2025 IB DP Mathematics - Analysis and Approaches Practice Paper with Answers

Let \(u = \mathrm{e}^{2x} + 3\). Then \(\mathrm{d}u = 2\mathrm{e}^{2x} \,\mathrm{d}x\), which gives \(\mathrm{e}^{2x} \,\mathrm{d}x = \frac{1}{2} \,\mathrm{d}u\). Next, find the new limits of integration: when \(x = 0\), \(u = \mathrm{e}^{0} + 3 = 4\), and when \(x = \ln 2\), \(u = \mathrm{e}^{2\ln 2} + 3 = 4 + 3 = 7\). Substituting these into the integral gives \(\frac{1}{2} \int_{4}^{7} \frac{1}{u} \,\mathrm{d}u\). Evaluating this results in \(\frac{1}{2} [\ln|u|]_{4}^{7} = \frac{1}{2} (\ln 7 - \ln 4) = \frac{1}{2} \ln\left(\frac{7}{4}\right)\).

(M1) for attempting substitution with \(u = \mathrm{e}^{2x} + 3\) or integration by inspection. (A1) for finding correct limits \(u = 4\) and \(u = 7\) (or for substituting back to \(x\) with the original limits). (A1) for the correct integrated expression \(\frac{1}{2} \ln(\mathrm{e}^{2x} + 3)\) or \(\frac{1}{2} \ln u\). (A1) for the correct final exact answer: \(\frac{1}{2}\ln\left(\frac{7}{4}\right)\) or \(\ln\left(\frac{\sqrt{7}}{2}\right)\).

To find the equation of the normal, we first find the coordinates of the point of contact.

When \(x = 0\):
\(f(0) = e^0 \cos(0) = 1 \cdot 1 = 1\)
So the point is \((0, 1)\).

Next, we find the derivative of \(f(x)\) using the product rule:
\(f'(x) = 2e^{2x} \cos(x) - e^{2x} \sin(x)\)

Now we evaluate the gradient of the tangent at \(x = 0\):
\(f'(0) = 2e^0 \cos(0) - e^0 \sin(0) = 2(1)(1) - 1(0) = 2\)

The gradient of the normal, \(m_n\), is the negative reciprocal of the gradient of the tangent:
\(m_n = -\frac{1}{2}\)

Using the point-slope formula for the normal line at \((0, 1)\):
\(y - 1 = -\frac{1}{2}(x - 0)\)

Multiply the entire equation by 2 to clear fractions:
\(2y - 2 = -x\)

Rearranging into the form \(ax + by + d = 0\):
\(x + 2y - 2 = 0\)

(M1) for finding the correct \(y\)-coordinate, \(f(0) = 1\).
(M1) for attempting to apply the product rule to differentiate \(f(x)\).
(A1) for the correct derivative \(f'(x) = 2e^{2x} \cos(x) - e^{2x} \sin(x)\).
(A1) for finding the correct normal gradient of \(-\frac{1}{2}\).
(A1) for the correct final equation in the required form: \(x + 2y - 2 = 0\) (or any integer multiple equivalent like \(-x - 2y + 2 = 0\)).

Let the terms of the arithmetic sequence be represented as:
\(u_1 = 3\)
\(u_3 = 3 + 2d\)
\(u_{11} = 3 + 10d\)

Since these three terms form a geometric sequence, the ratio between consecutive terms must be equal:
\(\frac{3+2d}{3} = \frac{3+10d}{3+2d}\)

Cross-multiplying to solve for \(d\):
\((3+2d)^2 = 3(3+10d)\)

Expanding both sides:
\(9 + 12d + 4d^2 = 9 + 30d\)

Subtracting 9 from both sides and rearranging terms:
\(4d^2 - 18d = 0\)

Factorising the quadratic equation:
\(2d(2d - 9) = 0\)

This yields two possible solutions:
\(d = 0\) or \(d = \frac{9}{2}\)

Since the question specifies that \(d \neq 0\), we conclude:
\(d = \frac{9}{2} = 4.5\)

(M1) for writing expressions for \(u_3\) and \(u_{11}\) in terms of \(d\) (i.e., \(3+2d\) and \(3+10d\)).
(M1) for setting up a correct geometric relation, e.g., \((3+2d)^2 = 3(3+10d)\).
(A1) for expanding and simplifying to obtain a correct quadratic equation, e.g., \(4d^2 - 18d = 0\).
(M1) for attempting to solve their quadratic equation.
(A1) for the correct value \(d = \frac{9}{2}\) (or \(4.5\)), with the explicit or implicit rejection of \(d = 0\).

(a) The acceleration \(a(t)\) is the derivative of the velocity \(v(t)\):
\(a(t) = v'(t) = 6t - 12\).
At \(t = 3\):
\(a(3) = 6(3) - 12 = 6\text{ m s}^{-2}\).

(b) To find the total distance, we determine where the velocity changes sign:
\(v(t) = 3(t^2 - 4t + 3) = 3(t-1)(t-3) = 0\), which gives \(t = 1\) or \(t = 3\).
For \(0 \le t < 1\), \(v(t) > 0\).
For \(1 < t < 3\), \(v(t) < 0\).

The total distance is given by:
\(\int_{0}^{3} |v(t)| dt = \int_{0}^{1} v(t) dt - \int_{1}^{3} v(t) dt\).

Let the displacement function be \(s(t) = \int v(t) dt = t^3 - 6t^2 + 9t\).
Evaluating \(s(t)\) at the boundaries:
\(s(0) = 0\)
\(s(1) = 1^3 - 6(1)^2 + 9(1) = 4\)
\(s(3) = 3^3 - 6(3)^2 + 9(3) = 27 - 54 + 27 = 0\).

The distance travelled in the first interval \([0, 1]\) is:
\(|s(1) - s(0)| = |4 - 0| = 4\).

The distance travelled in the second interval \([1, 3]\) is:
\(|s(3) - s(1)| = |0 - 4| = 4\).

Therefore, the total distance travelled is:
\(4 + 4 = 8\text{ m}\).

(a)
\(a(t) = 6t - 12\) (M1 for differentiating \(v(t)\))
\(a(3) = 6\text{ m s}^{-2}\) (A1)

(b)
Recognizing that \(v(t) = 0\) at \(t = 1\) (M1)
Attempt to split the integral into two intervals or use displacement at key points (M1)
For finding the distance in each part: \(\int_{0}^{1} v(t) dt = 4\) and \(\int_{1}^{3} v(t) dt = -4\) (A1)
Total distance = \(4 + |-4| = 8\text{ m}\) (A1)

(a) For an infinite geometric series to converge, we require \(|r| < 1\).
Since \(0 < \theta < \frac{\pi}{2}\), the value of \(\sin\theta\) satisfies \(0 < \sin\theta < 1\).
Thus, \(|r| = |\sin\theta| < 1\), meaning the sum to infinity always exists.

