Question

Consider $f ( x )=|1- x | 1 \leq x \leq 2 \text { and } $ $g ( x )= f ( x )+ b \sin \frac{\pi}{2} x , 1 \leq x \leq 2$ then which of the following is correct?

• A. Rolles theorem is applicable to both $f , g$ and $b =\frac{3}{2}$
• B. LMVT is not applicable to $f$ and Rolles theorem if applicable to $g$ with $b =\frac{1}{2}$
• C. LMVT is applicable to $f$ and Rolles theorem is applicable to $g$ with $b =1$
• D. Rolles theorem is not applicable to both $f , g$ for any real $b$.

Answer 1

Answer: (C)

$ f ( x )= x -1,1 \leq x \leq 2 $
$g ( x )= x -1+ b \sin \frac{\pi}{2} x , 1 \leq x \leq 2$
$f(1)=0 ; f(2)=1 \Rightarrow$ Rolle's theorem is not applicable to ' $f$ ' but LMVT is applicable to $f$.
$(\because x -1$ is continuous and differentiable in $[1,2]$ and $(1,2)$ respectively) Now $g(1)=b ; g(2)=1$ and
Function $x-1, \sin \frac{\pi}{2} x$ are both continuous in $[1,2]$ and $(1,2)$
$\therefore $ For Rolle's theorem to be applicable to $g$.
We must have $b=1$