Definite Integration Question 99
Question: Let $ a,b,c $ be non-zero real numbers such that $ \int_0^{3}{(3ax^{2}+2bx+c)dx}=\int_1^{3}{(3ax^{2}+2bx+c})dx, $ then
[BIT Ranchi 1991]
Options:
A) $ a+b+c=3 $
B) $ a+b+c=1 $
C) $ a+b+c=0 $
D) $ a+b+c=2 $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_0^{3}{(3ax^{2}+2bx+c)dx=\int_1^{3}{(3ax^{2}+2bx+c)dx}} $
therefore $ \int_0^{1}{(3ax^{2}+2bx+c)dx+\int_1^{3}{(3ax^{2}+2bx+c)dx}} $
$ =\int_1^{3}{(3ax^{2}+2bx+c)dx} $
therefore $ \int_0^{1}{(3ax^{2}+2bx+c)dx=0} $
therefore $ [ \frac{3ax^{3}}{3}+\frac{2bx^{2}}{2}+cx ]_0^{1}=0\Rightarrow a+b+c=0 $ .
BETA
