Definite Integration Ques 105
- 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 $(2, \infty)$
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Answer:
Correct Answer: 105.(b)
Solution:
- Here, $f(x)=(1-x)^{2} \cdot \sin ^{2} x+x^{2} \geq 0, \forall x$.
and $g(x)=\int _1^{x} \frac{2(t-1)}{t+1}-\log t \quad f(t) d t$
$\Rightarrow \quad g^{\prime}(x)=\frac{2(x-1)}{(x+1)}-\log x \cdot \underbrace{f(x)} _{+ve}$
For $g^{\prime}(x)$ to be increasing or decreasing,
let $\varphi(x)=\frac{2(x-1)}{(x+1)}-\log x$
$\varphi^{\prime}(x)=\frac{4}{(x+1)^{2}}-\frac{1}{x}=\frac{-(x-1)^{2}}{x(x+1)^{2}}$
$\varphi^{\prime}(x)<0$, for $x>1 \Rightarrow \varphi(x)<\varphi(1) \Rightarrow \varphi(x)<0$
From Eqs. (i) and (ii), we get
$$ g^{\prime}(x)<0 \text { for } x \in(1, \infty) $$
$\therefore g(x)$ is decreasing for $x \in(1, \infty)$.
BETA