(b) When \(\theta = \frac{\pi}{6}\):
\(u_1 = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\)
\(r = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\)

The sum to infinity is:
\(S_{\infty} = \frac{u_1}{1 - r} = \frac{\frac{\sqrt{3}}{2}}{1 - \frac{1}{2}} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}\).

(c) We are given \(S_{\infty} = 1 + \sqrt{2}\), so:
\(\frac{\cos\theta}{1 - \sin\theta} = 1 + \sqrt{2}\)
\ \cos\theta = (1 + \sqrt{2})(1 - \sin\theta)\)

Squaring both sides and using \(\cos^2\theta = 1 - \sin^2\theta\):
\(1 - \sin^2\theta = (1 + \sqrt{2})^2 (1 - \sin\theta)^2\)
\(1 - \sin^2\theta = (3 + 2\sqrt{2})(1 - \sin\theta)^2\)

Since \(\theta \in (0, \frac{\pi}{2})\), \(\sin\theta \neq 1\), so we can divide both sides by \(1 - \sin\theta\):
\(1 + \sin\theta = (3 + 2\sqrt{2})(1 - \sin\theta)\)
\(1 + \sin\theta = 3 + 2\sqrt{2} - (3 + 2\sqrt{2})\sin\theta\)
\((4 + 2\sqrt{2})\sin\theta = 2 + 2\sqrt{2}\)
\(\sin\theta = \frac{2 + 2\sqrt{2}}{4 + 2\sqrt{2}} = \frac{1 + \sqrt{2}}{2 + \sqrt{2}} = \frac{1 + \sqrt{2}}{\sqrt{2}(\sqrt{2} + 1)} = \frac{1}{\sqrt{2}}\).

Since \(0 < \theta < \frac{\pi}{2}\), the only solution is \(\theta = \frac{\pi}{4}\).

(a)
Recognizing that the condition for convergence is \(|r| < 1\) (M1)
Stating that for \(0 < \theta < \frac{\pi}{2}\), \(0 < \sin\theta < 1\), hence \(|r| < 1\) (R1)

(b)
Substituting \(\theta = \frac{\pi}{6}\) to find \(u_1 = \frac{\sqrt{3}}{2}\) and \(r = \frac{1}{2}\) (M1)
Correctly simplifying the fraction to show \(S_{\infty} = \sqrt{3}\) (A1)

(c)
Setting up the equation \(\frac{\cos\theta}{1 - \sin\theta} = 1 + \sqrt{2}\) and attempting to solve for \(\sin\theta\) (M1)
Obtaining \(\sin\theta = \frac{1}{\sqrt{2}}\) (or equivalent) (A1)
Identifying the correct value \(\theta = \frac{\pi}{4}\) (A1)

**(a)** To find the local maximum, we first differentiate \(f(x) = x e^{-x^2}\) using the product rule and chain rule:
\(f'(x) = (1)(e^{-x^2}) + (x)(-2x e^{-x^2})\)
\(f'(x) = e^{-x^2}(1 - 2x^2)\)

Setting \(f'(x) = 0\) for the local maximum:
\(e^{-x^2}(1 - 2x^2) = 0\)
Since \(e^{-x^2} > 0\) for all \(x\):
\(1 - 2x^2 = 0 \implies x^2 = \frac{1}{2}\)
Since \(x \ge 0\), we have:
\(x = \frac{1}{\sqrt{2}}\)

Now, find the \(y\)-coordinate by substituting \(x = \frac{1}{\sqrt{2}}\) into \(f(x)\):
\(y = f\left(\frac{1}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}} e^{-\left(\frac{1}{\sqrt{2}}\right)^2} = \frac{1}{\sqrt{2}} e^{-1/2} = \frac{1}{\sqrt{2e}}\)

So, the coordinates of the local maximum are \(\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2e}}\right)\).

**(b)** The area of the region \(R\) is given by the integral:
\(\text{Area} = \int_{0}^{k} x e^{-x^2} \, dx\)

Let \(u = -x^2\), which gives \(du = -2x \, dx \implies x \, dx = -\frac{1}{2} \, du\).
When \(x = 0\), \(u = 0\).
When \(x = k\), \(u = -k^2\).

Substituting these into the integral:
\(\text{Area} = \int_{0}^{-k^2} -\frac{1}{2} e^u \, du = \left[ -\frac{1}{2} e^u \right]_{0}^{-k^2} = -\frac{1}{2} e^{-k^2} - \left(-\frac{1}{2} e^0\right) = \frac{1}{2}(1 - e^{-k^2})\)

We are given that the area is \(\frac{e^9 - 1}{2e^9}\). Therefore:
\(\frac{1}{2}(1 - e^{-k^2}) = \frac{e^9 - 1}{2e^9}\)
\(\frac{1}{2}(1 - e^{-k^2}) = \frac{1}{2}\left(1 - e^{-9}\right)\)
\(1 - e^{-k^2} = 1 - e^{-9}\)
\(e^{-k^2} = e^{-9}\)
\(k^2 = 9\)
Since \(k > 0\), \(k = 3\).

**(a)**
* **M1** for attempting to use the product rule to find \(f'(x)\).
* **A1** for correct derivative \(f'(x) = e^{-x^2}(1 - 2x^2)\).
* **A1** for finding the correct \(x\)-coordinate: \(x = \frac{1}{\sqrt{2}}\).
* **A1** for finding the correct \(y\)-coordinate: \(y = \frac{1}{\sqrt{2e}}\).

**(b)**
* **M1** for attempting integration by substitution (recognizing the derivative of \(-x^2\) is present in some form).
* **A1** for obtaining the correct integrated expression \(\left[ -\frac{1}{2} e^{-x^2} \right]_0^k = \frac{1}{2}(1 - e^{-k^2})\).
* **A1** for setting up the equation and correctly solving to find \(k = 3\).

**(a)** To find the local maximum, we first differentiate \(f(x) = x e^{-x^2}\) using the product rule and chain rule:
\(f'(x) = (1)(e^{-x^2}) + (x)(-2x e^{-x^2})\)
\(f'(x) = e^{-x^2}(1 - 2x^2)\)

Setting \(f'(x) = 0\) for the local maximum:
\(e^{-x^2}(1 - 2x^2) = 0\)
Since \(e^{-x^2} > 0\) for all \(x\):
\(1 - 2x^2 = 0 \implies x^2 = \frac{1}{2}\)
Since \(x \ge 0\), we have:
\(x = \frac{1}{\sqrt{2}}\)

Now, find the \(y\)-coordinate by substituting \(x = \frac{1}{\sqrt{2}}\) into \(f(x)\):
\(y = f\left(\frac{1}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}} e^{-\left(\frac{1}{\sqrt{2}}\right)^2} = \frac{1}{\sqrt{2}} e^{-1/2} = \frac{1}{\sqrt{2e}}\)

So, the coordinates of the local maximum are \(\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2e}}\right)\).

