Let f(x)=3 √(x-2)+√(4-x) be a real valued function. If α and β are respectively the minimum and the maximum values of f, then α²+2 β² is equal to

Solution

Maths sir · 3 steps

Check the critical point and both ends of the domain, then compare the three values of f.

1. Find the domain. You need x - 2 ≥ 0 and 4 - x ≥ 0, so x ∈ [2, 4]. On a closed interval, the extreme values come from the critical points or the endpoints.

   f'(x) = 3/2√(x-2) - 1/2√(4-x) = 0 ⇒ 3√(4-x) = √(x-2)

2. Critical point. Squaring gives 9(4 - x) = x - 2, so 10x = 38 and x = 19/5, which lies inside [2, 4]. There, f = 3√(9/5) + √(1/5) = 9/√5 + 1/√5 = 2√5.

   x	f(x)	f(x)²
   2	0 + √2 = √2	2, smallest
   19/5	2√5	20, largest
   4	3√2 + 0 = 3√2	18

3. So. α = √2 and β = 2√5, so α² + 2β² = 2 + 2 × 20 = 42. The answer is (a).

The idea to keep. For a continuous function on a closed interval, list the critical points and both endpoints, and compare the values. The largest is the maximum, the smallest is the minimum.

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