Number of ordered pairs (a, b) satisfying sim | vedclass

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

Question

(B) Given the system of equations:$1$) $\int_{a}^{b} x^{3} dx = 0$$2$) $\int_{a}^{b} x^{2} dx = \frac{2}{3}$Solving the first equation:$\left[\frac{x^

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given the system of equations:$1$) $\int_{a}^{b} x^{3} dx = 0$$2$) $\int_{a}^{b} x^{2} dx = \frac{2}{3}$Solving the first equation:$\left[\frac{x^{4}}{4}\right]_{a}^{b} = 0 \Rightarrow \frac{b^{4}-a^{4}}{4} = 0 \Rightarrow b^{4} = a^{4}$This implies $b = a$ or $b = -a$.Case $1$: If $b = a$,then $\int_{a}^{a} x^{2} dx = 0$,which contradicts $\int_{a}^{b} x^{2} dx = \frac{2}{3}$.Case $2$: If $b = -a$,substitute into the second equation:$\int_{a}^{-a} x^{2} dx = \frac{2}{3}$$\left[\frac{x^{3}}{3}\right]_{a}^{-a} = \frac{2}{3}$$\frac{(-a)^{3} - a^{3}}{3} = \frac{2}{3}$$\frac{-a^{3} - a^{3}}{3} = \frac{2}{3}$$-2a^{3} = 2 \Rightarrow a^{3} = -1 \Rightarrow a = -1$Since $b = -a$,we get $b = 1$.Thus,there is only one ordered pair $(a, b) = (-1, 1)$ that satisfies the system.

Number of ordered pairs $(a, b)$ satisfying simultaneously the system of equations $\int_{a}^{b} x^{3} dx = 0$ and $\int_{a}^{b} x^{2} dx = \frac{2}{3}$ is

Similar Questions

$\int_5^{10} \frac{d x}{(x-1)(x-2)} = $

The limit of the area under the curve $y = e^{-x}$ from $x = 0$ to $x = h$ as $h \rightarrow \infty$ is:

What is the minimum value of $\int\limits_0^x {t{e^{ - {t^2}}}} dt$?

$\mathop \smallint \limits_0^\pi \sqrt {1 + 4{{\sin }^2}\frac{x}{2} - 4\sin \frac{x}{2}} \;dx = $

The value of the integral $\int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module