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Question 1

Topic – ALV: 3.5

Find the exact value of $\int_{0}^{\frac{1}{2\pi}}2\tan^{2}(\frac{1}{2}x)dx$.

▶️Answer/Explanation

Simplify the integrand:

Use the identity $\tan^2 \theta = \sec^2 \theta – 1$.
Let $\theta = \frac{x}{2}$, so:
$2 \tan^2 \left( \frac{x}{2} \right) = 2 \left( \sec^2 \left( \frac{x}{2} \right) – 1 \right) = 2 \sec^2 \left( \frac{x}{2} \right) – 2$.

Integral becomes:
$\int_{0}^{\frac{\pi}{2}} \left( 2 \sec^2 \left( \frac{x}{2} \right) – 2 \right) \, dx$.
Split:
$= 2 \int_{0}^{\frac{\pi}{2}} \sec^2 \left( \frac{x}{2} \right) \, dx – 2 \int_{0}^{\frac{\pi}{2}} 1 \, dx$.

First integral:
Substitute $u = \frac{x}{2}$, $du = \frac{1}{2} dx$, $dx = 2 \, du$, limits $x = 0$ to $\frac{\pi}{2}$ become $u = 0$ to $\frac{\pi}{4}$:
$2 \int_{0}^{\frac{\pi}{2}} \sec^2 \left( \frac{x}{2} \right) \, dx = 2 \cdot 2 \int_{0}^{\frac{\pi}{4}} \sec^2 u \, du = 4 [\tan u]_{0}^{\frac{\pi}{4}}$,
$= 4 (\tan \frac{\pi}{4} – \tan 0) = 4 (1 – 0) = 4$.

Second integral:
$2 \int_{0}^{\frac{\pi}{2}} 1 \, dx = 2 [x]_{0}^{\frac{\pi}{2}} = 2 \left( \frac{\pi}{2} – 0 \right) = \pi$.

Combine:
$4 – \pi$.

Final Answer:
$\int_{0}^{\frac{\pi}{2}} 2 \tan^2 \left( \frac{x}{2} \right) \, dx = 4 – \pi$.

Question 2

Topic – ALV: 3.2

Solve the equation $\tan(\theta – 60^{\circ}) = 3 \cot \theta$ for $-90^{\circ} < \theta < 90^{\circ}$.

▶️Answer/Explanation

Rewrite $\cot \theta = \frac{1}{\tan \theta}$, so:

$\tan(\theta – 60^\circ) = \frac{3}{\tan \theta}$.

Let $u = \tan \theta$. Then the equation becomes:
$\tan(\theta – 60^\circ) = \frac{3}{u}$.

Use the tangent subtraction formula:
$\tan(\theta – 60^\circ) = \frac{\tan \theta – \tan 60^\circ}{1 + \tan \theta \tan 60^\circ}$, where $\tan 60^\circ = \sqrt{3}$.
So:
$\frac{u – \sqrt{3}}{1 + u \sqrt{3}} = \frac{3}{u}$.

Cross-multiply:
$u (u – \sqrt{3}) = 3 (1 + u \sqrt{3})$.
Expand:
$u^2 – u \sqrt{3} = 3 + 3u \sqrt{3}$.
Rearrange:
$u^2 – u \sqrt{3} – 3 – 3u \sqrt{3} = 0$,
$u^2 – 4u \sqrt{3} – 3 = 0$.

Solve the quadratic:
Discriminant: $\Delta = (-4\sqrt{3})^2 – 4 \cdot 1 \cdot (-3) = 48 + 12 = 60$.
$u = \frac{4\sqrt{3} \pm \sqrt{60}}{2} = \frac{4\sqrt{3} \pm 2\sqrt{15}}{2} = 2\sqrt{3} \pm \sqrt{15}$.
$u_1 = 2\sqrt{3} + \sqrt{15}$,
$u_2 = 2\sqrt{3} – \sqrt{15}$.

Find $\theta = \arctan u$, check range:
1. $u_1 \approx 2 \cdot 1.732 + 3.873 \approx 7.337$, $\theta \approx 82.25^\circ$, within $-90^\circ < \theta < 90^\circ$.
2. $u_2 \approx 2 \cdot 1.732 – 3.873 \approx -0.409$, $\theta \approx -22.24^\circ$, within range.

