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 = 6cos^5(theta) - 5cos^3(theta) · (b) y = sqrt(2)x - sqrt(2)

9709/23 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 $\sin^2 x = \frac{1 - \cos 2x}{2}$ to rewrite the integrand as a sum of a constant and a cosine term, then integrate term by term.

Working

\int 6\sin^2 x\,dx = \int 6\left(\frac{1 - \cos 2x}{2}\right)dx

Simplify the integrand:

6\left(\frac{1 - \cos 2x}{2}\right) = 3 - 3\cos 2x

So

\int 6\sin^2 x\,dx = \int (3 - 3\cos 2x)\,dx

Integrate term by term:

\int 3\,dx = 3x

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

Therefore

\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

This integral looks like it needs a trigonometric identity because $\sin^2 x$ is not one of the basic integrands we know directly. The double-angle formula $\cos 2x = 1 - 2\sin^2 x$ can be rearranged to give $\sin^2 x = \frac{1 - \cos 2x}{2}$ . Substituting this into the integral turns $6\sin^2 x$ into $3 - 3\cos 2x$ , which is much easier to integrate.

Now integrate each term separately. The constant $3$ integrates to $3x$ . For $-3\cos 2x$ , recall that $\int \cos(ax+b)\,dx = \frac{1}{a}\sin(ax+b)+c$ , so with $a=2$ the integral is $-3\cdot \frac{1}{2}\sin 2x = -\frac{3}{2}\sin 2x$ . Adding the two results gives $3x - \frac{3}{2}\sin 2x + c$ .

Key Takeaways

The key idea is to rewrite unfamiliar powers of trigonometric functions using identities so that the integrand becomes a sum of terms with known antiderivatives. This question also reinforces the general rule $\int \cos(ax+b)\,dx = \frac{1}{a}\sin(ax+b)+c$ , where the factor $\frac{1}{a}$ appears because of the chain rule in reverse.

Common Mistakes

Using the wrong sign in the identity: $\sin^2 x = \frac{1+\cos 2x}{2}$ instead of $\frac{1-\cos 2x}{2}$ .
Forgetting the factor $\frac{1}{2}$ when substituting the identity, which would give $6\sin^2 x = 6 - 6\cos 2x$ incorrectly.
When integrating $-3\cos 2x$ , forgetting to divide by $2$ , giving $-3\sin 2x$ instead of $-\frac{3}{2}\sin 2x$ .
Omitting the constant of integration. The mark scheme condones its omission, but it is safer to always include $+c$ .

Things to Be Careful About

The coefficient $6$ must be multiplied by the $\frac{1}{2}$ from the identity, giving $3$ . Also, the derivative of $\sin 2x$ is $2\cos 2x$ , so the antiderivative of $\cos 2x$ must include the factor $\frac{1}{2}$ . Keep the constant of integration unless the question asks for a definite integral.

Techniques used

use the double-angle identity to rewrite sin^2 xintegrate term by termintegrate cos(2x) by dividing by the coefficient of x

Similar questions

Q23MMedium-EasyLogarithmic and Exponential Functions

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

DifficultyMedium-Easy

Worked solution

Approach

Let $y=e^{2x}$ so the equation becomes a quadratic in $y$ . Solve it by factorisation, then use the fact that $e^{2x}>0$ to discard the negative root and take natural logarithms to find $x$ .

Working

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

y(y-8)=48

Expanding:

y^2 - 8y = 48

So:

y^2 - 8y - 48 = 0

Factorising:

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

Hence:

y=12 \quad \text{or} \quad y=-4

Since $e^{2x}>0$ for all real $x$ , the root $y=-4$ is impossible. Therefore:

e^{2x}=12

Taking natural logarithms:

2x = \ln 12

So:

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

Answer

x = \frac{1}{2}\ln 12 \quad (\approx 1.24)

There is no other solution, because $e^{2x}=-4$ is impossible.

Final answer

x = (1/2) ln 12 ≈ 1.24, no other solution

Detailed explanation

Walkthrough

The equation contains the same exponential expression $e^{2x}$ twice, so it is natural to treat it as a single variable. Let $y=e^{2x}$ . Then the product $e^{2x}(e^{2x}-8)$ becomes $y(y-8)$ , and the equation is $y(y-8)=48$ . Expanding gives $y^2-8y=48$ , so $y^2-8y-48=0$ . This quadratic factorises as $(y-12)(y+4)=0$ , giving $y=12$ or $y=-4$ .

The key step is to remember that $e^{2x}$ is always positive for real $x$ . Therefore $y=-4$ cannot occur, and the only valid possibility is $y=12$ , i.e. $e^{2x}=12$ . To undo the exponential, take the natural logarithm of both sides: $2x=\ln 12$ . Then divide by 2 to get $x=\frac{1}{2}\ln 12$ , which is approximately 1.24.

Key Takeaways

This question tests the ability to recognise a quadratic structure inside an exponential equation, to solve that quadratic, and to use the positivity of the exponential function to reject an extraneous root. It also reinforces that $\ln(e^u)=u$ for any real $u$ , so taking natural logarithms is the correct inverse operation to solve for the exponent.

