Question

The product of all values of $ x $ satisfying the equation $ {\sin ^{ - 1}}\cos \left( {\dfrac{{2{x^2} + 10\left| x \right| + 4}}{{{x^2} + 5\left| x \right| + 3}}} \right) = \cot \left\{ {{{\cot }^{ - 1}}\left( {\dfrac{{2 - 18\left| x \right|}}{{9\left| x \right|}}} \right)} \right\} + \dfrac{\pi }{2} $ is
(A) $ 9 $
(B) $ - 9 $
(C) $ - 3 $
(D) $ - 1 $

Answer 1

Hint: This type of equation will be solved by the use of the basic concept of inverse trigonometric and also angle property of trigonometric.

Complete step-by-step answer:
 $ {\sin ^{ - 1}}\cos \left( {\dfrac{{2{x^2} + 10\left| x \right| + 4}}{{{x^2} + 5\left| x \right| + 3}}} \right) = \cot \left\{ {{{\cot }^{ - 1}}\left( {\dfrac{{2 - 18\left| x \right|}}{{9\left| x \right|}}} \right)} \right\} + \dfrac{\pi }{2} $ . . . . (given)
We have to find the product of real value of $ x $ which satisfies the equation.
So, the equation can be written as
 $\Rightarrow {\sin ^{ - 1}}\cos \left( {\dfrac{{2{x^2} + 10x + 4}}{{{x^2} + 5x + 3}}} \right) = \cot \left\{ {{{\cot }^{ - 1}}\left( {\dfrac{{2 - 18x}}{{9x}}} \right)} \right\} + \dfrac{\pi }{2} $ . . . . By $ \left( {\left| x \right| = x} \right) $
By changing $ \cos $ function to $ \sin $ function in left hand side of the above equation we get,
 $\Rightarrow {\sin ^{ - 1}}\sin \left[ {\dfrac{\pi }{2} - \left( {\dfrac{{2{x^2} + 10x4}}{{{x^2} + 5x + 3}}} \right)} \right] = \cot \left\{ {{{\cot }^{ - 1}}\left( {\dfrac{{2 - 18x}}{{9x}}} \right)} \right\} + \dfrac{\pi }{2} $ . . . . $ \cos \theta = \sin \left( {\dfrac{\pi }{2} - \theta } \right) $
Applying the inverse trigonometric property in the both side we get,
 $ \dfrac{\pi }{2} - \left( {\dfrac{{2{x^2} + 10x + 4}}{{{x^2} + 5x + 3}}} \right) = \dfrac{{2 - 18x}}{{9x}} + \dfrac{\pi }{2} $
On simplifying the above equation, we get,
 $ - \dfrac{{2{x^2} + 10x + 4}}{{{x^2} + 5x + 3}} = \dfrac{{2 - 18x}}{{9x}} $
On cross multiplying the above equation we get,
 $ - 18{x^3} - 90{x^2} - 36x = 2{x^2} + 10x + 6 - 18{x^3} - 90x - 54x $
By rearranging above equation we get,
 $\Rightarrow 2{x^2} - 8x + 6 = 0 $
Now, factorize the above equation to find the value of $ x $
 $\Rightarrow \left( {x - 1} \right)\left( {x - 3} \right) = 0 $
Since, there are two values of $ x $
 $ x = 1 $
 $ x = 3 $
Therefore, the product of two real roots $ = 3 \times 1 = 3 $
Hence the product of the real roots which satisfies the given equation is $ 3 $ .
Therefore from the above explanation the correct option is [] $ 3 $ .

Note: In this question we use inverse trigonometric $ \left[ {\cot .ca{t^{ - 1}}\left( \theta \right) = \theta \sin .{{\sin }^{ - 1}}\left( \theta \right)} \right] $
Also used $ \cos \theta = \sin \left[ {\dfrac{\pi }{2} - \theta } \right] $ , we must be careful on this.