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Inverse Trigonometric Functions

Section: Mathematics

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Question : 59 of 99

Marks: +1, -0

Let

x \cdot y = x^{2} + y^{3}

(x \cdot 1) \cdot 1 = x \cdot (1 \cdot 1)

. Then a value of

2\sin^{-1}\left(\frac{x^{4}+x^{2}-2}{x^{4}+x^{2}+2}\right)

[24-Jun-2022-Shift-2]

$\;\frac{\pi}{4}$
$\;\frac{\pi}{3}$
$\;\frac{\pi}{2}$
$\;\frac{\pi}{6}$

Solution: 👈: Video Solution

Given,

x \cdot y = x^{2} + y^{3}

\therefore x \cdot 1 = x^{2} + 1^{3} = x^{2} + 1

\;\text{ Now, }\;(x \cdot 1) \cdot 1 = (x^{2}+1) \cdot 1

\Rightarrow (x \cdot 1) \cdot 1 = (x^{2}+1)^{2} + 1^{3}

\Rightarrow (x \cdot 1) \cdot 1 = x^{4} + 1 + 2x^{2} + 1

\;\text{ Also, }\; x \cdot (1 \cdot 1)

= x \cdot (1^{2} + 1^{3})

= x \cdot 2

= x^{2} + 2^{3}

= x^{2} + 8

Given that,

(x \cdot 1) \cdot 1 = x \cdot (1 \cdot 1)

\therefore x^{4} + 1 + 2x^{2} + 1 = x^{2} + 8

\Rightarrow x^{4} + x^{2} - 6 = 0

\Rightarrow x^{4} + 3x^{2} - 2x^{2} - 6 = 0

\Rightarrow x^{2}(x^{2}+3) - 2(x^{3}+3) = 0

\Rightarrow (x^{2}+3)(x^{2}-2) = 0

\Rightarrow x^{2} = 2, -3

[x^{2} = -3.

not possible as square of anything should be always possible]

\therefore x^{2} = 2

\therefore

Now,

2\sin^{-1}\left(\frac{x^{4}+x^{2}-2}{x^{4}+x^{2}+2}\right)

= 2\sin^{-1}\left(\frac{2^{2}+2-2}{2^{2}+2+2}\right)

= 2\sin^{-1}\left(\frac{4}{8}\right)

= 2\sin^{-1}\left(\frac{1}{2}\right)

= 2 \times \frac{\pi}{6}

= \frac{\pi}{3}

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