Common Mistakes

Forgetting to reject $e^{2x}=-4$ ; since $e^{2x}$ is never negative, this root must be discarded.
Stopping at $e^{2x}=12$ without taking logarithms to find $x$ .
Making a sign error when rearranging to $y^2-8y-48=0$ .
Giving only the decimal answer without the exact form, or giving an incorrect approximation.

Things to Be Careful About

The natural logarithm is only defined for positive inputs, so the negative root must be removed before taking logs. Use $\ln 12$ as the exact answer, or give the decimal to an appropriate degree of accuracy, such as 1.24. There is no other solution, so do not include a second value for $x$ .

Techniques used

substitute a variable for the exponential termfactorise the resulting quadraticdiscard the negative root using positivity of the exponential functiontake natural logarithms and solve for x

Similar questions

Q3MediumAlgebraTrigonometry

(a)

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

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

For an equation of the form $|A|=|B|$ , either $A=B$ or $A=-B$ . Split the modulus equation into these two linear cases and solve each.

Working

2x-3=5x+2

-3x=5

x=-\frac{5}{3}

Now take the case where the two sides have opposite signs:

2x-3=-(5x+2)

2x-3=-5x-2

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 are solving $|2x-3|=|5x+2|$ . The absolute value of two quantities is equal when the quantities themselves are equal, or when one is the negative of the other. This gives two separate linear equations.

The first equation is $2x-3=5x+2$ . Rearranging gives $-3x=5$ , so $x=-\frac{5}{3}$ .

The second equation is $2x-3=-(5x+2)$ . Expanding the bracket gives $2x-3=-5x-2$ . Adding $5x$ to both sides gives $7x-3=-2$ , then adding $3$ gives $7x=1$ , so $x=\frac{1}{7}$ .

Both values can be checked by substituting back into the original modulus equation; each satisfies it.

Key Takeaways

A modulus equation of the form $|A|=|B|$ is equivalent to $A=B$ or $A=-B$ . This often produces two solutions. The method avoids considering separate sign cases for $x$ because both sides are already non-negative after applying the modulus.

Common Mistakes

Forgetting the second case $A=-B$ and giving only one solution.
Making a sign error when expanding $-(5x+2)$ .
If using the squaring method, failing to expand $(2x-3)^2=(5x+2)^2$ correctly or not showing the resulting quadratic equation.

The mark scheme requires an attempt at both cases: the first solution can be awarded directly, but the second requires showing the different-sign equation.

Things to Be Careful About

When solving $2x-3=-(5x+2)$ , the negative sign applies to both terms inside the bracket. Always check the final answers in the original equation. If using the alternative squaring method, the full quadratic solution must be shown.

Techniques used

split a modulus equation into two linear casessolve linear equationscheck candidate solutions

(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) with $x=\sec\theta$ . The solutions are $\sec\theta=-\frac{5}{3}$ or $\sec\theta=\frac{1}{7}$ . The second is impossible because it would require $\cos\theta=7$ , so only $\sec\theta=-\frac{5}{3}$ remains. Convert this to $\cos\theta=-\frac{3}{5}$ and find the angle in the interval $\pi<\theta<2\pi$ .

Working

From part (a), with $x=\sec\theta$ :

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

Since $\sec\theta=\frac{1}{7}$ would mean $\cos\theta=7$ , which is impossible, discard it.

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

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

For $\pi<\theta<2\pi$ , cosine is negative only in the third quadrant. Let $\alpha=\cos^{-1}\left(\frac{3}{5}\right)$ .

\alpha=0.927295\ldots

Thus

\theta=\pi+\alpha=4.068887\ldots

Answer

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

Final answer

θ = 4.07 rad (3 s.f.)

Detailed explanation

Walkthrough

The word 'Hence' tells us to reuse the result of 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 part (a). Therefore $\sec\theta$ must be one of the two values found in part (a).

The value $\sec\theta=\frac{1}{7}$ is impossible because $\sec\theta=\frac{1}{\cos\theta}$ , so it would mean $\cos\theta=7$ . Since $\cos\theta$ is always between $-1$ and $1$ , this cannot happen. So only $\sec\theta=-\frac{5}{3}$ remains.

Taking reciprocals gives $\cos\theta=-\frac{3}{5}$ . We need the solution in $\pi<\theta<2\pi$ . In this interval, cosine is negative only in the third quadrant, between $\pi$ and $\frac{3\pi}{2}$ . The reference angle is $\alpha=\cos^{-1}\left(\frac{3}{5}\right)\approx0.9273$ radians, so the third-quadrant angle is $\theta=\pi+\alpha\approx4.0689$ radians, which is $4.07$ to 3 significant figures.

Key Takeaways

This question combines modulus equations with reciprocal trigonometric functions. It shows how a result obtained for a general variable $x$ can be transferred to $\sec\theta$ . It also reinforces that $|\sec\theta|\ge1$ , so $\sec\theta=\frac{1}{7}$ is immediately invalid, and that solving $\cos\theta=c$ requires choosing the correct quadrant for the given interval.

