A curve has parametric equations $$ x = \tan\theta, \quad y = \sin\theta - 2\sin^3\theta, $$ for $0 < \theta < \frac{1}{2}\pi$. (a) Show that $\frac{dy}{dx} = 6\cos^5\theta - 5\cos^3\theta$. (b) Find the equation of the normal to the curve at the point where it crosses the $x$-axis. Give your answer in the form $y = mx + c$, where $m$ and $c$ are exact constants.

(a) dy/dx = 6 cos^5 θ - 5 cos^3 θ · (b) y = √2 x - √2

9709/25 Oct/Nov 2025 — Mathematics Solved Past Paper

Q13MMedium-EasyIntegration

Find $\int 6\sin^2 x\,dx$ .

DifficultyMedium-Easy

Worked solution

Approach

Use the double-angle identity to rewrite $\sin^2 x$ in terms of $\cos 2x$ , then integrate each term separately.

Working

\sin^2 x = \frac{1 - \cos 2x}{2}

Multiply by 6:

6\sin^2 x = 6 \cdot \frac{1 - \cos 2x}{2} = 3(1 - \cos 2x) = 3 - 3\cos 2x

Integrate term by term:

\int 6\sin^2 x\,dx = \int (3 - 3\cos 2x)\,dx = 3x - 3 \cdot \frac{\sin 2x}{2} + c

So:

\int 6\sin^2 x\,dx = 3x - \frac{3}{2}\sin 2x + c

Answer

3x - \frac{3}{2}\sin 2x + c

Final answer

3x - (3/2) sin 2x + c

Detailed explanation

Walkthrough

The integrand contains $\sin^2 x$ , which is not one of the basic integrals we know directly. We use the double-angle identity $\cos 2x = 1 - 2\sin^2 x$ to rewrite $\sin^2 x = \frac{1 - \cos 2x}{2}$ . This converts the integrand into a constant plus a cosine term, both of which we can integrate immediately.

Multiplying by 6 gives $6\sin^2 x = 3 - 3\cos 2x$ . Then integrating term by term:

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

Adding the constant of integration gives $3x - \frac{3}{2}\sin 2x + c$ .

Key Takeaways

This question tests the use of trigonometric identities to make an integrand integrable. The double-angle identity is essential for integrating $\sin^2 x$ or $\cos^2 x$ . It also tests the standard integral $\int \cos(ax+b)\,dx = \frac{1}{a}\sin(ax+b)+c$ .

Common Mistakes

Forgetting to divide by the coefficient of $x$ inside $\cos 2x$ when integrating, giving $3x - 3\sin 2x$ instead of $3x - \frac{3}{2}\sin 2x$ .
Using the wrong identity, e.g. writing $\sin^2 x = \frac{1+\cos 2x}{2}$ (that is for $\cos^2 x$ ).
Omitting the constant of integration; the mark scheme condones this, but it is safer to include it.

Things to Be Careful About

Ensure the identity is applied correctly: $\sin^2 x = \frac{1 - \cos 2x}{2}$ .
When integrating $\cos 2x$ , the factor $\frac{1}{2}$ must appear because of the chain rule in reverse.
The mark scheme allows omission of $+c$ , but including it is always correct.

Techniques used

apply the double-angle identity to rewrite sin^2 xintegrate term by term

Similar questions

Q23MMedium-EasyLogarithmic and Exponential Functions

Solve the equation $e^{2x}(e^{2x} - 8) = 48$ .

DifficultyMedium-Easy

Worked solution

Approach

Let $y = e^{2x}$ . Then the equation becomes a quadratic in $y$ . Solve the quadratic, use the fact that $e^{2x} > 0$ to keep the positive root, and take natural logarithms to find $x$ .

Working

Let $y = e^{2x}$ . Then

y(y-8) = 48

Expanding and rearranging to zero:

y^2 - 8y - 48 = 0

Factorising:

(y - 12)(y + 4) = 0

So $y = 12$ or $y = -4$ .

Since $y = e^{2x} > 0$ for all real $x$ , reject $y = -4$ . Therefore

e^{2x} = 12

Taking natural logarithms:

2x = \ln 12

Hence

x = \frac{1}{2}\ln 12

Numerically, $x \approx 1.24$ . There is no other real solution.

Answer

x = \frac{1}{2}\ln 12 \quad \text{or} \quad x \approx 1.24

Final answer

x = (1/2) ln 12 (≈ 1.24)

Detailed explanation

Walkthrough

We first notice that the variable $e^{2x}$ appears in both the factor outside and the bracket. Replacing it by a single letter, say $y = e^{2x}$ , turns the equation into a familiar quadratic. Expanding $y(y-8)=48$ gives $y^2 - 8y = 48$ , and moving everything to one side gives $y^2 - 8y - 48 = 0$ . This factorises as $(y-12)(y+4)=0$ , so the possible values are $y=12$ and $y=-4$ .

Now we return to the meaning of $y$ . Since $y = e^{2x}$ and an exponential function is always positive, $y=-4$ cannot occur for any real $x$ ; we ignore it. Thus $e^{2x}=12$ . Taking the natural logarithm of both sides uses the fact that $\ln(e^{2x}) = 2x$ , giving $2x = \ln 12$ . Dividing by 2 gives $x = \frac{1}{2}\ln 12 \approx 1.24$ . There is exactly one real solution.

Key Takeaways

Recognising a hidden quadratic by substitution is a powerful technique for equations of the form $A e^{kx} + B e^{2kx} + C = 0$ .
Exponentials are always positive, so only positive solutions of the quadratic can correspond to real values of $x$ .
To undo $e^{f(x)}$ , apply $\ln$ and use $\ln(e^{f(x)}) = f(x)$ .

Common Mistakes

Forgetting to reject $e^{2x} = -4$ and offering an impossible solution.
Factoring incorrectly because the equation was not rearranged to zero first.
Writing $x = \ln 12$ instead of $x = \frac{1}{2}\ln 12$ after $2x = \ln 12$ .
Stopping at $e^{2x} = 12$ without solving for $x$ , which would not earn the final mark.

Things to Be Careful About

The exponential function $e^{2x}$ is positive for all real $x$ ; this domain fact justifies rejecting the negative root.
Use the natural logarithm, not a logarithm of a different base, when solving equations involving $e$ .
The mark scheme requires an explicit final value of $x$ and no extra solutions; give both the exact form and, if useful, the decimal approximation.

