Consider f(x) = |1 - x| for 1 le x le 2 and g | vedclass

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(C) Given $f(x) = |1 - x|$. For $1 \le x \le 2$,$1 - x \le 0$,so $f(x) = x - 1$.$f(x)$ is continuous on $[1, 2]$ and differentiable on $(1, 2)$.$f(1)

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$LMVT$ is applicable to $f$,and Rolle's theorem is applicable to $g$ with $b = 1$.

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(C) Given $f(x) = |1 - x|$. For $1 \le x \le 2$,$1 - x \le 0$,so $f(x) = x - 1$.$f(x)$ is continuous on $[1, 2]$ and differentiable on $(1, 2)$.$f(1) = 0$ and $f(2) = 1$. Since $f(1) 
eq f(2)$,Rolle's theorem is not applicable to $f$. However,$LMVT$ is applicable to $f$ because it is continuous on $[1, 2]$ and differentiable on $(1, 2)$.Now,$g(x) = x - 1 + b \sin(\frac{\pi}{2}x)$.For Rolle's theorem to be applicable to $g(x)$ on $[1, 2]$,we must have $g(1) = g(2)$.$g(1) = 1 - 1 + b \sin(\frac{\pi}{2}) = b(1) = b$.$g(2) = 2 - 1 + b \sin(\pi) = 1 + b(0) = 1$.Setting $g(1) = g(2)$ gives $b = 1$.Thus,$LMVT$ is applicable to $f$,and Rolle's theorem is applicable to $g$ when $b = 1$.

Consider $f(x) = |1 - x|$ for $1 \le x \le 2$ and $g(x) = f(x) + b \sin(\frac{\pi}{2}x)$ for $1 \le x \le 2$. Which of the following is correct?

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