Question

Let $f\left(x\right)=\{\begin{array}{cc}1/\left|x\right|& for\left|x\right|\ge 1\\ a{x}^2+b& for\left|x\right|<1\end{array}$ The coefficients a and b so that f is continuous and differentiable at any point, are equal to

• A. $a=-1/2,b=3/2$
• B. $a=1/2,b=-3/2$
• C. $a=1,b=-1$
• D. none of these

Answer 1

The correct option is A $a=-1/2,b=3/2$

The given function is clearly continuous at all points excepts possibly
at $x=\pm 1$
For $f\left(x\right)$ to be continuous at $x=1$, we must have
$\underset{x\to 1-}{lim}f\left(x\right)=\underset{x\to 1+}{lim}f\left(x\right)=f\left(1\right)$
$\Rightarrow \underset{x\to 1}{lim}{ax}^2+b=\underset{x\to 1}{lim}\frac{1}{\left|x\right|}\Rightarrow a+b=1$ ...(1)
Now for $f\left(x\right)$ to be differentiable at $x=1$ we must have
$\underset{x\to 1}{lim}\frac{f\left(x\right)-f\left(1\right)}{x-1}=\underset{x\to 1+}{lim}\frac{f\left(x\right)-f\left(1\right)}{x-1}$
$\Rightarrow \underset{x\to 1}{lim}\frac{{ax}^2-a}{x-1}=\underset{x\to 1}{lim}\frac{\frac{1}{x}-1}{x-1}(\because a+b=1\therefore b-1=-a)$
$\Rightarrow \underset{x\to 1}{lim}a(x+1)=\underset{x\to 0}{lim}(-\frac{1}{x})\Rightarrow a=-\frac{1}{2}$
Substituting $a=-\frac{1}{2}$ in (1) we get $b=\frac{3}{2}$