Question

Consider $f(x)=|1-x|1\le x\le 2\text{ and }$ $g(x)=f(x)+b\sin \frac{\pi }{2}x,1\le x\le 2$ 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)

$f(x)=x-1,1\le x\le 2$
$g(x)=x-1+b\sin \frac{\pi }{2}x,1\le x\le 2$
$f(1)=0;f(2)=1\Rightarrow$ 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 $b=1$