Techniques used

substitute to form a quadratic equationfactorise a quadratic expressionuse positivity of exponentials to reject an invalid roottake natural logarithms to solve for the exponent

Similar questions

Q3MediumAlgebraTrigonometry

(a)

Solve the equation $|2x - 3| = |5x + 2|$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The equation $|2x - 3| = |5x + 2|$ is true when the expressions inside the modulus bars are equal, or when one is the negative of the other. Solve the two resulting linear equations.

Working

Case 1: $2x - 3 = 5x + 2$ .

2x - 3 = 5x + 2

Subtract $2x$ from both sides and subtract $2$ from both sides:

-3 - 2 = 5x - 2x

-5 = 3x

x = -\frac{5}{3}

Case 2: $2x - 3 = -(5x + 2)$ .

2x - 3 = -5x - 2

Add $5x$ to both sides and add $3$ to both sides:

7x = 1

x = \frac{1}{7}

Answer

x = -\frac{5}{3} \quad \text{or} \quad x = \frac{1}{7}

Final answer

x = -5/3 or x = 1/7

Detailed explanation

Walkthrough

We solve $|2x - 3| = |5x + 2|$ by removing the modulus signs. The absolute value of an expression equals the absolute value of another when the expressions are equal or when they are negatives of each other. This gives two linear equations.

For $2x - 3 = 5x + 2$ , rearranging gives $3x = -5$ , so $x = -\frac{5}{3}$ .

For $2x - 3 = -(5x + 2)$ , expanding the bracket gives $2x - 3 = -5x - 2$ , so $7x = 1$ and $x = \frac{1}{7}$ .

Both values satisfy the original equation, so the solution set is $x = -\frac{5}{3}$ or $x = \frac{1}{7}$ .

Key Takeaways

This question tests the ability to solve a modulus equation by considering the two possible sign cases. The key skill is recognising that $|A| = |B|$ is equivalent to $A = B$ or $A = -B$ .

Common Mistakes

A common mistake is to only solve $2x - 3 = 5x + 2$ and forget the case where the two sides have opposite signs, losing the solution $x = \frac{1}{7}$ . Another common mistake is to mishandle the negative sign when expanding $-(5x + 2)$ .

Things to Be Careful About

When using the case method, always solve both cases and check that each answer satisfies the original equation. If using the squaring method, remember that squaring can introduce extra solutions, so verify each result.

Techniques used

remove modulus signs by considering equal and opposite casessolve the resulting linear equations

(b)

Hence solve the equation $|2\sec\theta - 3| = |5\sec\theta + 2|$ for $\pi < \theta < 2\pi$ . Give your answer correct to 3 significant figures.

3M Similar

DifficultyMedium

Worked solution

Approach

Use the result of part (a) by substituting $x = \sec\theta$ . Solve the two possibilities for $\sec\theta$ , reject any impossible value, then convert to $\cos\theta$ and find the angle in the interval $\pi < \theta < 2\pi$ .

Working

From part (a), $|2x - 3| = |5x + 2|$ gives

x = -\frac{5}{3} \quad \text{or} \quad x = \frac{1}{7}

Putting $x = \sec\theta$ :

\sec\theta = -\frac{5}{3} \quad \text{or} \quad \sec\theta = \frac{1}{7}

Since $|\sec\theta| \geq 1$ , the value $\sec\theta = \frac{1}{7}$ is impossible. Therefore

\sec\theta = -\frac{5}{3}

Taking reciprocals:

\cos\theta = -\frac{3}{5}

Let the reference angle be $\alpha = \cos^{-1}\left(\frac{3}{5}\right)$ . In the interval $\pi < \theta < 2\pi$ , $\cos\theta$ is negative in the third quadrant, so

\theta = \pi + \alpha = \pi + \cos^{-1}\left(\frac{3}{5}\right)

\theta = 4.0688\ldots

Answer

\theta = 4.07 \text{ radians (3 s.f.)}

Final answer

theta = 4.07 radians (3 s.f.)

Detailed explanation

Walkthrough

Part (b) is designed to be solved using the answer to part (a). If we let $x = \sec\theta$ , then the equation $|2\sec\theta - 3| = |5\sec\theta + 2|$ has exactly the same form as $|2x - 3| = |5x + 2|$ . Therefore $\sec\theta$ must be one of the two values found in part (a): $-\frac{5}{3}$ or $\frac{1}{7}$ .

The value $\sec\theta = \frac{1}{7}$ is impossible because $\sec\theta = \frac{1}{\cos\theta}$ and $|\cos\theta| \leq 1$ , so $|\sec\theta| \geq 1$ . Hence only $\sec\theta = -\frac{5}{3}$ is possible.

Taking reciprocals gives $\cos\theta = -\frac{3}{5}$ . We need the angle in the interval $\pi < \theta < 2\pi$ where cosine is negative. In this interval, cosine is negative only in the third quadrant. The reference angle is $\alpha = \cos^{-1}\left(\frac{3}{5}\right) \approx 0.9273$ radians, so the third-quadrant angle is $\theta = \pi + \alpha \approx 4.0689$ radians, which is $4.07$ to 3 significant figures.

Key Takeaways

This question links a modulus equation with trigonometric functions. It shows how a substitution such as $x = \sec\theta$ can reuse an earlier algebraic result. It also reinforces the range of $\sec\theta$ and the importance of choosing the correct quadrant when solving $\cos\theta = k$ .

Common Mistakes

A common mistake is to keep $\sec\theta = \frac{1}{7}$ and try to solve $\cos\theta = 7$ , which has no solution. Another common mistake is to choose the second-quadrant angle $\pi - \alpha$ instead of the third-quadrant angle $\pi + \alpha$ , forgetting the interval $\pi < \theta < 2\pi$ . Some students also give the answer in degrees or fail to round to 3 significant figures.

Things to Be Careful About

The mark scheme allows the working to be done in degrees for the angle-finding method, but the final answer must be in radians because the interval is given in radians. Always check that the final angle lies in the required interval. Also remember that $\sec\theta$ cannot take values between $-1$ and $1$ , so one of the part (a) values must be rejected.

