Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

MathematicsInverse Trigonometric FunctionsIdentities, eqations and inequations involving ITFMedium2 minQB

MathematicsMediumsingle choice

The value of ' a ' for which $a x^{2} + sin^{- 1} (x^{2} - 2 x + 2) + cos^{- 1} (x^{2} - 2 x + 2) = 0$ has a real solution is

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Answer:

Solution:

For a real solution, $x^{2} - 2 x + 2 = 1$ , thus $x = 1$ . Substituting $x = 1$ into the equation gives $a + sin^{- 1} (1) + cos^{- 1} (1) = 0$ . Since $sin^{- 1} (1) = \frac{π}{2}$ and $cos^{- 1} (1) = 0$ , we have $a + \frac{π}{2} = 0$ . Therefore, $a = - \frac{π}{2}$ .

Stream:JEESubject:MathematicsTopic:Inverse Trigonometric FunctionsSubtopic:Identities, eqations and inequations involving ITF

⏱ 2mℹ️ Source: QB

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