiable at x = 1, then

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

Answer 1

Answer: (B)

Since f(x) is diff. at x = 1
$\therefore L{t}_{x\to 1-}\frac{f\left(x\right)-f\left(1\right)}{x-1}=L{t}_{x\to 1+}\frac{f\left(x\right)-f\left(1\right)}{x-1}$
$\therefore L{t}_{x\to 1-}\frac{a{x}^2-b-1}{x-1}=L{t}_{x\to 1+}\frac{\frac{1}{|x|}-1}{x-1}$
$\Rightarrow L{t}_{x\to 1}\frac{a\left(x^2-1\right)+a-b-1}{x-1}=L{t}_{x\to 1}\frac{1-x}{x\left(x-1\right)}....\left(1\right)$
$\Rightarrow a\left(1+1\right)+L{t}_{x\to 1}\frac{a-b-1}{x-1}=-1$
$\Rightarrow 2a+L{t}_{x\to 1}\frac{a-b-1}{x-1}=-1$
$\therefore a-b-1=0$
$\therefore$ (1) gives $-L{t}_{x\to 1}a\left(x+1\right)=-1$
$\Rightarrow 2a=-1\therefore a=-\frac{1}{2}$
$\therefore -\frac{1}{2}-b-1=0\therefore b=-3/2$