MHT CET202616 April 2026Evening ShiftMathematicsIndefinite IntegrationActual
If a > 0, b > 0 and 1 ax^2 + b dx = 1 6 ⁻¹ ( 2 x 3 ) + c , then 1 bx^2 + a dx =
Options
- A- 1 6 ⁻¹ ( 2 x 3 ) + c
- B1 6 ⁻¹ ( 3 x 2 ) + c
- C- 6 , ⁻¹ ( 2 x 3 ) + c
- D6 , ⁻¹ ( 3 x 2 ) + c
Correct answer
B. 1 6 ⁻¹ ( 3 x 2 ) + c
Step-by-step solution
We know that 1 ax^2 + b dx = 1 ab ⁻¹ ( a b x ) + c Comparing this with the given expression 1 6 ⁻¹ ( 2 x 3 ) + c , we get: 1 ab = 1 6 ab = 6 a b = 2 3 a b = 2 3 Multiplying the two equations, we get a^2 = 4 a = 2 (since a > 0 ). Substituting a = 2 in ab = 6 , we get b = 3 . Now, we need to evaluate 1 bx^2 + a dx : 1 3x^2 + 2 dx = 1 3 2 ⁻¹ ( 3 2 x ) + c = 1 6 ⁻¹ ( 3 x 2 ) + c Answer: 1 6 ⁻¹ ( 3 x 2 ) + c