Common Mistakes

Keeping $\sec\theta=\frac{1}{7}$ as a valid possibility instead of discarding it.
Choosing the wrong quadrant: $2\pi-\cos^{-1}\left(\frac{3}{5}\right)$ would give a positive cosine and is not a solution.
Giving the answer in degrees. The interval is given in radians, so the final answer should be in radians.
Rounding incorrectly: $4.0689$ rounds to $4.07$ , not $4.06$ or $4.08$ .

Things to Be Careful About

The mark scheme allows working in degrees for the method mark, but the final answer must be correct in radians to 3 significant figures. Remember that $\sec\theta=\frac{1}{\cos\theta}$ , and that $\cos\theta$ cannot equal $7$ . Within $\pi<\theta<2\pi$ , the third quadrant is the only place where $\cos\theta$ is negative.

Techniques used

substitute a previous result into a new variablediscard an impossible reciprocal valueidentify the correct quadrant for a trigonometric equationcompute a reference angle

Q45MMediumTrigonometry

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

DifficultyMedium

Worked solution

Approach

Let $t = \tan\theta$ . Rewrite $\cot\theta$ as $\frac{1}{\tan\theta}$ and use the compound angle formula for $\tan(\theta + 45^\circ)$ to express the equation in terms of $t$ . Then solve the resulting quadratic equation and take inverse tangent, keeping only values in $0^\circ < \theta < 90^\circ$ .

Working

Let $t = \tan\theta$ . Since $\cot\theta = \frac{1}{\tan\theta}$ ,

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

Since $\tan 45^\circ = 1$ ,

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

Multiply both sides by $t(1-t)$ :

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

Rearrange:

7t^2 - 6t + 1 = 0

Solve 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}

Thus

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

Taking inverse tangent:

\theta = \tan^{-1}\left(\frac{3 + \sqrt{2}}{7}\right) \approx 32.235^\circ

\theta = \tan^{-1}\left(\frac{3 - \sqrt{2}}{7}\right) \approx 12.764^\circ

Both values lie in $0^\circ < \theta < 90^\circ$ , and there are no others in the range.

Answer

\theta = 12.8^\circ \text{ or } 32.2^\circ

Final answer

θ = 12.8° and 32.2°

Detailed explanation

Walkthrough

The equation mixes $\cot\theta$ and $\tan(\theta + 45^\circ)$ . The key is to write everything in terms of $\tan\theta$ . Let $t = \tan\theta$ . Since $\cot\theta = \frac{1}{\tan\theta}$ , replace it by $\frac{1}{t}$ . For the compound angle, use

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

With $A = \theta$ , $B = 45^\circ$ , and $\tan 45^\circ = 1$ , this gives $\frac{t+1}{1-t}$ . Substituting gives $\frac{1}{t} \cdot \frac{t+1}{1-t} = 7$ .

Now multiply both sides by $t(1-t)$ . This is valid because $0^\circ < \theta < 90^\circ$ means $t > 0$ , and the roots will not make $t = 1$ . This gives $t + 1 = 7t - 7t^2$ , so $7t^2 - 6t + 1 = 0$ .

Apply the quadratic formula:

t = \frac{6 \pm \sqrt{36 - 28}}{14} = \frac{6 \pm \sqrt{8}}{14} = \frac{3 \pm \sqrt{2}}{7}.

Since both values are positive, each gives exactly one angle in the first quadrant. Taking inverse tangent gives approximately $32.235^\circ$ and $12.764^\circ$ , both in the required range $0^\circ < \theta < 90^\circ$ .

Key Takeaways

$\cot\theta = \frac{1}{\tan\theta}$ .
Compound angle formula for tangent: $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ .
A trigonometric equation can often be reduced to a quadratic by substituting $t = \tan\theta$ .
After solving for $\tan\theta$ , use inverse tangent and check the domain.

Common Mistakes

Using the wrong sign in the denominator of the tangent compound angle formula; it is $1 - \tan A \tan B$ , not $1 + \tan A \tan B$ .
Forgetting that $\tan 45^\circ = 1$ .
Making a sign error when rearranging $t + 1 = 7t - 7t^2$ ; the correct quadratic is $7t^2 - 6t + 1 = 0$ .
Stopping at values of $\tan\theta$ without converting to angles, or giving angles outside the specified range.
The mark scheme requires a method and a three-term quadratic; an unsupported answer is not enough.

Things to Be Careful About

Work in degrees throughout, since the angle is given as $45^\circ$ and the range is in degrees.
Check that multiplying by $t(1-t)$ is legitimate: in this domain $t \neq 0$ and the roots are not $t = 1$ , so no solution is lost or introduced.
The tangent function has period $180^\circ$ , so each value of $\tan\theta$ would normally give infinitely many angles; only the two in $0^\circ < \theta < 90^\circ$ are required.
Give final answers to an appropriate accuracy; the mark scheme accepts $12.8^\circ$ and $32.2^\circ$ , or greater accuracy $12.764\ldots^\circ$ and $32.235\ldots^\circ$ .

Techniques used

rewrite cotangent as reciprocal of tangentapply compound angle formula for tangentform and solve a quadratic equation in tan thetatake inverse tangent and check 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 $B$ is where the curve meets the $x$ -axis, so its $y$ -coordinate is $0$ . Set $y = 0$ in the equation of the curve and solve the resulting exponential equation for $x$ , using the laws of logarithms.