**(b)** The area of the region \(R\) is given by the integral:
\(\text{Area} = \int_{0}^{k} x e^{-x^2} \, dx\)

Let \(u = -x^2\), which gives \(du = -2x \, dx \implies x \, dx = -\frac{1}{2} \, du\).
When \(x = 0\), \(u = 0\).
When \(x = k\), \(u = -k^2\).

Substituting these into the integral:
\(\text{Area} = \int_{0}^{-k^2} -\frac{1}{2} e^u \, du = \left[ -\frac{1}{2} e^u \right]_{0}^{-k^2} = -\frac{1}{2} e^{-k^2} - \left(-\frac{1}{2} e^0\right) = \frac{1}{2}(1 - e^{-k^2})\)

We are given that the area is \(\frac{e^9 - 1}{2e^9}\). Therefore:
\(\frac{1}{2}(1 - e^{-k^2}) = \frac{e^9 - 1}{2e^9}\)
\(\frac{1}{2}(1 - e^{-k^2}) = \frac{1}{2}\left(1 - e^{-9}\right)\)
\(1 - e^{-k^2} = 1 - e^{-9}\)
\(e^{-k^2} = e^{-9}\)
\(k^2 = 9\)
Since \(k > 0\), \(k = 3\).

**(a)**
* **M1** for attempting to use the product rule to find \(f'(x)\).
* **A1** for correct derivative \(f'(x) = e^{-x^2}(1 - 2x^2)\).
* **A1** for finding the correct \(x\)-coordinate: \(x = \frac{1}{\sqrt{2}}\).
* **A1** for finding the correct \(y\)-coordinate: \(y = \frac{1}{\sqrt{2e}}\).

**(b)**
* **M1** for attempting integration by substitution (recognizing the derivative of \(-x^2\) is present in some form).
* **A1** for obtaining the correct integrated expression \(\left[ -\frac{1}{2} e^{-x^2} \right]_0^k = \frac{1}{2}(1 - e^{-k^2})\).
* **A1** for setting up the equation and correctly solving to find \(k = 3\).

(a) To find the local maximum, we find the first derivative \( f'(x) \) using the quotient rule:
\( f'(x) = \frac{\frac{1}{x} \cdot x^2 - \ln x \cdot 2x}{(x^2)^2} = \frac{1 - 2\ln x}{x^3} \).

Setting \( f'(x) = 0 \) gives:
\( 1 - 2\ln x = 0 \Rightarrow \ln x = \frac{1}{2} \Rightarrow x = e^{1/2} = \sqrt{e} \).

Substituting \( x = \sqrt{e} \) into \( f(x) \):
\( f(\sqrt{e}) = \frac{\ln(\sqrt{e})}{(\sqrt{e})^2} = \frac{\frac{1}{2}}{e} = \frac{1}{2e} \).

Thus, the local maximum is at \( \left(\sqrt{e}, \frac{1}{2e}\right) \).

(b) To find the points of inflection, we find the second derivative \( f''(x) \):
\( f''(x) = \frac{\left(-\frac{2}{x}\right)x^3 - (1 - 2\ln x)(3x^2)}{(x^3)^2} = \frac{-2x^2 - 3x^2 + 6x^2\ln x}{x^6} = \frac{6\ln x - 5}{x^4} \).

Setting \( f''(x) = 0 \) gives:
\( 6\ln x - 5 = 0 \Rightarrow \ln x = \frac{5}{6} \Rightarrow x = e^{5/6} \).

Since \( x^4 > 0 \) for all \( x > 0 \), the sign of \( f''(x) \) depends only on the numerator \( 6\ln x - 5 \).
For \( x < e^{5/6} \), \( f''(x) < 0 \), and for \( x > e^{5/6} \), \( f''(x) > 0 \).
Since \( f''(x) \) changes sign at \( x = e^{5/6} \), there is a point of inflection at \( x = e^{5/6} \).

(c) The curve crosses the x-axis when \( f(x) = 0 \), which gives \( \ln x = 0 \Rightarrow x = 1 \). The intercept is at \( (1, 0) \).

The gradient of the tangent at \( x = 1 \) is:
\( f'(1) = \frac{1 - 2\ln 1}{1^3} = 1 \).

The equation of the tangent is:
\( y - 0 = 1(x - 1) \Rightarrow y = x - 1 \).

(d) The region is bounded by \( x = 1 \) (where the curve crosses the x-axis) and \( x = \sqrt{e} \) (the x-coordinate of the local maximum).

The area \( A \) is:
\( A = \int_{1}^{\sqrt{e}} \frac{\ln x}{x^2} \, dx \).

Using integration by parts with \( u = \ln x \) and \( dv = x^{-2} \, dx \), we have \( du = \frac{1}{x} \, dx \) and \( v = -\frac{1}{x} \):
\( \int \frac{\ln x}{x^2} \, dx = -\frac{\ln x}{x} - \int \left(-\frac{1}{x^2}\right) \, dx = -\frac{\ln x + 1}{x} \).

Evaluating the definite integral:
\( A = \left[ -\frac{\ln x + 1}{x} \right]_{1}^{\sqrt{e}} = \left( -\frac{\ln(\sqrt{e}) + 1}{\sqrt{e}} \right) - \left( -\frac{\ln(1) + 1}{1} \right) = -\frac{1.5}{\sqrt{e}} + 1 = 1 - \frac{3}{2\sqrt{e}} \).

(a)
M1: Attempt to find \( f'(x) \) using the quotient or product rule.
A1: Correct derivative \( f'(x) = \frac{1 - 2\ln x}{x^3} \).
M1: Setting their \( f'(x) = 0 \) and solving for \( x \).
A1: Correct coordinates \( \left(\sqrt{e}, \frac{1}{2e}\right) \).

(b)
M1: Attempt to find \( f''(x) \) using the quotient or product rule.
A1: Correct simplified second derivative \( f''(x) = \frac{6\ln x - 5}{x^4} \).
A1: Setting \( f''(x) = 0 \) to find \( x = e^{5/6} \).
R1: Clear reasoning showing that \( f''(x) \) changes sign around \( x = e^{5/6} \).

