int_{0}^{1} (1 + |sin x|)(ax^2 + bx + c) dx = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the equation: $\int_{0}^{1} (1 + |\sin x|)(ax^2 + bx + c) dx = \int_{0}^{2} (1 + |\sin x|)(ax^2 + bx + c) dx$. We can rewrite the right side

Vedclass · Accepted Answer

At least one real root between $(1, 2)$

Vedclass · Answer

(A) Given the equation: $\int_{0}^{1} (1 + |\sin x|)(ax^2 + bx + c) dx = \int_{0}^{2} (1 + |\sin x|)(ax^2 + bx + c) dx$. We can rewrite the right side as: $\int_{0}^{1} (1 + |\sin x|)(ax^2 + bx + c) dx + \int_{1}^{2} (1 + |\sin x|)(ax^2 + bx + c) dx$. Subtracting the integral from $0$ to $1$ from both sides,we get: $\int_{1}^{2} (1 + |\sin x|)(ax^2 + bx + c) dx = 0$. Let $f(x) = ax^2 + bx + c$ and $g(x) = 1 + |\sin x|$. Since $g(x) > 0$ for all $x \in [1, 2]$,for the integral of the product $f(x)g(x)$ to be zero over the interval $[1, 2]$,the function $f(x)$ must change its sign at least once in the interval $(1, 2)$. By the Intermediate Value Theorem,if a continuous function changes sign in an interval,it must have at least one root in that interval. Therefore,$ax^2 + bx + c = 0$ has at least one real root between $(1, 2)$.

$\int_{0}^{1} (1 + |\sin x|)(ax^2 + bx + c) dx = \int_{0}^{2} (1 + |\sin x|)(ax^2 + bx + c) dx$. Then,the location of the roots of $ax^2 + bx + c = 0$ is:

Similar Questions

$\int_{-\pi}^\pi \frac{x \sin ^3 x}{4-\cos ^2 x} d x=$

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\tan^4 x} dx = $ . . . . . . .

$\int_{-\pi}^{\pi} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx = $

If $f$ is defined on $R$ such that $f(x) f(-x) = 9$, then find the value of $\int_{-23}^{23} \frac{dx}{3+f(x)}$.

$\int_0^{\frac{\pi}{2}} \frac{\sin^2 x}{\sin x + \cos x} dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module