Let x * y = x^2 + y^3 and (x * 1) * 1 = x * (1 * 1).

Question

Let $x * y = {x^2} + {y^3}$ and $(x * 1) * 1 = x * (1 * 1)$.

Then a value of $$2{\sin ^{ - 1}}\left( {{{{x^4} + {x^2} - 2} \over {{x^4} + {x^2} + 2}}} \right)$$ is :

Accepted Answer

The star "*" in this context represents a binary operation, similar to addition (+), subtraction (-), multiplication (\times ), and division (\div ). 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,

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

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

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

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

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

$$ = {\pi  \over 3}$$

Solutions · 1