How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?
Learn More

Rolle's Theorem | IIT JAM 2017 | Problem 10

Try this problem from IIT JAM 2017 exam (Problem 10).This problem needs the concept of Rolle's Theorem.

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

  • $(2,-1)$
  • $(-2,-1)$
  • $(-2,1)$
  • $(2,1)$

Key Concepts

Real Analysis

Continuity / Differentiability

 Mean-value theorem of differential calculus

Check the Answer

Answer: $(2,1)$

IIT JAM 2017 , Problem 10

Real Analysis : Robert G. Bartle

Try with Hints

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]

Subscribe to Cheenta at Youtube

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.