Question

Let \[f\left( x \right) = {\left( {1 - x} \right)^2}{\sin ^2}x + {x^2}\] for all \[x \in {\text{IR,}}\] and let \[g\left( x \right) = \mathop \smallint \nolimits_1^x \left( {\dfrac{{2\left( {t - 1} \right)}}{{t + 1}} - \ln t} \right)f\left( t \right)dt\] for all \[x \in (1,\;\infty ).\] 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 ) $

Answer 1

Hint: From the given options, it is clear that you have to find whether the function $ g(x) $ is increasing or decreasing on the given interval. So, to find whether the function is increasing or decreasing, find derivative of the function with help of Leibnitz rule and then you will get two functions, simplify them for their increasing or decreasing to eventually find for $ g(x) $

Complete step-by-step answer:
In this question, we have to find the slope of $ g(x) $ either increasing or decreasing in the given intervals. For this we will find its derivative with help of Leibnitz rule as follows
Leibnitz rule is used to find the derivative of the anti-derivative, this rule can be understood as consider there are two functions \[u\left( t \right){\text{ and }}v\left( t \right),\] which have the derivatives up to $ {n^{th}} $ order then their derivative will be given as
  $ {\left( {uv} \right)^n} = \sum\limits_{i = 0}^n {\left( {\begin{array}{*{20}{c}}
  n \\
  i
\end{array}} \right){u^{(n - i)}}{v^i}} $
Using this, finding the derivative of the given integral, we will get
\[ \Rightarrow g\left( x \right) = \mathop \smallint \nolimits_1^x \left( {\dfrac{{2\left( {t - 1} \right)}}{{t + 1}} - \ln t} \right)f\left( t \right)dt\]
Taking $ u(t) = \left( {\dfrac{{2\left( {t - 1} \right)}}{{t + 1}} - \ln t} \right)\;{\text{and}}\;v(t) = f(t) $ then $ g'(x) $ will be written as
  $ \Rightarrow g'\left( x \right) = \left( {\dfrac{{2\left( {x - 1} \right)}}{{x + 1}} - \ln x} \right)f\left( x \right) $
Here let \[h(x) = \left( {\dfrac{{2\left( {x - 1} \right)}}{{x + 1}} - \ln x} \right)\]
  $ \Rightarrow g'\left( x \right) = h(x)f\left( x \right) $
Now, in above equation, we can see that $ g'(x) $ is consist of two functions $ h(x)\;{\text{and}}\;f(x) $
And also we can see that $ f(x) $ is always positive in the interval $ (1,\;\infty ) $
So we have to check slope only for $ h(x) $
\[
   \Rightarrow h(x) = \left( {\dfrac{{2\left( {x - 1} \right)}}{{x + 1}} - \ln x} \right) \\
   \Rightarrow h(x) = \left( {\dfrac{{2\left( {x + 1} \right) - 4}}{{x + 1}} - \ln x} \right) \\
   \Rightarrow h(x) = \left( {2 - \dfrac{4}{{x + 1}} - \ln x} \right) \\
   \Rightarrow h'(x) = \left( {0 + \dfrac{4}{{{{(x + 1)}^2}}} - \dfrac{1}{x}} \right) \\
   \Rightarrow h'(x) = \left( {\dfrac{{4x - {{(x + 1)}^2}}}{{x{{(x + 1)}^2}}}} \right) \\
   \Rightarrow h'(x) = \left( {\dfrac{{4x - {x^2} - 1 - 2x}}{{x{{(x + 1)}^2}}}} \right) \\
   \Rightarrow h'(x) = \dfrac{{ - {{(x - 1)}^2}}}{{x{{(x + 1)}^2}}} \;
 \]
So we can see that numerator has square function and denominator has a square function that is always positive, so other denominator function will decide slope of the function,
We know that $ - \dfrac{1}{x} < 0,\;\forall x \in (1,\,\infty ) $
Therefore the function $ h(x) $ and hence $ g(x) $ is decreasing over the interval $ (1,\;\infty ) $
So, the correct answer is “Option B”.

Note: When finding the derivative of $ \dfrac{u}{v} $ type function like as in this question, try to simplify the $ \dfrac{u}{v} $ form to make the differentiation process easy. However it does not get affected or the result will vary if you differentiate directly or differentiate after simplifying.