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

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.