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Question: If in the expansion of \({\left( {\dfrac{1}{x} + x\tan x} \right)^5}\) the ratio of 4th term to 2nd ...

If in the expansion of (1x+xtan⁡x)5{\left( {\dfrac{1}{x} + x\tan x} \right)^5} the ratio of 4th term to 2nd term is 227π4\dfrac{2}{{27}}{\pi ^4}, then the value of x can be
(A) −π6 - \dfrac{\pi }{6}
(B) −π3 - \dfrac{\pi }{3}
(C) π3\dfrac{\pi }{3}
(D) π12\dfrac{\pi }{{12}}

Explanation

Solution

Hint: Start with using the binomial expansion of (x+y)n{(x + y)^n}. We know that (r+1)th{\left( {r + 1} \right)^{th}} term in its expansion is Tr+1=nCr(x)n−r(y)r{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 (1x+xtan⁡x)5{\left( {\dfrac{1}{x} + x\tan x} \right)^5}.
The above expression is of the form (x+y)n{(x + y)^n}. And we know that general term of expansion of (x+y)n{(x + y)^n} is given as:
⇒Tr+1=nCr(x)n−r(y)r\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:
⇒Tr+1=5Cr(1x)5−r.(xtan⁡x)r....(i)\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 is227π4\dfrac{2}{{27}}{\pi ^4}. So we have:
⇒T4T2=227π4.....(ii)\Rightarrow \dfrac{{{T_4}}}{{{T_2}}} = \dfrac{2}{{27}}{\pi ^4} .....(ii)
Now putting r=3r = 3 in equation (i)(i), we’ll get:
⇒T4=5C3(1x)5−3.(xtan⁡x)3 ⇒T4=5C3.xtan⁡3x  \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=1r = 1 in equation (i)(i), we’ll get:
⇒T2=5C1(1x)5−1.(xtan⁡x)1 ⇒T2=5C1.1x3tan⁡x  \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 T4{T_4} and T2{T_2} in equation (ii)(ii), we’ll get:

⇒5C3.xtan⁡3x5C1.1x3tan⁡x=227π4 ⇒5C35C1.x4tan⁡2x=227π4  \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 5C3=10^5{C_3} = 10 and 5C1=5^5{C_1} = 5. Putting its value, we’ll get:

⇒105.x4tan⁡2x=227π4 ⇒x4tan⁡2x=π427  \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:
⇒x4tan⁡2x=π481×3\Rightarrow {x^4}{\tan ^2}x = \dfrac{{{\pi ^4}}}{{81}} \times 3
Now 3 can be written as (3)2{\left( {\sqrt 3 } \right)^2}. Doing this, we’ll get:
⇒x4tan⁡2x=π481×(3)2\Rightarrow {x^4}{\tan ^2}x = \dfrac{{{\pi ^4}}}{{81}} \times {\left( {\sqrt 3 } \right)^2}
Comparing x4{x^4} with π481\dfrac{{{\pi ^4}}}{{81}} and tan⁡2x{\tan ^2}x with (3)2{\left( {\sqrt 3 } \right)^2}, we have:
⇒x4=π481\Rightarrow {x^4} = \dfrac{{{\pi ^4}}}{{81}} and tan⁡2x=(3)2{\tan ^2}x = {\left( {\sqrt 3 } \right)^2}
⇒x4=(π3)4\Rightarrow {x^4} = {\left( {\dfrac{\pi }{3}} \right)^4} and tan⁡x=±3\tan x = \pm \sqrt 3
⇒x=±π3\Rightarrow x = \pm \dfrac{\pi }{3} and x=±π3x = \pm \dfrac{\pi }{3}

Hence x=±π3x = \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 (x+y)n{\left( {x + y} \right)^n} as the given expression is in the same form. This expansion consists of a total of (n+1)\left( {n + 1} \right) terms. We were required to compare only the fourth and second term.