JEE 2023 — Mathematics PYQ

Humein function diya gaya hai: $f(x)=4{\sin }^{-1}\left(\frac{x^2}{x^2+1}\right)$.

Sabse pehle andar wali term $g(x)=\frac{x^2}{x^2+1}$ ko dekhte hain:

• Kyunki $x^2\ge 0$, isliye minimum value tab hogi jab $x=0$.

  $g(0)=\frac{0}{0+1}=0$.

• Maximum value ke liye hum ise aise likh sakte hain: $g(x)=\frac{x^2+1-1}{x^2+1}=1-\frac{1}{x^2+1}$.

  Jab $x\to \infty$, tab $\frac{1}{x^2+1}\to 0$, iska matlab $g(x)$ ki value $1$ ke bahut kareeb jayegi par kabhi $1$ nahi hogi.

• Isliye, $g(x)\in [0,1)$.

Step 2: Inverse Sine function apply karna

Ab hum jaante hain ki ${\sin }^{-1}(y)$ ek increasing function hai.

• Agar $y\in [0,1)$, toh ${\sin }^{-1}(y)$ ki value ${\sin }^{-1}(0)$ se ${\sin }^{-1}(1)$ tak jayegi.

• ${\sin }^{-1}(0)=0$

• ${\sin }^{-1}(1)=\frac{\pi }{2}$ (par $1$ include nahi hai, toh $\frac{\pi }{2}$ bhi include nahi hoga).

• Isliye, ${\sin }^{-1}\left(\frac{x^2}{x^2+1}\right)\in [0,\frac{\pi }{2})$.

Step 3: Final multiplication

Ab poore function ko $4$ se multiply karte hain:

• Lower bound: $4\times 0=0$

• Upper bound: $4\times \frac{\pi }{2}=2\pi$

• Range $\in [0,2\pi )$.

Correct Option: (3) $[0,2\pi )$.