Question

The function $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}ax(x-1)+b& x<1\end{array}\\ \begin{array}{cc}x-1& 1\le x\le 3\end{array}\\ \begin{array}{cc}p{x}^2+qx+2& x>3\end{array}\end{array}$

(i) $f\left(x\right)$ is continuous for all $x$

(ii) $f^{\prime }\left(1\right)$ does not exists

(iii) $f^{\prime }\left(x\right)$ is continuous at $x=3$

If $a,b,p,q$ are constant then

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

Answer 1

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

The correct options are
A $p=\frac{1}{3}$
B $a\in R-\left\{1\right\}$
C $b=0$
D $q=-1$

$f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}ax(x-1)+b& x<1\end{array}\\ \begin{array}{cc}x-1& 1\le x\le 3\end{array}\\ \begin{array}{cc}{px}^2+qx+2& x>3\end{array}\end{array}$

$f\left(x\right)$ is continuous at $x=1$

$\Rightarrow \underset{x\to 1^{-}}{lim}f\left(x\right)=f\left(1\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)$$\Rightarrow \underset{x\to 1^{-}}{lim}ax(x-1)+b=0$$\Rightarrow b=0$ and $a\in R$

Now, $f^{\prime }\left(1\right)=\underset{h\to 0}{lim}\frac{f(1+h)-f\left(1\right)}{h}$$=\{\begin{array}{c}\underset{h\to 0^{-}}{lim}\frac{a(1+h)(1+h-1)+b}{h}\\ \underset{h\to 0^{+}}{lim}\left(\frac{1+h-1}{h}\right)\end{array}=\{\begin{array}{c}{lim}_{h\to 0^{-}}\frac{a(1+h)h+0}{h}\\ \underset{h\to 0^{+}}{lim}\frac{h}{h}\end{array}$

$=\{\begin{array}{c}{lim}_{h\to 0^{-}}a(1+h)\\ 1\end{array}=\{\begin{array}{c}a\\ 1\end{array}$

$\because f^{\prime }\left(1\right)=$ does not exist

$\Rightarrow a\ne 1$

$\therefore a\in R-\left\{1\right\}$ and $b=0;f\left(x\right)$ is cont. at $x=3$

Further it is given that $f^{\prime }\left(x\right)$ is continuous at $x=3,$ and $f\left(x\right)$ is continuous $\forall x.$

In particular $f\left(x\right)$ is also continuous at $x=3$

$\Rightarrow \underset{x\to 3}{lim}f\left(x\right)=f\left(3\right)$

$\Rightarrow \underset{x\to 3}{lim}({px}^2+qx+2)=2$$\Rightarrow 9p+3q=2=2$$\Rightarrow 9p+3q=0$ ...(i)

$\because f^{\prime }\left(x\right)$ is continuous at $x=3,$ hence $f\left(x\right)$ is must be differentiable at $x=3$ as $f\left(x\right)$ is continuous at $x=3$

$\therefore f^{\prime }\left(3\right)=\underset{h\to 0}{lim}\frac{f(3+h)-f\left(3\right)}{h}$$=\{\begin{array}{c}\underset{h\to 0^{-}}{lim}\frac{h+3-1-2}{h}\\ \underset{h\to 0^{+}}{lim}\frac{p{(3+h)}^2+q(3+h)+2-2}{h}\end{array}$

$=\{\begin{array}{c}\underset{h\to 0^{-}}{lim}\frac{h}{h}\\ \underset{h\to 0^{+}}{lim}\frac{p{h}^2+6ph+qh+9p+3q}{h}\end{array}$$=\{\begin{array}{c}1\\ \underset{h\to 0^{+}}{lim}\frac{{ph}^2+6ph+qh}{h}\end{array}$

$[\because$ from equation (i)$9p+3q=0]$

$=\{\begin{array}{c}1\\ \underset{h\to 0^{+}}{lim}(ph+6p+q)\end{array}=\{\begin{array}{c}1\\ 6p+q\end{array}$

$\therefore f^{\prime }\left(3^{+}\right)=f^{\prime }\left(3^{-}\right)$$\Rightarrow 6p+q=1$ ...(ii)

Solving equation (i) and (ii) $p=\frac{1}{3},q=-1$

Thus $a\in R-\left\{1\right\};b=0;p=\frac{1}{3};q=-1$