Question

Consider $f\left(x\right)=|1-x|,1\le x\le 2$ and $g\left(x\right)=f\left(x\right)+b\sin \left(\frac{\pi }{2}x\right),1\le x\le 2.$ Then which of the following is CORRECT ?

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

Answer 1

The correct option is C LMVT is applicable to $f$ and Rolle's theorem is applicable to $g$ with $b=1$

$f\left(x\right)=x-1,1\le x\le 2$
$g\left(x\right)=x-1+b\sin \left(\frac{\pi }{2}x\right),1\le x\le 2$ $f\left(1\right)=0;f\left(2\right)=1$
$\because f\left(1\right)\ne f\left(2\right)$
$f^{\prime }\left(c\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}=1$
Hence, Rolle's theorem is not applicable to $f$ but LMVT is applicable to $f.$

$\because x-1$ is continuous and differentiable in $[1,2]\text{and}(1,2)$ respectively

Now $g\left(1\right)=b;g\left(2\right)=1$ and function $x-1$ and $\sin \left(\frac{\pi }{2}x\right)$ are both continuous in $[1,2]$
$\therefore$ For Rolle's theorem to be applicable to $g,$ we must have $b=1$