(c)
A1: Correctly finding the x-intercept at \( x = 1 \).
A1: Correct gradient of the tangent \( f'(1) = 1 \).
A1: Correct equation of the tangent \( y = x - 1 \).

(d)
M1: Formulating the correct area integral \( \int_{1}^{\sqrt{e}} \frac{\ln x}{x^2} \, dx \).
M1: Attempting integration by parts with correct substitutions.
A1: Correct antiderivative \( -\frac{\ln x + 1}{x} \).
A1: Correct final area \( 1 - \frac{3}{2\sqrt{e}} \) (or equivalent).

(a) To find \( f'(x) \), we use the product rule:
\( f'(x) = 2x e^{-kx} + x^2 (-k) e^{-kx} = x(2 - kx) e^{-kx} \).

To find the critical points, we set \( f'(x) = 0 \):
\( x(2 - kx) e^{-kx} = 0 \).
Since \( e^{-kx} > 0 \), we have \( x = 0 \) or \( 2 - kx = 0 \Rightarrow x = \frac{2}{k} \).

For \( x = 0 \):
\( f(0) = 0^2 e^{0} = 0 \).
Since \( f(x) \ge 0 \) for all \( x \ge 0 \), \( (0, 0) \) is a local minimum.

For \( x = \frac{2}{k} \):
\( f\left(\frac{2}{k}\right) = \left(\frac{2}{k}\right)^2 e^{-k\left(\frac{2}{k}\right)} = \frac{4}{k^2} e^{-2} = \frac{4}{e^2 k^2} \).
Since \( f'(x) > 0 \) for \( 0 < x < \frac{2}{k} \) and \( f'(x) < 0 \) for \( x > \frac{2}{k} \), the point \( \left(\frac{2}{k}, \frac{4}{e^2 k^2}\right) \) is a local maximum.

(b) To find the points of inflexion, we find the second derivative:
\( f''(x) = \frac{d}{dx} \left[ (2x - kx^2) e^{-kx} \right] = (2 - 2kx) e^{-kx} + (2x - kx^2)(-k) e^{-kx} \)
\( f''(x) = (2 - 2kx - 2kx + k^2 x^2) e^{-kx} = (k^2 x^2 - 4kx + 2) e^{-kx} \).

Set \( f''(x) = 0 \):
\( k^2 x^2 - 4kx + 2 = 0 \) (since \( e^{-kx} \neq 0 \)).
Using the quadratic formula:
\( x = \frac{4k \pm \sqrt{(-4k)^2 - 4(k^2)(2)}}{2k^2} = \frac{4k \pm \sqrt{16k^2 - 8k^2}}{2k^2} = \frac{4k \pm \sqrt{8k^2}}{2k^2} \).
Since \( k > 0 \), we have \( \sqrt{8k^2} = 2\sqrt{2}k \):
\( x = \frac{4k \pm 2\sqrt{2}k}{2k^2} = \frac{2 \pm \sqrt{2}}{k} \).
Since the quadratic expression \( k^2 x^2 - 4kx + 2 \) changes sign at these roots, the second derivative \( f''(x) \) changes sign, confirming these are points of inflexion.

(c) For \( k = 1 \), we have \( f(x) = x^2 e^{-x} \). The area of the region is:
\( A = \int_{0}^{2} x^2 e^{-x} \, dx \).

We use integration by parts: \( \int u \, dv = uv - \int v \, du \).
Let \( u = x^2 \) and \( dv = e^{-x} \, dx \Rightarrow du = 2x \, dx \) and \( v = -e^{-x} \).
\( A = \left[ -x^2 e^{-x} \right]_{0}^{2} - \int_{0}^{2} (-e^{-x})(2x) \, dx = -4e^{-2} + 2 \int_{0}^{2} x e^{-x} \, dx \).

We apply integration by parts again to \( \int_{0}^{2} x e^{-x} \, dx \):
Let \( u = x \) and \( dv = e^{-x} \, dx \Rightarrow du = dx \) and \( v = -e^{-x} \).
\( \int_{0}^{2} x e^{-x} \, dx = \left[ -x e^{-x} \right]_{0}^{2} - \int_{0}^{2} (-e^{-x}) \, dx = -2e^{-2} + \left[ -e^{-x} \right]_{0}^{2} \)
\( = -2e^{-2} - e^{-2} + e^{0} = 1 - 3e^{-2} \).

Substituting this back into the expression for \( A \):
\( A = -4e^{-2} + 2(1 - 3e^{-2}) = 2 - 10e^{-2} \).

**Part (a)**
- **M1**: Applying product rule to find \( f'(x) \).
- **A1**: Correct derivative \( f'(x) = x(2 - kx) e^{-kx} \) (or equivalent).
- **M1**: Setting \( f'(x) = 0 \) and solving for \( x \).
- **A1**: Correct coordinates for local minimum: \( (0, 0) \).
- **A1**: Correct coordinates for local maximum: \( \left(\frac{2}{k}, \frac{4}{e^2 k^2}\right) \) (accept \( \left(\frac{2}{k}, 4k^{-2}e^{-2}\right) \).

**Part (b)**
- **M1**: Applying product rule to find \( f''(x) \).
- **A1**: Correct simplified expression for \( f''(x) = (k^2 x^2 - 4kx + 2) e^{-kx} \).
- **M1**: Setting \( f''(x) = 0 \) and applying the quadratic formula.
- **A1**: Correctly obtaining \( x = \frac{2 \pm \sqrt{2}}{k} \) and noting a sign change occurs.

**Part (c)**
- **M1**: Writing down the correct integral \( \int_{0}^{2} x^2 e^{-x} \, dx \).
- **M1**: Attempting integration by parts (first time) with correct choice of \( u \) and \( dv \).
- **A1**: Correct expression after first integration by parts: \( \left[ -x^2 e^{-x} \right]_{0}^{2} + 2 \int_{0}^{2} x e^{-x} \, dx \).
- **M1**: Attempting integration by parts (second time) with correct choice of \( u \) and \( dv \).
- **A1**: Correct expression for the second integral: \( \left[ -x e^{-x} - e^{-x} \right]_{0}^{2} \) (or equivalent).
- **M1**: Correct substitution of limits \( 0 \) and \( 2 \).
- **A1**: Correct final area \( 2 - 10e^{-2} \) (or \( 2 - \frac{10}{e^2} \)).

(a) Using the identity \( \cos^2 x = 1 - \sin^2 x \):
\( f(x) = 2(1 - \sin^2 x) + 2\sin x = 2 - 2\sin^2 x + 2\sin x = -2\sin^2 x + 2\sin x + 2 \).

