UPSEE Solved Model Paper 7

Given:
f (x) = 3sin√

π2

9

−x2
We have to find the range of f (x)
Now,
f (x) = 3sin√

π2

9

−x2 exists if

π2

9

−x2 ≥ 0
[Using definition of square - root function]
⇒ −(x−

π

3

)(x+

π

3

) ≥ 0
⇒ (x−

π

3

)(x+

π

3

) ≤ 0
⇒ - π/3 ≤ x ≤ π/3
As

π2

9

−x2 ≥ 0
Let θ = √

π2

9

−x2 ≥ 0
So, 0 ≤ sin θ ≤

√3

2

[Because for 0 ≤ θ ≤

π

3

, 0 ≤ sin θ ≤

√3

2

]
⇒ 0 ≤ 3 sin √

π2

9

−x2 ≤

3√3

2

So, range of f (x) ∊ [0,

3√3

2

]
Hence, option 'C' is correct.