Theory of Equations 5 Question 8

8. Let a,b,c be non-zero real numbers such that

01(1+cos8x)(ax2+bx+c)dx=02(1+cos8x)(ax2+bx+c)dx

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Answer:

Correct Answer: 8. (b)

Solution:

  1. Consider,

f(x)=01(1+cos8x)(ax2+bx+c)dx

Obviously, f(x) is continuous and differentiable in the interval [1, 2]. Also,

f(1)=f(2)

[given]

By Rolle’s theorem, there exist atleast one point k(1,2), such that f(k)=0.

 Now, f(x)=(1+cos8x)(ax2+bx+c)f(k)=0(1+cos8k)(ak2+bk+c)=0ak2+bk+c=0[as(1+cos8k)0]x=k is root of ax2+bx+c=0,

where k(1,2)



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