Working

Set $y = 0$ in $y = 8e^{-\frac{1}{2}x} - 1$ :

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

Isolate the exponential term:

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

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

Take natural logarithms of both sides:

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

So $\frac{1}{2}x = \ln 8$ , giving $x = 2\ln 8$ .

Using the law of logarithms $\ln 8 = \ln 2^3 = 3\ln 2$ :

x = 2 \times 3\ln 2 = 6\ln 2

Answer

x = 6\ln 2

Final answer

x = 6 ln 2

Detailed explanation

Walkthrough

Point $B$ lies on the $x$ -axis, so its $y$ -coordinate is $0$ . We substitute $y = 0$ into the equation $y = 8e^{-\frac{1}{2}x} - 1$ to get an equation in $x$ alone. Rearranging gives $e^{-\frac{1}{2}x} = \frac{1}{8}$ . Because the variable $x$ is in the exponent, we take logarithms of both sides to bring it down. The rule $\ln(e^k) = k$ turns $-\frac{1}{2}x$ into a linear expression. The negative sign on both sides cancels, leaving $\frac{1}{2}x = \ln 8$ . Multiplying by $2$ gives $x = 2\ln 8$ , and applying the power law for logarithms $\ln 8 = \ln 2^3 = 3\ln 2$ yields the final form $x = 6\ln 2$ .

Key Takeaways

The $x$ -intercept of a curve is found by setting $y = 0$ and solving for $x$ .
To solve an equation of the form $e^{ax + b} = c$ , take natural logarithms.
The power law $\ln(k^n) = n\ln k$ is essential for simplifying.

Common Mistakes

Forgetting to take the negative of $\ln\frac{1}{8}$ (i.e. writing $\frac{1}{2}x = \ln 8$ but not handling the sign correctly).
Not explicitly showing the use of $\ln 8 = 3\ln 2$ — the mark scheme requires this detail to award the final mark.

Things to Be Careful About

The mark scheme insists on seeing the use of a logarithm property such as $\ln 8 = 3\ln 2$ or $64 = 2^6$ before awarding the conclusion mark.

Techniques used

set y = 0 to find x-interceptapply logarithm law to solve for xsimplify using ln 8 = 3 ln 2

(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 find the $y$ -coordinate of $A$ (where $x = 0$ ), so we know the triangle $OAB$ whose area can be computed. The shaded region is the area of triangle $OAB$ minus the area under the curve from $0$ to $6\ln 2$ . Integrate the curve, evaluate at the limits, and combine with the triangle area.

Working

Step 1: Find point $A$ .
Set $x = 0$ in $y = 8e^{-\frac{1}{2}x} - 1$ :

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

So $A = (0, 7)$ and from part (a), $B = (6\ln 2, 0)$ .

Step 2: Area of triangle $OAB$ .
With $O = (0,0)$ , $A = (0,7)$ , $B = (6\ln 2, 0)$ :

\text{Area}_{OAB} = \frac{1}{2} \times 6\ln 2 \times 7 = 21\ln 2

Step 3: Integrate the curve.

\int (8e^{-\frac{1}{2}x} - 1)\, dx = \frac{8}{-\frac{1}{2}}e^{-\frac{1}{2}x} - x = -16e^{-\frac{1}{2}x} - x

Step 4: Evaluate between $0$ and $6\ln 2$ .
At $x = 6\ln 2$ :

-16e^{-\frac{1}{2}(6\ln 2)} - 6\ln 2 = -16e^{-3\ln 2} - 6\ln 2 = -16 \cdot 2^{-3} - 6\ln 2 = -16 \cdot \frac{1}{8} - 6\ln 2 = -2 - 6\ln 2

At $x = 0$ :

-16e^{0} - 0 = -16

Area under the curve:

(-2 - 6\ln 2) - (-16) = 14 - 6\ln 2

Step 5: Shaded area = triangle $OAB$ − area under curve.

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

Answer

27\ln 2 - 14

Final answer

27 ln 2 - 14

Detailed explanation

Walkthrough

We start by reading point $A$ off the graph: it is the $y$ -intercept, obtained by setting $x = 0$ , which gives $y = 7$ . Together with $B = (6\ln 2, 0)$ and the origin $O$ , triangle $OAB$ is a right triangle with legs of length $7$ and $6\ln 2$ , so its area is $\frac{1}{2} \times 6\ln 2 \times 7 = 21\ln 2$ .

The shaded region sits between the straight line $AB$ and the curve. Because the curve is convex, it dips below the line, and the shaded region equals the area of triangle $OAB$ minus the area under the curve from $0$ to $6\ln 2$ .

To find the area under the curve we integrate $8e^{-\frac{1}{2}x} - 1$ . The antiderivative of $8e^{-\frac{1}{2}x}$ is obtained by dividing the coefficient by $-\frac{1}{2}$ , giving $-16e^{-\frac{1}{2}x}$ , and the antiderivative of $-1$ is $-x$ . So the antiderivative is $-16e^{-\frac{1}{2}x} - x$ .

