Answer:
Option C
Explanation:
We have ,
f(x) = $\begin{cases}ax+b, & if x\leq 1\\ax^{2}+c ,& if 1<x\leq2 \\ \frac{dx^{2}+1}{x},&if x\geq 2\end{cases}$
$\because$ f(x) is differentiable , hence f(x) must be continuous
$\therefore$ $\lim_{x \rightarrow 1-}f(x)=\lim_{x \rightarrow 1+}f(x)$
$\Rightarrow$ a+b=a+c $\Rightarrow$ b=c ........(i)
and $\lim_{x \rightarrow 2-}f(x)\lim_{x \rightarrow 2+}f(x)$
$\Rightarrow$ 4a+c = $\frac{4d+1}{2}$
$\Rightarrow$ 8a+2c=4d+1 ........(ii)
$\because$ f(x) is differentiable on R
$\therefore$ $f(x) = \begin{cases}a, & if x< 1\\2ax ,& if 1<x<2 \\d- \frac{1}{x^{2}},&if x> 2\end{cases}$
f(x) is differentiable at x=1
$\because$ a=2a $\Rightarrow$ a=0 .....(iii)
and f(x) is differentiable at x=2
$\because$ 4a=d-$\frac{1}{4} \Rightarrow$ d= $\frac{1}{4}$ .......(iv)
From Eqs.(ii) and (iv) ,we get
c=1=b
$\therefore$ ad-bc=$ 0 \times \frac{1}{4}-1$=-1
