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IIT JEE Advanced 2012 Paper 2

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Let f (x) =

(1-x)^2 \sin^2 x + x^2

for all x ∊ R, and let g (x) =

\int\limits_{1}^{x} \left( \frac{2(t-1)}{t+1} - \ln t \right)

f (t) dt for all x ∊ (1 , ∞).

Section: Mathematics

Question : 52 of 60

Marks: +1, -0

Solution:

g' (x) =

\left(\frac{2(x-1)}{x+1} - \ln x\right)

f (x). For x ∊ (1 , ∞) , f (x) > 0

Let h (x) =

\left(\frac{2(x-1)}{x+1} - \ln x\right)

⇒ h' (x)

\left(\frac{4}{(x+1)^2} - \frac{1}{x}\right)

\frac{-(x-1)^2}{(x+1)^2 x}

< 0

Also h(1) = 0. so, h(x) < 0 ∀ x > 1.

⇒ g(x) is decreasing on (1, ∞).

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