If φ(x)=1/√x ∫_(π/4)ˣ(4 √2 sin t-3 φ′(t)) d t, x>0,
then ∅′(π/4) is equal to :

Solution

Maths sir · 4 steps

Clear the 1/√x, differentiate, and use φ(π/4) = 0 at the lower limit.

1. Clear the fraction. Multiply both sides by √x to get √x φ(x) = ∫_(π/4)ˣ(4√2 sin t - 3φ'(t))dt. Put x = π/4. The limits are then equal, so the integral is 0 and φ(π/4) = 0.
2. Differentiate both sides. Use the product rule on the left and the Leibniz rule on the right. The upper limit is x, so the right side becomes the integrand at t = x.

   φ(x)/2√x + √x φ'(x) = 4√2 sin x - 3φ'(x)

3. Substitute the limit. Put x = π/4. The first term is 0 because φ(π/4) = 0. Also √(π/4) = √π/2 and 4√2 sin π/4 = 4, so the equation becomes √π/2φ' = 4 - 3φ', where φ' means φ'(π/4).
4. So. Collect the φ' terms to get φ'(3 + √π/2) = 4. This gives φ'(π/4) = 8/(6+√π), which is option (d).

The idea to keep. When the unknown function sits inside its own integral, differentiate with the Leibniz rule. Then put x equal to the lower limit: the integral is 0 there, so you get a value of the function for free.

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