Question

If in the expansion of ${\left( {\dfrac{1}{x} + x\tan x} \right)^5}$ the ratio of 4th term to 2nd term is $\dfrac{2}{{27}}{\pi ^4}$, then the value of x can be
(A) $ - \dfrac{\pi }{6}$
(B) $ - \dfrac{\pi }{3}$
(C) $\dfrac{\pi }{3}$
(D) $\dfrac{\pi }{{12}}$

Answer 1

Hint: Start with using the binomial expansion of ${(x + y)^n}$. We know that ${\left( {r + 1} \right)^{th}}$ term in its expansion is ${T_{r + 1}}{ = ^n}{C_r}{(x)^{n - r}}{(y)^r}$. Find out the 4th term and 2nd term and compare their ratio with the value given in the question.

Complete step by step answer:
According to the question, the given expression is ${\left( {\dfrac{1}{x} + x\tan x} \right)^5}$.
The above expression is of the form ${(x + y)^n}$. And we know that general term of expansion of ${(x + y)^n}$ is given as:
$ \Rightarrow {T_{r + 1}}{ = ^n}{C_r}{(x)^{n - r}}{(y)^r}$
Using this result, the general term of the expression in the question is:
$ \Rightarrow {T_{r + 1}}{ = ^5}{C_r}{\left( {\dfrac{1}{x}} \right)^{5 - r}}.{\left( {x\tan x} \right)^r} ....(i)$
From the information in the question, the ratio of 4th term to 2nd term is$\dfrac{2}{{27}}{\pi ^4}$. So we have:
$ \Rightarrow \dfrac{{{T_4}}}{{{T_2}}} = \dfrac{2}{{27}}{\pi ^4} .....(ii)$
Now putting $r = 3$ in equation $(i)$, we’ll get:
$
   \Rightarrow {T_4}{ = ^5}{C_3}{\left( {\dfrac{1}{x}} \right)^{5 - 3}}.{\left( {x\tan x} \right)^3} \\
   \Rightarrow {T_4}{ = ^5}{C_3}.x{\tan ^3}x \\
$
Again putting $r = 1$ in equation $(i)$, we’ll get:
$
   \Rightarrow {T_2}{ = ^5}{C_1}{\left( {\dfrac{1}{x}} \right)^{5 - 1}}.{\left( {x\tan x} \right)^1} \\
   \Rightarrow {T_2}{ = ^5}{C_1}.\dfrac{1}{{{x^3}}}\tan x \\
$
Putting these values of ${T_4}$ and ${T_2}$ in equation $(ii)$, we’ll get:
\[
   \Rightarrow \dfrac{{^5{C_3}.x{{\tan }^3}x}}{{^5{C_1}.\dfrac{1}{{{x^3}}}\tan x}} = \dfrac{2}{{27}}{\pi ^4} \\
   \Rightarrow \dfrac{{^5{C_3}}}{{^5{C_1}}}.{x^4}{\tan ^2}x = \dfrac{2}{{27}}{\pi ^4} \\
\]
We know that \[^5{C_3} = 10\] and \[^5{C_1} = 5\]. Putting its value, we’ll get:
\[
   \Rightarrow \dfrac{{10}}{5}.{x^4}{\tan ^2}x = \dfrac{2}{{27}}{\pi ^4} \\
   \Rightarrow {x^4}{\tan ^2}x = \dfrac{{{\pi ^4}}}{{27}} \\
\]
Multiplying and dividing by 3 on the right hand side, we’ll get:
\[ \Rightarrow {x^4}{\tan ^2}x = \dfrac{{{\pi ^4}}}{{81}} \times 3\]
Now 3 can be written as ${\left( {\sqrt 3 } \right)^2}$. Doing this, we’ll get:
\[ \Rightarrow {x^4}{\tan ^2}x = \dfrac{{{\pi ^4}}}{{81}} \times {\left( {\sqrt 3 } \right)^2}\]
Comparing \[{x^4}\] with \[\dfrac{{{\pi ^4}}}{{81}}\] and \[{\tan ^2}x\] with \[{\left( {\sqrt 3 } \right)^2}\], we have:
$ \Rightarrow {x^4} = \dfrac{{{\pi ^4}}}{{81}}$ and ${\tan ^2}x = {\left( {\sqrt 3 } \right)^2}$
\[ \Rightarrow {x^4} = {\left( {\dfrac{\pi }{3}} \right)^4}\] and $\tan x = \pm \sqrt 3 $
\[ \Rightarrow x = \pm \dfrac{\pi }{3}\] and \[x = \pm \dfrac{\pi }{3}\]

Hence \[x = \pm \dfrac{\pi }{3}\] is a valid solution. (B) is the correct option.

Note: Here we have solved the problem by using the general expansion of ${\left( {x + y} \right)^n}$ as the given expression is in the same form. This expansion consists of a total of $\left( {n + 1} \right)$ terms. We were required to compare only the fourth and second term.