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Let a ,b,c be non zero real number such that Definite integral under limit 0 to 1 (1+cos^8x)(ax^2+bx+c) =Definite integral under limit 0 to 2 (1+cos^8x)(ax^2+bx+c) , Then the quadratic equation ax^2+bx+c=0 has

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Solution: Let $f(x)=(1+cos^8x)(ax^2+bx+c)$

We are given

$Rightarrow$ $int_0^1f(x)hspace0.1cmdx=int_0^2f(x)hspace0.2cmdx$

$int_0^1f(x)hspace0.1cmdx=int_0^1f(x)hspace0.2cmdx+int_1^2f(x)hspace0.2cmdx$

$int_1^2f(x)hspace0.2cmdx=0$

If $f(x)> 0 (f(x)< 0)forall xin [1,2]$

Then $int_1^2f(x)hspace0.2cmdx> 0(int_1^2f(x)hspace0.2cmdx< 0)$

$herefore$ $f(x)$ is partly positive and partly negative on $[1,2]$

$Rightarrow$ There exist $alpha ,eta in [1,2]$ such that

As $f$ is continuous on $[1,2]$ there exists $gamma$ lying between $alpha$ and $eta$

$Rightarrow$ $(1+cos^8gamma )(agamma ^2+bgamma +c)=0$

$Rightarrow$ $agamma ^2+bgamma +c=0$ $[ecause 1+cos^8gamma geq 1]$

Thus , $ax^2+bx+c=0$ has at least one root in $[1,2]$

Posted by

Deependra Verma

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