Consider f (x) = | 1 - x | ,;,1 le x le | vedclass

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Question

$f (x) = x - 1, 1 \le x \le 2$

$g (x) = x - 1 + b sin \,\frac{\pi }{2}\,x, 1 \le x \le 2$

$f (1) = 0 ; f (2) = 1 $

==> Rolle's theorem

Vedclass · Accepted Answer

$LMVT$ is applicable to $f$ and Rolles theorem is applicable to $g$ with $b = 1$

Vedclass · Answer

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

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 ?

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