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f(x) = (1-x)^(2) cdot sin^(2) x + x^(2)...

`f(x) = (1-x)^(2) cdot sin^(2) x + x^(2) ge 0, AA x`
and `g(x) =int_(1)^(x) ((2(t-1))/((x+1))-log t ) f (t) dt` Which of the following is true? (A)g is increasing on `(1,infty)` (B)g is decreasing on `(1, infty)` (C) g is increasing on (1,2) and decreasing on `(2, infty)` (D) g is decreasing on (1,2) and increasing on `(2,infty)`

A

(a) g is increasing on `(1,infty)`

B

(b) g is decreasing on `(1, infty)`

C

(c) g is increasing on (1,2) and decreasing on `(2, infty)`

D

(d) g is decreasing on (1,2) and increasing on `(2,infty)`

Text Solution

Verified by Experts

The correct Answer is:
(b)
Here, `f(x) = (1-x)^(2) cdot sin^(2) x + x^(2) ge 0, AA x`
and `g(x) =int_(1)^(x) ((2(t-1))/((x+1))-log t ) f (t) dt`
`rArr g(x) = {(2(t-1))/((x+1))-log x } cdot f underbrace((x))_(+ ve) …(i)`
For g'(x) to be increasing of decreasing.
Let `phi (x) = (2(x-1))/(x+1 )-log x`
`phi' (x)=4/((x+1)^(2))-1/x = (-(x-1))/(x(x+)^(2)`
`phi' (x) lt0, AA xgt1`
`rArr phi (x) lt phi(1)`
`rArr phi (x) lt 0 …(ii)`
From Eqs. (i) and (ii), `g'(x)lt 0, x in (1,infty)`
`therefore g(x)` is decreasing on `x in (1, infty).`
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