IB Mathematics AHL 2.14 Odd and even functions AA HL Paper 3

(a)
The first graph is

\(S_1(x)=\sin x\).

The second graph is

\(S_2(x)=\sin(\sin x)\).

Both functions have the same \(x\)-intercepts because

\(\sin(\sin x)=0\)

when \(\sin x=0\). Therefore, the intercepts are:

\((0,0)\), \((\pi,0)\) and \((2\pi,0)\).

The \(y\)-intercept of both graphs is also \((0,0)\).

The graph of \(S_2\) has a smaller positive maximum and a less negative minimum than the graph of \(S_1\):

\(\max S_2(x)=\sin1\approx0.841\)

and

\(\min S_2(x)=-\sin1\approx-0.841\).

✅ The two curves have intercepts at \(x=0\), \(x=\pi\) and \(x=2\pi\), with \(S_2\) lying closer to the \(x\)-axis than \(S_1\).

(b)(i)
The maximum value of \(\sin x\) is \(1\), occurring at \(x=\dfrac{\pi}{2}\).

✅ Answer: \(\max S_1(x)=1\)

(b)(ii)
Since the maximum value of \(\sin x\) is \(1\):

\(\max S_2(x)=\sin1\)

\(\sin1=0.841470\ldots\)

✅ Answer: \(\max S_2(x)=0.841\)

(b)(iii)
The maximum value of \(S_3(x)=\sin(S_2(x))\) is obtained by applying sine to the maximum value of \(S_2\).

\(\max S_3(x)=\sin(\sin1)\)

\(\sin(\sin1)=0.745624\ldots\)

✅ Answer: \(\max S_3(x)=0.746\)

(c)
Let \(M_n\) denote the maximum value of \(S_n(x)\).

Since sine is increasing for inputs between \(0\) and \(1\):

\(M_{n+1}=\sin(M_n)\), with \(M_1=1\).

The maximum for \(S_5\) is still greater than \(0.6\), while the maximum for \(S_6\) is less than \(0.6\).

✅ Answer: \(n=6\)

(d)
For \(y=S_2(x)=\sin(\sin x)\), use the chain rule.

\(\dfrac{dy}{dx}=\cos(\sin x)\cos x\)

For a point of zero gradient:

\(\cos(\sin x)\cos x=0\)

First, consider:

\(\cos x=0\)

For \(0\leq x\leq2\pi\), this gives:

\(x=\dfrac{\pi}{2}\) or \(x=\dfrac{3\pi}{2}\).

Now consider:

\(\cos(\sin x)=0\)

Cosine is zero when its input has the form:

\(\sin x=\dfrac{(2k+1)\pi}{2}\), where \(k\in\mathbb{Z}\).

However, \(-1\leq\sin x\leq1\), whereas every value \(\dfrac{(2k+1)\pi}{2}\) has magnitude at least \(\dfrac{\pi}{2}>1\).

Therefore, \(\cos(\sin x)=0\) has no real solution.

✅ Hence, there are exactly two points of zero gradient, at \(x=\dfrac{\pi}{2}\) and \(x=\dfrac{3\pi}{2}\).

(e)
Since

\(S_3(x)=\sin(S_2(x))\),

the chain rule gives:

\(S_3′(x)=\cos(S_2(x))S_2′(x)\).

Also:

\(S_2′(x)=\cos(S_1(x))\cos x\).

Therefore:

\(S_3′(x)=\cos(S_2(x))\cos(S_1(x))\cos x\)

\(S_3′(x)=\cos(\sin(\sin x))\cos(\sin x)\cos x\).

✅ Hence, the required result is shown.

(f)
Base case: Let \(n=1\).

\(S_1(x)=\sin x\)

so

\(S_1′(x)=\cos x\).

This agrees with the stated formula for \(n=1\), where the product contains only \(\cos x\).

Inductive hypothesis: Assume the statement is true for \(n=k\), where \(k\in\mathbb{Z}^{+}\). Therefore:

\(S_k'(x)=\cos(S_{k-1}(x))\cos(S_{k-2}(x))\ldots\cos(S_1(x))\cos x\).

Inductive step: Since

\(S_{k+1}(x)=\sin(S_k(x))\),

differentiating using the chain rule gives:

\(S_{k+1}'(x)=\cos(S_k(x))S_k'(x)\).

Using the inductive hypothesis:

\(S_{k+1}'(x)=\cos(S_k(x))\cos(S_{k-1}(x))\cos(S_{k-2}(x))\ldots\cos(S_1(x))\cos x\).

This is exactly the required formula for \(n=k+1\).

Therefore, the statement is true for \(n=1\), and if it is true for \(n=k\), it is also true for \(n=k+1\).

✅ Hence, by mathematical induction, \(S_n'(x)=\cos(S_{n-1}(x))\cos(S_{n-2}(x))\ldots\cos(S_1(x))\cos x\) for all \(n\in\mathbb{Z}^{+}\).

(g)
As \(x\to0\), \(S_n(x)\to0\). Therefore, the limit has the indeterminate form \(\dfrac{0}{0}\), so l’Hôpital’s rule can be applied.

\(\displaystyle\lim_{x\to0}\dfrac{S_n(x)}{x}=\lim_{x\to0}\dfrac{S_n'(x)}{1}\)

Using the result from part (f):

\(\displaystyle\lim_{x\to0}\dfrac{S_n(x)}{x}=\lim_{x\to0}\left[\cos(S_{n-1}(x))\cos(S_{n-2}(x))\ldots\cos(S_1(x))\cos x\right]\)

As \(x\to0\):

\(S_1(x)\to0,\ S_2(x)\to0,\ldots,\ S_{n-1}(x)\to0\).

Therefore, every cosine factor approaches \(\cos0=1\).

\(\displaystyle\lim_{x\to0}\dfrac{S_n(x)}{x}=1\times1\times\cdots\times1\)

✅ Hence, \(\displaystyle\lim_{x\to0}\dfrac{S_n(x)}{x}=1\) for all \(n\in\mathbb{Z}^{+}\).