Question

Let a, b, c be non-zero real numbers such that ${\int }_0^3(3a{x}^2+2bx+c)dx={\int }_1^3(3a{x}^2+2bx+c)dx.$ Then

• A. a + b + c = 3
• B. a + b + c = 1
• C. a + b + c = 0
• D. a + b + c = 2

Answer 1

The correct option is C

a + b + c = 0

${\int }_0^3(3a{x}^2+2bx+c)dx={\int }_1^3(3a{x}^2+2bx+c)dx$

$\Rightarrow {\int }_0^1(3a{x}^2+2bx+c)dx+{\int }_1^3(3a{x}^2+2bx+c)dx={\int }_1^3(3a{x}^2+2bx+c)dx$

$\Rightarrow {\int }_0^1(3a{x}^2+2bx+c)dx=0$

$\Rightarrow [\frac{3a{x}^3}{3}+\frac{2b{x}^2}{2}+cx{]}_0^1=0$

$\Rightarrow a+b+c=0$