Mathematics 9709/22 — October/November 2024

Walkthrough

We are asked to differentiate $f(x) = 4\sin^2 3x$ and evaluate the derivative at $x = \frac{\pi}{4}$.

The function $f(x) = 4\sin^2 3x$ is a composite function. Breaking it down: we first compute $3x$, then take the sine of that, then square the result, then multiply by 4. To differentiate a composite function, we use the chain rule, working from the outside in.

The outermost operation is multiplication by 4 (a constant, which just carries through). Inside that, we have the square of $\sin 3x$. The chain rule says: differentiate the square first (bringing down the power 2 and reducing it by 1), then multiply by the derivative of $\sin 3x$, which is $3\cos 3x$ (the derivative of sine is cosine, and the chain rule on $3x$ gives the factor of 3).

So: $f'(x) = 4 \cdot 2(\sin 3x) \cdot 3\cos 3x = 24\sin 3x \cos 3x$.

An equivalent form uses the double-angle identity $\sin 6x = 2\sin 3x \cos 3x$, giving $f'(x) = 12\sin 6x$. Both forms are accepted by the mark scheme.

Next, substitute $x = \frac{\pi}{4}$. The argument $3x$ becomes $\frac{3\pi}{4}$. We need $\sin\left(\frac{3\pi}{4}\right)$ and $\cos\left(\frac{3\pi}{4}\right)$. The angle $\frac{3\pi}{4}$ is in the second quadrant, where sine is positive and cosine is negative. The reference angle is $\frac{\pi}{4}$, so $\sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2}$ and $\cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}$.

Multiplying: $24 \cdot \frac{\sqrt{2}}{2} \cdot \left(-\frac{\sqrt{2}}{2}\right) = 24 \cdot \left(-\frac{2}{4}\right) = 24 \cdot \left(-\frac{1}{2}\right) = -12$.

Key Takeaways

• The chain rule is the fundamental tool for differentiating composite functions like $\sin^2 3x$. You must differentiate each "layer" in turn and multiply the results.
• The double-angle identity $\sin 2\theta = 2\sin\theta\cos\theta$ can simplify derivative expressions and is often useful for checking answers.
• Exact trigonometric values for standard angles (multiples of $\frac{\pi}{4}$, $\frac{\pi}{6}$, etc.) must be known. Remember the signs in each quadrant: in the second quadrant, sine is positive but cosine and tangent are negative.

Common Mistakes

• Forgetting the factor of 3 from differentiating $3x$ inside the sine. The derivative of $\sin 3x$ is $3\cos 3x$, not $\cos 3x$.
• Forgetting the factor of 2 from differentiating the square. The derivative of $(\sin 3x)^2$ is $2(\sin 3x) \cdot 3\cos 3x$.
• Sign errors: $\cos\left(\frac{3\pi}{4}\right)$ is negative, so the final answer is negative.
• Substituting $x = \frac{\pi}{4}$ into $3x$ incorrectly — the argument becomes $\frac{3\pi}{4}$, not $\frac{\pi}{4}$.

Things to Be Careful About

• The mark scheme requires showing the differentiation step (M1) before the final answer — an unsupported answer would not earn full marks.
• When using the alternative form $12\sin 6x$, at $x = \frac{\pi}{4}$ we get $12\sin\left(\frac{3\pi}{2}\right) = 12 \cdot (-1) = -12$, which matches.
• Make sure your calculator (if used) is in radian mode, since the question uses $\pi$.

Walkthrough

We need to find the indefinite integral of $f(x) = 4\sin^2 3x$.

The integrand $\sin^2 3x$ cannot be integrated directly in its current form. The standard technique is to rewrite it using a trigonometric identity. The double-angle identity for cosine, $\cos 2\theta = 1 - 2\sin^2\theta$, can be rearranged to give $\sin^2\theta = \frac{1 - \cos 2\theta}{2}$.

Applying this with $\theta = 3x$:
$4\sin^2 3x = 4 \cdot \frac{1 - \cos 6x}{2} = 2(1 - \cos 6x) = 2 - 2\cos 6x$.

Now the integrand is a sum of two terms, each of which can be integrated using standard results:

• $\int 2\,dx = 2x$
• $\int -2\cos 6x\,dx = -2 \cdot \frac{\sin 6x}{6} = -\frac{1}{3}\sin 6x$

The general rule used here is $\int \cos(ax+b)\,dx = \frac{1}{a}\sin(ax+b) + C$. Here $a = 6$, so we divide by 6.

Combining the two results (and adding the constant of integration):
$\int f(x)\,dx = 2x - \frac{1}{3}\sin 6x + C$.

Key Takeaways

• To integrate $\sin^2$ or $\cos^2$ terms, always use the double-angle identities to convert them into linear combinations of $\cos 2\theta$ or $\sin 2\theta$.
• When integrating $\cos(ax+b)$ or $\sin(ax+b)$, remember to divide by the coefficient $a$.
• Always include the constant of integration $C$ for indefinite integrals.