Evaluating at $x = 6\ln 2$ uses $e^{-3\ln 2} = e^{\ln 2^{-3}} = 2^{-3} = \frac{1}{8}$ , which simplifies the exponential to $-2$ . Evaluating at $x = 0$ gives $-16$ . Subtracting gives $14 - 6\ln 2$ for the area under the curve.

Finally, shaded area $= 21\ln 2 - (14 - 6\ln 2) = 27\ln 2 - 14$ .

Key Takeaways

The area between a curve and a chord is often computed as (area of a convenient triangle or trapezoid) − (area under the curve).
When integrating $e^{ax}$ , divide the coefficient by $a$ in the antiderivative.
When evaluating at a logarithmic limit, use the rule $e^{n\ln k} = k^n$ .

Common Mistakes

Sign error when integrating $-1$ (forgetting the $-x$ term) or when integrating $8e^{-\frac{1}{2}x}$ (writing $+16e^{-\frac{1}{2}x}$ instead of $-16e^{-\frac{1}{2}x}$ ).
Forgetting to multiply by $2$ when converting $\frac{1}{2}x = \ln 8$ to $x = 2\ln 8$ in the evaluation step.
Adding the triangle area and the curve area instead of subtracting.
Miscomputing $\frac{1}{2} \times 6 \times 7 = 21$ — this is a common slip; it should be $21\ln 2$ , not $21$ .

Things to Be Careful About

The mark scheme allows decimals for the triangle-area step but requires an exact value for the final answer, so do not switch to decimals partway through.
The shaded area must be expressed in the form $p\ln 2 - q$ with both $p$ and $q$ positive integers — the answer $27\ln 2 - 14$ satisfies this.
The expression $e^{-3\ln 2}$ is a key simplification step; show it explicitly.

Techniques used

find y-interceptintegrate exponential-linear functionevaluate definite integral with logarithmic limitscompute area of right trianglesubtract to find bounded 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

We have parametric equations with parameter $\theta$ . To find $\frac{dy}{dx}$ , differentiate $x$ and $y$ with respect to $\theta$ , then use

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

Then simplify using $\sin^2\theta + \cos^2\theta = 1$ to express the result only in terms of $\cos\theta$ .

Working

Differentiate $x = \tan\theta$ :

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

Differentiate $y = \sin\theta - 2\sin^3\theta$ :

\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} = \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(1-\cos^2\theta)\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 = 6cos^5(theta) - 5cos^3(theta)

Detailed explanation

Walkthrough

We are given a curve in parametric form, so $x$ and $y$ are both functions of the parameter $\theta$ . To find the gradient $\frac{dy}{dx}$ , we cannot differentiate $y$ directly with respect to $x$ . Instead we use the chain rule in the form

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

First differentiate $x = \tan\theta$ . The derivative of $\tan\theta$ is $\sec^2\theta$ . Then differentiate $y = \sin\theta - 2\sin^3\theta$ term by term. The derivative of $\sin\theta$ is $\cos\theta$ . For $-2\sin^3\theta$ , use the chain rule: the derivative of $\sin^3\theta$ is $3\sin^2\theta\cos\theta$ , so the derivative is $-6\sin^2\theta\cos\theta$ . Hence $\frac{dy}{d\theta} = \cos\theta - 6\sin^2\theta\cos\theta$ .

Dividing by $\sec^2\theta$ is the same as multiplying by $\cos^2\theta$ . This gives $\cos^3\theta(1 - 6\sin^2\theta)$ . The question wants the answer only in terms of $\cos\theta$ , so replace $\sin^2\theta$ using $\sin^2\theta = 1 - \cos^2\theta$ . Expanding gives $6\cos^5\theta - 5\cos^3\theta$ .

Key Takeaways

This question tests parametric differentiation: differentiating both coordinates with respect to the parameter and then dividing. It also tests the chain rule for a power of a trigonometric function and the Pythagorean identity used to rewrite an expression in a single trigonometric function.

Common Mistakes

Forgetting to divide by $\frac{dx}{d\theta}$ and instead writing $\frac{dy}{dx} = \frac{dy}{d\theta}$ .
Differentiating $\sin^3\theta$ incorrectly, e.g. writing $3\sin^2\theta$ without the extra $\cos\theta$ factor from the chain rule.
Stopping at $\cos^3\theta(1 - 6\sin^2\theta)$ without using the identity to express the answer in terms of $\cos\theta$ only. The mark scheme says AG (answer given), so all necessary detail must be shown.

Things to Be Careful About

Since the question says 'Show that', the final line must be reached explicitly; an unsupported answer is not enough.
Keep the parameter domain $0<\theta<\frac{\pi}{2}$ in mind, although it is not needed in part (a) except to ensure the curve is defined.
Use radians; the parameter is measured in radians.

Techniques used

differentiate parametric equations with respect to the parameterapply the chain rule to a power of a trigonometric functiondivide the parametric derivatives to form dy/dxuse the Pythagorean identity to rewrite in terms of cos 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.

5M Similar

DifficultyMedium

Worked solution

Approach

The curve crosses the $x$ -axis where $y=0$ . Solve this for $\theta$ , substitute into $x=\tan\theta$ to find the point, and use the result from part (a) to find the gradient of the curve there. The normal has gradient equal to the negative reciprocal of this gradient.

