Question

$f\left(x\right)$ is defined as under:$f\left(x\right)=\{\begin{array}{cc}ax(x-1)+b,& x<1\\ x-1,& 1\le x\le 3\\ c{x}^2+dx+2,& x>3\end{array}$

$f^{\prime }\left(x\right)$ is discontinuous at $x=3$. Then $a\ne k,b=m,c=\frac{1}{h},d=-p.$ Find $k+m+h+p$?

Answer 1

$f\left(x\right)=\{\begin{array}{cc}ax(x-1)+b,& x<1\\ x-1,& 1\le x\le 3\\ c{x}^2+dx+2,& x>3\end{array}$
$f^{\prime }\left(x\right)=\{\begin{array}{cc}2ax-a,& x<1\\ 1,& 1\le x\le 3\\ 2cx+d,& x>3\end{array}$
$f\left(x\right)$ is differentiable at $x=1,3$

$L{f}^{\prime }\left(1\right)=R{f}^{\prime }\left(1\right)$

$\Rightarrow 2a-a=1$

$\Rightarrow a=1$
Also, $L{f}^{\prime }\left(3\right)=R{f}^{\prime }\left(3\right)$

$\Rightarrow 1=6c+d$ ....(1)
Since, $f\left(x\right)$ is differentiable at $x=1,3$

$\Rightarrow$ f(x) is contnuous at $x=1,3$

$LHL=RHL=f\left(1\right)$

${lim}_{h\to 0}f(1-h)=0$

$\Rightarrow {lim}_{h\to 0}a(1-h{)}^2-a(1-h)+b=0$

$\Rightarrow b=0$
Also,$LHL=RHL=f\left(3\right)$

${lim}_{h\to 0}f(3+h)=0$

$\Rightarrow {lim}_{h\to 0}c(3+h{)}^2+d(3+h)+2=2$

$\Rightarrow 3c+d=0$ ....(2)
Solving (1) and (2), we get

$c=\frac{1}{3},d=-1,m=0,$
On comparing with given values ,
$p=1,h=3,k=1,m=0$
$\Rightarrow k+m+h+p=5$