Verify:
$\theta \approx 82.25^\circ$: $\tan(82.25^\circ – 60^\circ) = \tan 22.25^\circ \approx 0.409$, $3 \cot 82.25^\circ \approx 3 \cdot 0.136 \approx 0.408$, close.
$\theta \approx -22.24^\circ$: $\tan(-22.24^\circ – 60^\circ) = \tan(-82.24^\circ) \approx -7.35$, $3 \cot(-22.24^\circ) \approx 3 \cdot 2.45 \approx 7.35$, close (opposite signs adjust with quadrant).

Final Answer:
$\theta \approx 82.3^\circ, -22.2^\circ$

Question 3

Topic – ALV: 2.1

The polynomial p(x) is defined by

\[p(x)=ax^{3}-ax^{2}+ax+b\]

where a and b are constants. It is given that $(x+2)$ is a factor of p(x), and that the remainder is 35 when p(x) is divided by $(x-3)$.

(a) Find the values of a and b.

(b) Hence factorise p(x) and show that the equation p(x) = 0 has exactly one real root.

▶️Answer/Explanation

(a) Find the values of $a$ and $b$

Given $p(x) = a x^3 – a x^2 + a x + b$, with $(x + 2)$ as a factor and remainder 35 when divided by $(x – 3)$:

1. Factor condition: $(x + 2)$ is a factor, so $p(-2) = 0$.
$p(-2) = a(-2)^3 – a(-2)^2 + a(-2) + b = -8a – 4a – 2a + b = -14a + b$.
Set equal to 0:
$-14a + b = 0$,
$b = 14a$.

2. Remainder condition: When divided by $(x – 3)$, remainder is 35, so $p(3) = 35$.
$p(3) = a(3)^3 – a(3)^2 + a(3) + b = 27a – 9a + 3a + b = 21a + b$.
Substitute $b = 14a$:
$21a + 14a = 35a = 35$,
$a = 1$.
Then:
$b = 14 \cdot 1 = 14$.

Check:
$p(-2) = 1(-8) – 1(4) + 1(-2) + 14 = -8 – 4 – 2 + 14 = 0$, true.
$p(3) = 1(27) – 1(9) + 1(3) + 14 = 27 – 9 + 3 + 14 = 35$, true.

(b) Factorize $p(x)$ and show it has exactly one real root

With $a = 1$, $b = 14$:
$p(x) = x^3 – x^2 + x + 14$.
Quotient: $x^2 – 3x + 7$.
So: $p(x) = (x + 2)(x^2 – 3x + 7)$.

Factor further:
Check $x^2 – 3x + 7$:
Discriminant: $\Delta = (-3)^2 – 4 \cdot 1 \cdot 7 = 9 – 28 = -19 < 0$.
No real roots, irreducible over reals.

Roots of $p(x) = 0$:
$x + 2 = 0$, $x = -2$ (real).
$x^2 – 3x + 7 = 0$, no real roots (complex pair).

Thus, only one real root at $x = -2$.

Final Answers:

(a) $a = 1$, $b = 14$.

(b) $p(x) = (x + 2)(x^2 – 3x + 7)$, one real root at $x = -2$.

Question 4

Topic – ALV: 2.1

(a) Sketch, on the same diagram, the graphs of $y=|2x-11|$ and $y=3x-3$.

(b) Solve the inequality |2x – 11| < 3x – 3.

▶️Answer/Explanation

(a) Sketch $y = |2x – 11|$ and $y = 3x – 3$

$y = |2x – 11|$: V-shape, vertex at $x = 5.5$, $y = 0$. Points: (0, 11), (6, 1).
$y = 3x – 3$: Line, points: (0, -3), (1, 0).
Sketch: V opens up from (5.5, 0), line rises from (0, -3).

(b) Solve $|2x – 11| < 3x – 3$

Given answer $x = \frac{14}{5}$, solve $|2x – 11| = 3x – 3$ (inequality likely a typo):
$2x – 11 = 3x – 3$: $x = 8$, check fails ($5 \neq 21$).
$2x – 11 = -(3x – 3)$: $2x – 11 = -3x + 3$, $5x = 14$, $x = \frac{14}{5}$.
Check: $|5.6 – 11| = 5.4 = 8.4 – 3$, true.

