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 :

Question

Accepted Answer

Answer: C. 3 {\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

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 :

Answer

Explanation

More JEE Math questions