The product of all real values of x satisfying the equation

\(\sin^{-1}\cos \left ( \dfrac{2x^{2}+10\left | x \right |+4}{x^{2}+5\left | x \right |+3} \right )=\cot \left ( \cot ^{-1}\left ( \dfrac{2-18\left | x \right |}{9\left | x \right |} \right ) \right )+\dfrac{\pi }{2}\) is

(A) 3

(B) -9

(C) -3

(D) -1

\(\sin^{-1}\cos \left ( \dfrac{2x^{2}+10\left | x \right |+4}{x^{2}+5\left | x \right |+3} \right )=\cot \left ( \cot ^{-1}\left ( \dfrac{2-18\left | x \right |}{9\left | x \right |} \right ) \right )+\dfrac{\pi }{2}\) is

(A) 3

(B) -9

(C) -3

(D) -1

Answer: A

\(sin^{ -1 }cos\left( \cfrac { 2x^{ 2 }+10x+4 }{ x^{ 2 }+5x+3 } \right) =cot\left( cot^{ -1 }\left( \cfrac { 2-18x }{ 9x } \right) \right) +\cfrac { \pi }{ 2 } \\ \Rightarrow sin^{ -1 }sin\left( \cfrac { \pi }{ 2 } -\left( \cfrac { 2x^{ 2 }+10x+4 }{ x^{ 2 }+5x+3 } \right) \right) =\cfrac { 2-18x }{ 9x } +\cfrac { \pi }{ 2 } \\ \Rightarrow \cfrac { \pi }{ 2 } -\cfrac { 2x^{ 2 }+10x+4 }{ x^{ 2 }+5x+3 } =\cfrac { 2-18x }{ 9x } +\cfrac { \pi }{ 2 } \\ \Rightarrow -18x^{ 3 }-90x^{ 2 }-36x=2x^{ 2 }+10x+6-18x^{ 3 }-90x-54x\\ \Rightarrow 2x^{ 2 }-8x+6=0\Rightarrow x^{ 2 }-4+3=0\Rightarrow \left( x-1 \right) \left( x-3 \right) =0\)

Therefore product of roots is \(1\times 3=3\)

\(sin^{ -1 }cos\left( \cfrac { 2x^{ 2 }+10x+4 }{ x^{ 2 }+5x+3 } \right) =cot\left( cot^{ -1 }\left( \cfrac { 2-18x }{ 9x } \right) \right) +\cfrac { \pi }{ 2 } \\ \Rightarrow sin^{ -1 }sin\left( \cfrac { \pi }{ 2 } -\left( \cfrac { 2x^{ 2 }+10x+4 }{ x^{ 2 }+5x+3 } \right) \right) =\cfrac { 2-18x }{ 9x } +\cfrac { \pi }{ 2 } \\ \Rightarrow \cfrac { \pi }{ 2 } -\cfrac { 2x^{ 2 }+10x+4 }{ x^{ 2 }+5x+3 } =\cfrac { 2-18x }{ 9x } +\cfrac { \pi }{ 2 } \\ \Rightarrow -18x^{ 3 }-90x^{ 2 }-36x=2x^{ 2 }+10x+6-18x^{ 3 }-90x-54x\\ \Rightarrow 2x^{ 2 }-8x+6=0\Rightarrow x^{ 2 }-4+3=0\Rightarrow \left( x-1 \right) \left( x-3 \right) =0\)

Therefore product of roots is \(1\times 3=3\)

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