Question

Let $I_r(x) = \int \frac{x^r dx}{x^4 + 4} (x > 0)$

If $I_1(\sqrt{2}) = \frac{9\pi}{16}$, then $\lim_{x\to 0} \frac{\cos(I_1(x))}{x^2}$ is equal to -

• A. -$\frac{1}{4}$
• B. -$\frac{1}{8}$
• C. $\frac{1}{8}$
• D. 1

Answer 1

Answer: (B)

-$\frac{1}{8}$

Answer 2

1. Find $I_1(x)$: $I_1(x) = \int \frac{x dx}{x^4 + 4}$ Let $u = x^2$, then $du = 2x dx$, so $x dx = \frac{1}{2} du$. $I_1(x) = \int \frac{\frac{1}{2} du}{u^2 + 4} = \frac{1}{2} \int \frac{du}{u^2 + 2^2}$ Using the standard integral $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$: $I_1(x) = \frac{1}{2} \cdot \frac{1}{2} \tan^{-1}(\frac{u}{2}) + C = \frac{1}{4} \tan^{-1}(\frac{x^2}{2}) + C$.

2. Determine the constant $C$ using $I_1(\sqrt{2}) = \frac{9\pi}{16}$: Substitute $x = \sqrt{2}$ into $I_1(x)$: $I_1(\sqrt{2}) = \frac{1}{4} \tan^{-1}(\frac{(\sqrt{2})^2}{2}) + C = \frac{1}{4} \tan^{-1}(\frac{2}{2}) + C = \frac{1}{4} \tan^{-1}(1) + C$ $I_1(\sqrt{2}) = \frac{1}{4} \cdot \frac{\pi}{4} + C = \frac{\pi}{16} + C$. Given $I_1(\sqrt{2}) = \frac{9\pi}{16}$, we have: $\frac{\pi}{16} + C = \frac{9\pi}{16} \implies C = \frac{8\pi}{16} = \frac{\pi}{2}$. So, $I_1(x) = \frac{1}{4} \tan^{-1}(\frac{x^2}{2}) + \frac{\pi}{2}$.

3. Evaluate the limit $\lim_{x\to 0} \frac{\cos(I_1(x))}{x^2}$: First, find $I_1(0)$: $I_1(0) = \frac{1}{4} \tan^{-1}(\frac{0^2}{2}) + \frac{\pi}{2} = \frac{1}{4} \tan^{-1}(0) + \frac{\pi}{2} = 0 + \frac{\pi}{2} = \frac{\pi}{2}$. As $x \to 0$, $I_1(x) \to \frac{\pi}{2}$. The limit is of the form $\frac{\cos(\pi/2)}{0} = \frac{0}{0}$, so we can apply L'Hopital's Rule. Let $L = \lim_{x\to 0} \frac{\cos(I_1(x))}{x^2}$. Differentiate numerator and denominator with respect to $x$: $L = \lim_{x\to 0} \frac{-\sin(I_1(x)) \cdot I_1'(x)}{2x}$. From the definition $I_r(x) = \int \frac{x^r dx}{x^4 + 4}$, by the Fundamental Theorem of Calculus, $I_1'(x) = \frac{x}{x^4 + 4}$. Substitute $I_1'(x)$: $L = \lim_{x\to 0} \frac{-\sin(I_1(x)) \cdot \frac{x}{x^4 + 4}}{2x}$ $L = \lim_{x\to 0} \frac{-\sin(I_1(x))}{2(x^4 + 4)}$. Now, substitute $x=0$: $L = \frac{-\sin(I_1(0))}{2(0^4 + 4)} = \frac{-\sin(\pi/2)}{2(4)} = \frac{-1}{8}$.

The final answer is $\boxed{\text{-}\frac{1}{8}}$.