JEE MAIN - Mathematics (2021 - 17th March Evening Shift - No. 5)
The number of solutions of the equation
$${\sin ^{ - 1}}\left[ {{x^2} + {1 \over 3}} \right] + {\cos ^{ - 1}}\left[ {{x^2} - {2 \over 3}} \right] = {x^2}$$, for x$$\in$$[$$-$$1, 1], and [x] denotes the greatest integer less than or equal to x, is :
$${\sin ^{ - 1}}\left[ {{x^2} + {1 \over 3}} \right] + {\cos ^{ - 1}}\left[ {{x^2} - {2 \over 3}} \right] = {x^2}$$, for x$$\in$$[$$-$$1, 1], and [x] denotes the greatest integer less than or equal to x, is :
0
Infinite
2
4
Explanation
There are three cases possible for $$x \in [ - 1,1]$$
Case I : $$x \in \left[ { - 1, - \sqrt {{2 \over 3}} } \right)$$
$$ \therefore $$ $${\sin ^{ - 1}}(1) + {\cos ^{ - 1}}(0) = {x^2}$$
$$ \Rightarrow {x^2} = {\pi \over 2} + {\pi \over 2} = \pi $$
$$ \Rightarrow x = \pm \sqrt \pi $$ $$ \to $$ (Reject)
Case II : $$x \in \left( { - \sqrt {{2 \over 3}} ,\sqrt {{2 \over 3}} } \right)$$
$$ \therefore $$ $${\sin ^{ - 1}}(0) + {\cos ^{ - 1}}( - 1) = {x^2}$$
$$ \Rightarrow 0 + \pi = {x^2}$$
$$ \Rightarrow x = \pm \sqrt x $$ $$ \to $$ (Reject)
Case III : $$x \in \left( {\sqrt {{2 \over 3}} ,1} \right)$$
$$ \therefore $$ $${\sin ^{ - 1}}(0) + {\cos ^{ - 1}}(0) = {x^2}$$
$$ \Rightarrow {x^2} - \pi \Rightarrow x - \pm \sqrt x $$ (Reject)
$$ \therefore $$ No solution. There, the correct answer is (1).
Case I : $$x \in \left[ { - 1, - \sqrt {{2 \over 3}} } \right)$$
$$ \therefore $$ $${\sin ^{ - 1}}(1) + {\cos ^{ - 1}}(0) = {x^2}$$
$$ \Rightarrow {x^2} = {\pi \over 2} + {\pi \over 2} = \pi $$
$$ \Rightarrow x = \pm \sqrt \pi $$ $$ \to $$ (Reject)
Case II : $$x \in \left( { - \sqrt {{2 \over 3}} ,\sqrt {{2 \over 3}} } \right)$$
$$ \therefore $$ $${\sin ^{ - 1}}(0) + {\cos ^{ - 1}}( - 1) = {x^2}$$
$$ \Rightarrow 0 + \pi = {x^2}$$
$$ \Rightarrow x = \pm \sqrt x $$ $$ \to $$ (Reject)
Case III : $$x \in \left( {\sqrt {{2 \over 3}} ,1} \right)$$
$$ \therefore $$ $${\sin ^{ - 1}}(0) + {\cos ^{ - 1}}(0) = {x^2}$$
$$ \Rightarrow {x^2} - \pi \Rightarrow x - \pm \sqrt x $$ (Reject)
$$ \therefore $$ No solution. There, the correct answer is (1).
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