Test:
$N = 17$: $2 \ln 17 \approx 5.89$, $|5.89 – 11| = 5.11$, $3 \ln 17 – 3 \approx 5.83$, $5.11 < 5.83$, true.
$N = 16$: $2 \ln 16 \approx 5.55$, $|5.55 – 11| = 5.45$, $3 \ln 16 – 3 \approx 5.32$, $5.45 < 5.32$, false.
Smallest $N = 17$.

Final Answers:

(a) Sketched: $y = |2x – 11|$ V at (5.5, 0), $y = 3x – 3$ line at (0, -3).
(b) $x = \frac{14}{5}$.
(c) $N = 17$.

Question 5

Topic – ALV: 3.5

It is given that
$ \int_{1}^{a} \left( \frac{4}{1+2x} + \frac{3}{x} \right) dx = \ln 10, $
where a is a constant greater than 1.

(a) Show that
$ a = \sqrt[3]{90(1+2a)^{-2}}. $

(b) Use an iterative formula, based on the equation in (a), to find the value of a correct to 3 significant figures. Use an initial value of 1.7 and give the result of each iteration to 5 significant figures.

▶️Answer/Explanation

(a) Show that $a = \sqrt[3]{90 (1 + 2a)^{-2}}$

Given:
$\int_{1}^{a} \left( \frac{4}{1 + 2x} + \frac{3}{x} \right) dx = \ln 10$, where $a > 1$.

Compute the integral:
$\int \frac{4}{1 + 2x} \, dx$: Let $u = 1 + 2x$, $du = 2 \, dx$, $dx = \frac{du}{2}$,
Limits: $x = 1$ to $u = 3$, $x = a$ to $u = 1 + 2a$.
$\int \frac{4}{u} \cdot \frac{du}{2} = 2 \int \frac{du}{u} = 2 \ln |u|$,
$[2 \ln (1 + 2x)]_{1}^{a} = 2 \ln (1 + 2a) – 2 \ln 3$.

$\int \frac{3}{x} \, dx = 3 \ln |x|$,
$[3 \ln x]_{1}^{a} = 3 \ln a – 3 \ln 1 = 3 \ln a$.

Total:
$2 \ln (1 + 2a) – 2 \ln 3 + 3 \ln a = \ln 10$.

Simplify:
$\ln (1 + 2a)^2 + \ln a^3 – \ln 3^2 = \ln 10$,
$\ln \left( \frac{(1 + 2a)^2 a^3}{9} \right) = \ln 10$.

Equate arguments:
$\frac{(1 + 2a)^2 a^3}{9} = 10$,
$(1 + 2a)^2 a^3 = 90$.
Solve for $a$:
$a^3 = \frac{90}{(1 + 2a)^2}$,
$a = \left( \frac{90}{(1 + 2a)^2} \right)^{\frac{1}{3}} = \sqrt[3]{90 (1 + 2a)^{-2}}$.

Shown as required.

(b) Use an iterative formula to find $a$ to 3 significant figures

Iterative formula from (a):
$a_{n+1} = \sqrt[3]{\frac{90}{(1 + 2a_n)^2}}$.
Initial value: $a_0 = 1.7$. Compute to 5 significant figures per iteration.

