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Question 1

Topic-2.2 – Logarithmic and Exponential Functions

Use logarithms to show that the equation $5^{8y}=6^{7x}$ can be expressed in the form $y=kx$. Give the value of the constant k correct to 3 significant figures

▶️Answer/Explanation

Take the natural logarithm of both sides:

\[ \ln(5^{8y}) = \ln(6^{7x}) \]

Apply the logarithm power rule ($\ln(a^b) = b \ln a$):

\[ 8y \ln 5 = 7x \ln 6 \]

Solve for $y$:

\[ 8y = 7x \frac{\ln 6}{\ln 5} \]
\[ y = \frac{7 \ln 6}{8 \ln 5} x \]

This is in the form $y = kx$, where:

\[ k = \frac{7 \ln 6}{8 \ln 5} \]

Now, calculate $k$ to 3 significant figures:
$\ln 6 \approx 1.791759$,
$\ln 5 \approx 1.609438$,
$7 \ln 6 \approx 7 \cdot 1.791759 = 12.542313$,
$8 \ln 5 \approx 8 \cdot 1.609438 = 12.875504$,
$k = \frac{12.542313}{12.875504} \approx 0.97411$.

To 3 significant figures, $k \approx 0.974$.

Final Answer:

The equation becomes $y = kx$, where $k = 0.974$.

Question 2

(a) Topic-2.4 – Differentiation

(b) Topic-2.3 – Trigonometry

Let $f(x) = 4sin^{2}3x.$

(a) Find the value of $f'(\frac{1}{4}\pi)$.

(b) Find $\int f(x)dx.$

▶️Answer/Explanation

Part (a): Find $f'(\frac{1}{4}\pi)$

First, compute the derivative $f'(x)$:
$f(x) = 4 \sin^2 3x$,
Use the chain rule: $\frac{d}{dx} [\sin^2 3x] = 2 \sin 3x \cdot \frac{d}{dx} [\sin 3x]$,
$\frac{d}{dx} [\sin 3x] = \cos 3x \cdot 3 = 3 \cos 3x$,
So, $f'(x) = 4 \cdot 2 \sin 3x \cdot 3 \cos 3x = 24 \sin 3x \cos 3x$.

Use the double-angle identity: $2 \sin \theta \cos \theta = \sin 2\theta$:
$f'(x) = 24 \sin 3x \cos 3x = 12 \cdot 2 \sin 3x \cos 3x = 12 \sin 6x$.

Evaluate at $x = \frac{1}{4}\pi$:
$6x = 6 \cdot \frac{1}{4}\pi = \frac{3}{2}\pi$,
$\sin \frac{3}{2}\pi = \sin 270^\circ = -1$,
$f'(\frac{1}{4}\pi) = 12 \cdot (-1) = -12$.

Part (b): Find $\int f(x) \, dx$

Integrate $f(x) = 4 \sin^2 3x$:
Use the identity $\sin^2 \theta = \frac{1 – \cos 2\theta}{2}$,
$f(x) = 4 \cdot \frac{1 – \cos 6x}{2} = 2 – 2 \cos 6x$,
$\int (2 – 2 \cos 6x) \, dx = 2x – 2 \cdot \frac{\sin 6x}{6} + C = 2x – \frac{1}{3} \sin 6x + C$.

Final Answer:

(a) $f'(\frac{1}{4}\pi) = -12$

(b) $\int f(x) \, dx = 2x – \frac{1}{3} \sin 6x + C$

Question 3

Topic-2.4- Differentiation

A curve has equation $6e^{-x}y^{2}+e^{2x}-12y+7=0.$

Find the gradient of the curve at the point $(\ln 3, 2).$

▶️Answer/Explanation

Define the equation as $F(x, y) = 6e^{-x} y^2 + e^{2x} – 12y + 7 = 0$. The gradient is $\frac{dy}{dx}$, found by:

\[ \frac{d}{dx} [F(x, y)] = 0 \]

Differentiate each term:
$\frac{d}{dx} [6e^{-x} y^2] = 6 \left[ e^{-x} \cdot 2y \frac{dy}{dx} + (-e^{-x}) y^2 \right] = 12 e^{-x} y \frac{dy}{dx} – 6 e^{-x} y^2$,
$\frac{d}{dx} [e^{2x}] = 2 e^{2x}$,
$\frac{d}{dx} [-12y] = -12 \frac{dy}{dx}$,
$\frac{d}{dx} [7] = 0$.

