JEE Main 2022 · 24 Jun · Shift 2 · Mathematics Q1

MathematicsSingle correct+4 / -1Question 1 of 90

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

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

\dfrac\pi4

\dfrac\pi3

\dfrac\pi2

\dfrac\pi6

Answer and worked solution

Official answer: B

Use the rule. $x*1=x^2+1$ , so $(x*1)*1=(x^2+1)^2+1$ . Also $1*1=2$ , so $x*(1*1)=x^2+8$ .

Solve. $x^4+2x^2+2=x^2+8$ gives $x^4+x^2-6=0$ , so $x^2=2$ .

Evaluate. $\dfrac{x^4+x^2-2}{x^4+x^2+2}=\dfrac48=\dfrac12$ , and $2\sin^{-1}\frac12=\dfrac\pi3$ .

(B)

Topics:Types of functions Inverse trigonometric functions

JEE Main 2022 · 24 Jun · Shift 2, Mathematics question 1