(b) Setting \( f(x) = 2 \):
\( -2\sin^2 x + 2\sin x + 2 = 2 \Rightarrow -2\sin^2 x + 2\sin x = 0 \)
\( 2\sin x(1 - \sin x) = 0 \).
This gives \( \sin x = 0 \) or \( \sin x = 1 \).
For \( 0 \le x \le 2\pi \):
- \( \sin x = 0 \Rightarrow x = 0, \pi, 2\pi \)
- \( \sin x = 1 \Rightarrow x = \frac{\pi}{2} \).

Thus, the solutions are \( x = 0, \frac{\pi}{2}, \pi, 2\pi \).

(c) Let \( u = \sin x \) where \( -1 \le u \le 1 \). The function becomes a quadratic function:
\( g(u) = -2u^2 + 2u + 2 \).

To find the stationary points, we can differentiate \( g(u) \) or find the vertex:
\( g'(u) = -4u + 2 = 0 \Rightarrow u = \frac{1}{2} \).

Evaluating \( g(u) \) at \( u = \frac{1}{2} \):
\( g\left(\frac{1}{2}\right) = -2\left(\frac{1}{4}\right) + 2\left(\frac{1}{2}\right) + 2 = -\frac{1}{2} + 1 + 2 = \frac{5}{2} \).
Since the coefficient of \( u^2 \) is negative, this represents the maximum value.
We find the corresponding values of \( x \):
\( \sin x = \frac{1}{2} \Rightarrow x = \frac{\pi}{6}, \frac{5\pi}{6} \).

To find the minimum value, we evaluate \( g(u) \) at the boundaries of its domain \( u = -1 \) and \( u = 1 \):
- For \( u = 1 \) (which corresponds to \( x = \frac{\pi}{2} \)), \( g(1) = 2 \).
- For \( u = -1 \) (which corresponds to \( x = \frac{3\pi}{2} \)), \( g(-1) = -2(-1)^2 + 2(-1) + 2 = -2 - 2 + 2 = -2 \).

Comparing the values:
- The absolute maximum is \( \frac{5}{2} \), which occurs at \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \).
- The absolute minimum is \( -2 \), which occurs at \( x = \frac{3\pi}{2} \).

(d) We want to solve \( f(x) > 2 \):
\( -2\sin^2 x + 2\sin x + 2 > 2 \Rightarrow 2\sin x(1 - \sin x) > 0 \).
Since \( 1 - \sin x \ge 0 \) for all real \( x \), the inequality \( 2\sin x(1 - \sin x) > 0 \) is satisfied if and only if:
\( \sin x > 0 \) and \( 1 - \sin x \neq 0 \).

- \( \sin x > 0 \Rightarrow 0 < x < \pi \).
- \( 1 - \sin x \neq 0 \Rightarrow \sin x \neq 1 \Rightarrow x \neq \frac{\pi}{2} \).

Thus, the solution is \( 0 < x < \frac{\pi}{2} \) or \( \frac{\pi}{2} < x < \pi \).

**Part (a)**
- **M1**: Substituting \( \cos^2 x = 1 - \sin^2 x \) into the expression for \( f(x) \).
- **A1**: Correct algebraic simplification showing \( f(x) = -2\sin^2 x + 2\sin x + 2 \).

**Part (b)**
- **M1**: Setting the expression equal to 2 and simplifying to \( 2\sin x(1 - \sin x) = 0 \) (or equivalent quadratic form).
- **M1**: Identifying the two cases \( \sin x = 0 \) and \( \sin x = 1 \).
- **A1**: Correct solutions from \( \sin x = 0 \): \( x = 0, \pi, 2\pi \).
- **A1**: Correct solution from \( \sin x = 1 \): \( x = \frac{\pi}{2} \).

**Part (c)**
- **M1**: Finding the critical value of \( \sin x \) (either by setting \( g'(u) = 0 \) to get \( \sin x = \frac{1}{2} \) or using \( f'(x) = 2\cos x(1 - 2\sin x) = 0 \)).
- **A1**: Finding the maximum value \( \frac{5}{2} \) (or 2.5).
- **A1**: Finding the corresponding \( x \)-values for the maximum: \( x = \frac{\pi}{6}, \frac{5\pi}{6} \).
- **M1**: Checking boundaries/critical points for the minimum (e.g. \( u = -1 \), \( u = 1 \), or \( x = \frac{3\pi}{2} \)).
- **A1**: Finding the minimum value \( -2 \).
- **A1**: Finding the corresponding \( x \)-value for the minimum: \( x = \frac{3\pi}{2} \).

**Part (d)**
- **M1**: Setting up the inequality \( 2\sin x(1 - \sin x) > 0 \).
- **R1**: Recognizing that \( 1 - \sin x \ge 0 \) implies we need \( \sin x > 0 \) and \( \sin x \neq 1 \).
- **A1**: Identifying \( 0 < x < \pi \) (from \( \sin x > 0 \)).
- **A1**: Excluding \( x = \frac{\pi}{2} \) to give final answer: \( 0 < x < \frac{\pi}{2} \cup \frac{\pi}{2} < x < \pi \) (or equivalent notation).

(a) Let \(X\) be the mass of an apple, so \(X \sim N(150, \sigma^2)\). We are given \(P(X > 180) = 0.12\), which means \(P(X \le 180) = 0.88\). Using the standard normal distribution, the z-score corresponding to a cumulative probability of 0.88 is \(z \approx 1.175\). Setting up the equation: \(\frac{180 - 150}{\sigma} = 1.175\). Solving for \(\sigma\) gives \(\sigma = \frac{30}{1.175} \approx 25.5\) (or 25.532). (b) We first find the probability that a single apple weighs more than 160 grams: \(P(X > 160)\) using \(X \sim N(150, 25.532^2)\). Using a GDC, we find \(P(X > 160) \approx 0.3477\). Let \(Y\) be the number of apples that weigh more than 160 grams in a sample of 5. Then \(Y \sim B(5, 0.3477)\). Using the binomial probability formula or a GDC, we find \(P(Y = 2) = \binom{5}{2} (0.3477)^2 (1 - 0.3477)^3 \approx 0.335\).

(a) M1 for attempting to use the standard normal distribution (e.g., finding the z-score of 1.175). M1 for setting up the equation \(\frac{180 - 150}{\sigma} = 1.175\). A1 for \(\sigma \approx 25.5\) (accept 25.532). (b) M1 for finding the probability of a single apple weighing more than 160g, \(p \approx 0.348\). A1 for using binomial distribution to find \(P(Y = 2) \approx 0.335\) (accept 0.336 if using rounded value of \(\sigma = 25.5\)).

