Applications Of Derivatives Question 190
Question: In which of the following function is rolle’s theorem applicable-
Options:
A) $ f(x)= \begin{matrix} x,, \\ 0,, \\ \end{matrix} .,\begin{matrix} ,0\le x<1 \\ x=1 \\ \end{matrix} $ on [0, 1]
B) $ f(x)= \begin{matrix} \frac{\sin x}{x},, \\ 1,, \\ \end{matrix} .\begin{matrix} -\pi \le x<0 \\ x=0 \\ \end{matrix} $ on [- $ \pi $ ,0]
C) $ f(x)=\frac{x^{2}-x-6}{x-1} $ on [-2, 3]
D) $ f(x)= \begin{matrix} \frac{x^{3}-2x^{2}-5x+6}{x-1},,ifx\ne 1, \\ -,6,if,x=1 \\ \end{matrix} . $ on [-2, 3]
Show Answer
Answer:
Correct Answer: D
Solution:
[d] (1) Discontinuous at $ x=1\Rightarrow $ not applicable. (2) $ F(x) $ is not continuous (jump discontinuity) at x=0. (3) Discontinuity (missing point) at $ x=1\Rightarrow $ not applicable. (4) Notice that $ x^{3}-2x^{2}-5x+6=(x-1)(x^{2}-x-6). $
Hence, $ f(x)=x^{2}-x-6 $ if $ x\ne 1 $ and $ f(1)=-6. $ Thus, $ f $ is continuous at x=1. So, $ f(x)=x^{2}-x-6 $ is continuous in the interval $ [ -,2,3 ] $ . Also, note that $ f(-2)=f(3)=0. $
Hence, Rolle-s Theorem implies $ f’(x)=2x-1. $ Setting $ f’(x)=0 $ , we obtain x=1/2 which lies between -2 and 3.
BETA
