Question

the number of solutions of the equation \[{\tan ^{ - 1}}\left( {\dfrac{1}{{2x + 1}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{1}{{4x + 1}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{2}{{{x^2}}}} \right)\] is
(A) 0
(B) 1
(C) 2
(D) 3

Answer 1

Hint: here, we will be using the formula for the sum of two inverse tangent functions.

Complete step-by-step answer:
Given, the equation \[{\tan ^{ - 1}}\left( {\dfrac{1}{{2x + 1}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{1}{{4x + 1}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{2}{{{x^2}}}} \right)\] 
The sum of two inverse tangent functions \[{\tan ^{ - 1}}a\] and\[{\tan ^{ - 1}}b\] is given by \[{\tan ^{ - 1}}a + {\tan ^{ - 1}}b = {\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)\]
Now if we apply this formula to the given equation, then the equation is given by 
\[{\tan ^{ - 1}}\left( {\dfrac{{\dfrac{1}{{2x + 1}} + \dfrac{1}{{4x + 1}}}}{{1 - \dfrac{1}{{2x + 1}} \times \dfrac{1}{{4x + 1}}}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{2}{{{x^2}}}} \right)\]
\[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{4x + 1 + 2x + 1}}{{\left( {2x + 1} \right)\left( {4x + 1} \right)}}}}{{\dfrac{{\left( {2x + 1} \right)\left( {4x + 1} \right) - 1}}{{\left( {2x + 1} \right)\left( {4x + 1} \right)}}}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{2}{{{x^2}}}} \right)\]
\[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{6x + 2}}{{8{x^2} + 4x + 2x + 1 - 1}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{2}{{{x^2}}}} \right)\]
\[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{6x + 2}}{{8{x^2} + 6x}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{2}{{{x^2}}}} \right)\]
\[ \Rightarrow \dfrac{{6x + 2}}{{8{x^2} + 6x}} = \dfrac{2}{{{x^2}}}\]
\[ \Rightarrow \left( {6x + 2} \right){x^2} = 2\left( {8{x^2} + 6x} \right)\]
\[ \Rightarrow 6{x^3} + 2{x^2} = 16{x^2} + 12x\]
\[ \Rightarrow 6{x^3} - 14{x^2} - 12x = 0\]
\[ \Rightarrow 2x(3{x^2} - 7x - 6) = 0\]
\[ \Rightarrow 2x(3{x^2} - 9x + 2x - 6) = 0\]
\[ \Rightarrow 2x(3x(x - 3) + 2(x - 3)) = 0\]
\[ \Rightarrow 2x(3x + 2)(x - 3) = 0\]
\[ \Rightarrow x = 0,x = \dfrac{{ - 2}}{3},x = 3\]
\[ \Rightarrow \] the given equation has 3 solutions \[x = 0,x = \dfrac{{ - 2}}{3}\] and \[x = 3\]
Therefore, (D) 3 is the required solution.

Note: these types of questions always require the formula for sum of two inverse tangent functions