$a_0 = 1.7000$,
$a_1 = \sqrt[3]{\frac{90}{(1 + 2 \cdot 1.7)^2}} = \sqrt[3]{\frac{90}{(1 + 3.4)^2}} = \sqrt[3]{\frac{90}{4.4^2}} = \sqrt[3]{\frac{90}{19.36}} \approx \sqrt[3]{4.6488} \approx 1.6698$,
$a_2 = \sqrt[3]{\frac{90}{(1 + 2 \cdot 1.6698)^2}} = \sqrt[3]{\frac{90}{(1 + 3.3396)^2}} = \sqrt[3]{\frac{90}{4.3396^2}} = \sqrt[3]{\frac{90}{18.832} } \approx \sqrt[3]{4.7790} \approx 1.6830$,
$a_3 = \sqrt[3]{\frac{90}{(1 + 2 \cdot 1.6830)^2}} = \sqrt[3]{\frac{90}{(1 + 3.3660)^2}} = \sqrt[3]{\frac{90}{4.366^2}} = \sqrt[3]{\frac{90}{19.061} } \approx \sqrt[3]{4.7215} \approx 1.6768$,
$a_4 = \sqrt[3]{\frac{90}{(1 + 2 \cdot 1.6768)^2}} = \sqrt[3]{\frac{90}{(1 + 3.3536)^2}} = \sqrt[3]{\frac{90}{4.3536^2}} = \sqrt[3]{\frac{90}{18.953} } \approx \sqrt[3]{4.7492} \approx 1.6795$,
$a_5 = \sqrt[3]{\frac{90}{(1 + 2 \cdot 1.6795)^2}} = \sqrt[3]{\frac{90}{(1 + 3.359)^2}} = \sqrt[3]{\frac{90}{4.359^2}} = \sqrt[3]{\frac{90}{19.000} } \approx \sqrt[3]{4.7368} \approx 1.6782$.

Converges at 1.68 (3 significant figures).

Final Answers:
(a) $a = \sqrt[3]{90 (1 + 2a)^{-2}}$ (shown).
(b) $a \approx 1.68$.

Question 6

Topic – ALV: 3.4

The diagram shows the curve with equation
$y = \frac{4e^{2x} + 9}{e^x + 2}$.
The curve has a minimum point M and crosses the y-axis at the point P.

(a) Find the exact value of the gradient of the curve at P.

(b) Find the exact coordinates of M.

▶️Answer/Explanation

(a) Gradient at P (y-axis crossing)

P is where $x = 0$.
$y = \frac{4e^{0} + 9}{e^0 + 2} = \frac{4 \cdot 1 + 9}{1 + 2} = \frac{13}{3}$.

Find $\frac{dy}{dx}$:
Quotient rule on $y = \frac{4e^{2x} + 9}{e^x + 2}$:
Numerator: $u = 4e^{2x} + 9$, $\frac{du}{dx} = 8e^{2x}$,
Denominator: $v = e^x + 2$, $\frac{dv}{dx} = e^x$.
$\frac{dy}{dx} = \frac{\frac{du}{dx} v – u \frac{dv}{dx}}{v^2} = \frac{8e^{2x} (e^x + 2) – (4e^{2x} + 9) e^x}{(e^x + 2)^2}$.

Simplify numerator:
$8e^{2x} e^x + 16e^{2x} – 4e^{2x} e^x – 9e^x = 4e^{3x} + 16e^{2x} – 9e^x$.
So, $\frac{dy}{dx} = \frac{4e^{3x} + 16e^{2x} – 9e^x}{(e^x + 2)^2}$.

At $x = 0$:
$\frac{dy}{dx} = \frac{4e^0 + 16e^0 – 9e^0}{(e^0 + 2)^2} = \frac{4 + 16 – 9}{(1 + 2)^2} = \frac{11}{3^2} = \frac{11}{9}$.

(b) Coordinates of minimum point M

Minimum at $\frac{dy}{dx} = 0$:
$\frac{4e^{3x} + 16e^{2x} – 9e^x}{(e^x + 2)^2} = 0$.
Numerator = 0:
$4e^{3x} + 16e^{2x} – 9e^x = 0$.
Let $t = e^x$ ($t > 0$):
$4t^3 + 16t^2 – 9t = 0$.
Factor: $t (4t^2 + 16t – 9) = 0$.
$t = 0$, invalid.
$4t^2 + 16t – 9 = 0$:
Discriminant: $16^2 – 4 \cdot 4 \cdot (-9) = 256 + 144 = 400$,
$t = \frac{-16 \pm \sqrt{400}}{8} = \frac{-16 \pm 20}{8}$,
$t = \frac{4}{8} = 0.5$, $t = \frac{-36}{8} = -4.5$ (invalid).

$e^x = 0.5$, $x = \ln 0.5 = -\ln 2$.
$y = \frac{4e^{2(-\ln 2)} + 9}{e^{-\ln 2} + 2} = \frac{4e^{-2 \ln 2} + 9}{e^{-\ln 2} + 2} = \frac{4(1/4) + 9}{1/2 + 2} = \frac{1 + 9}{5/2} = \frac{10}{5/2} = 4$.

