Question

Given the continuous function $y=f(x)=\{\begin{array}{ll}{x}^2+10x+8,& x\le -2\\ a{x}^2+bx+c& -2<x<0,a\ne 0\\ x^2+2x& x\ge 0\end{array}$ 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)

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