IBDP Maths analysis and approaches Topic: SL 2.9 : The function ax and its graph. HL Paper 1

Detail Solution

(a)  Step 1: Define an even function.
A function \( f(x) \) is considered even if \( f(-x) = f(x) \) for all \( x \) in the domain of the function.

Step 2: Calculate \( f(-x) \).
We start by substituting \(-x\) into the function:
\[
f(-x) = \frac{\sin^2(k(-x))}{(-x)^2}
\]
Using the property of the sine function, \(\sin(-\theta) = -\sin(\theta)\), we have:
\[
f(-x) = \frac{\sin^2(-kx)}{x^2} = \frac{(-\sin(kx))^2}{x^2} = \frac{\sin^2(kx)}{x^2}
\]

Step 3: Compare \( f(-x) \) and \( f(x) \).
Now we compare \( f(-x) \) with \( f(x) \):
\[
f(x) = \frac{\sin^2(kx)}{x^2}
\]
Since both \( f(-x) \) and \( f(x) \) yield the same expression:
\[
f(-x) = f(x)
\]

Step 4: Conclude that \( f \) is an even function.
Since \( f(-x) = f(x) \) holds true for all \( x \neq 0 \), we conclude that the function \( f(x) = \frac{\sin^2(kx)}{x^2} \) is indeed an even function.

(b) 

 Step 1: Identify the limit of the function as \( x \) approaches 0.
We need to evaluate the limit:
\[
\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin^{2}(kx)}{x^{2}}.
\]

Step 2: Apply the limit property of sine.
Using the fact that \(\lim_{u \to 0} \frac{\sin u}{u} = 1\), we can rewrite the limit:
\[
\lim_{x \to 0} \frac{\sin^{2}(kx)}{x^{2}} = \lim_{x \to 0} \left(\frac{\sin(kx)}{kx}\right)^{2} \cdot \frac{k^{2}}{1} = k^{2} \cdot \lim_{x \to 0} \left(\frac{\sin(kx)}{kx}\right)^{2} = k^{2} \cdot 1 = k^{2}.
\]

Step 3: Set the limit equal to 16.
According to the problem, we have:
\[
k^{2} = 16.
\]

Step 4: Solve for \( k \).
Taking the square root of both sides gives:
\[
k = 4 \quad \text{(since \( k \in \mathbb{R}^+ \))}.
\]

————Markscheme—————–

(a) \( f(-x) = \frac{\sin^2(-kx)}{(-x)^2} = \frac{\sin^2(kx)}{x^2} = f(x) \), so \( f \) is even.
(b) Using L’Hôpital’s rule:
\( \lim_{x \to 0} \frac{\sin^2(kx)}{x^2} = \lim_{x \to 0} \frac{2k \sin(kx) \cos(kx)}{2x} = \lim_{x \to 0} \frac{k \sin(2kx)}{2x} = k^2 \)
Given \( k^2 = 16 \), so \( k = 4 \).