SATHEE: Inverse Circular Functions 2 Question 3

Inverse Circular Functions 2 Question 3

3. If $\cos^{- 1} \frac{2}{3 x} + \cos^{- 1} \frac{3}{4 x} = \frac{π}{2} x > \frac{3}{4}$ , then $x$ is equal to

(a) $\frac{\sqrt{145}}{10}$

(b) $\frac{\sqrt{146}}{12}$

(d) $\frac{\sqrt{145}}{11}$

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Answer:

Correct Answer: 3. (c)

Solution:

Key Idea Use the formula,

$\cos^{- 1} x + \cos^{- 1} y = \cos^{- 1} (x y - \sqrt{1 - x^{2}} \sqrt{1 - y^{2}})$

We have, $\cos^{- 1} \frac{2}{3 x} + \cos^{- 1} \frac{3}{4 x} = \frac{π}{2}$

$\Rightarrow \cos^{- 1} \frac{2}{3 x} \cdot \frac{3}{4 x} - \sqrt{1 - \frac{4}{9 x^{2}}} \sqrt{1 - \frac{9}{16 x^{2}}} = \frac{π}{2}$

$[∵ \cos^{- 1} x + \cos^{- 1} y = \cos^{- 1} (x y - \sqrt{1 - x^{2}} \sqrt{1 - y^{2}})]$

$\Rightarrow \cos^{- 1} \frac{1}{2 x^{2}} - \frac{\sqrt{9 x^{2} - 4} \sqrt{16 x^{2} - 9}}{12 x^{2}} = \frac{π}{2}$

$\Rightarrow \frac{6 - \sqrt{9 x^{2} - 4} \sqrt{16 x^{2} - 9}}{12 x^{2}} = \cos \frac{π}{2} = 0$

$\Rightarrow \sqrt{9 x^{2} - 4} \sqrt{16 x^{2} - 9} = 6$

On squaring both sides,

$\Rightarrow (9 x^{2} - 4) (16 x^{2} - 9) = 36$

$\Rightarrow 144 x^{4} - 81 x^{2} - 64 x^{2} + 36 = 36$

$\Rightarrow 144 x^{4} - 145 x^{2} = 0$

$\Rightarrow x^{2} (144 x^{2} - 145) = 0$

$\Rightarrow x = 0$ or $x = \pm \sqrt{\frac{145}{144}} = \pm \frac{\sqrt{145}}{12}$

But $x > \frac{3}{4}$ ,

$x = \frac{\sqrt{145}}{12}$

Inverse Circular Functions 2 Question 3

3. If $\cos^{- 1} \frac{2}{3 x} + \cos^{- 1} \frac{3}{4 x} = \frac{π}{2} x > \frac{3}{4}$ , then $x$ is equal to

Answer:

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Inverse Circular Functions 2 Question 3

3. If cos−1⁡23x+cos−1⁡34x=π2x>34, then x is equal to

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3. If $\cos^{- 1} \frac{2}{3 x} + \cos^{- 1} \frac{3}{4 x} = \frac{π}{2} x > \frac{3}{4}$ , then $x$ is equal to