JEE Main 2022 — Inverse Trigonometric Functions Question with Solution

The star "*" in this context represents a binary operation, similar to addition (+), subtraction (-), multiplication (×), and division (÷). It is a custom operation defined by the problem statement, and the specific rules of the operation are provided in the problem.

In this case, the operation "*" is defined by the equation $x\ast y=x^2+y^3$, which means if you have two numbers $x$ and $y$, then the result of applying the "*" operation to them is $x^2+y^3$.

The problem also specifies an additional rule for this operation: $(x\ast 1)\ast 1=x\ast (1\ast 1)$, which needs to be taken into account when solving the problem. This is a type of "associativity" condition.

Given,

$$x\ast y=x^2+y^3$$

$$\therefore$$ $$x\ast 1=x^2+1^3=x^2+1$$

Now, $$(x\ast 1)\ast 1=(x^2+1)\ast 1$$

$$\Rightarrow (x\ast 1)\ast 1=(x^2+1{)}^2+1^3$$

$$\Rightarrow (x\ast 1)\ast 1=x^4+1+2x^2+1$$

Also, $$x\ast (1\ast 1)$$

$$=x\ast (1^2+1^3)$$

$$=x\ast 2$$

$$=x^2+2^3$$

$$=x^2+8$$

Given that,

$$(x\ast 1)\ast 1=x\ast (1\ast 1)$$

$$\therefore$$ $$x^4+1+2x^2+1=x^2+8$$

$$\Rightarrow x^4+x^2-6=0$$

$$\Rightarrow x^4+3x^2-2x^2-6=0$$

$$\Rightarrow x^2(x^2+3)-2(x^3+3)=0$$

$$\Rightarrow (x^2+3)(x^2-2)=0$$

$$\Rightarrow x^2=2,-3$$

[$$x^2=-3$$ not possible as square of anything should be always positive]

$$\therefore$$ $$x^2=2$$

$$\therefore$$ Now,

$$2{\sin }^{-1}\left(\frac{x^4+x^2-2}{x^4+x^2+2}\right)$$

$$=2{\sin }^{-1}\left(\frac{2^2+2-2}{2^2+2+2}\right)$$

$$=2{\sin }^{-1}\left(\frac{4}{8}\right)$$

$$=2{\sin }^{-1}\left(\frac{1}{2}\right)$$

$$=2\times \frac{\pi }{6}$$

$$=\frac{\pi }{3}$$