Discussion Forum

1)

If f(x)  = $\begin{cases}ax+b, & if x\leq 1\\ax^{2}+c ,& if 1<x\leq2 \\ \frac{dx^{2}+1}{x},&if x\geq 2\end{cases}$

is differentiable on R, then ad-bc=

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