If `f(x) = {{:(e ^(x),,"," 0 le x lt 1 ,, ""), (2- e^(x - 1),,"," 1 lt x le 2,, and g(x) = int_(0)^(x) f(t ) dt","),( x- e,,"," 2lt x le 3 ,, ""):}`
` x in [ 1, 3 ] `, then

Options

A. g(x) has a local maxima at `x=1 + log_e` 2 and local minima at x=e
B. f(x) has a local maxima at x=1 and local minima at x=2
C. f(x) and f(x) have same points of local maxima and local minima
D. none of these

Answer

C

Explanation

It is given that
`f(x)={{:(e^x","0 le x le 1),(2-e^(x+1)","1 lt x le 2 and g(x)= underset(0)overset(x)int f(t) dt "," x in [1,3]),(x-e "," 2 lt le 3):}`
`therefore g(x) =f(x) = {{:(2-e^(x-1) 1 le x le 2 ),(x-e 2 le x le 3 ):}`
`so g(x)=0 rArr x =1 + log _e 2 and x =e `
Also `g' (x) gt 0 for x rarr (1,1 + log_e 2) and g'(x) lt 0 ` for `x in (1+ log _e) 2,2 `
So g(x) attains a local maximum at `x=1 + log_e 2` Similay
`g(x) le 0 for 2 lt x lt e "and " g(x) gt 0 for e lt x lt 3`
`rArr g(x) ` attains local minimum at x=e
we have

Clearly `f(x) gt 0 for x in (0,1) lt 0 for x in (1,2) gt 0 for x in (2,3) `
So, f(x) attains local maximum at x=1 and local minimum at x=2
Hence option ( c) is incorrect .