Question

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

Answer: (C)

$f\left(x\right)=x-1,1\le x\le 2$
$g\left(x\right)=x-1+bsin\frac{\pi }{2}x$,
$1\le x\le 2$
$f\left(1\right)=0$ ; $f\left(2\right)=1$
$\Rightarrow$ Rolle’s theorem is not applicable to ${}^{\prime }{f}^{\prime }$ but LMVT is applicable to $f$.
[$\therefore x-1$ is continuous and differentiable in $\left[1,2\right]$ and $\left(1,2\right)$ respectively].
Now, $g\left(1\right)=b$ ; $g\left(2\right)=1$ and function $x-1$, $sin\frac{\pi }{2}x$ are both continuous in $\left[1,2\right]$ and $\left[1,2\right]$.
$\therefore$ for Rolle’s theorem to be applicable to $g$.
We must have $b=1$