All values of a for which the inequality frac | vedclass

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

Question

(C) Given inequality: $\frac{1}{\sqrt{a}} \int_{1}^{a} (\frac{3}{2} \sqrt{x} + 1 - \frac{1}{\sqrt{x}}) dx < 4$First, evaluate the integral: $\int_{1}^

Vedclass · Accepted Answer

$(0, 4)$

Vedclass · Answer

(C) Given inequality: $\frac{1}{\sqrt{a}} \int_{1}^{a} (\frac{3}{2} \sqrt{x} + 1 - \frac{1}{\sqrt{x}}) dx First, evaluate the integral: $\int_{1}^{a} (\frac{3}{2} x^{1/2} + 1 - x^{-1/2}) dx$$= [\frac{3}{2} \cdot \frac{x^{3/2}}{3/2} + x - \frac{x^{1/2}}{1/2}]_{1}^{a}$$= [x^{3/2} + x - 2x^{1/2}]_{1}^{a}$$= (a^{3/2} + a - 2a^{1/2}) - (1^{3/2} + 1 - 2(1)^{1/2})$$= a^{3/2} + a - 2a^{1/2} - (1 + 1 - 2) = a^{3/2} + a - 2a^{1/2}$Now substitute back into the inequality:$\frac{1}{a^{1/2}} (a^{3/2} + a - 2a^{1/2}) $a + a^{1/2} - 2 $a + a^{1/2} - 6 Let $t = a^{1/2}$, where $t > 0$:$t^2 + t - 6 $(t + 3)(t - 2) Since $t > 0$, $t + 3$ is always positive, so we must have $t - 2 Thus, $a^{1/2} Since the integral is defined for $a > 0$, the interval is $(0, 4)$.

All values of $a$ for which the inequality $\frac{1}{\sqrt{a}} \int_{1}^{a} (\frac{3}{2} \sqrt{x} + 1 - \frac{1}{\sqrt{x}}) dx < 4$ is satisfied, lie in the interval

Similar Questions

$\int_0^{\pi / 4} \sqrt{1-\sin 2 x} \,d x =$

$\int_{0}^{\frac{\pi}{4}} (\tan^n x + \tan^{n-2} x) d(x - [x])$ is : (where $[.]$ denotes the greatest integer function)

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

The integral $\int_{0}^{1} \frac{1}{7^{\left[\frac{1}{x}\right]}} dx$,where $[.]$ denotes the greatest integer function,is equal to:

If $\int \limits_{-0.15}^{0.15} |100 x^2 - 1| dx = \frac{k}{3000}$,then $k$ is equal to $..........$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module