Question

Let $f\left(x\right)=$$\{\begin{array}{c}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 $\left[x\right]$ denotes the greatest integer function. Then which of the following options is correct

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

Given, $f\left(x\right)=\{\begin{array}{c}4x^2+2\left[x\right]x,-\frac{1}{2}\le x<0\\ a{x}^2-bx,0\le x<\frac{1}{2}\end{array}$
$\Rightarrow f\left(x\right)=\{\begin{array}{c}4x^2-2x,-\frac{1}{2}\le x<0\\ a{x}^2-bx,0\le x<\frac{1}{2}\end{array}$

Clearly, $f\left(x\right)$ is continuous in $(\frac{-1}{2},\frac{1}{2})\because f\left(0^{+}\right)=f\left(0^{-}\right)=f\left(0\right)=0$

$f^{\prime }\left(x\right)=$$\{\begin{array}{c}8x-2,-\frac{1}{2}\le x<0\\ 2ax-b,0<x<\frac{1}{2}\end{array}$
For $f\left(x\right)$ to be differentiable , At $x=0,L.H.D$ must be equal to $R.H.D$.
$\Rightarrow 8\left(0\right)-2=2\left(a\right)\left(0\right)-b$
$\Rightarrow b=2,a\in R$