Question

For $0 < {x_1} < {x_2} < \dfrac{\pi }{2}$,
A. $\dfrac{{\tan {x_2}}}{{\tan {x_1}}} < \dfrac{{{x_2}}}{{{x_1}}}$
B. $\dfrac{{\tan {x_2}}}{{\tan {x_1}}} > \dfrac{{{x_2}}}{{{x_1}}}$
C. $\dfrac{{\tan {x_2}}}{{\tan {x_1}}} = \dfrac{{{x_2}}}{{{x_1}}}$
D. None of these

Answer 1

Hint: To solve the question, we will first imagine a function that will help us find a suitable option. Then we will differentiate the function to find whether the function is increasing or decreasing in a given domain. Once we obtain the trend of the function we can use the given condition to find which one is the suitable option for the question.

Complete step by step answer:
We can see that all the options have $\tan x$ and $x$ involved in them.
So, let us imagine a function such as $f(x) = \dfrac{{\tan x}}{x}$.
Now, on differentiating the function with respect to $x$, we get,
$f'(x) = \dfrac{d}{{dx}}\left( {\dfrac{{\tan x}}{x}} \right)$
Using the quotient rule of differentiation, we get,
$ \Rightarrow f'(x) = \dfrac{{x.{{\sec }^2}x - \tan x}}{{{x^2}}}$
Taking, ${\sec ^2}x$ common from Right Hand Side, we get,
$ \Rightarrow f'(x) = {\sec ^2}x\left( {\dfrac{{x - \sin x.\cos x}}{{{x^2}}}} \right)$

Using the trigonometric identity $\sin 2x = 2\sin x\cos x$, we get,
$ \Rightarrow f'(x) = {\sec ^2}x\left( {\dfrac{{x - \dfrac{{\sin 2x}}{2}}}{{{x^2}}}} \right)$
Taking LCM in the numerator, we get,
Now, we know, $\sec x = \dfrac{1}{{\cos x}}$, using this property, we get,
$ \Rightarrow f'(x) = \left( {\dfrac{{x - \dfrac{{\sin 2x}}{2}}}{{{x^2}{{\cos }^2}x}}} \right)$
Simplifying the expression further, we get,
$ \Rightarrow f'(x) = \dfrac{{2x - \sin 2x}}{{2{x^2}{{\cos }^2}x}}$
We are given, that, $0 < {x_1} < {x_2} < \dfrac{\pi }{2}$, that is any value of x lies between $\left( {0,\dfrac{\pi }{2}} \right)$.
Therefore, $0 < x < \dfrac{\pi }{2}$
Multiplying all the sides of the inequality by $2$,
$ \Rightarrow 0 < 2x < \pi $
We know, in $\left( {0,\pi } \right)$, the value of $2x > \sin 2x$. Therefore $2x - \sin 2x > 0$.

Also, we know that any square term is always non-negative.
Therefore, the numerator and denominator of the derivative of the function $f(x) = \dfrac{{\tan x}}{x}$ is positive. So, the overall derivative function is positive.
Since, $f'(x) > 0$, that means $f(x)$ is an increasing function.
That is, for ${x_1} < {x_2}$.
$ \Rightarrow f({x_1}) < f({x_2})$
$ \Rightarrow \dfrac{{\tan {x_1}}}{{{x_1}}} < \dfrac{{\tan {x_2}}}{{{x_2}}}$
Now, changing the sides over the inequality to get $\tan x$terms on one side and $x$terms on one side, we get,
$ \Rightarrow \dfrac{{{x_2}}}{{{x_1}}} < \dfrac{{\tan {x_2}}}{{\tan {x_1}}}$
$ \therefore \dfrac{{\tan {x_2}}}{{\tan {x_1}}} > \dfrac{{{x_2}}}{{{x_1}}}$

Therefore, the correct option is B.

Note: We must keep one thing in mind that, when we assume a function, it must be continuous as well as differentiable, to satisfy the fundamental conditions of differentiability.Sometimes it may arise that the assumed function is continuous but not differentiable, which will not work to solve the problem, so we must keep that in mind. We must have a good grip over derivatives to tackle such questions.