Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
MHT CET202613 April 2026Evening ShiftMathematicsDefinite IntegrationActual

If ₀^a (x^2 - 4x + 1)dx = 6 , then the real value of a is

Options

  1. A4
  2. B6
  3. C3
  4. D-3

Correct answer

B. 6

Step-by-step solution

Evaluating the given definite integral: ₀^a (x^2 - 4x + 1)dx = [ x^3 3 - 2x^2 + x ]₀^a Substituting the limits, we get: a^3 3 - 2a^2 + a = 6 Multiplying by 3 on both sides: a^3 - 6a^2 + 3a - 18 = 0 Factorizing the polynomial: a^2(a - 6) + 3(a - 6) = 0 (a^2 + 3)(a - 6) = 0 Since a is a real number, a^2 + 3 0 . Therefore, a - 6 = 0 a = 6 . Answer: 6

Practice Definite Integration on Quantrex Academy →

More from Definite Integration

₀^1 2 x+5 x^2+3 x+2 ~d x= 2025₀^1 x^ 5 / 2 (1-x)^ 3 / 2 ~d x= 2025_ n [ 1 n^2 ^2 1 n^2 + 2 n^2 ^2 4 n^2 + 3 n^2 ^2 9 n^2 + + 1 n^2 ^2 1 ]= 2025₀^1 x Sin ⁻¹ x d x= 2025_ - 2 ^ 2 (x-[x]) d x= 2025₀^2 x^2(2-x)^5 d x= 2025If f(x)= Max x^3-4, x^4-4 , and g(x)= Min x^2, x^3 , then _ -1 ^1(f(x)-g(x)) d x= 2025_ n 2 n [ 2 n + 2 2 n + 3 2 n + + 2 ]= 2025 Full Definite Integration list All MHT CET PYQs