So:

\[ 12 e^{-x} y \frac{dy}{dx} – 6 e^{-x} y^2 + 2 e^{2x} – 12 \frac{dy}{dx} = 0 \]

Solve for $\frac{dy}{dx}$:
\[ (12 e^{-x} y – 12) \frac{dy}{dx} = 6 e^{-x} y^2 – 2 e^{2x} \]
\[ \frac{dy}{dx} = \frac{6 e^{-x} y^2 – 2 e^{2x}}{12 e^{-x} y – 12} \]

Evaluate at $(\ln 3, 2)$:
$x = \ln 3$, $e^x = 3$, $e^{-x} = \frac{1}{3}$, $e^{2x} = e^{2 \ln 3} = 9$,
$y = 2$,
Numerator: $6 \cdot \frac{1}{3} \cdot 2^2 – 2 \cdot 9 = 6 \cdot \frac{1}{3} \cdot 4 – 18 = 8 – 18 = -10$,
Denominator: $12 \cdot \frac{1}{3} \cdot 2 – 12 = 12 \cdot \frac{2}{3} – 12 = 8 – 12 = -4$,
$\frac{dy}{dx} = \frac{-10}{-4} = \frac{5}{2}$.

Final Answer:

Gradient at $(\ln 3, 2) = \frac{5}{2}$.

Question 4

(a) Topic-2.2- Logarithmic and exponential functions

(b) Topic-2.2- Logarithmic and exponential functions

(a) Sketch the graphs of $y=1+e^{2x}$ and $y=|x-4|$ on the same diagram.

(b) The two graphs meet at the point P.

Show that the x-coordinate of P satisfies the equation $x=\frac{1}{2}ln(3-x)$.

(c) Use an iterative formula, based on the equation in part (b), to find the x-coordinate of P correct to 3 significant figures. Use an initial value of 0.45 and give the result of each iteration to 5 significant figures.

▶️Answer/Explanation

Part (a): Sketch the graphs

$y = 1 + e^{2x}$:
Exponential growth, minimum at $x = -\infty$ where $y \to 1$, increases rapidly as $x$ increases.
At $x = 0$, $y = 1 + e^0 = 2$.

$y = |x – 4|$:
V-shape, vertex at $(4, 0)$, symmetric about $x = 4$.
At $x = 0$, $y = |0 – 4| = 4$; at $x = 4$, $y = 0$.

Sketch: The graphs intersect once to the left of $x = 4$, since $1 + e^{2x}$ starts below $|x – 4|$ and grows faster.

Part (b): Show the x-coordinate of P satisfies $x = \frac{1}{2} \ln (3 – x)$

At intersection point P, $y = 1 + e^{2x} = |x – 4|$. Since P is left of $x = 4$ (from the sketch), $x < 4$, so $|x – 4| = 4 – x$:

\[ 1 + e^{2x} = 4 – x \]
\[ e^{2x} = 4 – x – 1 \]
\[ e^{2x} = 3 – x \]
\[ 2x = \ln (3 – x) \]
\[ x = \frac{1}{2} \ln (3 – x) \]

This is the required equation (noting $x < 4$, and $3 – x > 0$).

Part (c): Iterative solution for $x$ to 3 significant figures

Use the iterative formula:

\[ x_{n+1} = \frac{1}{2} \ln (3 – x_n) \]

Initial value: $x_0 = 0.45$. Compute to 5 significant figures:

$x_0 = 0.45$
$3 – 0.45 = 2.55$
$\ln 2.55 \approx 0.93609$
$x_1 = \frac{1}{2} \cdot 0.93609 = 0.46805$

$x_1 = 0.46805$
$3 – 0.46805 = 2.53195$
$\ln 2.53195 \approx 0.92882$
$x_2 = \frac{1}{2} \cdot 0.92882 = 0.46441$

$x_2 = 0.46441$
$3 – 0.46441 = 2.53559$
$\ln 2.53559 \approx 0.93036$
$x_3 = \frac{1}{2} \cdot 0.93036 = 0.46518$

$x_3 = 0.46518$
$3 – 0.46518 = 2.53482$
$\ln 2.53482 \approx 0.93005$
$x_4 = \frac{1}{2} \cdot 0.93005 = 0.46502$

$x_4 = 0.46502$
$3 – 0.46502 = 2.53498$
$\ln 2.53498 \approx 0.93011$
$x_5 = \frac{1}{2} \cdot 0.93011 = 0.46506$

Converges to $x \approx 0.465$ (3 significant figures).

Final Answer:

(a) Sketch: $y = 1 + e^{2x}$ (exponential, starts at $y = 1$), $y = |x – 4|$ (V at $(4, 0)$), intersect left of $x = 4$.

(b) $x = \frac{1}{2} \ln (3 – x)$ (shown).

Question 5

(a) Topic- 2.1- Algebra

(b) Topic- 2.1- Algebra

The polynomial $p(x)$ is defined by

$p(x)=ax^{3}+bx^{2}-ax+8,$

where a and b are constants. It is given that $(x+2)$ is a factor of $p(x)$, and that the remainder is 24 when
$p(x)$ is divided by $(x-2)$.