Working

Set $y=0$ :

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

\sin\theta(1 - 2\sin^2\theta) = 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}.

Hence

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

At this point,

x = \tan\frac{\pi}{4} = 1.

So the curve crosses the $x$ -axis at $(1,0)$ .

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

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

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

\cos^3\frac{\pi}{4} = \frac{1}{2\sqrt{2}}, \qquad \cos^5\frac{\pi}{4} = \frac{1}{4\sqrt{2}}.

Thus

\frac{dy}{dx} = \frac{6}{4\sqrt{2}} - \frac{5}{2\sqrt{2}} = \frac{3}{2\sqrt{2}} - \frac{5}{2\sqrt{2}} = -\frac{1}{\sqrt{2}}.

The normal gradient is the negative reciprocal:

m = \sqrt{2}.

Using the point $(1,0)$ :

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

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

Answer

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

Final answer

y = sqrt(2)x - sqrt(2)

Detailed explanation

Walkthrough

The curve crosses the $x$ -axis where $y=0$ . Substitute $y = \sin\theta - 2\sin^3\theta = 0$ . Factor as $\sin\theta(1 - 2\sin^2\theta)=0$ . Since $0<\theta<\frac{\pi}{2}$ , $\sin\theta$ is not zero, so $1 - 2\sin^2\theta=0$ . This gives $\sin^2\theta = \frac{1}{2}$ , and within the given interval $\theta = \frac{\pi}{4}$ .

At this parameter value, $x = \tan\frac{\pi}{4} = 1$ , so the point is $(1,0)$ .

To find the gradient of the curve at this point, substitute $\theta = \frac{\pi}{4}$ into the result from part (a):

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

Since $\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$ , this evaluates to $-\frac{1}{\sqrt{2}}$ . The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of the tangent gradient, which is $\sqrt{2}$ .

Finally, use the point-slope form of a straight line through $(1,0)$ with gradient $\sqrt{2}$ :

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

so $y = \sqrt{2}x - \sqrt{2}$ .

Key Takeaways

This part combines solving a trigonometric equation, evaluating a parametric derivative at a specific point, and using the perpendicular-gradient condition for a normal. It shows how to move from a parameter value to a Cartesian point and line equation.

Common Mistakes

Dividing by $\sin\theta$ without noting that $\sin\theta$ could be zero; here the domain rules it out, but the reason should be stated.
Forgetting to find the $x$ -coordinate of the crossing point.
Using the tangent gradient instead of taking the negative reciprocal for the normal.
Using degrees for the final answer; the mark scheme allows degrees when solving for $\theta$ initially, but the parameter value must be in radians, and the final line should use exact values.

Things to Be Careful About

The mark scheme says 'Allow if degrees used' for the first method mark of solving $y=0$ , but the final $\theta$ must be $\frac{\pi}{4}$ in radians.
The gradient from part (a) must be evaluated exactly, not as a decimal.
The normal equation must be given in the requested form $y=mx+c$ with exact constants $m=\sqrt{2}$ and $c=-\sqrt{2}$ .

Techniques used

solve a trigonometric equation for the parameterevaluate the parametric gradient at the crossing pointtake the negative reciprocal to find the normal gradientform the equation of a straight line through a given 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 of $p(x)$ , we must have $p(-2)=0$ . Substitute $x=-2$ into $p(x)$ and solve the resulting linear equation for $k$ .

Working

Substitute $x=-2$ into $p(x)$ :

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

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

32-8k+4k-34+18=0

Simplify:

16-4k=0

4k=16

k=4

Answer

k=4

Final answer

k = 4

Detailed explanation

Walkthrough

The factor theorem says that if $(x+2)$ is a factor of $p(x)$ , then $x=-2$ is a root, so $p(-2)=0$ . This turns the problem into a simple substitution.

Substitute $x=-2$ into every term of $p(x)$ :

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

Evaluate each power:

32-8k+4k-34+18=0

Combine the constant terms $32-34+18=16$ and the $k$ -terms $-8k+4k=-4k$ :

16-4k=0

Solve:

k=4

Key Takeaways

The factor theorem links a linear factor $x-a$ to the condition $p(a)=0$ .
Substituting a known root into a polynomial with an unknown coefficient gives a linear equation for that coefficient.

Common Mistakes

Forgetting to evaluate all powers correctly, especially $(-2)^4=16$ and $(-2)^3=-8$ .
Sign errors when combining $-8k+4k$ and the constants.
Not setting $p(-2)=0$ before solving.

Things to Be Careful About

Use brackets when substituting negative values: $(-2)^2=4$ , not $-4$ .
The factor theorem applies only when the polynomial is exactly divisible by the linear factor, so the remainder is zero.

Techniques used

apply factor theoremsubstitute x = -2solve linear equation for k

(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

With $k=4$ , write down $p(x)$ . Since $x+2$ is a factor, one real root is $x=-2$ , so $\alpha=-2$ . Divide $p(x)$ by $x+2$ to find the remaining cubic factor. The non-integer real root $\beta$ satisfies this cubic equation, which can be rearranged into the required form.

