Question

The $f\left( x \right) = \dfrac{1}{2}\sin x\tan x - \log \left( {\sec x} \right)$ is increasing in:
1) $\left( { - \dfrac{\pi }{2},0} \right)$
2) $\left( {\pi ,\dfrac{{3\pi }}{2}} \right)$
3) $\left( {0,\dfrac{\pi }{2}} \right)$
4) $\left( {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right)$

Answer 1

Hint: We will take the differentiation of the given function using the product rule, $\dfrac{{du\left( x \right)v\left( x \right)}}{{dx}} = u'\left( x \right)v\left( x \right) + u\left( x \right)v'\left( x \right)$ and chain rule, ${\left( {u\left( {v\left( x \right)} \right)} \right)^\prime } = u'\left( {v\left( x \right)} \right)v'\left( x \right)$. Then, simplify the expression and determine the interval where $f'\left( x \right) > 0$.

Complete step-by-step answer:
We have to find the interval where the function $f\left( x \right) = \dfrac{1}{2}\sin x\tan x - \log \left( {\sec x} \right)$ is increasing.
We will differentiate the function with respect to $x$
We will use product rules to differentiate the first term. Product rule states that $\dfrac{{du\left( x \right)v\left( x \right)}}{{dx}} = u'\left( x \right)v\left( x \right) + u\left( x \right)v'\left( x \right)$
And we will use chain rule to differentiate the second term.
Chain rule states that ${\left( {u\left( {v\left( x \right)} \right)} \right)^\prime } = u'\left( {v\left( x \right)} \right)v'\left( x \right)$
\[f'\left( x \right) = \dfrac{1}{2}\left( {\dfrac{{d\left( {\sin x} \right)}}{{dx}}\tan x + \dfrac{{d\left( {\tan x} \right)}}{{dx}}\sin x} \right) - \dfrac{{d\left( {\log \left( {\sec x} \right)} \right)}}{{dx}}\dfrac{d}{{dx}}\left( {\sec x} \right)\]
Now, we know that \[\dfrac{{d\left( {\sin x} \right)}}{{dx}} = \cos x\], \[\dfrac{{d\left( {\tan x} \right)}}{{dx}} = {\sec ^2}x\], \[\dfrac{d}{{dx}}\left( {\sec x} \right) = \sec x\tan x\], and \[\dfrac{{d\left( {\log x} \right)}}{{dx}} = \dfrac{1}{x}\]
On substituting the values in the expression, we will get,
$
  f'\left( x \right) = \dfrac{1}{2}\left( {\cos x\tan x + {{\sec }^2}x\sin x} \right) - \dfrac{1}{{\sec x}}\tan x\sec x \\
   \Rightarrow f'\left( x \right) = \dfrac{1}{2}\left( {\cos x\tan x + {{\sec }^2}x\sin x} \right) - \tan x \\
$
We know that $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and $\sec \theta = \dfrac{1}{{\cos \theta }}$
Then,
$
  f'\left( x \right) = \dfrac{1}{2}\left( {\dfrac{{\cos x\sin x}}{{\cos x}} + \dfrac{{\tan x\cos x}}{{{{\cos }^2}x}}} \right) - \tan x \\
   \Rightarrow f'\left( x \right) = \dfrac{1}{2}\left( {\sin x + \dfrac{{\tan x}}{{\cos x}}} \right) - \tan x \\
   \Rightarrow f'\left( x \right) = \dfrac{{\sin x}}{2} + \dfrac{{\tan x\sec x}}{2} - \tan x \\
   \Rightarrow f'\left( x \right) = \dfrac{{\sin x}}{2} + \dfrac{{\tan x}}{2}\left( {\sec x - 1} \right) \\
   \Rightarrow f'\left( x \right) = \dfrac{1}{2}\left( {\sin x + \tan x\left( {\sec x - 1} \right)} \right) \\
$
For the function to be increasing $f'\left( x \right) > 0$
\[\dfrac{1}{2}\left( {\sin x + \tan x\left( {\sec x - 1} \right)} \right) > 0\]
\[\sin x - \tan x\left( {\sec x - 1} \right) > 0\]
\[ \Rightarrow \sin x > \tan x\left( {\sec x - 1} \right)\]
This function is increasing only when $\left( {0,\dfrac{\pi }{2}} \right)$
Hence, option 3 is correct.

Note: The points at which the differentiation of the function is greater than 0, then it is an increasing function and if the differentiation of the function is less than 0, then the function is a decreasing function.