Rolle's Theorem | IIT JAM 2017 | Problem 10

$$f(x)=\left\{\begin{array}{ll}1+x & \text { if } x<0 \\ (1-x)(p x+q) & \text { if } x \geq 0\end{array}\right.$$

satisfies the assumptions of Rolle's theorem in the interval $[-1,1],$ then the ordered pair $(p, q)$ is

Rolle's Theorem :

Let a function $f:[a, b] \rightarrow R$ be such that

1. $f$ is continuous on $[a, b]$
2. $f$ is differentiable at every point of $(a, b)$
3. $f(a)=f(b)$

Then there exists at least one point $c \in(a, b)$ such that $f^{\prime}(c)=0$

We can easily see that $3^{rd}$ assumption of Rolle's theorem is satisfied for $f(x)$ irrespective of the values of $p,q$.

Since $f(-1)=0=f(1)\quad \forall p,q$

Since $f(x)$ satisfies $1^{st}$ assumption, then

$\begin{aligned}& \quad \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)\\&\text { ie, } \lim _{x \rightarrow 0^{-}}(1+x)=\lim _{x \rightarrow 0^{+}}(1-x)(px+q)=q\\&\Rightarrow 1=q\end{aligned}$

$L f^{\prime}(0)=R f^{\prime}(0) \cdots \cdots(*)$

$\begin{aligned} \text{Now, } L f^{\prime}(0) &=\lim _{h \rightarrow 0^{-}} \frac{f(0+h)-f(0)}{h} \\&=\lim _{h \rightarrow 0^{-}} \frac{(1+h)-q}{h} \\ &=\lim _{h \rightarrow 0^{-}} \frac{1+h-1}{h}[\text{because } q=1] \\&=\lim_{h \to 0^{-}} \frac hh\\&=1\end{aligned}$

$\begin{aligned}\text{and, } R f^{\prime}(0)&=\displaystyle\lim _{h \rightarrow 0^{+}} \frac{f(0+h)-f(0)}{h}\\&=\lim _{h \rightarrow 0^{+}} \frac{(1-h)\left(ph+q\right)-q}{h}\\&=\lim _{h \rightarrow 0^{+}} \frac{(1-h)(ph +1)-1}{h}\quad[\text{because } q=1]\\&=\lim _{h \rightarrow 0^{+}}\frac{ph+1- ph^{2}-h-1}{h}\\&=\lim _{h \rightarrow 0^{+}} \frac{h(p-ph-1)}{h}\\&=\lim_{h \ to 0^{+}} (p-ph-1)\\&=p-1\end{aligned}$

Then by $(*) \text{we have}, \quad P-1=1 \Rightarrow P=2$

Then order pair $(p,q)\equiv (2,1)$ [ANS]