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If f(x)={{:(x",",0lexle1),(2-e^(x-1)",",...

If `f(x)={{:(x",",0lexle1),(2-e^(x-1)",",1ltxle2),(x-e",",2ltxle3):}` and `g'(x)=f(x), x in [1,3]`, then```

A

`g(x)` has local maximum at ` x = 1 + log _e 2 ` and local minima at ` x = e `

B

` f(x) ` has local maxima at ` x =1 ` and local minima at ` x = 2 `

C

`g(x) ` has no local minima

D

`f(x) ` has no local maxima

Text Solution

Verified by Experts

The correct Answer is:
A, B
Given,
`" " f(x) = {{:( e ^(x),, ",", if 0 le x le 1 ) , ( 2- e^(x - 1),, ",", if 1 lt x le2), ( x - e ,, ",", if 2 lt x le 3 ):}`
and ` " " g(x) = int _( 0) ^(x) f (t) dt `
`rArr " " g ' (x) = f(x) `
Put ` " " g ' (x) = 0 rArr x = 1 + log_e 2 and x = e`.
Also, ` " " g''(x) = {{:( e ^(x)",",, if 0 le x le 1), (- e ^( x -1 )",",, if 1 lt x le 2), ( 1",",, if 2 lt x le 3 ):}`
At `" " x = 1 +log _e 2`,
`g'' ( 1+ log_e 2) = - e ^( log_e 2) lt 0 , g(x)` has a local maximum.
Also, at ` x = e`,
`g '' (e) = 1 gt 0, g (x)` has a local, minima.
`because f(x)` is discontinuous at ` x = 1`, then we get local maxima at ` x = 1 ` and local minima at ` x = 2`.
Hence (a) and (b) are correct answers.
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