If m and n respectively are the number of local maximum and local minimum points of the function f(x) = ∫₀^(x²) (t² - 5t + 4)/(2 + eᵗ)dt, then the ordered pair (m, n) is equal to

Solution

Maths sir · 3 steps

Differentiate with Leibniz first, and do not lose the factor 2x from the upper limit x².

1. Differentiate the integral. By the Leibniz rule, put t = x² into the integrand and multiply by the derivative of the upper limit, which is 2x.

   f'(x) = 2x(x⁴ - 5x² + 4)/(2 + e^(x²)) = 2x(x-1)(x+1)(x-2)(x+2)/(2 + e^(x²))

2. Critical points. The denominator is always positive, so the sign of f' is the sign of x(x-1)(x+1)(x-2)(x+2). It is zero at x = -2, -1, 0, 1, 2. For x > 2 every factor is positive, and the sign flips at each simple root as you move left.

   Point	Sign of f' left to right	Type
   x = -2	- to +	local minimum
   x = -1	+ to -	local maximum
   x = 0	- to +	local minimum
   x = 1	+ to -	local maximum
   x = 2	- to +	local minimum

3. So. There are 2 local maxima (x = ± 1) and 3 local minima (x = 0, ± 2), so (m, n) = (2, 3), option (b).

The idea to keep. For ∫₀ᵍ⁽ˣ⁾ h(t) dt, the derivative is h(g(x)) g'(x). The factor g'(x) can add its own critical points, so always include it in the sign check.

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