The true solution set of the inequality sqrt{ | vedclass

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

Question

(D) First,evaluate the integrals:$\int_{0}^{x} dz = [z]_{0}^{x} = x$$\int_{0}^{\pi} \sin^2 x dx = \int_{0}^{\pi} \frac{1 - \cos 2x}{2} dx = \left[ \fr

Vedclass · Accepted Answer

$(2, 3)$

Vedclass · Answer

(D) First,evaluate the integrals:$\int_{0}^{x} dz = [z]_{0}^{x} = x$$\int_{0}^{\pi} \sin^2 x dx = \int_{0}^{\pi} \frac{1 - \cos 2x}{2} dx = \left[ \frac{x}{2} - \frac{\sin 2x}{4} \right]_{0}^{\pi} = \frac{\pi}{2} - 0 = \frac{\pi}{2}$Substitute these into the inequality:$\sqrt{5x - 6 - x^2} + \frac{\pi}{2} x > x \left( \frac{\pi}{2} \right)$$\sqrt{5x - 6 - x^2} + \frac{\pi x}{2} > \frac{\pi x}{2}$Subtract $\frac{\pi x}{2}$ from both sides:$\sqrt{5x - 6 - x^2} > 0$For the square root to be defined and greater than zero,the expression inside must be strictly positive:$5x - 6 - x^2 > 0$$-(x^2 - 5x + 6) > 0$$x^2 - 5x + 6 $(x - 2)(x - 3) Thus,the solution set is $x \in (2, 3)$.

The true solution set of the inequality $\sqrt{5x-6-x^2} + \left( \frac{\pi}{2} \int_{0}^{x} dz \right) > x \int_{0}^{\pi} \sin^2 x dx$ is:

Similar Questions

If $I_{n}=\int_{0}^{\pi / 4} \tan ^{n} x d x$,where $n$ is a positive integer,then $I_{10}+I_{8}$ is

If $\int_0^b \frac{dx}{1+x^2} = \int_b^{\infty} \frac{dx}{1+x^2}$,then $b$ is equal to

The value of the integral $\int_{-1}^2 \log _e\left(x+\sqrt{x^2+1}\right) d x$ is:

If $f(x) = \int_{-1}^{x} |t| dt$,then for any $x \geq 0$,$f(x)$ is equal to

If $f(x) = \begin{cases} 4x + 3, & 1 \le x \le 2 \\ 3x + 5, & 2 < x \le 4 \end{cases}$,then $\int_1^4 f(x) \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module