Question

Consider function $f\left(x\right)=\{\begin{array}{cc}ax(x-1)+b,& x<1\\ x-1,& 1\le x\le 3.\\ p{x}^2+qx+2,& x>3\end{array}$ If $f\left(x\right)$ satisfies the following conditions
$a\left)f\right(x)$ is continuous for all $x$.
$b\left)f^{\prime }\right(1)$ does not exist.
$c\left)f^{\prime }\right(x)$ is continuous at $x=3$. Then the value of $p+\frac{q}{3}+1$ is

Answer 1

Sol. $f\left(x\right)$ is continuous $\forall x\in R$.
Hence, it must be continuous at $x=1,3$.

$f\left(1^{-}\right)=\underset{x\to 1}{lim}ax(x-1)+b=b$
$f\left(1^{+}\right)=\underset{x\to 1}{lim}(x-1)=0$
Now $f\left(1^{-}\right)=f\left(1^{+}\right)=f\left(1\right)$ (for continuity at $x=1$ )
$\Rightarrow b=0$
$f\left(3^{-}\right)=\underset{x\to 3}{lim}(x-1)=2$
$f\left(3^{+}\right)=\underset{x\to 3}{lim}(p{x}^2+qx+2)=9p+3q+2$
Now $f\left(3^{-}\right)=f\left(3^{+}\right)=f\left(3\right)$ (for continuity at $x=3$ )
$\Rightarrow 9p+3q=0\cdots \left(1\right)$
$f^{\prime }\left(x\right)=\{\begin{array}{cc}2ax-a,& x<1\\ 1,& 1\le x\le 3\\ 2px+q,& x>3\end{array}$
Now given that $f^{\prime }\left(1\right)$ does not exist
$\Rightarrow f^{\prime }\left(1^{+}\right)\ne f^{\prime }\left(1^{-}\right)$
$\Rightarrow 1\ne 2a-a$
$\Rightarrow a\ne 1$
Also given that $f^{\prime }\left(3\right)$ exists.
$\Rightarrow f^{\prime }\left(3^{-}\right)=f^{\prime }\left(3^{+}\right)$
$\Rightarrow 1=6p+q\cdots \left(2\right)$
Solving equations $\left(1\right)$ and $\left(2\right)$ for $p$ and $q$, we get $p=1/3,q=-1.$
Thus $p+\frac{q}{3}+1=1$