Question

Given the continuous function $y=f(x)=\begin{cases}x^2+10 x+8, & x \leq-2 \\a x^2+b x+c & -2< x< 0, a \neq 0 \\x^2+2 x & x \geq 0\end{cases}$ If a line $L$ touches the graph of $y=f(x)$ at three points then
If $y=f(x)$ is differentiable at $x=0$ then the value of $b$

• A. is -1
• B. is 2
• C. is 4
• D. can not be determined

Answer 1

Answer: (B)

$f(x)=\begin{cases}x^2+10 x+8, & x \leq-2 \\ ax ^2+b x+c & -2< x< 0, a \neq 0 \\ x^2+2 x & x \geq 0\end{cases}$
$\text { for continuous at } x =0 \Rightarrow c =0 $
$\text { continuous at } x=-2 \Rightarrow 4-20+8=4 a-2 b$
$-8=4 a-2 b $
$2 a - b =-4$....(1)
now let the line $y=m x+p$ is tangent to all the 3 curves solving $y=m x+p$ and $y=x^2+2 x$
$x^2+2 x=m x+p$
$x^2+(2-m) x-p=0$
$D =0$
$(2-m)^2+4 p=0$....(2)
again solving $y=m x+p$ and $y=x^2+10 x+8$
$x^2+10 x+8=m x+p$
$x ^2+(10- m ) x +8- p $
$(10- m )^2-4(8- p )=0$
$(10- m )^2-32+4 p =0$
$(10- m )^2-(2- m )^2=32$
$(100-20 m )-(4-4 m )=32 \left( m ^2 \text { cancels out }\right) $
$96-16 m =32 \Rightarrow 64=16 m $
$m =4 \text { and } p =-1$
hence equation of line tangent to $1^{\text {st }}$ and last curves is
$y=4 x-1$
$\text { now solving this with } y=a^2+b x (a s c=0) $
$a^2+b x=4 x-1 $
$D =0$
$(b-4)^2+(b-4) x+1=0$
$(b-4)^2=4 a$
Also
$b =2 a +4 \text { (from 1) }$
$\therefore 4 a^2=4 a \Rightarrow a \neq 0 \Rightarrow a=1 \text { and } b=6$
$f ^{\prime}\left(0^{-}\right)=\underset{x \rightarrow 0}{\text{Lim}} 2 ax + b = b ; f ^{\prime}\left(0^{+}\right)=\underset{x \rightarrow 0}{\text{Lim}} 2 x +2=2 \Rightarrow b =2$