Let f(x) = x + a/(π ² - 4)sin x + b/(π ² - 4)cos x,x ∈ R be a function which
satisfies f(x) = x + ∫₀^(π /2) sin (x + y)f(y)dy. then (a+b) is equal to

Solution

Maths sir · 4 steps

Expand sin(x+y) and the integral becomes two unknown constants. Name them and solve for them.

1. Name the constants. Write sin(x+y) = sin x cos y + cos x sin y. Then f(x) = x + A sin x + B cos x, where A = ∫₀^(π/2) f(y)cos y dy and B = ∫₀^(π/2) f(y)sin y dy. Comparing with the given form, a = A(π²-4) and b = B(π²-4).
2. Evaluate the integrals. Put f(y) = y + A sin y + B cos y back in. On [0, π/2] you need ∫ y cos y dy = π/2-1, ∫ y sin y dy = 1, ∫ sin y cos y dy = 1/2 and ∫ cos² y dy = ∫ sin² y dy = π/4. So A = π/2-1 + A/2 + (π B)/4 and B = 1 + (π A)/4 + B/2.

   2A - π B = 2π - 4 ; -π A + 2B = 4

3. Solve the system. The determinant is 4-π². Cramer's rule gives A = (8π-8)/(4-π²) and B = (2π²-4π+8)/(4-π²). Multiplying by π²-4 gives a = 8-8π and b = -2π²+4π-8.
4. So. a+b = 8-8π-2π²+4π-8 = -2π²-4π = -2π(π+2), which is option (a).

The idea to keep. When a definite integral with constant limits sits inside a functional equation, it is just a number. Call it a constant, substitute f back in and solve the linear equations you get.

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