NDA Mathematics Question (2026) — Solution

Step-by-step solution

Let I = \,d (2+ )(3+4 ) . Substitute t = , which gives dt = - \,d . The integral becomes: I = -dt (2+t)(3+4t) Using partial fractions, we can write: -1 (2+t)(3+4t) = A' 2+t + B' 3+4t -1 = A'(3+4t) + B'(2+t) Putting t = -2, we get -1 = A'(3 - 8) A' = 1 5 . Putting t = - 3 4 , we get -1 = B' (2 - 3 4 ) -1 = B' ( 5 4 ) B' = - 4 5 . So, the integral is: I = ( 1/5 2+t - 4/5 3+4t ) dt I = 1 5 |2+t| - 4 5 1 4 |3+4t| + C I = 1 5 |2+t| - 1 5 |3+4t| + C Substituting back t = : I = 1 5 |2+ | - 1 5 |3+4 | + C Comparing this with the given expression A |2+ | + B |3+4 |, we get: B = - 1 5 Answer: - 1 5