(a) To find the point of intersection, we set \(f(x) = g(x)\), which gives \(\mathrm{e}^{0.5x} = 4 \cos(x)\). Using a GDC to find the intersection in the interval \([0, \pi]\), we get \(x \approx 1.11818...\), which rounds to \(1.12\) (to 3 significant figures). (b) The region is bounded by the two curves and the \(y\)-axis (\(x = 0\)). On the interval \([0, 1.11818]\), the graph of \(g(x)\) lies above the graph of \(f(x)\). The area \(A\) of this region is given by \(A = \int_{0}^{1.11818} (g(x) - f(x)) \, \mathrm{d}x = \int_{0}^{1.11818} (4 \cos(x) - \mathrm{e}^{0.5x}) \, \mathrm{d}x\). Evaluating this integral using a GDC gives \(A \approx 2.10\).

(a) M1 for equating the functions or attempting to find the intersection on a GDC. A1 for \(x \approx 1.12\) (accept 1.118). (b) M1 for expressing the area as an integral with correct limits (0 and their answer from part a). M1 for the correct integrand \(4 \cos(x) - \mathrm{e}^{0.5x}\). A1 for \(2.10\) (accept 2.099).

(a) Let \(X\) be the mass of a randomly chosen apple, where \(X \sim N(150, 18^2)\).
Using a GDC with lower bound \(130\), upper bound \(165\), mean \(\mu = 150\), and standard deviation \(\sigma = 18\):
\(P(130 < X < 165) \approx 0.664\)

(b) We require \(P(X < w) = 0.12\).
Using the inverse normal function on a GDC with area \(0.12\), \(\mu = 150\), and \(\sigma = 18\):
\(w \approx 129\text{ g}\) (to 3 significant figures, or \(128.9\))

(c) Let \(Y\) be the number of small apples in the sample of \(10\). Then \(Y \sim B(10, 0.12)\).
We want to find \(P(Y \ge 2)\):
\(P(Y \ge 2) = 1 - P(Y \le 1)\)
Using a GDC for binomial cumulative probability with \(n = 10\), \(p = 0.12\), and \(x = 1\):
\(P(Y \le 1) \approx 0.6583\)
Therefore,
\(P(Y \ge 2) = 1 - 0.6583 \approx 0.342\)

(a)
[M1] for attempting to use normal cumulative distribution with correct parameters.
[A1] for \(0.664\) (accept \(0.66444...\)).

(b)
[M1] for setting up the equation \(P(X < w) = 0.12\) or using the inverse normal function.
[A1] for \(w = 129\) (accept \(128.9\) or \(128.85...\)).

(c)
[M1] for identifying binomial distribution with parameters \(n = 10\) and \(p = 0.12\), and expressing the required probability as \(P(Y \ge 2)\) or \(1 - P(Y \le 1)\).
[A1] for \(0.342\) (accept \(0.34172...\)).

(a) On the interval \([0, 1.5]\), \(g(x) > f(x)\).
The area \(A\) of region \(R\) is given by the integral:
\(A = \int_{0}^{1.5} (g(x) - f(x)) \, dx = \int_{0}^{1.5} (\cos(x) + 1.5 - (e^{-0.5x^2} + 1)) \, dx\)
\(A = \int_{0}^{1.5} (\cos(x) + 0.5 - e^{-0.5x^2}) \, dx\)
Using a GDC to evaluate this numerical integral:
\(A \approx 0.662\) (to 3 significant figures, or \(0.6616...\))

(b) The volume \(V\) of the solid of revolution formed by rotating \(R\) about the \(x\)-axis is given by:
\(V = \pi \int_{0}^{1.5} \left( [g(x)]^2 - [f(x)]^2 \right) \, dx\)
\(V = \pi \int_{0}^{1.5} \left( (\cos(x) + 1.5)^2 - (e^{-0.5x^2} + 1)^2 \right) \, dx\)
Using a GDC to evaluate the integral and multiplying by \(\pi\):
\(V \approx 8.25\) (to 3 significant figures, or \(8.246...\))

(a)
[M1] for setting up the correct integral representing the area, including correct limits.
[A2] for \(0.662\) (award [A1] for \(0.66\) or for the unrounded answer \(0.6616...\)).

(b)
[M1] for attempting to use the volume of revolution formula \(\pi \int (y_1^2 - y_2^2) \, dx\).
[A1] for the correct integral expression \(\pi \int_{0}^{1.5} \left( (\cos(x) + 1.5)^2 - (e^{-0.5x^2} + 1)^2 \right) \, dx\).
[A1] for \(8.25\) (accept \(8.246...\)).

Part (a): Let \(M\) represent the mass of a melon, where \(M \sim N(1200, 100^2)\). We want to find \(P(1150 < M < 1350)\). Using a graphic display calculator (GDC) with lower bound 1150, upper bound 1350, mean 1200, and standard deviation 100, we find: \(P(1150 < M < 1350) \approx 0.624655...\). To three significant figures, this is \(0.625\). Part (b): Let \(Y\) be the number of melons in the sample of 8 that have a mass between 1150 g and 1350 g. The variable \(Y\) follows a binomial distribution, \(Y \sim B(8, p)\), where \(p \approx 0.624655...\). We want to find \(P(Y \ge 6)\). Using a GDC, we find: \(P(Y \ge 6) = 1 - P(Y \le 5)\). With the accurate value \(p = 0.624655...\), we get \(P(Y \ge 6) \approx 0.369\). If the rounded value \(p = 0.625\) is used, we get \(P(Y \ge 6) \approx 0.370\).

Part (a) [3 Marks]: (M1) for setting up the normal distribution model or writing \(P(1150 < M < 1350)\). (M1) for attempting to use GDC normal CDF (or standardizing to find \(P(-0.5 < Z < 1.5)\)). (A1) for \(0.625\) (or \(0.624655...\)). Part (b) [4 Marks]: (M1) for identifying the binomial distribution model \(Y \sim B(8, p)\). (M1) for recognizing the need to find \(P(Y \ge 6)\) (e.g., writing \(1 - P(Y \le 5)\) or \(P(Y=6) + P(Y=7) + P(Y=8)\)). (A1) for correct GDC binomial inputs or intermediate probability terms. (A1) for \(0.369\) (accept \(0.370\) from using the rounded value \(p = 0.625\)).

(a)
Let \(V(t)\) be the volume of water in the tank at time \(t\).
The rate of change of the volume is given by:
\(V'(t) = I(t) - O(t)\)

At \(t = 4\):
\(V'(4) = 150 \sin^2(0.8) - 80 e^{0.2}\)
\(V'(4) \approx 77.195 - 97.712 \approx -20.517...\)

So, the rate of change is \(-20.5\text{ litres per hour}\) (to 3 s.f.).
Since \(V'(4) < 0\), the volume of water is decreasing.

