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Let f(x) = e^(x)- x and g(x) = x^(2) - ...

Let `f(x) = e^(x)- x and g(x) = x^(2) - x , AA x in R`. Then, the set of all `x in R`, when the function `h(x)= (fog)(x)` is increasing, is

A

`[0, (1)/(2)] uu [1, oo)`

B

`[-1, (-1)/(2)] uu [ (1)/(2), oo)`

C

`[0, oo)`

D

`[(-1)/(2), 0] uu [1, oo)`

Text Solution

Verified by Experts

The correct Answer is:
A
The given functions are
`" " f(x) = e ^(x) - x `,
`and " " g (x) = x ^(2) - x, AA x in R`
Then, `h(x) = ( fog) (x) = f(g(x))`
Now, `h'(x) = f'(g(x)) * g'(x)`
`" " = ( e^(g(x)) - 1) * ( 2x - 1) = (e^("("x^(2) - x")") - 1) ( 2x - 1)`
`" " = (e ^(x( x - 1)) - 1)(2x - 1)`
`because ` It is given that `h(x)` is an increasing function, so `h'(x) ge 0`
`rArr " " (e^(x(x- 1)) - 1) ( 2x - 1) ge 0`
Case I `( 2x - 1) ge 0 and (e ^(x( x- 1)) -1 ge 0 `
`rArr " " x ge (1) /(2) and x (x - 1) ge 0`
`rArr x in [ 1//2, oo) and x in ( - oo, 0], uu[ 1, oo), "so " x in [1, oo)`
Case II `(2x -1) le 0 and [ e^(x(x -1 ) ) - 1] le 0`
`rArr x le (1)/(2) and x (x - 1) le 0 rArr x in (-oo, (1) /(2)] and x in [0, 1]`
So, `" " x in [ 0, (1) /(2)]`
From, the above cases, ` x in [0, (1)/(2)] uu [ 1, oo)`
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