Consider `f(x)=|1-x|,1 le xle2 and g(x)=f(x)+b sin.(pi)/(2)x, 1 le xle 2` then which of the following is correct?

Options

A. Rolle's theorem is applicable to both f and g with `b=(3)/(2)`.
B. LMVT is not applicable to f and Rolle's theorem is applicable to g with `b=(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 and g for any real b.

Answer

C

Explanation

`f(x)=x-1,1lexle2`
`g(x)=x-1+b sin.(pi)/(2)x, 1le xle 2`
`f(1)=0,f(2)=1 rArr `Rolle's theorem is not applicable to 'f' but LMVT is application to f
`(because" "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.(pi)/(2)` are both continuuous in `[1,2]` and (1,2)`
`therefore" For Rolle's theorem to be applicable to g, we must have b = 1