Techniques used

substitute sec theta for x using part (a)reject impossible secant value using its rangefind reference angle and identify the correct quadrant

Q45MMediumTrigonometry

Solve the equation $\cot\theta\tan(\theta + 45^\circ) = 7$ for $0^\circ < \theta < 90^\circ$ .

DifficultyMedium

Worked solution

Approach

Let $t = \tan\theta$ . Rewrite $\cot\theta$ as $\frac{1}{\tan\theta}$ and use the compound angle formula for $\tan(\theta + 45^\circ)$ . This gives an equation in $t$ only, which simplifies to a quadratic. Solve the quadratic, then find the angles in the given range.

Working

Let $t = \tan\theta$ . Since $\tan 45^\circ = 1$ , the compound angle formula gives

\tan(\theta + 45^\circ) = \frac{\tan\theta + \tan 45^\circ}{1 - \tan\theta\tan 45^\circ} = \frac{t + 1}{1 - t}

Also, $\cot\theta = \frac{1}{\tan\theta} = \frac{1}{t}$ . Substituting into the equation:

\frac{1}{t} \cdot \frac{t + 1}{1 - t} = 7

Multiplying by $t(1-t)$ :

t + 1 = 7t(1 - t) = 7t - 7t^2

Rearranging:

7t^2 - 6t + 1 = 0

Using the quadratic formula:

t = \frac{6 \pm \sqrt{36 - 28}}{14} = \frac{6 \pm 2\sqrt{2}}{14} = \frac{3 \pm \sqrt{2}}{7}

So

\tan\theta = \frac{3 - \sqrt{2}}{7} \quad \text{or} \quad \tan\theta = \frac{3 + \sqrt{2}}{7}

Both values are positive and less than $1$ , so both give angles in $0^\circ < \theta < 45^\circ$ , which is inside the required range. Therefore

\theta = 12.764\ldots^\circ \quad \text{or} \quad \theta = 32.235\ldots^\circ

There are no other solutions in $0^\circ < \theta < 90^\circ$ .

Answer

\theta = 12.8^\circ \text{ or } 32.2^\circ \quad (12.764\ldots^\circ, 32.235\ldots^\circ)

Final answer

θ = 12.8° or 32.2° (12.764...°, 32.235...°)

Detailed explanation

Walkthrough

We need to solve an equation that mixes $\cot\theta$ and $\tan(\theta + 45^\circ)$ . The key is to express everything in terms of one trigonometric ratio, $\tan\theta$ . Let $t = \tan\theta$ .

First, $\cot\theta = \frac{1}{\tan\theta} = \frac{1}{t}$ . This is the definition of cotangent.

Next, use the compound angle formula for tangent:

\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}

With $A = \theta$ and $B = 45^\circ$ , and knowing $\tan 45^\circ = 1$ , this becomes

\tan(\theta + 45^\circ) = \frac{t + 1}{1 - t}

Substitute both expressions into the original equation:

\frac{1}{t} \cdot \frac{t + 1}{1 - t} = 7

Multiplying by $t(1-t)$ removes the denominators. It is important to multiply correctly: $t+1 = 7t(1-t)$ . Expanding gives $t+1 = 7t - 7t^2$ , and rearranging gives

7t^2 - 6t + 1 = 0

Now solve the quadratic in $t$ . Using the quadratic formula,

t = \frac{6 \pm \sqrt{(-6)^2 - 4(7)(1)}}{2(7)} = \frac{6 \pm \sqrt{8}}{14} = \frac{3 \pm \sqrt{2}}{7}

Both roots are positive and less than $1$ , so both are valid for $\theta$ in the range $0^\circ < \theta < 90^\circ$ ; in fact both lie in $0^\circ < \theta < 45^\circ$ . Taking inverse tangent gives the two final angles.

Key Takeaways

Cotangent can always be rewritten as $\frac{1}{\tan\theta}$ .
The compound angle formula for tangent is essential when an angle like $\theta + 45^\circ$ appears.
After substitution, many trigonometric equations reduce to a quadratic in $\tan\theta$ .
Always check that each solution lies in the given interval and that no denominator is zero at the solution.

Common Mistakes

Forgetting that $\tan 45^\circ = 1$ and therefore miswriting the compound angle formula.
Making a sign error when expanding $7t(1-t)$ or rearranging the quadratic.
Stopping after solving for $\tan\theta$ instead of finding $\theta$ .
Omitting one of the two solutions because the quadratic has two positive roots.
Not showing the initial attempt to express the equation in terms of $\tan\theta$ only; the mark scheme requires this method mark.

Things to Be Careful About

The denominator $1 - t$ must not be zero; $t=1$ would make $\tan(\theta+45^\circ)$ undefined, but neither solution is $t=1$ .
Since the range is $0^\circ < \theta < 90^\circ$ , both positive roots are valid. If the range were different, extra solutions from adding $180^\circ$ might also need consideration.
Use degrees mode on your calculator when finding $\theta$ from $\tan\theta$ .
Give final answers to the accuracy requested; the mark scheme accepts $12.8$ and $32.2$ , or greater accuracy.

Techniques used

express cotangent in terms of tangentapply the compound angle formula for tangentform and solve a quadratic in tan thetatake inverse tangent and verify the range

Similar questions

Q5MediumLogarithmic and Exponential FunctionsIntegration

Fig. 1

The diagram shows the curve with equation $y = 8e^{-\frac{1}{2}x} - 1$ . The curve meets the axes at the points $A$ and $B$ . The shaded region is bounded by the curve and the line segment $AB$ .

(a)

Show that the $x$ -coordinate of $B$ is $6\ln 2$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Point $B$ lies on the $x$ -axis, so its $y$ -coordinate is $0$ . Setting $y = 0$ in $y = 8e^{-\frac{1}{2}x} - 1$ gives an exponential equation which we solve for $x$ using logarithm laws.

Working

8e^{-\frac{1}{2}x} - 1 = 0

8e^{-\frac{1}{2}x} = 1

e^{-\frac{1}{2}x} = \frac{1}{8}

Take the natural logarithm of both sides:

-\frac{1}{2}x = \ln\frac{1}{8}

-\frac{1}{2}x = -\ln 8

Using $\ln 8 = \ln 2^3 = 3\ln 2$ :

-\frac{1}{2}x = -3\ln 2

x = 6\ln 2

Answer

x = 6\ln 2

Final answer

x = 6 ln 2

Detailed explanation

Walkthrough

Since point $B$ is where the curve meets the $x$ -axis, its $y$ -coordinate must be $0$ . We substitute $y = 0$ into the equation of the curve to get an equation purely in $x$ .

