The range of the function f(x)=√(3-x)+√(2+x) is :

Solution

Maths sir · 4 steps

Square the function: the cross term becomes one quadratic under a root, and you can track that easily.

1. Find the domain. Both roots need non-negative insides: 3-x ≥ 0 and 2+x ≥ 0. So x ∈ [-2, 3], and f(x) ≥ 0 everywhere on it.

   f(x)² = (3-x)+(2+x)+2√(3-x)(2+x) = 5 + 2√(25/4-(x-1/2)²)

2. Range of the root. On [-2, 3] the quadratic (3-x)(2+x) is 0 at both ends and reaches its peak of 25/4 at x = 1/2. So √(3-x)(2+x) runs from 0 to 5/2.
3. Range of f squared. So f² runs from 5 + 0 = 5 up to 5 + 2 × 5/2 = 10. Check: f(-2) = f(3) = √5 and f(1/2) = 2√2.5 = √10.
4. So. Since f ≥ 0, take the positive root: the range is [√5, √10], option (d).

The idea to keep. For √(a-x)+√(b+x), square it. The sum inside is constant, so the range depends only on the product term, which is a downward parabola on the domain.

Answer from the NTA final answer key. Solution by PrepWiser.