Question

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

$f\left(x\right) = x-1, 1 \le x \le 2$
$g\left(x\right) = x - 1 + b\,sin \frac{\pi}{2} x$,
$1 \le x \le 2$
$f\left(1\right) = 0$ ; $f\left(2\right) = 1$
$\Rightarrow $ Rolle’s theorem is not applicable to $'f'$ 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 $b = 1$