After moving the $-1$ to the right-hand side, we divide by $8$ to isolate the exponential term $e^{-x/2}$ . At this point, the equation is in the form $e^{kx} = c$ where $c$ is a constant, so we take the natural logarithm of both sides. The key identity is $\ln(e^u) = u$ , which lets us remove the exponential on the left.

The result $-\frac{1}{2}x = \ln\frac{1}{8} = -\ln 8$ simplifies to $\frac{1}{2}x = \ln 8$ . We use the logarithm power law $\ln(a^n) = n\ln a$ to write $\ln 8 = \ln 2^3 = 3\ln 2$ . Multiplying both sides by $2$ gives the required $x = 6\ln 2$ .

Key Takeaways

A point on the $x$ -axis has $y = 0$ , so substituting $y = 0$ into the curve equation gives the $x$ -intercept.
Equations of the form $e^{kx} = c$ (with $c > 0$ ) are solved by taking natural logarithms.
The logarithm power law $\ln(a^n) = n\ln a$ is used to express $\ln 8$ in terms of $\ln 2$ .

Common Mistakes

Forgetting to apply the negative sign when moving $-\frac{1}{2}x$ across the equals sign.
Writing $\ln\frac{1}{8}$ as $\frac{1}{\ln 8}$ instead of $-\ln 8$ (a misuse of logarithm laws).
Forgetting to multiply by $2$ at the final step, leaving the answer as $3\ln 2$ rather than $6\ln 2$ .

Things to Be Careful About

The question is a "show that", so every step must be clearly visible. In particular, the mark scheme requires explicit use of $\ln 8 = 3\ln 2$ for the A1 mark.
The exponential $e^{-x/2}$ and the negative coefficient both matter; a sign error will give an incorrect final sign.

Techniques used

set y equal to zeroisolate the exponentialapply natural logarithmsimplify using logarithm laws

(b)

Find the area of the shaded region. Give your answer in the form $p\ln 2 - q$ , where $p$ and $q$ are positive integers.

5M Similar

DifficultyMedium

Worked solution

Approach

First identify the coordinates of $A$ and $B$ from the curve and the axes. The shaded region lies between the curve and the line segment $AB$ , so its area equals the area of triangle $OAB$ minus the area under the curve from $0$ to $6\ln 2$ .

Working

Coordinates of $A$ and $B$ :

At $A$ (where the curve meets the $y$ -axis), $x = 0$ :

y = 8e^{0} - 1 = 8 - 1 = 7

So $A = (0, 7)$ .

From part (a), $B = (6\ln 2, 0)$ .

Area of triangle $OAB$ :

\text{Area}(\triangle OAB) = \frac{1}{2} \times 6\ln 2 \times 7 = 21\ln 2

Area under the curve from $0$ to $6\ln 2$ :

Integrate term by term:

\int \left(8e^{-\frac{1}{2}x} - 1\right) dx = -16e^{-\frac{1}{2}x} - x + C

Apply the limits:

\begin{aligned} \left[\,-16e^{-\frac{1}{2}x} - x\,\right]_0^{6\ln 2} &= \left(-16e^{-\frac{1}{2}\cdot 6\ln 2} - 6\ln 2\right) - \left(-16e^{0} - 0\right) \\ &= \left(-16e^{-3\ln 2} - 6\ln 2\right) - (-16) \\ &= -16 \cdot e^{\ln(1/8)} - 6\ln 2 + 16 \\ &= -16 \cdot \frac{1}{8} - 6\ln 2 + 16 \\ &= -2 - 6\ln 2 + 16 \\ &= 14 - 6\ln 2 \end{aligned}

Shaded area:

\begin{aligned} \text{Shaded area} &= 21\ln 2 - (14 - 6\ln 2) \\ &= 21\ln 2 - 14 + 6\ln 2 \\ &= 27\ln 2 - 14 \end{aligned}

Answer

27\ln 2 - 14

Final answer

27 ln 2 - 14

Detailed explanation

Walkthrough

The shaded region is bounded above by the line segment $AB$ and below by the curve. A useful technique when the bounding line is straight is to find the area of the larger triangle (here $\triangle OAB$ ) and subtract the area under the curve.

First we identify point $A$ by setting $x = 0$ in the curve equation, which gives $y = 7$ . The point $B$ is the $x$ -intercept, already found in part (a) to be $6\ln 2$ . With $O$ at the origin, triangle $OAB$ has legs of length $6\ln 2$ (along the $x$ -axis) and $7$ (along the $y$ -axis).

To integrate the curve, recall the rule $\int e^{ax}\,dx = \frac{1}{a}e^{ax}$ . With $a = -\frac{1}{2}$ and coefficient $8$ , the antiderivative of $8e^{-x/2}$ is $8 \cdot \frac{1}{-1/2}e^{-x/2} = -16e^{-x/2}$ . The antiderivative of $-1$ is $-x$ .

When evaluating at $x = 6\ln 2$ , the term $e^{-\frac{1}{2}\cdot 6\ln 2} = e^{-3\ln 2} = e^{\ln 2^{-3}} = 2^{-3} = \frac{1}{8}$ . This is the key step that produces a clean numerical answer.

At $x = 0$ , the antiderivative is $-16 \cdot 1 - 0 = -16$ , so subtracting this value adds $16$ to the result.

Finally, the area under the curve $14 - 6\ln 2$ is subtracted from the triangle's area $21\ln 2$ to give the shaded region. Combining the $\ln 2$ terms gives $21\ln 2 + 6\ln 2 = 27\ln 2$ , and the constant becomes $-14$ .

Key Takeaways

The area between a straight line and a curve can be found by subtracting the area under the curve from the area of the corresponding triangle.
For an integrand of the form $e^{ax}$ , the antiderivative is $\frac{1}{a}e^{ax}$ , including the sign of $a$ .
The relationship $e^{n\ln a} = a^n$ is essential for evaluating such integrals at limits that involve logarithms.

