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

$$\int_{0}^{1} (1+\cos ^8x)(ax^2+bx+c)dx=\int_{0}^{2} (1+\cos ^8x)(ax^2+bx+c)dx$$, then the equation $ax^2+bx+c=0$ has (choose the correct option)

$1.$ No root in $(0,2)$

$2.$ Atleast one root in $(0,2)$

$3.$ Both roots in $(0,2)$

Could someone give me slight hint to solve this problem?

  • 2
    First, note that the above can be written as $\int_1^2 p(x) q(x) dx$ where $p(x) \ge 1$ and $q$ is the quadratic in question. If $q$ has no roots in $(0,2)$ then it does not change sign there. – 2017-02-04
  • 0
    @copper.hat SO for integral to be $0$ quadratic will have atleast one root in $(1,2)$ . Am I correct? – 2017-02-04
  • 1
    Correct, otherwise there could not be equality. – 2017-02-04
  • 0
    Since$ \int_{1}^{2} (1+\cos ^8x)(ax^2+bx+c)dx=0$ and since $1+\cos ^8x>0$ on $(1,2)$ then $ax^2+bx+c$ must have at least one root in $(1,2)\subset(0,2)$. – 2017-02-04

1 Answers 1

2

As it has been noted above, the conditions imply that \begin{equation*} \tag{$*$} \int_1^2 (1+\cos^8 x)(ax^2+bx+c)\,dx=0. \end{equation*} Since $1+\cos^8 x \ge 1$ in $[1,2],$ there must be a root of the quadratic function $ax^2+bx+c$ in $(1,2).$ On the other hand, $(*)$ implies that $$ Aa+Bb+Cc=0 $$ where $A,B,C$ are real numbers with $C \ne 0.$ Hence $$ c=\alpha a+\beta b $$ for some $\alpha,\beta \in \mathbf R.$ In effect, given any $a,b$ such that $a \ne 0,$ the function $ax^2+bx+c$ satisfies $(*)$ for a suitable $c.$ Now take positive $a,b,$ and define $c$ accordingly. As we now know, the corresponding function $q(x)=ax^2+bx+c$ has a root in $(1,2),$ hence its discriminant $\Delta$ is positive. But then $q(x)$ has also a negative root, namely $$ \frac{-b-\sqrt\Delta}{2a} < 0. $$ Thus it is impossible in general to have both roots in $(0,2),$ whence 2 is the correct answer.