(b)
To find the minimum volume, we look for critical points where \(V'(t) = 0\):
\(150 \sin^2(0.2t) - 80 e^{0.05t} = 0\)

Using a GDC to find the roots in the interval \([0, 12]\) yields:
\(t_1 \approx 4.8450...\) and \(t_2 \approx 9.7185...\)

We evaluate \(V(t) = 500 + \int_0^t (I(u) - O(u)) \, du\) at these critical points and the endpoints:
- At \(t = 0\):
\(V(0) = 500\text{ litres}\)

- At \(t \approx 4.845\):
\(V(4.845) = 500 + \int_0^{4.845} (150 \sin^2(0.2u) - 80 e^{0.05u}) \, du\)
\(V(4.845) \approx 500 - 250.10 \approx 249.90\text{ litres}\)

- At \(t = 12\):
\(V(12) = 500 + \int_0^{12} (150 \sin^2(0.2u) - 80 e^{0.05u}) \, du \approx 271.39\text{ litres}\)

Comparing these values, the absolute minimum volume of water is approximately \(250\text{ litres}\), which occurs at \(t \approx 4.85\text{ hours}\).

(c)
To find the maximum volume, we check the endpoints and the local maximum.
The local maximum occurs at \(t_2 \approx 9.7185\) (since \(V'(t)\) changes sign from positive to negative there).

- At \(t \approx 9.719\):
\(V(9.719) = 500 + \int_0^{9.719} (150 \sin^2(0.2u) - 80 e^{0.05u}) \, du \approx 355.34\text{ litres}\)

Comparing this with the endpoint values:
\(V(0) = 500\text{ litres}\)
\(V(9.719) \approx 355\text{ litres}\)
\(V(12) \approx 271\text{ litres}\)

Therefore, the absolute maximum volume of water in the tank is \(500\text{ litres}\) (which occurs at the start, \(t = 0\)).

(d)
The total amount of water that flowed out of the tank is given by:
\(\int_0^{12} O(t) \, dt = \int_0^{12} 80 e^{0.05t} \, dt\)

Evaluating the integral:
\(\int_0^{12} 80 e^{0.05t} \, dt = \left[ \frac{80}{0.05} e^{0.05t} \right]_0^{12} = 1600 (e^{0.6} - 1) \approx 1315.39\text{ litres}\)

To 3 significant figures, this is \(1320\text{ litres}\) (or \(1315\text{ litres}\) if written to 4 s.f.).

(a)
- (M1) for setting up the rate equation \(V'(t) = I(t) - O(t)\).
- (A1) for \(-20.5\text{ litres per hour}\) (accept \(-20.5\) or \(-20.6\) due to minor rounding).
- (A1) for stating that the volume is decreasing because the rate of change is negative.

(b)
- (M1) for recognizing that critical points occur when \(I(t) - O(t) = 0\).
- (A1) for finding both critical times \(t \approx 4.85\) and \(t \approx 9.72\).
- (M1) for evaluating or comparing the volume at \(t \approx 4.85\) with other key points (endpoints \(t = 0\) and \(t = 12\)).
- (R1) for verifying it is a local/global minimum (e.g., noting derivative changes sign from negative to positive, or comparing values: \(V(0) = 500\), \(V(4.85) \approx 250\), \(V(12) \approx 271\)).
- (A1) for the final time \(t \approx 4.85\text{ hours}\) (accept \(4.84\)).

(c)
- (M1) for realizing they must check the local maximum at \(t \approx 9.72\) and the endpoints.
- (A1) for finding the volume at \(t \approx 9.72\) is \(355\text{ litres}\).
- (R1) for comparing \(V(0) = 500\), \(V(9.72) \approx 355\), and \(V(12) \approx 271\).
- (A1) for identifying the maximum volume as \(500\text{ litres}\).

(d)
- (M1) for writing down the correct definite integral \(\int_0^{12} 80 e^{0.05t} \, dt\).
- (A1) for \(1320\text{ litres}\) (or \(1315\text{ litres}\)).

(a) Given that \(X \sim N(\mu, \sigma^2)\):
\(P(X < 100) = 0.10 \implies z_1 = -1.28155...\)
\(P(X > 160) = 0.05 \implies P(X < 160) = 0.95 \implies z_2 = 1.64485...\)

Setting up the standardisation equations:
\(\frac{100 - \mu}{\sigma} = -1.28155\)
\(\frac{160 - \mu}{\sigma} = 1.64485\)

Subtracting the first equation from the second:
\(\frac{60}{\sigma} = 2.92640\)
\(\sigma = \frac{60}{2.92640} \approx 20.503\text{ g}\)

Substituting \(\sigma\) back into the first equation:
\(100 - \mu = -1.28155 \times 20.503\)
\(\mu = 100 + 26.276 \approx 126.276\text{ g}\)

To three significant figures:
\(\mu = 126\) and \(\sigma = 20.5\).

(b) \(P(100 < X < 160) = 1 - P(X < 100) - P(X > 160) = 1 - 0.10 - 0.05 = 0.85\).

(c) Let \(Y\) be the number of Premium apples in a sample of 10. Then \(Y \sim B(10, 0.85)\).
(i) \(P(Y = 8) = \binom{10}{8} (0.85)^8 (0.15)^2 \approx 0.276\).
(ii) \(P(Y \ge 8) = 1 - P(Y \le 7) \approx 0.820\).

(d) Let \(W\) be the number of Premium apples in a box of 40. Then \(W \sim B(40, 0.85)\).
A box is rejected if \(W < 31\) (i.e., \(W \le 30\)).
Using the GDC for cumulative binomial distribution:
\(P(W \le 30) \approx 0.0898\).

(e) Let \(p\) be the new probability of an apple being Premium. We require:
\(P(W \le 30) < 0.07\), where \(W \sim B(40, p)\).
Using GDC trial or solver to find the boundary value for \(p\):
For \(p = 0.853\), \(P(W \le 30) \approx 0.0748\)
For \(p = 0.854\), \(P(W \le 30) \approx 0.0699\)
Thus, we require \(p \ge 0.8538...\) (or \(p \ge 0.854\)).

Now, we require \(P(100 < X < 160) \ge 0.8538...\) where \(X \sim N(\mu, 20.5^2)\).
Using GDC numerical solver to find the boundary values of \(\mu\) for which \(\text{normalcdf}(100, 160, \mu, 20.5) = 0.8538...\):
\(\mu_1 \approx 127.4\text{ g}\)
\(\mu_2 \approx 132.6\text{ g}\)

Therefore, the required range is \(127.4 \le \mu \le 132.6\) grams.
*(Note: If the rounded value \(0.854\) is used, \(127.5 \le \mu \le 132.5\) is obtained. Both are accepted.)*

**Part (a)**
- **M1**: Attempting to find standard normal z-values for both boundaries (e.g., \(-1.28\) and \(1.64\)).
- **M1**: Setting up standardisation equations with \(\mu\) and \(\sigma\).
- **A1**: Finding a correct unrounded value for \(\sigma\) (e.g., \(20.503...\)).
- **A1**: Finding a correct unrounded value for \(\mu\) (e.g., \(126.276...\)) and concluding with the 3 s.f. shown values.

