Let the inverse trigonometric functions take principal values. Th… — Solution | JEE Mathematics

MCQMediumJEE 2024Inverse Trigonometric Functions

Let the inverse trigonometric functions take principal values. The number of real solutions of the equation $2\sin^{-1}(x) + 3\cos^{-1}(x) = \frac{2\pi}{5}$ is:

A
$0$
B
$1$
C
$2$
D
Infinite

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: $2\sin^{-1}x + 3\cos^{-1}x = \frac{2\pi}{5}$

Find: The number of real solutions.

Let $a = \sin^{-1}x$ and $b = \cos^{-1}x$ . Using the identity

a + b = \frac{\pi}{2}

we have

2a + 3b = \frac{2\pi}{5}

Using the principal value ranges

Substitute $b = \frac{\pi}{2} - a$ into the given equation:

2a + 3\left(\frac{\pi}{2} - a\right) = \frac{2\pi}{5}

2a + \frac{3\pi}{2} - 3a = \frac{2\pi}{5}

-a + \frac{3\pi}{2} = \frac{2\pi}{5}

-a = \frac{2\pi}{5} - \frac{3\pi}{2} = \frac{4\pi}{10} - \frac{15\pi}{10} = -\frac{11\pi}{10}

a = \frac{11\pi}{10}

Range check trick

Since $a = \sin^{-1}x$ , its principal value must satisfy

-\frac{\pi}{2} \le a \le \frac{\pi}{2}

But the calculation gives $a = \frac{11\pi}{10}$ , which lies outside this range. Therefore no real value of $x$ can satisfy the equation.

So, the number of real solutions is $0$ , and the correct option is A.

Common mistakes

Using the identity incorrectly as $\sin^{-1}x = \cos^{-1}x$ . This is wrong because the correct relation is $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ . Always convert one inverse function in terms of the other before solving.
Ignoring the principal value range of $\sin^{-1}x$ . Even if algebra gives a value for $a$ , it must satisfy $-\frac{\pi}{2} \le a \le \frac{\pi}{2}$ . Check the range before concluding that a solution exists.
Making an error while simplifying $\frac{2\pi}{5} - \frac{3\pi}{2}$ . A wrong common denominator can change the final conclusion. Convert both terms to denominator $10$ first.

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Step-by-step solution

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