Working

With $k=4$ ,

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

Since $x+2$ is a factor, $\alpha=-2$ .

Divide $p(x)$ by $x+2$ using synthetic division with divisor $-2$ :

\begin{array}{c|ccccc} -2 & 2 & 4 & 4 & 17 & 18\\ & & -4 & 0 & -8 & -18\\ \hline & 2 & 0 & 4 & 9 & 0 \end{array}

So

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

The equation $p(x)=0$ therefore has roots from $x+2=0$ and from

2x^3+4x+9=0

The integer root is $\alpha=-2$ , and the non-integer real root $\beta$ satisfies the cubic.

Rearrange the cubic:

2x^3+4x+9=0

2x^3=-4x-9

x^3=-2x-4.5

Taking cube roots gives

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

Hence $\beta$ satisfies the required equation.

Answer

\alpha=-2,\qquad \beta=\sqrt[3]{-2\beta-4.5}

Final answer

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

Detailed explanation

Walkthrough

First use the value $k=4$ found in part (a). The factor $(x+2)$ tells us immediately that $x=-2$ is a root, and because $\alpha$ is the integer real root, $\alpha=-2$ .

To find the other factor, divide the quartic by $x+2$ . Synthetic division with $-2$ gives quotient $2x^3+4x+9$ and remainder $0$ . Therefore

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

The non-integer root $\beta$ must come from the cubic factor, so it satisfies $2x^3+4x+9=0$ . Rearranging isolates $x^3$ on one side:

2x^3=-4x-9

x^3=-2x-4.5

Taking the cube root of both sides gives exactly the required form. This form is useful because it can be used as an iteration formula in part (d).

Key Takeaways

A factor of a polynomial gives a root directly.
Polynomial division reduces a quartic to a cubic and a linear factor.
Rearranging an equation into the form $x=F(x)$ prepares it for iterative solution.

Common Mistakes

Dividing incorrectly and missing the $4x$ term in the quotient.
Forgetting that $\alpha=-2$ is the integer root.
Rearranging the cubic incorrectly, e.g. dividing only one term by $2$ .

Things to Be Careful About

The quotient is $2x^3+4x+9$ , not $2x^3+4x^2+9$ ; the $x^2$ term is zero.
When taking cube roots, no sign ambiguity arises: the cube root is defined for negative values too.
The equation $x=\sqrt[3]{-2x-4.5}$ is an AG (answer given) form, so show the rearrangement clearly.

Techniques used

state factor rootdivide polynomial by x + 2rearrange cubic equation to iterative form

(c)

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

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the cubic $f(x)=2x^3+4x+9$ . Evaluate $f$ at the endpoints $-1.4$ and $-1.0$ . If the signs are different, the continuous function has a root between them; since $\beta$ is the real root of this cubic, this locates $\beta$ .

Working

Let

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

Evaluate at $x=-1$ :

f(-1)=2(-1)^3+4(-1)+9=-2-4+9=3

Evaluate at $x=-1.4$ :

f(-1.4)=2(-1.4)^3+4(-1.4)+9=-5.488-5.6+9=-2.088

There is a sign change:

f(-1.4)=-2.088<0,\qquad f(-1)=3>0

Since $f$ is continuous, its root lies between $-1.4$ and $-1.0$ . Hence

-1.4<\beta<-1.0

Answer

-1.4<\beta<-1.0

Final answer

-1.4 < beta < -1.0

Detailed explanation

Walkthrough

The cubic factor is $f(x)=2x^3+4x+9$ . A continuous function that changes sign between two points must cross zero somewhere between them. So evaluate $f$ at the two proposed bounds.

At $x=-1$ :

f(-1)=2(-1)^3+4(-1)+9=3

At $x=-1.4$ :

f(-1.4)=2(-1.4)^3+4(-1.4)+9=-2.088

The signs are opposite, so the root $\beta$ lies in $(-1.4,-1.0)$ .

Key Takeaways

Sign change on a continuous function is a standard way to locate a root.
The cubic has only one real root, so the sign change identifies $\beta$ specifically.

Common Mistakes

Evaluating $(-1.4)^3$ incorrectly (it is negative).
Using the quartic $p(x)$ instead of the cubic factor.
Forgetting to state the conclusion that the sign change implies the root is in the interval.

Things to Be Careful About

The function must be continuous; polynomials are continuous everywhere.
Use enough decimal places to make the sign clear.
A sign change is sufficient for at least one root; here the cubic has exactly one real root, so it is $\beta$ .

Techniques used

evaluate cubic at interval endpointsapply sign-change argument

(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 obtained in part (b):

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

Start with a value in the interval $(-1.4,-1.0)$ , for example $x_0=-1.2$ . Iterate until successive values agree to 5 significant figures, then round the final value to 3 significant figures.

Working

Take $x_0=-1.2$ . The first iteration is

x_1=\sqrt[3]{-2(-1.2)-4.5}=\sqrt[3]{-2.1}=-1.2806

Continuing:

\begin{array}{c|c} n & x_n\\ \hline 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 \end{array}

To 5 significant figures, the iteration has converged to $-1.2570$ . Therefore, correct to 3 significant figures,

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

Answer

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

Final answer

beta = -1.26 (3 s.f.)

