Differential Calculus-1
Differential Calculus-1
DAILY_CHALLENGE
Grade 12
Question:
Let $f$ be a function defined on $(-\pi/2, \pi/2)$ as follows: $f(x) = \begin{cases} \frac{2^{[1/n]} - [x] - \frac{[x]}{[n2-1]}}{x\tan x} & x \neq 0 \\ k & x = 0 \end{cases}$. The value of $k$ so that $f$ is continuous at $x = 0$ is:
$\frac{1}{2}(ln2)^2 - \frac{1}{2}ln2 + 1$
$(ln2)^2 + \frac{1}{2}(ln2) + 1$
$(ln2)^2 + (ln2) + \frac{1}{2}$
$\frac{1}{2}(ln2)^2 + ln2 + \frac{1}{2}$
Step-by-Step Solution
Key Concept: Apply multi-variable Taylor expansion carefully and identify the leading order term in both numerator and denominator.
We find $k = \lim_{x \to 0^+} \frac{(2e^y)^x - x - x \ln 2 - 1}{x^2 \frac{\tan x}{x}}$. Using Taylor expansion of $(2e^x)^x = 1 + x \ln(2e) + \frac{x^2}{2!}(\ln 2e)^2 + \ldots$, the numerator becomes $x \ln(2e) + \frac{x^2}{2}(\ln 2e)^2 + \ldots - x - x\ln 2 = \frac{x^2}{2}(\ln 2)^2 + \ldots$. Thus $k = \frac{1}{2}(\ln 2 + 1)^2 = \frac{1}{2}(\ln 2)^2 + \ln 2 + \frac{1}{2}$.
Correct Answer: 4