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

If ${I_n} = \int_0^\infty {{e^{ - x}}{x^{n - 1}}dx,} $ then $\int_0^\infty {{e^{ - \lambda x}}{x^{n - 1}}dx = } $

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^2 x \, dx = $ . . . . . . .

The value of $\sum\limits_{r = 2}^{16} {\int\limits_r^{r + 1} {\frac{{dx}}{{\left( {2r - x} \right)\left( {2r + 2 - x} \right)}}} }$ is equal to

$\int_0^{16} \frac{\sqrt{x}}{1+\sqrt{x}} d x=$

Evaluate the definite integral $\int_{0}^{\frac{\pi}{2}} \cos ^{2} x \,d x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module