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 the value of $b$ for which $f\left(x\right)$ is differentiable in $(-\frac{1}{2},\frac{1}{2})$ is:

Answer 1

$f\left(x\right)=\{\begin{array}{cc}4x^2-2x,& -\frac{1}{2}\le x<0\\ a{x}^2-bx,& 0\le x<\frac{1}{2}\end{array}$

$f^{\prime }\left(x\right)=\{\begin{array}{cc}8x-2,& -\frac{1}{2}<x<0\\ 2ax-b,& 0<x<\frac{1}{2}\end{array}$
since $\left[x\right]=-1$ when $-\frac{1}{2}<x<0$
Now, at $x=0$, for differentiablity of $f\left(x\right)$
$f^{\prime }\left(0^{-}\right)=f^{\prime }\left(0^{+}\right)$
$\Rightarrow 8\left(0\right)-2=2a\left(0\right)-b\Rightarrow b=2$ and $a\in R$