(a) Find the values of a and b.

(b) Factorise $p(x)$ and hence show that the equation $p(x)=0$ has exactly one real root.

▶️Answer/Explanation

Part (a): Find $a$ and $b$

Given $(x + 2)$ is a factor:
$p(-2) = 0$.
$p(-2) = a(-2)^3 + b(-2)^2 – a(-2) + 8 = -8a + 4b + 2a + 8 = -6a + 4b + 8 = 0$.
$-6a + 4b + 8 = 0 \implies -3a + 2b + 4 = 0 \quad (1)$.

Remainder is 24 when divided by $(x – 2)$:
$p(2) = 24$.
$p(2) = a(2)^3 + b(2)^2 – a(2) + 8 = 8a + 4b – 2a + 8 = 6a + 4b + 8 = 24$.
$6a + 4b + 8 = 24 \implies 6a + 4b = 16 \implies 3a + 2b = 8 \quad (2)$.

Solve (1) and (2):
$-3a + 2b = -4$
$3a + 2b = 8$
Add: $4b = 4 \implies b = 1$.
Substitute into (2): $3a + 2 \cdot 1 = 8 \implies 3a + 2 = 8 \implies 3a = 6 \implies a = 2$.

So, $a = 2$, $b = 1$.

Part (b): Factorize $p(x)$ and show one real root

With $a = 2$, $b = 1$:
$p(x) = 2x^3 + x^2 – 2x + 8$.

Since $(x + 2)$ is a factor, use synthetic division with $x = -2$:
Coefficients: $2, 1, -2, 8$.
\[
\begin{array}{r|rrrr}
-2 & 2 & 1 & -2 & 8 \\
& & -4 & 6 & -8 \\
\hline
& 2 & -3 & 4 & 0 \\
\end{array}
\]
Quotient: $2x^2 – 3x + 4$, remainder 0.
$p(x) = (x + 2)(2x^2 – 3x + 4)$.

Check $2x^2 – 3x + 4$ for roots:
Discriminant: $\Delta = (-3)^2 – 4 \cdot 2 \cdot 4 = 9 – 32 = -23 < 0$. No real roots.

Roots of $p(x) = 0$:
$(x + 2)(2x^2 – 3x + 4) = 0 \implies x + 2 = 0 \implies x = -2$.
The quadratic has no real roots, so $x = -2$ is the only real root.

Part (c): Solve $p\left(\frac{1}{2} \csc \theta\right) = 0$ for $-90^\circ < \theta < 90^\circ$

$p\left(\frac{1}{2} \csc \theta\right) = 0$.
Since $p(x) = 0$ at $x = -2$:
$\frac{1}{2} \csc \theta = -2 \implies \csc \theta = -4 \implies \sin \theta = -\frac{1}{4}$.

For $-90^\circ < \theta < 90^\circ$:
$\sin \theta = -\frac{1}{4} \implies \theta = -\arcsin\left(\frac{1}{4}\right) \approx -14.48^\circ$, which is in the interval.
$\sin \theta$ is positive at $\arcsin\left(\frac{1}{4}\right) \approx 14.48^\circ$, but we need the negative case.

Final Answer:

(a) $a = 2$, $b = 1$

(b) $p(x) = (x + 2)(2x^2 – 3x + 4)$, one real root at $x = -2$

Question 6

(a) Topic-2.6 – Numerical solution of equations

(b) Topic-2.5 – Integration

(d) Topic-2.5 – Integration

The diagram shows the curves with equations
$y = \sqrt[3]{5x^{2} + 7}$
and
$y = \frac{27}{2x + 5}$
for
$x \geq 0$.

The curves meet at the point (2, 3).

Region A is bounded by the curve
$y = \sqrt[3]{5x^{2} + 7}$
and the straight lines
$x = 0$ ,
$x = 2$
and
$y = 0$

Region B is bounded by the two curves and the straight line
$x = 0.$

(a) Use the trapezium rule with two intervals to find an approximation to the area of region A. Give
your answer correct to 3 significant figures.

(b) Find the exact total area of regions A and B. Give your answer in the form klnm, where k and m are constants.

(d) State, with a reason, whether your answer to part (c) is an over-estimate or an under-estimate of
the area of region B.

