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

Correct answer: (B) (π)/3. Given x×y=x^2+y^3 Now x×1× 1=x×1×1 x^2+1^3×1=x×1^2+1^3 x2+1^2+1^3=x^2+2^3 x^4+1+2x^2+1=x^2+8 x^4+x^2−6=0 Let x^2=t t^2+t-6=0 t+3t−2=0 t≠−3 so t=2 or x^2=2 Now 2sin^(-1)(x^4+x^2-2)/(x^4+x^2+2)=2sin^(-1)(4+2-2)/(4+2+2) =2sin^(-1)1/2=(π)/6×2=(π)/3

Q31 / 90 · JEE Main 2022 (24 Jun Shift 2) · Mathematics

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

A
$\frac{π}{4}$
Correct answer
B
$\frac{π}{3}$
C
$\frac{π}{6}$
D
$π$

Solution

Given $x \times y = x^{2} + y^{3}$

Now $(x \times 1) \times 1 = x \times (1 \times 1)$

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

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

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

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

Let $x^{2} = t$

$t^{2} + t - 6 = 0$

$(t + 3) (t - 2) = 0$

$t \neq - 3$ so $t = 2$ or $x^{2} = 2$

Now $2 \sin^{- 1} (\frac{x^{4} + x^{2} - 2}{x^{4} + x^{2} + 2}) = 2 \sin^{- 1} (\frac{4 + 2 - 2}{4 + 2 + 2})$
$= 2 \sin^{- 1} (\frac{1}{2}) = \frac{π}{6} \times 2 = \frac{π}{3}$

Hence, the correct answer is option: 'B'.

Let x×y=x2+y3 and x×1×1=x ×1×1. Then a value of 2sin-1x4+x2-2x4+x2+2 is

Solution

Relevant Questions

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