UPSEE Solved Model Paper 7

Given: f (x) = $3\sin\sqrt{\frac{\pi^2}{9} - x^2}$ We have to find the range of f (x) Now, f (x) = $3\sin\sqrt{\frac{\pi^2}{9} - x^2}$ exists if $\frac{\pi^2}{9} - x^2$ ≥ 0 [Using definition of square - root function] ⇒ $-\left(x-\frac{\pi}{3}\right)\left(x+\frac{\pi}{3}\right)$ ≥ 0 ⇒ $\left(x-\frac{\pi}{3}\right)\left(x+\frac{\pi}{3}\right)$ ≤ 0 ⇒ - π/3 ≤ x ≤ π/3 As $\frac{\pi^2}{9} - x^2$ ≥ 0 Let θ = $\sqrt{\frac{\pi^2}{9} - x^2}$ ≥ 0 So, 0 ≤ sin θ ≤ $\frac{\sqrt{3}}{2}$ [Because for 0 ≤ θ ≤ $\frac{\pi}{3}$ , 0 ≤ sin θ ≤ $\frac{\sqrt{3}}{2}$] ⇒ 0 ≤ 3 sin $\sqrt{\frac{\pi^2}{9} - x^2}$ ≤ $\frac{3\sqrt{3}}{2}$ So, range of f (x) ∊ $\left[0,\frac{3\sqrt{3}}{2}\right]$ Hence, option 'C' is correct.