Let f(x)=(1-x)^ 2 ^ 2 x+x^ 2 for all x ∈ I R and let g(x)=∫_ 1 ^ x ((2(t-1))/(t+1)- t) f(t) d t for all x ∈(1, ∞) . Question: Which of the following is true?

Correct answer: B. g is decreasing on (1, ∞) g(x)=∫_ 1 ^ x ((2(t-1))/(t+1)- t) f(t) d t , g^ (x)=[(2(x-1))/(x+1)- x] f(x) Here f(x)>0, x ∈(1, ∞) Let h(x)=(2(x-1))/(x+1)- x h^ (x)= 4 (x+1)^ 2 -(1)/(x)= -(x-1)^ 2 (x+1)^ 2 x 1, h(x) 1 g^ (x) < 0 x ∈(1, ∞) Therefore, g(x) is decreasing on (1, ∞) .

JEE Advanced 2012Paper 2

Exam: JEE Advanced
Year: 2012
Paper: Paper 2
Subject: Mathematics
Chapter: Application of Derivatives

JEE Advanced 2012 · Paper 2 · Mathematics · Application of Derivatives

Let f(x)=(1-x)^ 2 ^ 2 x+x^ 2 for all x ∈ I R and let g(x)=∫_ 1 ^ x...

Let

f(x)=(1-x)^{2} \sin ^{2} x+x^{2}

for all

x \in I R

and let

g(x)=\int_{1}^{x}\left(\frac{2(t-1)}{t+1}-\ln t\right) f(t) d t

for all

x \in(1, \infty)

. Question: Which of the following is true?

Options

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 $(2, \infty)$

Correct answer

B. $g$ is decreasing on $(1, \infty)$

Step-by-step solution

g(x)=\int_{1}^{x}\left(\frac{2(t-1)}{t+1}-\ln t\right) f(t) d t

\therefore \quad g^{\prime}(x)=\left[\frac{2(x-1)}{x+1}-\ln x\right] f(x)

Here

f(x)>0, \forall x \in(1, \infty)

Let

h(x)=\frac{2(x-1)}{x+1}-\ln x

\therefore \quad h^{\prime}(x)=\frac{4}{(x+1)^{2}}-\frac{1}{x}=\frac{-(x-1)^{2}}{(x+1)^{2} x} < 0, x \in(1, \infty)

\Rightarrow h(x)

is decreasing function. Hence, for

x>1, h(x) < h(1) \Rightarrow h(x) < 0 \forall x>1

\Rightarrow g^{\prime}(x) < 0 \forall x \in(1, \infty)

Therefore,

g(x)

is decreasing on

(1, \infty)

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