Detailed explanation

Walkthrough

The equation $x=\sqrt[3]{-2x-4.5}$ is already in the form $x=F(x)$ , so we can iterate:

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

Choose $x_0=-1.2$ , which lies in the interval found in part (c). Substitute into the formula to get $x_1$ , then use $x_1$ to get $x_2$ , and so on. The table shows the sequence settling down. When two consecutive values agree to 5 significant figures, the iteration has converged to that accuracy. The stable value is $-1.2570$ to 5 s.f., so rounding to 3 s.f. gives $-1.26$ .

Key Takeaways

An equation rewritten as $x=F(x)$ can be solved by repeated substitution.
Iterations should be continued until successive approximations agree to the required precision.
Rounding should be done only at the final stage, not to the intermediate iterates.

Common Mistakes

Rounding each iterate too early; the question asks for 5 significant figures for each result, so keep enough digits.
Using the wrong iterative formula, e.g. forgetting the minus signs.
Stopping after one or two iterations, which is not enough to justify the final answer.

Things to Be Careful About

The cube root of a negative number is negative, so the iterates stay negative.
The final answer must be given to exactly 3 significant figures: $-1.26$ , not $-1.257$ .
Show enough iterations to justify convergence; the mark scheme requires evidence such as agreement to 5 s.f. or a sign change in the interval $[-1.265,-1.255]$ .

Techniques used

iterate fixed-point formularound results to significant figuresconfirm convergence to required accuracy

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

Write the square root as a power, differentiate using the product rule, then set the derivative to zero and solve for $x$ . Finally substitute back into $y$ to find the exact $y$ -coordinate.

Working

Let

y = 4e^{1-2x}(3x-1)^{\frac{1}{2}}

Differentiate using the product rule:

\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

Factor out the common non-zero factor $e^{1-2x}(3x-1)^{-\frac{1}{2}}$ :

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

So

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

x = \frac{7}{12}

Substitute $x = \frac{7}{12}$ into $y$ :

y = 4e^{1 - \frac{14}{12}}\sqrt{\frac{21}{12} - 1}

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

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

Answer

The stationary point is

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

Final answer

(7/12, 2√3e^(-1/6))

Detailed explanation

Walkthrough

We need the stationary point, so we must differentiate $y$ with respect to $x$ and set the derivative equal to zero.

First, rewrite the square root as a power: $\sqrt{3x-1} = (3x-1)^{\frac{1}{2}}$ . This lets us apply the power rule cleanly.

The expression is a product of $4e^{1-2x}$ and $(3x-1)^{\frac{1}{2}}$ , so we use the product rule:

\frac{dy}{dx} = 4e^{1-2x} \cdot \frac{d}{dx}\left[(3x-1)^{\frac{1}{2}}\right] + \frac{d}{dx}\left[4e^{1-2x}\right] \cdot (3x-1)^{\frac{1}{2}}

The derivative of $4e^{1-2x}$ is $-8e^{1-2x}$ because the chain rule multiplies by $-2$ . The derivative of $(3x-1)^{\frac{1}{2}}$ is $\frac{3}{2}(3x-1)^{-\frac{1}{2}}$ because the chain rule multiplies by $3$ .

This gives

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

Setting this equal to zero, we factor out $e^{1-2x}(3x-1)^{-\frac{1}{2}}$ , which is never zero for $x > \frac{1}{3}$ . This leaves

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

Solving gives $3x - 1 = \frac{3}{4}$ , so $x = \frac{7}{12}$ . Finally, substitute this value into the original equation to obtain the exact $y$ -coordinate:

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

Key Takeaways

This question tests the product rule, the chain rule for exponential functions, and the chain rule for powers of linear expressions. It also requires solving an equation where a common non-zero factor can be cancelled, and then evaluating a composite expression exactly.

Common Mistakes

Forgetting the chain rule when differentiating $e^{1-2x}$ : the derivative must include the factor $-2$ .
Forgetting the chain rule when differentiating $(3x-1)^{\frac{1}{2}}$ : the derivative must include the factor $3$ .
Applying the product rule incorrectly, for example differentiating the whole product as a single power.
Dividing by $e^{1-2x}(3x-1)^{-\frac{1}{2}}$ without noting it is non-zero for $x > \frac{1}{3}$ . This is acceptable here because the stationary point is not on the boundary of the domain.
Leaving the $x$ -coordinate as $\frac{3}{4}$ instead of solving fully for $x$ .

Things to Be Careful About

The domain of the original curve is $3x - 1 \geq 0$ , i.e. $x \geq \frac{1}{3}$ . The derivative is undefined at $x = \frac{1}{3}$ because of the negative power, but the stationary point found is $x = \frac{7}{12}$ , which is inside the domain.

The mark scheme allows unsimplified derivatives, but the final coordinates must be exact. Be careful to simplify $\sqrt{\frac{3}{4}}$ to $\frac{\sqrt{3}}{2}$ and to combine it correctly with the factor $4$ .

Techniques used

apply the product ruleuse the chain rule for exponential and power functionsset the derivative equal to zerosolve the resulting equation for the stationary pointevaluate the original function at the stationary x-coordinate

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