Common Mistakes

• Forgetting to divide by 6 when integrating $\cos 6x$. The integral of $\cos 6x$ is $\frac{1}{6}\sin 6x$, not $\sin 6x$.
• Incorrectly applying the identity: $4\sin^2 3x = 2 - 2\cos 6x$, not $4 - 4\cos 6x$. The factor of 4 divided by 2 gives 2.
• Omitting the constant of integration $C$ — this would lose a mark.
• Confusing the identity: $\sin^2\theta = \frac{1 - \cos 2\theta}{2}$, not $\frac{1 + \cos 2\theta}{2}$ (that's for $\cos^2\theta$).

Things to Be Careful About

• The mark scheme requires showing the identity step (M1) before integrating — you must demonstrate that you rewrote the integrand.
• The final answer can be written in equivalent forms, but $2x - \frac{1}{3}\sin 6x + C$ is the standard form.
• Check your answer by differentiating: $\frac{d}{dx}\left(2x - \frac{1}{3}\sin 6x\right) = 2 - 2\cos 6x$, which matches the rewritten integrand.

Walkthrough

Start by sketching the exponential graph. The basic curve $y = \mathrm{e}^{2x}$ has $y \to 0$ as $x \to -\infty$ and passes through $(0,1)$. Adding $1$ shifts it up, so the asymptote becomes $y = 1$ and the curve passes through $(0,2)$. It is always above $y=1$ and increases without bound as $x$ increases.

Next sketch the modulus graph. For $x \leq 4$, $|x-4| = 4-x$, a line of slope $-1$; for $x \geq 4$, $|x-4| = x-4$, a line of slope $1$. This gives a V shape with vertex $(4,0)$ and $y$-intercept $(0,4)$.

The exponential curve starts near $y=1$ on the left and rises; the modulus graph starts at $(0,4)$ and falls until $(4,0)$. Because the exponential value at $x=0$ is $2$ and the modulus value at $x=0$ is $4$, while at $x=4$ the exponential value is huge and the modulus value is $0$, the two curves cross exactly once between $x=0$ and $x=4$.

Key Takeaways

Recognise the standard shapes of exponential and modulus graphs. An exponential of the form $y = a + \mathrm{e}^{kx}$ has a horizontal asymptote at $y=a$. A modulus graph $y=|x-c|$ is V-shaped with vertex at $(c,0)$.

Common Mistakes

• Drawing the exponential graph below $y=1$ or without its horizontal asymptote.
• Drawing the modulus vertex at $x=-4$ instead of $x=4$.
• Showing more than one intersection, or placing the intersection on the wrong branch.

Things to Be Careful About

The exponential curve must stay above its asymptote $y=1$. The modulus graph must be V-shaped and symmetric about $x=4$, crossing the $y$-axis at $(0,4)$. The single intersection $P$ lies on the decreasing branch $y=4-x$.

Walkthrough

At the intersection point $P$, the two graphs have the same $y$-coordinate, so we equate their equations. The modulus graph $y=|x-4|$ has two branches: $y=4-x$ for $x<4$ and $y=x-4$ for $x>4$. From the sketch, $P$ lies between $x=0$ and $x=4$, so we use the left branch $y=4-x$.

This gives $1+\mathrm{e}^{2x}=4-x$. Subtract $1$ to isolate the exponential term: $\mathrm{e}^{2x}=3-x$. Taking natural logarithms uses the fact that $\ln$ and $\mathrm{e}$ are inverse functions, so $\ln(\mathrm{e}^{2x})=2x$. Finally divide by $2$ to obtain the required form.

Key Takeaways

To solve an equation involving a modulus, first decide which branch applies. To solve an exponential equation, isolate the exponential term before taking logarithms. The identity $\ln(\mathrm{e}^{u})=u$ is essential.

Common Mistakes

• Using $|x-4|=x-4$ instead of $4-x$ for the intersection.
• Taking logarithms before isolating the exponential term, e.g. writing $\ln(1+\mathrm{e}^{2x})=\ln(4-x)$ and not simplifying.
• Forgetting to divide by $2$ at the end.
• Not showing enough working for a 'show that' question.

Things to Be Careful About

Because this is an AG (answer given) question, every step must be shown clearly. The logarithm is only defined when $3-x>0$, which is true at the intersection. Use $\ln(\mathrm{e}^{2x})=2x$, not $\ln(\mathrm{e}^{2x})=\mathrm{e}^{2x}\ln 1$ or similar.

Walkthrough

The equation from part (b) is $x=\frac{1}{2}\ln(3-x)$. To use iteration, write it as $x_{n+1}=\frac{1}{2}\ln(3-x_n)$. Start with $x_0=0.45$ and substitute repeatedly.

For example, $x_1=\frac{1}{2}\ln(3-0.45)=\frac{1}{2}\ln 2.55=0.46805$ to 5 significant figures. Repeating gives the table of values. The values move from $0.46805$ down to about $0.46450$, then settle near $0.46508$. Since $x_6$ and $x_7$ are both $0.46508$ to 5 significant figures, the sequence has converged enough. Because $0.46508$ lies between $0.4645$ and $0.4655$, the root is $0.465$ correct to 3 significant figures.