Common Mistakes

A common error is integrating $8e^{-x/2}$ as $8e^{-x/2}$ itself, forgetting the $\frac{1}{a} = -2$ factor.
Forgetting the negative sign on the constant term $-1$ when integrating; its antiderivative is $-x$ , not $x$ .
Sign errors when applying the limits of a definite integral: the lower-limit value is subtracted from the upper-limit value.
Computing $e^{-3\ln 2}$ incorrectly; this equals $\frac{1}{8}$ , not $8$ .

Things to Be Careful About

The mark scheme awards a mark for attempting the area of the triangle minus the area under the curve; it is acceptable to use a decimal approximation for the triangle area ( $14.56$ ) only at this step, and the $21\ln 2$ form should be used in the final answer.
The final answer is of the form $p\ln 2 - q$ ; ensure both $p$ and $q$ are positive integers ( $p = 27$ , $q = 14$ ).

Techniques used

find the y-intercept of the curveuse the x-intercept from part (a)integrate the exponential termevaluate the definite integralcompute the triangle areasubtract to obtain the shaded area

Q6MediumDifferentiation

A curve has parametric equations

x = \tan\theta, \quad y = \sin\theta - 2\sin^3\theta,

for $0 < \theta < \frac{1}{2}\pi$ .

(a)

Show that $\frac{dy}{dx} = 6\cos^5\theta - 5\cos^3\theta$ .

4M Similar

DifficultyMedium

Worked solution

Approach

For parametric equations, use

\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}.

Differentiate $x = \tan\theta$ and $y = \sin\theta - 2\sin^3\theta$ with respect to $\theta$ , then divide and simplify using $\sec^2\theta = \frac{1}{\cos^2\theta}$ and $\sin^2\theta + \cos^2\theta = 1$ .

Working

\frac{dx}{d\theta} = \sec^2\theta

Differentiate $y$ term by term. The derivative of $\sin\theta$ is $\cos\theta$ , and by the chain rule the derivative of $-2\sin^3\theta$ is $-6\sin^2\theta\cos\theta$ .

\frac{dy}{d\theta} = \cos\theta - 6\sin^2\theta\cos\theta

Therefore

\frac{dy}{dx} = \frac{\cos\theta - 6\sin^2\theta\cos\theta}{\sec^2\theta} = \left(\cos\theta - 6\sin^2\theta\cos\theta\right)\cos^2\theta

= \cos^3\theta - 6\sin^2\theta\cos^3\theta = \cos^3\theta\left(1 - 6\sin^2\theta\right)

Using $\sin^2\theta = 1 - \cos^2\theta$ :

\frac{dy}{dx} = \cos^3\theta\left(1 - 6\left(1 - \cos^2\theta\right)\right) = \cos^3\theta\left(6\cos^2\theta - 5\right)

= 6\cos^5\theta - 5\cos^3\theta

Answer

\frac{dy}{dx} = 6\cos^5\theta - 5\cos^3\theta

Final answer

dy/dx = 6 cos^5 θ - 5 cos^3 θ

Detailed explanation

Walkthrough

We are given parametric equations, so the slope of the curve is found by dividing the derivative of $y$ with respect to the parameter by the derivative of $x$ with respect to the parameter.

First differentiate $x = \tan\theta$ : the derivative is $\sec^2\theta$ . Then differentiate $y = \sin\theta - 2\sin^3\theta$ . The first term gives $\cos\theta$ . For the second term, use the chain rule: the derivative of $\sin^3\theta$ is $3\sin^2\theta\cos\theta$ , so the derivative of $-2\sin^3\theta$ is $-6\sin^2\theta\cos\theta$ .

Now divide $\frac{dy}{d\theta}$ by $\frac{dx}{d\theta}$ . Since $\frac{dx}{d\theta} = \sec^2\theta = \frac{1}{\cos^2\theta}$ , dividing by $\sec^2\theta$ is the same as multiplying by $\cos^2\theta$ . This makes the expression depend only on $\cos\theta$ after using $\sin^2\theta = 1 - \cos^2\theta$ . Expand and simplify to obtain the required result.

Key Takeaways

For parametric curves, $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ .
The chain rule is needed for powers of trigonometric functions.
The identity $\sin^2\theta + \cos^2\theta = 1$ lets us rewrite everything in terms of $\cos\theta$ .
When a question says "show that", every simplification step must be shown.

Common Mistakes

Forgetting to divide by $\frac{dx}{d\theta}$ , or dividing by it incorrectly.
Differentiating $\sin^3\theta$ as $3\sin^2\theta$ without multiplying by $\cos\theta$ .
Stopping before fully simplifying to the target expression.
The mark scheme says AG - necessary detail needed, so an unsupported answer is not enough.

Things to Be Careful About

Keep the parameter $\theta$ in radians throughout.
Remember $\sec^2\theta = \frac{1}{\cos^2\theta}$ , so dividing by $\sec^2\theta$ multiplies by $\cos^2\theta$ .
Watch the sign when expanding $-6(1 - \cos^2\theta)$ : it becomes $6\cos^2\theta - 6$ , so $1 - 6(1 - \cos^2\theta) = 6\cos^2\theta - 5$ .

Techniques used

differentiate parametric equations with respect to the parameterapply the chain rule to a trigonometric powerdivide derivatives to obtain dy/dxsimplify using the Pythagorean identity

(b)

Find the equation of the normal to the curve at the point where it crosses the $x$ -axis. Give your answer in the form $y = mx + c$ , where $m$ and $c$ are exact constants.

5M Similar

DifficultyMedium

Worked solution

Approach

The curve crosses the $x$ -axis where $y = 0$ . Find the corresponding value of $\theta$ , then evaluate $\frac{dy}{dx}$ at that point using part (a). The normal has gradient equal to the negative reciprocal of the tangent gradient, and passes through the crossing point.

