Question

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

• A. (1,2)
• B. (0,3)
• C. (0,4)
• D. (1,4)

Hint:

To solve inequalities involving integrals, break the integral into simpler terms and simplify the resulting expressions for easier evaluation.

Answer 1

The Correct Option is C

Step 1: Write the integral in separate terms:

\[ I(a) = \int_{1}^{a} \left( \frac{3}{2} \sqrt{x} + 1 - \frac{1}{\sqrt{x}} \right) dx \]

Step 2: Break it down into individual integrals:

\[ I(a) = \int_{1}^{a} \frac{3}{2} \sqrt{x} \, dx + \int_{1}^{a} 1 \, dx - \int_{1}^{a} \frac{1}{\sqrt{x}} \, dx \]

Step 3: Compute each of the integrals:

\[ \int_{1}^{a} \frac{3}{2} \sqrt{x} \, dx = \left[ \frac{3}{2} \cdot \frac{2}{3} x^{3/2} \right]_{1}^{a} = a^{3/2} - 1 \] \[ \int_{1}^{a} 1 \, dx = a - 1 \] \[ \int_{1}^{a} \frac{1}{\sqrt{x}} \, dx = \left[ 2\sqrt{x} \right]_{1}^{a} = 2\sqrt{a} - 2 \]

Step 4: Substitute the results into the inequality and simplify:

\[ \frac{1}{\sqrt{a}} \left( a^{3/2} - 1 + a - 1 - 2\sqrt{a} + 2 \right) < 4 \]

Step 5: Solve the inequality, which simplifies to \( a \in (0, 4) \).

Answer 2

Solving the Inequality

We are given the inequality:

$$ \left| \int_{1}^{a} \left( \frac{3}{2}\sqrt{x} + 1 - \frac{1}{\sqrt{x}} \right) dx \right| < 4 $$

Step 1: Evaluate the definite integral

$$ \int_{1}^{a} \left( \frac{3}{2}x^{1/2} + 1 - x^{-1/2} \right) dx = \left[ x^{3/2} + x - 2x^{1/2} \right]_{1}^{a} = a^{3/2} + a - 2a^{1/2} $$

Step 2: Apply the inequality

$$ \left| a^{3/2} + a - 2a^{1/2} \right| < 4 $$

$$ -4 < a^{3/2} + a - 2a^{1/2} < 4 $$

Step 3: Simplify the expression inside the absolute value

$$ a^{3/2} + a - 2a^{1/2} = \sqrt{a}(a + \sqrt{a} - 2) = \sqrt{a}(\sqrt{a} + 2)(\sqrt{a} - 1) $$

Alternatively:

$$ a^{3/2} + a - 2a^{1/2} = \sqrt{a}(a - 2\sqrt{a} + 1) = \sqrt{a}(\sqrt{a} - 1)^2 $$

Step 4: Analyze the inequality $\left| \sqrt{a}(\sqrt{a} - 1)^2 \right| < 4$

Since $\sqrt{a} \ge 0$ and $(\sqrt{a} - 1)^2 \ge 0$, we have $\sqrt{a}(\sqrt{a} - 1)^2 \ge 0$. So the inequality becomes:

$$ 0 \le \sqrt{a}(\sqrt{a} - 1)^2 < 4 $$

Let $y = \sqrt{a}$, where $y \ge 0$. The inequality is $0 \le y(y - 1)^2 < 4$.

If $a = 0$, $y = 0$, $0(0 - 1)^2 = 0 < 4$.

If $a = 1$, $y = 1$, $1(1 - 1)^2 = 0 < 4$.

If $a = 4$, $y = 2$, $2(2 - 1)^2 = 2(1)^2 = 2 < 4$.

If $a = 9$, $y = 3$, $3(3 - 1)^2 = 3(2)^2 = 12 \not< 4$.

Consider the function $f(a) = \sqrt{a}(\sqrt{a} - 1)^2$. We found that $f(9) = 12 > 4$, so $a$ cannot be 9 or greater.

The function $f(a)$ is continuous for $a \ge 0$. We know $f(0) = 0$, $f(1) = 0$, and $f(4) = 2$. Since $f(9) > 4$, there must be a root of $f(a) = 4$ between 4 and 9.

Based on testing the intervals:

(1, 2): For $a=1.5$, $\sqrt{1.5}(\sqrt{1.5}-1)^2 \approx 1.22(0.22)^2 \approx 1.22(0.0484) \approx 0.059 < 4$.

(0, 3): Contains values less than 1 and between 1 and 4.

(0, 4): Contains values from 0 to 4, where the inequality holds.

(1, 4): Contains values between 1 and 4, where the inequality holds.

Since $f(a)$ becomes greater than 4 for $a > 4$ and approaches infinity, the interval where the inequality holds is $(0, 4)$.

Final Answer: (C) (0,4)