Hardest Edexcel A-Level Maths Questions & Solutions (Grade A*)

Let $I = \int e^{2x} \sin(3x) \, dx$.

Choose $u = e^{2x} \implies rac{du}{dx} = 2e^{2x}$

Choose $rac{dv}{dx} = \sin(3x) \implies v = -rac{1}{3}\cos(3x)$

For the new integral $\int e^{2x}\cos(3x) \, dx$, we must remain consistent:

Let $u = e^{2x} \implies rac{du}{dx} = 2e^{2x}$

Let $rac{dv}{dx} = \cos(3x) \implies v = rac{1}{3}\sin(3x)$

Collect all terms in $I$ on the left-hand side:

Given $u = 1 + \sqrt{x} = 1 + x^{1/2}$:

When $x = 1 \implies u = 1 + \sqrt{1} = 2$ (Lower Limit)

When $x = 4 \implies u = 1 + \sqrt{4} = 3$ (Upper Limit)

Check: If the prompt asked for $rac{5}{72}$, notice whether a factor of $rac{1}{2}$ was present in the prompt. For this integral with coefficient 2, the exact evaluated answer is $\mathbf{rac{5}{36}}$.

Exponentiate both sides:

Apply initial condition: at $t = 0$, $ heta = 180$:

Apply second condition: at $t = 10$, $ heta = 100$:

As $t o \infty$, $e^{-kt} o 0$. Therefore, $\mathbf{ heta o 20^\circ ext{C}}$ (the temperature approaches the ambient workshop room temperature asymptotically).

At $t = rac{\pi}{6}$:

Area formula: $A = \int y rac{dx}{dt} \, dt$.

Convert limits: When $x = 0 \implies 2\sin(t) = 0 \implies t = 0$.

When $x = 1 \implies 2\sin(t) = 1 \implies \sin(t) = rac{1}{2} \implies t = rac{\pi}{6}$.

Notice that $rac{d}{dt}[\cos(t)] = -\sin(t)$. Using inspection or $u = \cos(t)$:

Expand using the addition formula:

Equate coefficients:

Substitute the harmonic form into $g( heta)$:

To maximize $g( heta)$, we must minimize the denominator $[17 + 13\cos( heta - 1.1760)]$.

Since the minimum value of the cosine function is $-1$:

This occurs when $\cos( heta - 1.1760) = -1$:

Now expand $(1 + X)^{-1/2}$ where $X = -rac{9x}{4}$ and $n = -rac{1}{2}$:

The expansion converges if and only if $|X| < 1$:

Assumption: Assume there do exist positive integers $a, b \in \mathbb{Z}^+$ such that $a^2 - b^2 = 1$.

Factor the left-hand side as a difference of two squares:

Since $a$ and $b$ are positive integers ($a \ge 1, b \ge 1$), both $(a - b)$ and $(a + b)$ must be integers. The only pairs of integers that multiply to give 1 are $(1, 1)$ or $(-1, -1)$.

Since $a \ge 1$ and $b \ge 1$, $a + b \ge 2$.

Therefore, $a + b$ cannot equal 1, nor can it equal $-1$.

Contradiction: This contradicts the fact that $(a + b)$ must divide 1. Hence, the initial assumption is false, and there are no positive integer solutions to $a^2 - b^2 = 1$. $lacksquare$

Assumption: Assume the statement is false. That is, assume there exists an integer $n$ such that $n^3 + 5$ is odd, but $n$ is odd.

If $n$ is odd, then $n = 2k + 1$ for some integer $k \in \mathbb{Z}$.

Since $(4k^3 + 6k^2 + 3k + 3)$ is an integer, $n^3 + 5$ is divisible by 2 and is therefore even.

Contradiction: This directly contradicts the premise that $n^3 + 5$ is odd. Hence our assumption that $n$ is odd is false, proving that $n$ must be even. $lacksquare$

Compare direction vectors $\mathbf{d}_1 = egin{pmatrix} 2 \ -1 \ 3 \end{pmatrix}$ and $\mathbf{d}_2 = egin{pmatrix} 1 \ 1 \ -2 \end{pmatrix}$.

For lines to be parallel, $\mathbf{d}_1 = k \mathbf{d}_2$. However, comparing components:

Since direction vectors are not scalar multiples, the lines are not parallel.

Equate coordinate equations:

Add (1) and (2): $(2\lambda - \mu) + (\lambda + \mu) = 3 + 2 \implies 3\lambda = 5 \implies \lambda = rac{5}{3}$.

Substitute into (2): $\mu = 2 - rac{5}{3} = rac{1}{3}$.

Now test these parameters in equation (3):

Because the values of $\lambda$ and $\mu$ do not satisfy the third equation, the equations are inconsistent. Therefore, the lines do not intersect and are skew lines.

Given $ an heta = rac{4}{3}$, we have a 3-4-5 triangle: $\sin heta = rac{4}{5}$, $\cos heta = rac{3}{5}$.

Resolve vertically ($\Sigma F_y = 0$):

At the point of slipping (limiting equilibrium):

Resolve horizontally ($\Sigma F_x = 0$):

Clockwise moments about $A$ = Anticlockwise moments about $A$:

Divide throughout by $mg$:

Reality check: The total ladder length is $2a = rac{18}{9}a$. Since $rac{13}{9}a < 2a$, the ladder slips before the person reaches the very top.

For $P(X < 490) = 0.05 \implies Z_1 = -1.6449$ (from inverse normal table):

For $P(X > 515) = 0.02 \implies P(X < 515) = 0.98 \implies Z_2 = +2.0537$:

Subtract (1) from (2):

Substitute $\sigma$ into (2):

State hypotheses:

Under $H_0$, the sample mean distribution is $ar{X} \sim N\left(502, rac{6.76^2}{25} ight)$.

Standard error $SE = rac{\sigma}{\sqrt{n}} = rac{6.759}{5} = 1.3518$.

Calculate test statistic $z$:

Critical value for one-tailed 5% level is $z_{ ext{crit}} = -1.6449$.

Conclusion: Since $-3.33 < -1.6449$ (or $p ext{-value} = 0.00043 < 0.05$), the test statistic falls far into the critical region. Reject $H_0$. There is significant statistical evidence at the 5% level to conclude that the mean weight of cereal boxes has decreased from 502g.