Question

Find \[{\tan ^{ - 1}}\left( {\dfrac{1}{{2x + 1}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{1}{{4x + 1}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{2}{{{x^2}}}} \right)\] has
\[1)\] one solution
\[2)\] two solutions
\[3)\] three solutions
\[4)\] no solution

Answer 1

Hint: This is an inverse trigonometric function question. Here in the LHS we will use the formula of the $tan^{-1}A + tan^{-1}B$ formula. Then we will simplify it. After this we will multiply tan on both sides. Then we will get a cubic equation. We will solve the equation to get our required answers.

Complete step-by-step answer:
Given problem is \[{\tan ^{ - 1}}\left( {\dfrac{1}{{2x + 1}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{1}{{4x + 1}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{2}{{{x^2}}}} \right)\]
we find how many solutions have this problem. It means we find how many \[x\] values have this problem.
Given a problem to use the formulae is,
\[{\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)\,,\,\,xy < 1\]
Here \[x = \dfrac{1}{{2x + 1}}\] and \[y = \dfrac{1}{{4x + 1}}\]
Now apply the formulae to a given problem
\[{\tan ^{ - 1}}\left( {\dfrac{1}{{2x + 1}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{1}{{4x + 1}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{1}{{2x + 1}} + \dfrac{1}{{4x + 1}}}}{{1 - \left( {\dfrac{1}{{2x + 1}}} \right)\left( {\dfrac{1}{{4x + 1}}} \right)}}} \right)\,\]
Take least common multiply on numerator and denominator
\[ = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{4x + 1 + 2x + 1}}{{\left( {2x + 1} \right)\left( {4x + 1} \right)}}}}{{\dfrac{{\left( {2x + 1} \right)\left( {4x + 1} \right) - 1}}{{\left( {2x + 1} \right)\left( {4x + 1} \right)}}}}} \right)\]
\[ = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{6x + 2}}{{\left( {2x + 1} \right)\left( {4x + 1} \right)}}}}{{\dfrac{{8{x^2} + 2x + 4x + 1 - 1}}{{\left( {2x + 1} \right)\left( {4x + 1} \right)}}}}} \right)\]
\[ = {\tan ^{ - 1}}\left( {\dfrac{{6x + 2}}{{8{x^2} + 6x}}} \right)\]
And problem to equating,
\[{\tan ^{ - 1}}\left( {\dfrac{{6x + 2}}{{8{x^2} + 6x}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{2}{{{x^2}}}} \right)\]
Both sides multiply on \[\tan \]then \[{\tan ^{ - 1}}\]cancels,
\[\dfrac{{6x + 2}}{{8{x^2} + 6x}} = \dfrac{2}{{{x^2}}}\]
Left-hand side numerator and denominator take common \[2\] and cancel that
\[\dfrac{{2\left( {3x + 1} \right)}}{{2\left( {4{x^2} + 3x} \right)}} = \dfrac{2}{{{x^2}}}\]
The equating the values,
\[
  {x^2}\left( {3x + 1} \right) = 2\left( {4{x^2} + 3x} \right) \\
  3{x^3} + {x^2} = 8{x^2} + 6x \\
  3{x^3} + {x^2} - 8{x^2} - 6x = 0 \\
  3{x^3} - 7{x^2} - 6x = 0 \\
 \]
Take common on \[x\]the term,
\[\left( x \right)\left( {3{x^2} - 7x - 6} \right) = 0\]
And equating to zero. Now use factorize method,
\[3{x^2} - 7x - 6 = 0\]
\[ - 7x\] term separate to \[ - 9x\] and \[2x\]
\[3{x^2} - 9x + 2x - 6 = 0\]
Take the first two terms to common on \[3x\]
Then \[3x\left( {x - 3} \right)\]
And last two terms to common on \[2\]
Then \[2\left( {x - 3} \right)\]
And join both terms,
\[3x\left( {x - 3} \right) + 2\left( {x - 3} \right) = 0\]
Take common on \[\left( {x - 3} \right)\]a term to both parts,
Now equating both terms to 0
\[\left( {x - 3} \right)\left( {3x + 2} \right) = 0\]
\[x - 3 = 0,\,3x + 2 = 0\]
\[x - 3 = 0\] in this term to add both sides on \[3\]
\[x = 3\]
\[3x + 2 = 0\]in this term to subtract both sides on \[2\]
\[3x = - 2\]
Now divide both sides on \[3\]
\[x = \dfrac{{ - 2}}{3}\]
We get \[x = 3,x = \dfrac{{ - 2}}{3}\]
Finally\[x\], values are,
\[x = 0,x = 3,x = \dfrac{{ - 2}}{3}\]
So, \[x\] have \[3\] values. And our problems have \[3\] solutions.

So, the correct answer is “Option (3)”.

Note: In formulae on a given problem is \[{\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)\,,\,\,xy < 1\]
The least common multiple means to take common multiply least value on given terms . Then equating zero on getting the polynomial. Then find the \[x\] terms to get \[x\] values on getting the polynomial. Factorize means, given a polynomial to find the factors. In the used factorize method carefully find the factors on a given problem.