Question

If in the expression of \[{{\left( \dfrac{1}{x}+x\tan x \right)}^{5}}\], the ratio of 4th term to the second 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:First of all use the formula for the general term as \[{{T}_{r+1}}=n{{C}_{r}}{{x}^{n-r}}{{y}^{r}}\] by using r = 1 and r = 3 for the second and fourth term respectively. Now, equate the ratio of the fourth term and the second term to \[\dfrac{2}{27}{{\pi }^{4}}\] and get an equation. In this equation, substitute the options and check which satisfies the equation.

Complete step-by-step answer:
We are given that in the expression of \[{{\left( \dfrac{1}{x}+x\tan x \right)}^{5}}\], the ratio of 4th term to the second term is \[\dfrac{2}{27}{{\pi }^{4}}\]. We have to find the value of x. Let us take the expression given in the question.
\[E={{\left( \dfrac{1}{x}+x\tan x \right)}^{5}}\]
We know that the general term as (r + 1)th term in the expression of \[{{\left( x+y \right)}^{n}}\] is given by \[{{T}_{r+1}}=n{{C}_{r}}{{x}^{n-r}}{{y}^{r}}\]. So, by using this, we get the general term in the expansion of the above expression as,
\[{{T}_{r+1}}=5{{C}_{r}}{{\left( \dfrac{1}{x} \right)}^{5-r}}{{\left( x\tan x \right)}^{r}}....\left( i \right)\]
By substituting the value of r = 1 in equation (i), we get, the second term in the expansion as
\[{{T}_{1+1}}={{T}_{2}}=5{{C}_{1}}{{\left( \dfrac{1}{x} \right)}^{5-1}}{{\left( x\tan x \right)}^{1}}\]
\[{{T}_{2}}=5{{C}_{1}}{{\left( \dfrac{1}{x} \right)}^{4}}\left( x\tan x \right)\]
\[{{T}_{2}}=\dfrac{5{{C}_{1}}\tan x}{{{x}^{3}}}....\left( ii \right)\]
By substituting the value of r = 3 in equation (i), we get, the fourth term in the expansion as,
\[{{T}_{3+1}}={{T}_{4}}=5{{C}_{3}}{{\left( \dfrac{1}{x} \right)}^{5-3}}{{\left( x\tan x \right)}^{3}}\]
\[{{T}_{4}}=5{{C}_{3}}{{\left( \dfrac{1}{x} \right)}^{2}}{{\left( x\tan x \right)}^{3}}\]
\[{{T}_{4}}=\dfrac{5{{C}_{3}}{{x}^{3}}{{\left( \tan x \right)}^{3}}}{{{x}^{2}}}\]
\[{{T}_{4}}=5{{C}_{3}}x{{\left( \tan x \right)}^{3}}....\left( iii \right)\]
By dividing equation (iii) with equation (ii), we get,
\[\dfrac{\text{4th term}}{\text{2nd term}}=\dfrac{5{{C}_{3}}x{{\left( \tan x \right)}^{3}}}{\dfrac{5{{C}_{1}}\tan x}{{{x}^{3}}}}.....\left( iv \right)\]
We are given that,
\[\dfrac{\text{4th term}}{\text{2nd term}}=\dfrac{2}{27}{{\pi }^{4}}....\left( v \right)\]
So, from equation (iv) and (v), we get,
\[\dfrac{5{{C}_{3}}x{{\left( \tan x \right)}^{3}}}{\dfrac{5{{C}_{1}}\tan x}{{{x}^{3}}}}=\dfrac{2}{27}{{\pi }^{4}}\]
We know that,
\[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
So, we get,
\[\dfrac{\dfrac{5!}{3!2!}{{\left( \tan x \right)}^{2}}{{x}^{4}}}{\dfrac{5!}{1!4!}}=\dfrac{2}{27}{{\pi }^{4}}\]
\[=\dfrac{4!}{3!2!}{{\left( \tan x \right)}^{2}}{{x}^{4}}=\dfrac{2}{27}{{\pi }^{4}}\]
\[=\dfrac{4\times 3!}{3!2!}{{\left( \tan x \right)}^{2}}{{x}^{4}}=\dfrac{2}{27}{{\pi }^{4}}\]
\[=2{{\left( \tan x \right)}^{2}}{{x}^{4}}=\dfrac{2}{27}{{\pi }^{4}}\]
