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Question

If $$\cos^{-1} \left (\dfrac {2}{3x}\right ) + \cos^{-1} \left (\dfrac {3}{4x}\right ) = \dfrac {\pi}{2}\left (x > \dfrac {3}{4}\right )$$ then $$x$$ is equal to


A
14512
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B
14510
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C
14612
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D
14511
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Solution

The correct option is A $$\dfrac {\sqrt {145}}{12}$$
$$\cos^{-1}\left (\dfrac {2}{3x}\right ) + \cos^{-1} \left (\dfrac {3}{4x}\right ) = \dfrac {\pi}{2}\left (x > \dfrac {3}{4}\right )$$
$$\Rightarrow \cos^{-1} \left (\dfrac {3}{4x}\right ) = \dfrac {\pi}{2} - \cos^{-1} \left (\dfrac {2}{3x}\right )$$
$$\Rightarrow \cos^{-1} \left (\dfrac {3}{4x}\right ) = \sin^{-1} \left (\dfrac {2}{3x}\right )$$
$$\Rightarrow \cos \left (\cos^{-1} \left (\dfrac {3}{4x}\right )\right ) = \cos \left (\sin^{-1}\dfrac {2}{3x}\right )$$
$$\Rightarrow\dfrac {3}{4x} = \dfrac {\sqrt {9x^{2} - 4}}{3x}$$
$$\Rightarrow\dfrac {81}{16} + 4 = 9x^{2}$$
$$\Rightarrow x^{2} = \dfrac {145}{16 \times 9}\Rightarrow x = \dfrac {\sqrt {145}}{12}$$.

Mathematics

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