Working

Set $y = 0$ :

\sin\theta - 2\sin^3\theta = 0

\sin\theta\left(1 - 2\sin^2\theta\right) = 0

Since $0 < \theta < \frac{\pi}{2}$ , $\sin\theta \neq 0$ , so

1 - 2\sin^2\theta = 0 \quad \Rightarrow \quad \sin^2\theta = \frac{1}{2}

Thus

\theta = \frac{\pi}{4}

and the crossing point is

x = \tan\frac{\pi}{4} = 1, \quad y = 0

From part (a), the gradient of the tangent at $\theta = \frac{\pi}{4}$ is

\frac{dy}{dx} = 6\cos^5\frac{\pi}{4} - 5\cos^3\frac{\pi}{4}

Since $\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$ ,

\frac{dy}{dx} = 6\left(\frac{1}{\sqrt{2}}\right)^5 - 5\left(\frac{1}{\sqrt{2}}\right)^3 = \frac{6}{4\sqrt{2}} - \frac{5}{2\sqrt{2}} = -\frac{1}{\sqrt{2}}

So the gradient of the normal is the negative reciprocal:

m = \sqrt{2}

The normal passes through $(1, 0)$ , so its equation is

y - 0 = \sqrt{2}(x - 1)

y = \sqrt{2}x - \sqrt{2}

Answer

y = \sqrt{2}x - \sqrt{2}

Final answer

y = √2 x - √2

Detailed explanation

Walkthrough

The curve crosses the $x$ -axis where $y = 0$ . Substitute $y = \sin\theta - 2\sin^3\theta$ and factor. Since $\theta$ is in the first quadrant, $\sin\theta$ is not zero, so the factor $1 - 2\sin^2\theta$ must be zero. This gives $\sin^2\theta = \frac{1}{2}$ , and because $\theta$ is acute, $\theta = \frac{\pi}{4}$ .

At that point, $x = \tan\frac{\pi}{4} = 1$ , so the point is $(1, 0)$ . Use the derivative from part (a) to find the tangent gradient. Substitute $\theta = \frac{\pi}{4}$ : $\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$ , so the gradient is $-\frac{1}{\sqrt{2}}$ .

The normal is perpendicular to the tangent, so its gradient is the negative reciprocal: $m = \sqrt{2}$ . Using the point $(1, 0)$ , the equation is $y = \sqrt{2}(x - 1)$ , which simplifies to $y = \sqrt{2}x - \sqrt{2}$ .

Key Takeaways

A curve crosses the $x$ -axis where $y = 0$ .
The normal gradient is the negative reciprocal of the tangent gradient.
To find the equation of a line, use the point-gradient form $y - y_1 = m(x - x_1)$ .

Common Mistakes

Forgetting to find the $x$ -coordinate of the crossing point before writing the normal equation.
Using the tangent gradient instead of its negative reciprocal for the normal.
Answering with $\theta$ in degrees; the mark scheme requires radians here.
Not simplifying the gradient expression correctly when substituting $\cos\frac{\pi}{4}$ .

Things to Be Careful About

The domain $0 < \theta < \frac{\pi}{2}$ means $\sin\theta > 0$ , so the solution $\theta = \frac{\pi}{4}$ is unique; other solutions such as $\frac{3\pi}{4}$ are outside the domain.
The final answer must have exact constants; leave $\sqrt{2}$ rather than a decimal.
The normal passes through the point where the curve crosses the $x$ -axis, not through the origin.

Techniques used

solve a trigonometric equation for the parameterevaluate the parametric derivative at a pointtake the negative reciprocal for the normal gradientwrite the equation of a line through a point

Q7MediumAlgebraNumerical Solution of Equations

The polynomial $p(x)$ is defined by

p(x) = 2x^4 + kx^3 + kx^2 + 17x + 18,

where $k$ is a constant. It is given that $(x + 2)$ is a factor of $p(x)$ .

(a)

Find the value of $k$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the factor theorem: since $(x+2)$ is a factor, $p(-2)=0$ . Substitute $x=-2$ into $p(x)$ , set the result equal to zero and solve for $k$ .

Working

Substitute $x=-2$ :

p(-2) = 2(-2)^4 + k(-2)^3 + k(-2)^2 + 17(-2) + 18

Simplify:

p(-2) = 32 - 8k + 4k - 34 + 18

p(-2) = 16 - 4k

Since $(x+2)$ is a factor, $p(-2)=0$ :

16 - 4k = 0

k = 4

Answer

k = 4

Final answer

k = 4

Detailed explanation

Walkthrough

Because $(x+2)$ is a factor of $p(x)$ , the factor theorem says $x=-2$ is a root, so $p(-2)=0$ . Substitute $x=-2$ into the polynomial. Each term simplifies: $2(-2)^4=32$ , $k(-2)^3=-8k$ , $k(-2)^2=4k$ , $17(-2)=-34$ , and the constant is $18$ . Adding gives $16-4k$ . Setting this equal to zero gives $k=4$ .

Key Takeaways

The factor theorem links a linear factor to a root: if $(x+2)$ is a factor, then substituting $x=-2$ must give zero. This turns a polynomial condition into an equation for an unknown constant.

Common Mistakes

Forgetting to apply the factor theorem and trying to divide first.
Sign errors when substituting negative values, especially $(-2)^3=-8$ and $(-2)^4=16$ .
Arithmetic slips when combining $32$ , $-34$ and $18$ .

Things to Be Careful About

Powers of negative numbers: even powers are positive, odd powers are negative.
The factor theorem requires the polynomial to equal zero at $x=-2$ , not at $x=2$ .

Techniques used

apply the factor theoremsubstitute a root into the polynomialsolve a linear equation for a constant

(b)

It is given that the equation $p(x) = 0$ has exactly two real roots, denoted by $\alpha$ and $\beta$ , where $\alpha$ is an integer and $\beta$ is not an integer.

State the value of $\alpha$ and show that $\beta$ satisfies the equation $x = \sqrt[3]{-2x - 4.5}$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Use the factor theorem to identify the integer root $\alpha=-2$ . Then divide $p(x)$ by $x+2$ to obtain the cubic factor, set it equal to zero and rearrange to the required form.

Working

With $k=4$ :

p(x)=2x^4+4x^3+4x^2+17x+18

Since $(x+2)$ is a factor, $x=-2$ is a root, so

\alpha = -2

Divide $2x^4+4x^3+4x^2+17x+18$ by $x+2$ :

p(x) = (x+2)(2x^3+4x+9)

Therefore $p(x)=0$ gives $x=-2$ or

2x^3+4x+9=0

The non-integer real root $\beta$ satisfies this cubic. Rearrange:

2x^3 = -4x - 9

x^3 = -2x - 4.5

x = \sqrt[3]{-2x - 4.5}

Answer

\alpha = -2, \quad \beta \text{ satisfies } x = \sqrt[3]{-2x - 4.5}

Final answer

alpha = -2 and beta satisfies x = cube root(-2x - 4.5)

Detailed explanation

Walkthrough

With $k=4$ , the polynomial is $p(x)=2x^4+4x^3+4x^2+17x+18$ . Since $(x+2)$ is a factor, one real root is $x=-2$ , so $\alpha=-2$ . Divide $p(x)$ by $x+2$ . Synthetic division with $-2$ gives quotient $2x^3+4x+9$ and remainder $0$ , so $p(x)=(x+2)(2x^3+4x+9)$ . The other real root $\beta$ must satisfy $2x^3+4x+9=0$ . Rearrange: $2x^3=-4x-9$ , divide by 2, then take cube roots: $x=\sqrt[3]{-2x-4.5}$ .