**Part (b)**
- **A1**: Correct probability of \(0.85\) (or equivalent).

**Part (c)**
- **(i) A1**: Correct probability \(0.276\).
- **(ii) M1**: Setting up correct binomcdf expression \(1 - P(Y \le 7)\).
- **A1**: Correct probability \(0.820\).

**Part (d)**
- **M1**: Recognising that "fewer than 31" corresponds to \(W \le 30\) for \(W \sim B(40, 0.85)\).
- **A1**: Correct probability \(0.0898\).

**Part (e)**
- **M1**: Setting up the inequality for the binomial parameter \(P(W \le 30) < 0.07\).
- **A1**: Correct critical probability \(p \ge 0.8538\) (or \(p \ge 0.854\)).
- **M1**: Setting up normal cumulative probability equal to this critical value \(P(100 < X < 160) = 0.8538...\).
- **M1**: Finding two boundary values for \(\mu\) using symmetry or GDC numerical solver.
- **A1**: Correct range \(127.4 \le \mu \le 132.6\) (accept \(127.5 \le \mu \le 132.5\)).

(a) Let \(X\) be the weight of a randomly chosen apple, where \(X \sim N(150, 12^2)\).
Using a GDC to find \(\text{P}(140 \le X \le 165)\):
\(\text{P}(140 \le X \le 165) \approx 0.692022\)
To three significant figures, the probability is 0.692.

(b) We are given \(\text{P}(X < w) = 0.15\).
Using the inverse normal function on a GDC with mean 150 and standard deviation 12:
\(w \approx 137.563\) grams.
To three significant figures, \(w = 138\) grams.

(c) Let \(Y\) be the number of small apples in a box of 20. Then \(Y \sim \text{B}(20, 0.15)\).

(i) We want to find \(\text{P}(Y = 4)\).
Using GDC (binompdf):
\(\text{P}(Y = 4) = \binom{20}{4} (0.15)^4 (0.85)^{16} \approx 0.18214\)
To three significant figures, \(\text{P}(Y = 4) = 0.182\).

(ii) We want to find \(\text{P}(Y \ge 3)\).
\(\text{P}(Y \ge 3) = 1 - \text{P}(Y \le 2)\)
Using GDC (binomcdf):
\(\text{P}(Y \le 2) \approx 0.40490\)
\(\text{P}(Y \ge 3) = 1 - 0.40490 = 0.59510\)
To three significant figures, \(\text{P}(Y \ge 3) = 0.595\).

(d) (i) Let \(V\) be the number of small apples in a bag of 6. Then \(V \sim \text{B}(6, 0.15)\).
A bag is rejected if \(V > 1\).
\(\text{P}(\text{Rejected}) = \text{P}(V \ge 2) = 1 - \text{P}(V \le 1) = 1 - (\text{P}(V = 0) + \text{P}(V = 1))\)
\(\text{P}(V = 0) = (0.85)^6 \approx 0.377150\)
\(\text{P}(V = 1) = 6 \times 0.15 \times (0.85)^5 \approx 0.399335\)
\(\text{P}(V \ge 2) = 1 - (0.377150 + 0.399335) = 1 - 0.776485 = 0.223515\)
To three significant figures, this is 0.224.

(ii) Let \(W\) be the number of rejected bags in 100 bags.
Using the rounded probability \(p = 0.224\), \(W \sim \text{B}(100, 0.224)\).
We want \(\text{P}(W < 20) = \text{P}(W \le 19)\).
Using GDC (binomcdf):
\(\text{P}(W \le 19) \approx 0.24176\)
To three significant figures, the probability is 0.242.

*Note: If using the unrounded probability \(p = 0.223515...\), then \(W \sim \text{B}(100, 0.223515...)\).*
Using GDC (binomcdf):
\(\text{P}(W \le 19) \approx 0.24647\)
To three significant figures, the probability is 0.246.

(e) Let \(V_p \sim \text{B}(6, p)\) be the number of small apples in a bag of 6.
We require \(\text{P}(V_p \ge 2) < 0.1\).
This is equivalent to \(1 - \text{P}(V_p \le 1) < 0.1 \implies \text{P}(V_p \le 1) > 0.9\).
\(\text{P}(V_p \le 1) = (1-p)^6 + 6p(1-p)^5\).
We set up the equation \((1-p)^6 + 6p(1-p)^5 = 0.9\).
Using a GDC numerical solver or by plotting \(y = (1-p)^6 + 6p(1-p)^5 - 0.9\) and finding the root:
\(p \approx 0.09258\).
Thus, the maximum possible value of \(p\) is 0.0926 (or 9.26%).

(a)
**M1** for attempting to use normal cumulative distribution function (sketch or written probability expression).
**A1** for 0.692.

(b)
**M1** for setting up a correct equation or inverse normal statement (e.g. \(\text{P}(X < w) = 0.15\)).
**A1** for 138 (accept 137.6 or 137.56...).

(c)
(i) **M1** for identifying binomial distribution with parameters \(n=20\) and \(p=0.15\).
**A1** for 0.182.
(ii) **M1** for attempting to find \(1 - \text{P}(Y \le 2)\).
**A1** for 0.595.

(d)
(i) **M1** for identifying binomial distribution with \(n=6\) and \(p=0.15\).
**M1** for expressing the rejection probability as \(1 - (\text{P}(V = 0) + \text{P}(V = 1))\) or equivalent.
**A1** for obtaining 0.223515... and showing it rounds to 0.224.

(ii) **M1** for identifying binomial distribution with \(n=100\) and \(p = 0.224\) (or unrounded 0.223515...).
**M1** for identifying the need to calculate \(\text{P}(W \le 19)\).
**A1** for 0.242 (if using 0.224) OR 0.246 (if using 0.223515...).

(e)
**M1** for formulating the inequality/equation in terms of \(p\), e.g., \((1-p)^6 + 6p(1-p)^5 > 0.9\) or \(1 - (1-p)^6 - 6p(1-p)^5 < 0.1\).
**M1** for a valid method to solve the equation/inequality using GDC (e.g. graphing or numerical solver).
**A1** for \(p = 0.0926\) (accept 9.26%).