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.