Let f be a twice differentiable function on ℝ. If f′(0)=4 and f(x) + ∫₀ˣ (x - t)f'(t)dt = ( e²ˣ + e⁻²ˣ )cos 2x + 2/ax , then (2 a+1)⁵ a² is equal to _______________.

Solution

Maths sir · 3 steps

Differentiate once and put x = 0: the integral term vanishes there, and you get a directly.

1. Differentiate the integral. By the Leibniz rule, the derivative of ∫₀ˣ (x-t)f'(t) dt is (x-x)f'(x) + ∫₀ˣ f'(t) dt. The first term is zero, and the second is f(x) - f(0).

   f'(x) + f(x) - f(0) = 2(e²ˣ - e⁻²ˣ)cos 2x - 2(e²ˣ + e⁻²ˣ)sin 2x + 2/a

2. Put x = 0. The left side becomes f'(0) + f(0) - f(0) = f'(0) = 4. On the right, e⁰ - e⁰ = 0 and sin 0 = 0, so only 2/a survives. This gives 2/a = 4, so a = 1/2.
3. So. (2a+1)⁵ a² = 2⁵ × 1/4 = 32/4 = 8.

The common slip: 12500. Solving 2/a = 4 as a = 2 gives 5⁵ × 4 = 12500, not 8. The equation gives a = 2/4 = 1/2.

The idea to keep. When the integrand has x inside it, like (x-t), differentiate it with the Leibniz rule: the boundary term plus the integral of ∂/∂ x. Then plug in the point where the integral vanishes.

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