Question

Consider a 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
$\left(\text{i}\right)f\left(x\right)$ is continuous for all $x$.
$\left(\text{ii}\right)f\left(x\right)$ is differentiable at $x=3$.
Then which of the following option(s) is (are) correct?

• A. $a\in R$
• B. $p=-\frac{1}{3}$
• C. $q=-1$
• D. $b=0$

Answer 1

Answer: (A), (C), (D)

$b=0$

$f\left(x\right)$ 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 $x=1$ )
$\Rightarrow b=0,a\in R$

$f\left(3^{-}\right)=\underset{x\to 3}{lim}(x-1)=2$
$f\left(3^{+}\right)=\underset{x\to 3}{lim}(p{x}^2+qx+2)=9p+3q+2$
Now, $f\left(3^{-}\right)=f\left(3^{+}\right)=f\left(3\right)$ (for continuity at $x=3$ )
$\Rightarrow 9p+3q=0\cdots \left(1\right)$

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
$p=\frac{1}{3},q=-1.$