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Question: the number of solutions of the equation \[{\tan ^{ - 1}}\left( {\dfrac{1}{{2x + 1}}} \right) + {\tan...

the number of solutions of the equation tan⁡−1(12x+1)+tan⁡−1(14x+1)=tan⁡−1(2x2){\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

Explanation

Solution

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(12x+1)+tan⁡−1(14x+1)=tan⁡−1(2x2){\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⁡−1a{\tan ^{ - 1}}a andtan⁡−1b{\tan ^{ - 1}}b is given by tan⁡−1a+tan⁡−1b=tan⁡−1(a+b1−ab){\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(12x+1+14x+11−12x+1×14x+1)=tan⁡−1(2x2){\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)
⇒tan⁡−1(4x+1+2x+1(2x+1)(4x+1)(2x+1)(4x+1)−1(2x+1)(4x+1))=tan⁡−1(2x2)\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)
⇒tan⁡−1(6x+28x2+4x+2x+1−1)=tan⁡−1(2x2)\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{6x + 2}}{{8{x^2} + 4x + 2x + 1 - 1}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{2}{{{x^2}}}} \right)
⇒tan⁡−1(6x+28x2+6x)=tan⁡−1(2x2)\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{6x + 2}}{{8{x^2} + 6x}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{2}{{{x^2}}}} \right)
⇒6x+28x2+6x=2x2\Rightarrow \dfrac{{6x + 2}}{{8{x^2} + 6x}} = \dfrac{2}{{{x^2}}}
⇒(6x+2)x2=2(8x2+6x)\Rightarrow \left( {6x + 2} \right){x^2} = 2\left( {8{x^2} + 6x} \right)
⇒6x3+2x2=16x2+12x\Rightarrow 6{x^3} + 2{x^2} = 16{x^2} + 12x
⇒6x3−14x2−12x=0\Rightarrow 6{x^3} - 14{x^2} - 12x = 0
⇒2x(3x2−7x−6)=0\Rightarrow 2x(3{x^2} - 7x - 6) = 0
⇒2x(3x2−9x+2x−6)=0\Rightarrow 2x(3{x^2} - 9x + 2x - 6) = 0
⇒2x(3x(x−3)+2(x−3))=0\Rightarrow 2x(3x(x - 3) + 2(x - 3)) = 0
⇒2x(3x+2)(x−3)=0\Rightarrow 2x(3x + 2)(x - 3) = 0
⇒x=0,x=−23,x=3\Rightarrow x = 0,x = \dfrac{{ - 2}}{3},x = 3
⇒\Rightarrow the given equation has 3 solutions x=0,x=−23x = 0,x = \dfrac{{ - 2}}{3} and x=3x = 3
Therefore, (D) 3 is the required solution.

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