QJEEM21Maf03a: JEE Main 2021 Mathematics - Class 12 JEEM21Maf03a Solution: Given that the inverse trigonometric functions take principal…

Question

Verified Solution • 2025–26 Session moderate

Given that the inverse trigonometric functions take principal values only. Then, the number of real values of x which satisfy

{\sin ^{ - 1}}\left( {{{3x} \over 5}} \right) + {\sin ^{ - 1}}\left( {{{4x} \over 5}} \right) = {\sin ^{ - 1}}x

is equal to :

Accepted Answer

{\sin ^{ - 1}}{{3x} \over 5} + {\sin ^{ - 1}}{{4x} \over 5} = {\sin ^{ - 1}}x

{\sin ^{ - 1}}\left( {{{3x} \over 5}\sqrt {1 - {{16{x^2}} \over {25}}} + {{4x} \over 5}\sqrt {1 - {{9{x^2}} \over {25}}} } \right) = {\sin ^{ - 1}}x

{{3x} \over 5}\sqrt {1 - {{16{x^2}} \over {25}}} + {{4x} \over 5}\sqrt {1 - {{9{x^2}} \over {25}}} = x

x = 0

or

3\sqrt {25 - 16{x^2}} + 4\sqrt {25 - 9{x^2}} = 25

4\sqrt {25 - 9{x^2}} = 25 - 3\sqrt {25 - 16{x^2}}

Squaring we get

16(25 - 9{x^2}) = 625 - 9(25 - 16{x^2}) - 150\sqrt {25 - 16{x^2}}

400 = 625 + 225 - 150\sqrt {25 - 16{x^2}}

\sqrt {25 - 16{x^2}} = 3 \Rightarrow 25 - 16{x^2} = 9

\Rightarrow {x^2} = 1

Put x = 0, 1,

-

1 in the original equation

We see that all values satisfy the original equation.

Number of solution = 3

Step-by-Step Verified Solution