Key Takeaways

Polynomial division reduces a quartic to a cubic once a linear factor is known. The remaining roots come from the quotient. Rearranging a cubic into the form $x=F(x)$ is the basis for an iterative formula.

Common Mistakes

Forgetting to include the zero $x^2$ term when dividing; the quotient must be $2x^3+0x^2+4x+9$ .
Stopping after finding $\alpha=-2$ and not deriving the cubic equation for $\beta$ .
Rearranging incorrectly: from $2x^3+4x+9=0$ , divide by 2 before taking cube roots.

Things to Be Careful About

The problem says the equation has exactly two real roots; $\alpha$ is the integer root and $\beta$ is the non-integer root of the cubic.
The equation $x=\sqrt[3]{-2x-4.5}$ is equivalent to the cubic only for real cube roots.

Techniques used

apply the factor theorem to identify a rootdivide a polynomial by a linear factorrearrange a cubic equation to an iterative form

(c)

Show by calculation that $-1.4 < \beta < -1.0$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Define $h(x)=x-\sqrt[3]{-2x-4.5}$ . Since $\beta$ satisfies $x=\sqrt[3]{-2x-4.5}$ , $\beta$ is a root of $h$ . Evaluate $h$ at $x=-1.4$ and $x=-1.0$ and look for a sign change.

Working

At $x=-1.4$ :

\sqrt[3]{-2(-1.4)-4.5} = \sqrt[3]{-1.7} \approx -1.1935

h(-1.4) = -1.4 - (-1.1935) = -0.2065 < 0

At $x=-1.0$ :

\sqrt[3]{-2(-1.0)-4.5} = \sqrt[3]{-2.5} \approx -1.3572

h(-1.0) = -1.0 - (-1.3572) = 0.3572 > 0

Since $h$ is continuous and changes sign between $-1.4$ and $-1.0$ , its root $\beta$ lies in that interval.

Answer

-1.4 < \beta < -1.0

Final answer

-1.4 < beta < -1.0

Detailed explanation

Walkthrough

Since $\beta$ satisfies $x=\sqrt[3]{-2x-4.5}$ , define $h(x)=x-\sqrt[3]{-2x-4.5}$ . Then $\beta$ is a zero of $h$ . Evaluate $h$ at the two endpoints. For $x=-1.4$ , the cube root is $\sqrt[3]{-2(-1.4)-4.5}=\sqrt[3]{-1.7}\approx -1.1935$ , so $h(-1.4)\approx -0.2065<0$ . For $x=-1.0$ , $\sqrt[3]{-2(-1)-4.5}=\sqrt[3]{-2.5}\approx -1.3572$ , so $h(-1.0)\approx 0.3572>0$ . A continuous function changing sign between $-1.4$ and $-1.0$ has a root in that interval.

Key Takeaways

A sign change over an interval locates a root of a continuous equation. The function chosen can be the rearranged equation $x-F(x)=0$ or the original cubic.

Common Mistakes

Evaluating only $x$ or only the cube root instead of the difference.
Using the wrong interval endpoints or substituting $x=1.4$ instead of $-1.4$ .
Not giving numerical values, so no sign change is demonstrated.

Things to Be Careful About

Cube roots of negative numbers are negative.
Need both endpoint values with opposite signs and a conclusion that a root lies between them.

Techniques used

evaluate a function at interval endpointsdetect a sign change to locate a root

(d)

Use an iterative formula, based on the equation in part (b), to find the value of $\beta$ correct to 3 significant figures. Give the result of each iteration to 5 significant figures.

3M Similar

DifficultyMedium

Worked solution

Approach

Use the iterative formula $x_{n+1}=\sqrt[3]{-2x_n-4.5}$ with starting value $x_0=-1.2$ . Iterate, recording each value to 5 significant figures, until consecutive values agree to that precision.

Working

x_{n+1} = \sqrt[3]{-2x_n - 4.5}, \quad x_0 = -1.2

$n$	$x_n$ (5 s.f.)
0	-1.2000
1	-1.2806
2	-1.2469
3	-1.2612
4	-1.2552
5	-1.2577
6	-1.2567
7	-1.2571
8	-1.2569
9	-1.2570
10	-1.2570

Since $x_9$ and $x_{10}$ are both $-1.2570$ to 5 significant figures, the iteration has converged. Thus

\beta \approx -1.2570

Correct to 3 significant figures:

\beta = -1.26

Answer

\beta = -1.26 \text{ (3 s.f.)}

Final answer

beta = -1.26

Detailed explanation

Walkthrough

Use the iterative formula from part (b): $x_{n+1}=\sqrt[3]{-2x_n-4.5}$ . Starting from $x_0=-1.2$ , each iteration produces a new approximation. The table shows values rounded to 5 significant figures. After several iterations the values settle at $-1.2570$ to 5 significant figures. Therefore $\beta\approx -1.2570$ , which is $-1.26$ correct to 3 significant figures.

Key Takeaways

Iterative formulae of the form $x_{n+1}=F(x_n)$ can be used to refine approximations. Convergence is shown when consecutive values agree to the required precision. Intermediate values should be kept to more significant figures than the final answer.

Common Mistakes

Rounding every intermediate value too early, causing the iteration to drift.
Giving the final answer as $-1.257$ instead of $-1.26$ (correct to 3 significant figures).
Stopping after one or two iterations without showing convergence.

Things to Be Careful About

The question asks for each iteration to 5 significant figures, so display intermediate values to 5 s.f.
The final answer must be to exactly 3 significant figures: $-1.26$ .
Use full precision in the calculation and round only when recording values.

