Question

Let $f\left(x\right)=\{\begin{array}{cc}4x^2+2\left[x\right]x,& -\frac{1}{2}\le x<0\\ a{x}^2-bx,& 0\le x<\frac{1}{2}\end{array},$
where $[.]$ denotes the greatest integer function. Then

• A. $f\left(x\right)$ is continuous and differentiable in $(-\frac{1}{2},\frac{1}{2})$ for all real $a$, provided $b=2.$
• B. $f\left(x\right)$ is continuous and differentiable in $(-\frac{1}{2},\frac{1}{2})$ if $f\left(a\right)=4,b=2.$
• C. $f\left(x\right)$ is continuous and differentiable in $(-\frac{1}{2},\frac{1}{2})$ if $a=4$ and $b=2.$
• D. For no choice of $a$ and $b,f\left(x\right)$ is differentiable in $(-\frac{1}{2},\frac{1}{2}).$

Answer 1

The correct option is A $f\left(x\right)$ is continuous and differentiable in $(-\frac{1}{2},\frac{1}{2})$ for all real $a$, provided $b=2.$

Clearly, $f\left(x\right)$ is continuous in $(\frac{-1}{2},\frac{1}{2})$ since $f\left(0^{+}\right)=f\left(0^{-}\right)=f\left(0\right)=0.$
Now, to check differentiablity,
$f^{\prime }\left(x\right)=$$\{\begin{array}{c}8x-2;-\frac{1}{2}<x<0\\ 2ax-b;0<x<\frac{1}{2}\end{array}$
since $\left[x\right]=-1$ when $\frac{-1}{2}\le x<0$
Now at $x=0,$
$L.H.D.$ must be equal to $R.H.D.$
$\Rightarrow 8\left(0\right)-2=2\left(a\right)\left(0\right)-b$
$\Rightarrow b=2$ and $a\in R$