QJEEM22Maf969: JEE Main 2022 Mathematics - Class 12 JEEM22Maf969 Solution: Let and . Then a value of is :

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 $x * y = x^2 + y^3$ , 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: $(x * 1) * 1 = x * (1 * 1)$ , which needs to be taken into account when solving the problem. This is a type of "associativity" condition.

Given,

$x\, * \,y = {x^2} + {y^3}$

$\therefore$ $x\, * \,1 = {x^2} + {1^3} = {x^2} + 1$

Now, $(x\, * \,1)\, * \,1 = ({x^2} + 1)\, * \,1$

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

$\Rightarrow (x\, * \,1)\, * \,1 = {x^4} + 1 + 2{x^2} + 1$

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

$= x\, * \,({1^2} + {1^3})$

$= x\, * \,2$

$= {x^2} + {2^3}$

$= {x^2} + 8$

Given that,

$(x\, * \,1)\, * \,1 = x\, * \,(1\, * \,1)$

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

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

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

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

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

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

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

$\therefore$ ${x^2} = 2$

$\therefore$ Now,

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

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

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

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

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

$= {\pi \over 3}$