\[={{\left( \tan x \right)}^{2}}{{x}^{4}}=\dfrac{{{\pi }^{4}}}{27}....\left( vi \right)\]
As the above equation is not a standard equation, so now, we will use the options.
(a) By substituting \[x=\dfrac{-\pi }{6}\] in equation (vi), we get,
\[={{\left[ \tan \left( \dfrac{-\pi }{6} \right) \right]}^{2}}{{\left( \dfrac{-\pi }{6} \right)}^{4}}=\dfrac{{{\pi }^{4}}}{27}\]
We know that \[\tan \left( \dfrac{-\pi }{6} \right)=\dfrac{-1}{\sqrt{3}}\], so we get,
\[{{\left( \dfrac{-1}{\sqrt{3}} \right)}^{2}}{{\left( \dfrac{\pi }{6} \right)}^{4}}=\dfrac{{{\pi }^{4}}}{27}\]
\[\dfrac{{{\pi }^{4}}}{3888}\ne \dfrac{{{\pi }^{4}}}{27}\]
\[LHS\ne RHS\]
So, this value of x is not correct.
(b) By substituting \[x=\dfrac{-\pi }{3}\] in equation (vi), we get,
\[={{\left[ \tan \left( \dfrac{-\pi }{3} \right) \right]}^{2}}{{\left( \dfrac{-\pi }{3} \right)}^{4}}=\dfrac{{{\pi }^{4}}}{27}\]
We know that \[\tan \left( \dfrac{-\pi }{3} \right)=\sqrt{3}\], so we get,
\[{{\left( -\sqrt{3} \right)}^{2}}{{\left( \dfrac{\pi }{3} \right)}^{4}}=\dfrac{{{\pi }^{4}}}{27}\]
\[3\times \dfrac{{{\pi }^{4}}}{81}=\dfrac{{{\pi }^{4}}}{27}\]
\[\dfrac{{{\pi }^{4}}}{27}=\dfrac{{{\pi }^{4}}}{27}\]
\[LHS=RHS\]
So, this value of x is correct.
(c) By substituting \[x=\dfrac{\pi }{3}\] in equation (vi), we get,
\[={{\left[ \tan \left( \dfrac{\pi }{3} \right) \right]}^{2}}{{\left( \dfrac{\pi }{3} \right)}^{4}}=\dfrac{{{\pi }^{4}}}{27}\]
We know that \[\tan \left( \dfrac{\pi }{3} \right)=\sqrt{3}\], so we get,
\[{{\left( \sqrt{3} \right)}^{2}}{{\left( \dfrac{\pi }{3} \right)}^{4}}=\dfrac{{{\pi }^{4}}}{27}\]
\[3\times \dfrac{{{\pi }^{4}}}{81}=\dfrac{{{\pi }^{4}}}{27}\]
\[\dfrac{{{\pi }^{4}}}{27}=\dfrac{{{\pi }^{4}}}{27}\]
\[LHS=RHS\]
So, this value of x is correct.
(d) By substituting \[x=\dfrac{\pi }{12}\] in equation (vi), we get,
\[={{\left[ \tan \left( \dfrac{\pi }{12} \right) \right]}^{2}}{{\left( \dfrac{\pi }{12} \right)}^{4}}=\dfrac{{{\pi }^{4}}}{27}\]
We know that \[\tan \left( \dfrac{\pi }{12} \right)=\left( 2-\sqrt{3} \right)\], so we get,
\[{{\left( 2-\sqrt{3} \right)}^{2}}{{\left( \dfrac{\pi }{12} \right)}^{4}}=\dfrac{{{\pi }^{4}}}{27}\]
\[\left( 4+3-4\sqrt{3} \right)\dfrac{{{\pi }^{4}}}{{{12}^{4}}}=\dfrac{{{\pi }^{4}}}{27}\]
\[LHS\ne RHS\]
So, this value of x is not correct.
 Hence, option (b) and (c) are the correct answers.

Note: Students must note that in insolvable questions or equations, we use the options and check if it satisfies the equation that is LHS = RHS or not. Also in this question, many students make this mistake of taking r = 2 and r = 4 for finding second and fourth term which is wrong because we know the formula of \[\left( {{T}_{r+1}} \right)th\] term and not \[\left( {{T}_{r}} \right)th\] term. So, for the second term, we should use r = 1 to make \[{{T}_{r+1}}={{T}_{2}}\] , and for the fourth term, we should use r = 3 to make \[{{T}_{r+1}}={{T}_{4}}\] and so on.