Question

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

On the basis of above information, answer the following questions : 21. If $I_1(\sqrt{2}) = \frac{9\pi}{16}$, then $lim_{x\to0} \frac{cos(I_1(x))}{x^2}$ 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. Calculate $I_1(x)$: $I_1(x) = \int \frac{x dx}{x^4 + 4}$. Let $u = x^2$, so $du = 2x dx$. $I_1(x) = \int \frac{1/2 du}{u^2 + 2^2} = \frac{1}{2} \cdot \frac{1}{2} \tan^{-1}\left(\frac{u}{2}\right) + C = \frac{1}{4} \tan^{-1}\left(\frac{x^2}{2}\right) + C$.

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

3. Evaluate the limit: We need to find $L = \lim_{x\to0} \frac{\cos(I_1(x))}{x^2}$. As $x \to 0$, $I_1(x) \to \frac{1}{4} \tan^{-1}(0) + \frac{\pi}{2} = \frac{\pi}{2}$. Thus, as $x \to 0$, $\cos(I_1(x)) \to \cos(\frac{\pi}{2}) = 0$. The limit is of the form $\frac{0}{0}$. Using L'Hopital's Rule: $L = \lim_{x\to0} \frac{-\sin(I_1(x)) \cdot I_1'(x)}{2x}$. From the definition, $I_1'(x) = \frac{x}{x^4 + 4}$. $L = \lim_{x\to0} \frac{-\sin(I_1(x)) \cdot \frac{x}{x^4 + 4}}{2x} = \lim_{x\to0} \frac{-\sin(I_1(x))}{2(x^4 + 4)}$. Substitute $x=0$: $L = \frac{-\sin(\pi/2)}{2(0^4+4)} = \frac{-1}{8}$.