JEE MAIN - Mathematics (2021 - 16th March Evening Shift - No. 13)
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 :
$${\sin ^{ - 1}}\left( {{{3x} \over 5}} \right) + {\sin ^{ - 1}}\left( {{{4x} \over 5}} \right) = {\sin ^{ - 1}}x$$ is equal to :
2
0
3
1
Explanation
$${\sin ^{ - 1}}{{3x} \over 5} + {\sin ^{ - 1}}{{4x} \over 5} = {\sin ^{ - 1}}x$$
$${\sin ^{ - 1}}\left( {{{3x} \over 5}\sqrt {1 - {{16{x^2}} \over {25}}} + {{4x} \over 5}\sqrt {1 - {{9{x^2}} \over {25}}} } \right) = {\sin ^{ - 1}}x$$
$${{3x} \over 5}\sqrt {1 - {{16{x^2}} \over {25}}} + {{4x} \over 5}\sqrt {1 - {{9{x^2}} \over {25}}} = x$$
$$x = 0 $$ 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
$${\sin ^{ - 1}}\left( {{{3x} \over 5}\sqrt {1 - {{16{x^2}} \over {25}}} + {{4x} \over 5}\sqrt {1 - {{9{x^2}} \over {25}}} } \right) = {\sin ^{ - 1}}x$$
$${{3x} \over 5}\sqrt {1 - {{16{x^2}} \over {25}}} + {{4x} \over 5}\sqrt {1 - {{9{x^2}} \over {25}}} = x$$
$$x = 0 $$ 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
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