- Home
- Standard 12
- Mathematics
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 ?
Rolles theorem is applicable to both $f, g$ and $b =\frac{3}{2}\,$
$LMVT$ is not applicable to $f$ and Rolles theorem if applicable to $g$ with $b =\frac{1}{2}\,$
$LMVT$ is applicable to $f$ and Rolles theorem is applicable to $g$ with $b = 1$
Rolles theorem is not applicable to both $f, g$ for any real $b.$
Solution
$f (x) = x – 1, 1 \le x \le 2$
$g (x) = x – 1 + b sin \,\frac{\pi }{2}\,x, 1 \le x \le 2$
$f (1) = 0 ; f (2) = 1 $
==> Rolle's theorem is not applicable to $' f '$ but $LMVT$ is applicable to $f.$
$( x – 1$ 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 $(1, 2)$
For Rolle's theorem to be applicable to $g.$
We must have $b = 1$