▶️Answer/Explanation

Solution :-

(a) Use y-values $\sqrt[3]{7}$, $\sqrt[3]{12}$ and $\sqrt[3]{27}$, or decimal equivalents

Use correct formula, or equivalent, with $h=1$

Obtain 4.75

(b) Integrate to obtain form $k~ln(2x+5)$

Obtain correct $\frac{27}{2}ln(2x+5)$

Apply limits 0 and 2 to obtain $\frac{27}{2}ln~9$ or exact equivalent of required form and
no extra terms

(d) State under-estimate …

… because (a) is over-estimate due to tops of trapezia being above curve

Question 7

(a) Topic-2.3 -Trigonometry

(b) Topic-2.3 -Trigonometry

(a) Express $4~sin~\theta~sin(\theta+60^{\circ})$ in the form

$a+R~sin(2\theta-\alpha),$

where a and R are positive integers and $0^{\circ}<\alpha<90^{\circ}$.

(b) Hence find the smallest positive value of $\theta$ satisfying the equation

$\frac{1}{5}+4~sin~\theta~sin(\theta+60^{\circ})=0.$

▶️Answer/Explanation

Part (a): Express $4 \sin \theta \sin(\theta + 60^\circ)$ as $a + R \sin(2\theta – \alpha)$

Use the identity $\sin A \sin B = \frac{1}{2} [\cos(A – B) – \cos(A + B)]$:
$A = \theta$, $B = \theta + 60^\circ$,
$4 \sin \theta \sin(\theta + 60^\circ) = 4 \cdot \frac{1}{2} [\cos(\theta – (\theta + 60^\circ)) – \cos(\theta + \theta + 60^\circ)] = 2 [\cos(-60^\circ) – \cos(2\theta + 60^\circ)]$,
$\cos(-60^\circ) = \cos 60^\circ = \frac{1}{2}$,
$= 2 \left[ \frac{1}{2} – \cos(2\theta + 60^\circ) \right] = 1 – 2 \cos(2\theta + 60^\circ)$.

Rewrite $-\cos(2\theta + 60^\circ) = \sin(2\theta + 60^\circ – 90^\circ) = \sin(2\theta – 30^\circ)$:
$4 \sin \theta \sin(\theta + 60^\circ) = 1 – 2 \cos(2\theta + 60^\circ) = 1 + 2 \sin(2\theta – 30^\circ)$.

Matches form $a + R \sin(2\theta – \alpha)$:
$a = 1$, $R = 2$, $\alpha = -30^\circ$.
But $\alpha$ must be in $(0^\circ, 90^\circ)$. Adjust:
$\sin(2\theta – 30^\circ) = -\sin(2\theta – 30^\circ – 180^\circ) = -\sin(2\theta – 210^\circ)$,
$1 + 2 \sin(2\theta – 30^\circ) = 1 – 2 \sin(2\theta – 210^\circ)$,
Still negative $\alpha$. Test: Original fits with $\alpha = 30^\circ$, but let’s correct:
Use identity correctly: Should be $1 + 2 \sin(2\theta + 30^\circ)$, adjust phase:
Final check: $1 – 2 \cos(2\theta + 60^\circ)$, use $-\cos x = \sin(x – 90^\circ)$, but let’s derive properly: Correct form is:
$R \sin(2\theta – \alpha) = 2 \sin(30^\circ – 2\theta)$, so adjust:
Final: $4 \sin \theta \sin(\theta + 60^\circ) = 1 + 2 \sin(2\theta – 30^\circ)$, $\alpha = 30^\circ$ after correction.

Part (b): Smallest positive $\theta$ for $\frac{1}{5} + 4 \sin \theta \sin(\theta + 60^\circ) = 0$

\[ \frac{1}{5} + 1 + 2 \sin(2\theta – 30^\circ) = 0 \]
\[ 1 + \frac{1}{5} + 2 \sin(2\theta – 30^\circ) = 0 \]
\[ \frac{6}{5} + 2 \sin(2\theta – 30^\circ) = 0 \]
\[ 2 \sin(2\theta – 30^\circ) = -\frac{6}{5} \]
\[ \sin(2\theta – 30^\circ) = -\frac{3}{5} \]

Solve for smallest positive $\theta$:
$2\theta – 30^\circ = \arcsin\left(-\frac{3}{5}\right) \approx -36.87^\circ$ or $180^\circ – (-36.87^\circ) = 216.87^\circ$,
$2\theta – 30^\circ = -36.87^\circ + 360k^\circ$,
$2\theta = -6.87^\circ + 360k^\circ$,
$\theta = -3.435^\circ + 180k^\circ$.
Smallest positive: $k = 1$, $\theta = -3.435^\circ + 180^\circ = 176.565^\circ$.

Final Answer:

(a) $4 \sin \theta \sin(\theta + 60^\circ) = 1 + 2 \sin(2\theta – 30^\circ)$, $a = 1$, $R = 2$, $\alpha = 30^\circ$

(b) $\theta = 176.565^\circ$ (smallest positive)

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