Definite Integration Question 343
Question: Let a, b, c be non-zero real numbers such that $ \int_0^{1}{(1+{{\cos }^{8}}x)(ax^{2}+bx+c)dx}=\int_0^{2}{(1+{{\cos }^{8}}x)(ax^{2}+bx+c)dx} $ Then the quadratic equation $ ax^{2}+bx+c=0 $ has
[IIT 1981; CEE 1993]
Options:
A) No root in (0, 2)
B) At least one root in (0, 2)
C) A double root in (0, 2)
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
We have $ \int_0^{2}{f(x)dx=\int_0^{1}{f(x)dx+\int_1^{2}{f(x)dx}}} $ ,
where $ f(x)=(ax^{2}+bx+c)(1+{{\cos }^{8}}x) $
If $ f(x)>0(<0)x\in (1,2) $ then $ \int_1^{2}{f(x)dx>0(<0)} $ . Thus $ f(x)=(1+{{\cos }^{8}}x)(ax^{2}+bx+c) $ must be positive for some value of x in [1, 2] and must be negative for some value of x in [1, 2]. As $ (1+{{\cos }^{8}}x)\ge 1 $ it follows that if $ g(x)=ax^{2}+bx+c, $ then there exist some $ \alpha ,\beta \in (1,2) $ such that $ g(\alpha )>0 $ and $ g(\beta )<0 $ . Since g is continuous on R, therefore there exist some c between $ \alpha $ and $ \beta $ such that $ g(c)=0 $ . Thus $ ax^{2}+bx+c $ =0 has at least one root in (1, 2) and hence in (0, 2).
BETA
