Let x*y=x^2+y^3 and (x*1)*1=x*(1*1) . Then a value of 2sin^-1 (fracx^4+x^2-2x^4+x^2+2 ) is

Sets, Relations and Function: Let x*y=x^2+y^3 and (x*1)*1=x*(1*1) . Then a value of 2sin^-1… | JEE Main 2022

JEE Main 2022 (Online) 24th June Evening ShiftQuestion 1

Let $x*y=x^{2}+y^{3}$ and $(x*1)*1=x*(1*1)$. Then a value of $2\sin^{-1}\left(\frac{x^{4}+x^{2}-2}{x^{4}+x^{2}+2}\right)$ is

Sets, Relations and Function

$\frac{\pi}{4}$$\frac{\pi}{3}$✓ Correct answer$\frac{\pi}{2}$$\frac{\pi}{6}$

Solution

$(x*1)*1=(x^2+1)*1=(x^2+1)^2+1$; $x*(1*1)=x*2=x^2+8$. $(x^2+1)^2+1=x^2+8\Rightarrow x^4+x^2-6=0\Rightarrow x^2=2$ (real). $\dfrac{x^4+x^2-2}{x^4+x^2+2}=\dfrac{4}{8}=\dfrac12$, so $2\sin^{-1}\tfrac12=\dfrac{\pi}{3}$. Option B.

Sets, Relations and Function - JEE Mathematics question (JEE Main 2022 (Online) 24th June Evening Shift)

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