Let g(x) = cos(x^2),f(x) = sqrt{x},and alpha, | vedclass

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

Question

(D) Given the quadratic equation $18x^2 - 9\pi x + \pi^2 = 0$. Factoring the quadratic: $(3x - \pi)(6x - \pi) = 0$. Thus,the roots are $\alpha = \frac

Vedclass · Accepted Answer

$\frac{1}{2}(\sqrt{3} - 1)$

Vedclass · Answer

(D) Given the quadratic equation $18x^2 - 9\pi x + \pi^2 = 0$. Factoring the quadratic: $(3x - \pi)(6x - \pi) = 0$. Thus,the roots are $\alpha = \frac{\pi}{6}$ and $\beta = \frac{\pi}{3}$ (since $\alpha Next,we find the composite function $(g \circ f)(x) = g(f(x)) = \cos((\sqrt{x})^2) = \cos(x)$. The area $A$ bounded by $y = \cos(x)$,$x = \frac{\pi}{6}$,$x = \frac{\pi}{3}$,and $y = 0$ is given by the integral: $A = \int_{\pi/6}^{\pi/3} \cos(x) \, dx$. Evaluating the integral: $A = [\sin(x)]_{\pi/6}^{\pi/3} = \sin(\frac{\pi}{3}) - \sin(\frac{\pi}{6})$. Substituting the values: $A = \frac{\sqrt{3}}{2} - \frac{1}{2} = \frac{\sqrt{3} - 1}{2}$ sq. units.

Let $g(x) = \cos(x^2)$,$f(x) = \sqrt{x}$,and $\alpha, \beta$ (where $\alpha < \beta$) be the roots of the quadratic equation $18x^2 - 9\pi x + \pi^2 = 0$. Then,the area (in sq. units) bounded by the curve $y = (g \circ f)(x)$ and the lines $x = \alpha$,$x = \beta$,and $y = 0$ is:

Similar Questions

The area enclosed between the curve ${y^2}(2a - x) = {x^3}$ and the line $x = 2a$ above the $x$-axis is

The area bounded by the parabola $y = 4x^2$,the $y$-axis,and the lines $y = 1$ and $y = 4$ is:

Find the area of the region bounded by $x^{2}=4y$,$y=2$,$y=4$ and the $y-$axis in the first quadrant.

The area of the region enclosed by the curve $f(x) = \max \{\sin x, \cos x\}$,$-\pi \leq x \leq \pi$ and the $x$-axis is

The area bounded by the $X$-axis and the curve $y=x(x-2)(x+1)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module