Key Takeaways

An equation of the form $x=F(x)$ can be solved numerically by repeated substitution. Keep enough accuracy during iteration and round only when reporting. To justify a final answer to 3 significant figures, show iterations until consecutive values agree within the required interval.

Common Mistakes

• Using $x_{n+1}=\ln(3-x_n)$ without the factor $\frac{1}{2}$.
• Rounding each iteration to too few figures, which can change the convergence.
• Stopping after one or two iterations.
• Giving the final answer as $0.46508$ instead of rounding to 3 significant figures.
• Not showing enough iterations to justify the 3 significant figure answer.

Things to Be Careful About

Use the initial value $0.45$ exactly. Report each iteration to 5 significant figures as requested. The final answer must be to 3 significant figures, so $0.465$ is correct, not $0.46508$. The interval $[0.4645,0.4655]$ is the range of values that round to $0.465$.

Walkthrough

The factor theorem states that $(x - c)$ is a factor of a polynomial $p(x)$ if and only if $p(c) = 0$. Here the factor is $(x + 2)$, which we write as $(x - (-2))$, so $c = -2$ and we must have $p(-2) = 0$. Substituting $x = -2$ into $p(x) = ax^3 + bx^2 - ax + 8$ gives $-8a + 4b + 2a + 8 = -6a + 4b + 8$, and setting this equal to zero produces the first linear equation in $a$ and $b$.

The remainder theorem states that the remainder when $p(x)$ is divided by $(x - c)$ equals $p(c)$. Here we divide by $(x - 2)$, so the remainder is $p(2)$, which is given as 24. Substituting $x = 2$ gives $8a + 4b - 2a + 8 = 6a + 4b + 8$, and equating this to 24 gives the second equation $6a + 4b - 16 = 0$.

We now have two linear equations. Adding them eliminates $a$ because the coefficients of $a$ are $-6$ and $+6$; this directly yields $b = 1$. Substituting $b = 1$ back into either equation gives $a = 2$. These are the required values.

Key Takeaways

• The factor theorem: $(x - c)$ is a factor of $p(x)$ iff $p(c) = 0$.
• The remainder theorem: the remainder of $p(x) \div (x - c)$ is $p(c)$.
• Setting up and solving simultaneous linear equations from polynomial conditions.

Common Mistakes

• Sign errors when substituting $x = -2$: note $(-2)^3 = -8$, $(-2)^2 = 4$, and $-a(-2) = +2a$.
• Equating $p(2)$ to 0 instead of to the given remainder 24.
• Arithmetic slips when collecting like terms in each substitution.

Things to Be Careful About

• The mark scheme awards M1 for each substitution step, so show both $p(-2)$ and $p(2)$ explicitly.
• The A1 marks require the simplified equations $-6a + 4b + 8 = 0$ and $6a + 4b - 16 = 0$ (or equivalent).
• Check your final values by verifying both original conditions are satisfied.

Walkthrough

First substitute the found values $a = 2$, $b = 1$ into $p(x)$ to obtain the explicit cubic $2x^3 + x^2 - 2x + 8$. Since $(x + 2)$ is a known factor, we divide the cubic by it. The division is performed term by term: $2x^3 \div x = 2x^2$; multiplying back by $(x+2)$ gives $2x^3 + 4x^2$, which we subtract, leaving $-3x^2 - 2x + 8$. Then $-3x^2 \div x = -3x$, and subtracting $-3x^2 - 6x$ leaves $4x + 8$. Finally $4x \div x = 4$, and subtracting $4x + 8$ leaves remainder 0. The quotient is $2x^2 - 3x + 4$, confirming the factorisation $p(x) = (x+2)(2x^2 - 3x + 4)$.

To show there is exactly one real root, examine the quadratic factor. A quadratic $Ax^2 + Bx + C$ has real roots only when its discriminant $B^2 - 4AC \geq 0$. Here the discriminant is $(-3)^2 - 4(2)(4) = 9 - 32 = -23 < 0$, so $2x^2 - 3x + 4$ has no real roots. Therefore the only real root of $p(x) = 0$ comes from $x + 2 = 0$, i.e. $x = -2$. Hence the equation has exactly one real root.

Key Takeaways

• Polynomial division by a linear factor, using the known factor to simplify the cubic.
• The discriminant determines whether a quadratic has real roots.
• A cubic factorised as (linear)(quadratic) has real roots exactly where its factors do; if the quadratic has none, the cubic has exactly one.

Common Mistakes

• Sign errors when subtracting during polynomial division.
• Omitting the discriminant calculation and simply asserting the quadratic has no real roots.
• Forgetting to state that the root from the linear factor is $x = -2$.

Things to Be Careful About

• The mark scheme awards M1 for dividing at least as far as the $x$ term; show the division steps.
• A1 is for obtaining $(x+2)(2x^2 - 3x + 4)$.
• The final A1 requires reference to the root $-2$ AND the discriminant $-23$ with the conclusion that there is no further real root.