Let I(x)=∫ dx/(x-11)^(11/13)(x+15)^(15/13). If I(37)-I(24)=1/4(1/b^(1/13)-1/c^(1/13)),b,c∈ N, then 3( b+c) is equal to

Solution

Maths sir · 5 steps

The powers add up to 2, so divide by (x+15)² and substitute t = (x-11)/(x+15).

1. Spot the pattern. The exponents add to 11/13 + 15/13 = 2. So rewrite the integrand as 1/((x-11)/(x+15))^(11/13)(x+15)².
2. Substitute. Let t = (x-11)/(x+15). Then dt = ((x+15)-(x-11))/(x+15)² dx = 26/(x+15)² dx, which absorbs the (x+15)² exactly.

   I(x) = 1/26∫ t^(-11/13) dt = 1/26×13/2 t^(2/13) + C = 1/4((x-11)/(x+15))^(2/13) + C

3. Put in limits. At x = 37, t = 26/52 = 1/2. At x = 24, t = 13/39 = 1/3. The constant C cancels in the difference.
4. Match the form. I(37) - I(24) = 1/4((1/2)^(2/13) - (1/3)^(2/13)) = 1/4(1/4^(1/13) - 1/9^(1/13)). So b = 4 and c = 9.
5. So. 3(b + c) = 3 × 13 = 39, which is option (a).

The idea to keep. For ∫ dx/(x-a)ᵐ(x+b)ⁿ with m + n = 2, divide by (x+b)² and put t = (x-a)/(x+b). The derivative of t supplies the (x+b)².

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