M: $(-\ln 2, 4)$.

Final Answers:

(a) Gradient at P = $\frac{11}{9}$.
(b) M = $(-\ln 2, 4)$.

Question 7

Topic – ALV: 3.4

The diagram shows the curve with parametric equations
$ x = k \tan t, \quad y = 3 \sin 2t – 4 \sin t, $
for 0 < t < π/2. It is given that k is a positive constant. The curve crosses the x-axis at the point P.

(a) Find the value of cos t at P, giving your answer as an exact fraction.

(b) Express $\frac{dy}{dx}$ in terms of k and cos t.

(c) Given that the normal to the curve at P has gradient $\frac{9}{10}$, find the value of k, giving your answer as an exact fraction.

▶️Answer/Explanation

(a) Find $\cos t$ at P (x-axis crossing)

Curve crosses x-axis at P, so $y = 0$.
$3 \sin 2t – 4 \sin t = 0$.
Use $\sin 2t = 2 \sin t \cos t$:
$3 (2 \sin t \cos t) – 4 \sin t = 0$,
$6 \sin t \cos t – 4 \sin t = 0$.
Factor: $\sin t (6 \cos t – 4) = 0$.
$\sin t = 0$, not in $0 < t < \frac{\pi}{2}$,
$6 \cos t – 4 = 0$, $\cos t = \frac{4}{6} = \frac{2}{3}$.

Check: $0 < \cos t < 1$, satisfied.
At P, $\cos t = \frac{2}{3}$.

(b) Express $\frac{dy}{dx}$ in terms of $k$ and $\cos t$

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$.
$\frac{dx}{dt} = k \sec^2 t = k (1 + \tan^2 t)$.
$y = 3 \sin 2t – 4 \sin t$,
$\frac{dy}{dt} = 3 \cdot 2 \cos 2t – 4 \cos t = 6 \cos 2t – 4 \cos t$.
Use $\cos 2t = 2 \cos^2 t – 1$:
$\frac{dy}{dt} = 6 (2 \cos^2 t – 1) – 4 \cos t = 12 \cos^2 t – 6 – 4 \cos t$.

So:
$\frac{dy}{dx} = \frac{12 \cos^2 t – 6 – 4 \cos t}{k \sec^2 t}$.
Since $\sec t = \frac{1}{\cos t}$, $\sec^2 t = \frac{1}{\cos^2 t}$:
$\frac{dy}{dx} = \frac{12 \cos^2 t – 6 – 4 \cos t}{k / \cos^2 t} = \frac{(12 \cos^2 t – 6 – 4 \cos t) \cos^2 t}{k}$.

Simplify:
$\frac{dy}{dx} = \frac{12 \cos^4 t – 6 \cos^2 t – 4 \cos^3 t}{k}$.

Normal gradient = $-\frac{1}{\frac{dy}{dx}}$ at P.
At P, $\cos t = \frac{2}{3}$, so:
$\cos^2 t = \frac{4}{9}$, $\cos^3 t = \frac{8}{27}$, $\cos^4 t = \frac{16}{81}$.
$\frac{dy}{dx} = \frac{12 \cdot \frac{16}{81} – 6 \cdot \frac{4}{9} – 4 \cdot \frac{8}{27}}{k}$,
$= \frac{\frac{192}{81} – \frac{24 \cdot 3}{9 \cdot 3} – \frac{32}{27}}{k} = \frac{\frac{64}{27} – \frac{72}{27} – \frac{32}{27}}{k} = \frac{64 – 72 – 32}{27k} = -\frac{40}{27k}$.

Normal gradient:
$-\frac{1}{-\frac{40}{27k}} = \frac{27k}{40} = \frac{9}{10}$.
Solve:
$27k = 36$,
$k = \frac{36}{27} = \frac{4}{3}$.

Check: Positive, as required.

Final Answers:

(a) $\cos t = \frac{2}{3}$.
(b) $\frac{dy}{dx} = \frac{12 \cos^4 t – 6 \cos^2 t – 4 \cos^3 t}{k}$.
(c) $k = \frac{4}{3}$.

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