Multiple choice

Let a, b, c be non-zero real numbers such the : $\displaystyle \int_{0}^{1}\left ( 1+\cos ^{8}x \right )\left ( ax^{2}+bx+c \right )dx=\int_{0}^{2}\left ( 1+\cos ^{8}x \right )\left ( ax^{2}+bx+c \right )dx,$ then the quadratic equation $\displaystyle ax^{2}+bx+c=0$ has

  1. no root in $\displaystyle \left ( 0,2 \right )$
  2. atleast one root in $\displaystyle \left ( 0,2 \right )$
  3. a double root in $\displaystyle \left ( 0,2 \right )$

  4. none
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let f(x) = ax^2 + bx + c and g(x) = 1 + cos^8(x). The equation is integral from 0 to 1 of f(x)g(x) dx = integral from 0 to 2 of f(x)g(x) dx. This implies the integral from 1 to 2 of f(x)g(x) dx = 0. Since g(x) is always positive, f(x) must change sign in the interval (1, 2), which is within (0, 2). By the Intermediate Value Theorem, f(x) must have at least one root in (0, 2).

AI explanation

Let the integrand be h(x) = (1 + cos^8 x)(ax^2 + bx + c). The given equation can be written as the integral from 0 to 1 of h(x) minus the integral from 0 to 1 of h(x) equals the integral from 1 to 2 of h(x). This simplifies to the integral from 0 to 2 of h(x) = 0. By the mean value theorem for definite integrals, since the function is continuous, it must equal zero at some point within the interval. Because the term 1 + cos^8 x is always strictly positive, the quadratic ax^2 + bx + c must be zero at that same point. Therefore, the quadratic equation has at least one root in (0, 2).

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