Techniques used

iterate a rearrangement of the equationround intermediate values to a required precisionidentify convergence from repeated values

Q86MMediumDifferentiation

The equation of a curve is $y = 4e^{1-2x}\sqrt{3x-1}$ .

Find the exact coordinates of the stationary point of the curve.

DifficultyMedium

Worked solution

Approach

Use the product rule to differentiate $y = 4e^{1-2x}(3x-1)^{\frac{1}{2}}$ . Set $\frac{dy}{dx}=0$ , divide out the non-zero exponential factor, solve for $x$ , then substitute back to find the exact $y$ -coordinate.

Working

Let

u = 4e^{1-2x}, \quad v = (3x-1)^{\frac{1}{2}}

Then

\frac{du}{dx} = -8e^{1-2x}, \qquad \frac{dv}{dx} = \frac{3}{2}(3x-1)^{-\frac{1}{2}}

By the product rule,

\frac{dy}{dx} = \frac{du}{dx}v + u\frac{dv}{dx}

\frac{dy}{dx} = -8e^{1-2x}(3x-1)^{\frac{1}{2}} + 4e^{1-2x}\cdot \frac{3}{2}(3x-1)^{-\frac{1}{2}}

\frac{dy}{dx} = -8e^{1-2x}(3x-1)^{\frac{1}{2}} + 6e^{1-2x}(3x-1)^{-\frac{1}{2}}

At a stationary point, $\frac{dy}{dx} = 0$ :

-8e^{1-2x}(3x-1)^{\frac{1}{2}} + 6e^{1-2x}(3x-1)^{-\frac{1}{2}} = 0

Since $e^{1-2x} \neq 0$ , divide by it:

-8(3x-1)^{\frac{1}{2}} + 6(3x-1)^{-\frac{1}{2}} = 0

Multiply through by $(3x-1)^{\frac{1}{2}}$ :

-8(3x-1) + 6 = 0

3x - 1 = \frac{6}{8} = \frac{3}{4}

x = \frac{1 + \frac{3}{4}}{3} = \frac{7}{12}

Substitute into $y$ :

y = 4e^{1 - 2\cdot \frac{7}{12}}\sqrt{3\cdot\frac{7}{12} - 1}

= 4e^{-\frac{1}{6}}\sqrt{\frac{3}{4}}

= 4e^{-\frac{1}{6}}\cdot \frac{\sqrt{3}}{2} = 2\sqrt{3}e^{-\frac{1}{6}}

Answer

\left(\frac{7}{12},\ 2\sqrt{3}e^{-\frac{1}{6}}\right)

Final answer

x = 7/12, y = 2√3 e^(-1/6)

Detailed explanation

Walkthrough

This question asks for the stationary point, so the key step is to find where the gradient is zero. The curve is a product of two functions, $4e^{1-2x}$ and $(3x-1)^{\frac{1}{2}}$ , so we need the product rule. Each factor also needs the chain rule.

First write the square root as a power: $\sqrt{3x-1} = (3x-1)^{\frac{1}{2}}$ . Let $u = 4e^{1-2x}$ and $v = (3x-1)^{\frac{1}{2}}$ . The derivative of $u$ is $4e^{1-2x}\cdot(-2) = -8e^{1-2x}$ , because the derivative of $1-2x$ is $-2$ . The derivative of $v$ is $\frac{1}{2}(3x-1)^{-\frac{1}{2}}\cdot 3 = \frac{3}{2}(3x-1)^{-\frac{1}{2}}$ , because the derivative of $3x-1$ is $3$ .

Then apply the product rule: $\frac{dy}{dx} = u'v + uv'$ . This gives the two terms shown in the working. At a stationary point the gradient is zero, so set the whole expression equal to zero.

Because $e^{1-2x}$ is never zero, it can be divided out safely. This leaves an equation with powers of $3x-1$ . Multiplying by $(3x-1)^{\frac{1}{2}}$ clears the negative fractional power and gives a simple linear equation, $3x-1 = \frac{3}{4}$ . Solving gives $x = \frac{7}{12}$ .

Finally substitute this value back into the original equation for $y$ . The exponent simplifies to $-\frac{1}{6}$ , and the square root becomes $\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$ . Multiplying by $4$ gives $2\sqrt{3}e^{-\frac{1}{6}}$ .

Key Takeaways

The product rule is essential when a function is written as a product of two simpler functions.
Composite functions such as $e^{1-2x}$ and $(3x-1)^{\frac{1}{2}}$ require the chain rule.
The exponential function is never zero, so it can be cancelled from equations safely.
Stationary points are found by solving $\frac{dy}{dx}=0$ , then substituting to find the corresponding $y$ -coordinate.
Exact answers should be left in simplified surd/exponential form.

Common Mistakes

Forgetting the inner derivative when differentiating $e^{1-2x}$ or $(3x-1)^{\frac{1}{2}}$ .
Mixing up the order or signs in the product rule.
Attempting to divide by $(3x-1)^{-\frac{1}{2}}$ without handling the fractional power correctly.
Finding $x$ but failing to substitute back to obtain the $y$ -coordinate.
The mark scheme requires the derivative to be written in the form $k_1 e^{1-2x}(3x-1)^{\frac{1}{2}} + k_2 e^{1-2x}(3x-1)^{-\frac{1}{2}}$ ; omitting either term loses credit.

Things to Be Careful About

The domain is $x \geq \frac{1}{3}$ because of the square root. The stationary point $x = \frac{7}{12}$ lies in this domain, so it is valid.
The factor $(3x-1)^{\frac{1}{2}}$ is positive for $x > \frac{1}{3}$ , so multiplying the equation by it does not change the solution set at the stationary point.
Simplify $\sqrt{\frac{3}{4}}$ correctly to $\frac{\sqrt{3}}{2}$ , not $\frac{3}{2}$ or $\sqrt{3}$ .
Keep the answer exact: do not replace $e^{-\frac{1}{6}}$ or $\sqrt{3}$ with decimals unless asked.
Check that the exponent $1 - 2x$ is computed correctly: at $x = \frac{7}{12}$ , it is $1 - \frac{7}{6} = -\frac{1}{6}$ .

Techniques used

apply the product ruledifferentiate composite exponential and radical functionsset the derivative equal to zero and solve for xsubstitute back to find the exact y-coordinate

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