Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade None
Question:
If $\int \frac{dx}{\sqrt{9x^2 + 4x + 6}}$ to evaluate $I$, one of the most proper substitution could be:
$\sqrt{9x^2 + 4x + 6} = u \pm 3x$
$\sqrt{9x^2 + 4x + 6} = 3u \pm x$
$x = \frac{1}{t}$
$9x^2 + 4x + 6 = \frac{1}{t}$
Step-by-Step Solution
Key Concept: Analyzing the limit condition forces constraints on the degree and leading coefficient of polynomial $f(x)$, uniquely determining it as $x^4$.
Since $\lim_{x \to 0}\left(1 + \frac{f(x)}{x^3}\right)^{1/x}$ exists, we must have $\lim_{x \to 0} \frac{f(x)}{x^3} = 0$, implying $f(x) = a_4x^4 + a_5x^5 + \cdots$ where $a_4 \neq 0$. Thus $f(x)$ is at least degree 4, so $f(x) = a_4x^4$. From $\lim_{x \to 0}\left(1 + \frac{f(x)}{x^3}\right)^{1/x} = e$, we get $a_4 = 1$, so $f(x) = x^4$. The area between $y = x^4$ and $x^2 + y^2 = 2$ is $2\int_0^1\left(\sqrt{2-x^2} - x^4\right)dx = \frac{\pi}{2} - \frac{3}{5}$.
Correct Answer: 1