Let f(x)=∫₀^(x²) (t²-8 t+15)/eᵗ dt, x ∈ R. Then the numbers of local maximum and local minimum points of f, respectively, are :

Solution

Maths sir · 3 steps

Differentiate with the chain rule. The extra factor 2x adds x = 0 as a critical point.

1. Differentiate. By the Leibniz rule, put t = x² into the integrand and multiply by the derivative of the upper limit, which is 2x. The quadratic factors as t² - 8t + 15 = (t-3)(t-5).

   f'(x) = 2x e^(-x²)(x²-3)(x²-5)

2. Critical points. e^(-x²) is always positive, so f'(x) = 0 only at x = 0, ±√3, ±√5. These are five simple roots, so the sign of f' flips at each one. For x > √5 every factor is positive, so f' > 0 there.

   Point	Sign of f' before	Sign of f' after	Type
   -√5	-	+	local min
   -√3	+	-	local max
   0	-	+	local min
   √3	+	-	local max
   √5	-	+	local min

3. So. f has local maxima at ±√3 and local minima at -√5, 0 and √5. That is 2 maxima and 3 minima, option (c).

The idea to keep. When the upper limit of an integral is a function of x, the derivative of that limit multiplies the integrand. Its own zeros, like 2x = 0 here, are critical points too.

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