Question 1
(a) Show that$$ \sum_{r=1}^{n} (3r+2)=\frac{n}{2}(3n+7) $$
Click to view the solution
Solution:
(a) Show that $ \sum_{r=1}^n (3r + 2) = \frac{n}{2}(3n + 7) $
We start by expanding the sum:
$ \sum_{r=1}^n (3r + 2) = \sum_{r=1}^n 3r + \sum_{r=1}^n 2 $
Step 1: Evaluate $ \sum_{r=1}^n 3r $
Factor out the constant $3$:
$ \sum_{r=1}^n 3r = 3 \sum_{r=1}^n r $
The sum of the first $n$ integers is given by the formula:
$ \sum_{r=1}^n r = \frac{n(n+1)}{2} $
Thus:
$ \sum_{r=1}^n 3r = 3 \cdot \frac{n(n+1)}{2} = \frac{3n(n+1)}{2} $
Step 2: Evaluate $ \sum_{r=1}^n 2 $
Since $2$ is a constant:
$ \sum_{r=1}^n 2 = 2n $
Step 3: Combine the two sums
Now we can combine both results:
$ \sum_{r=1}^n (3r + 2) = \frac{3n(n+1)}{2} + 2n $
Step 4: Simplify the expression
To simplify, express $2n$ with a denominator of 2:
$ \sum_{r=1}^n (3r + 2) = \frac{3n(n+1)}{2} + \frac{4n}{2} $
Now combine the two terms:
$ \sum_{r=1}^n (3r + 2) = \frac{3n^2 + 3n + 4n}{2} = \frac{3n^2 + 7n}{2} $
Finally, factor the result:
$ \sum_{r=1}^n (3r + 2) = \frac{n}{2}(3n + 7) $
(b) Hence, or otherwise, evaluate $ \sum_{r=10}^{40} (3r + 2) $
We can use the result from part (a) to calculate the sum from $r = 10$ to $r = 40$:
$ \sum_{r=10}^{40} (3r + 2) = \sum_{r=1}^{40} (3r + 2) - \sum_{r=1}^{9} (3r + 2) $
Step 1: Evaluate $ \sum_{r=1}^{40} (3r + 2) $
Using the formula from part (a):
$ \sum_{r=1}^{40} (3r + 2) = \frac{40}{2}(3 \cdot 40 + 7) = 20 \cdot (120 + 7) = 20 \cdot 127 = 2540 $
Step 2: Evaluate $ \sum_{r=1}^{9} (3r + 2) $
Using the formula from part (a):
$ \sum_{r=1}^{9} (3r + 2) = \frac{9}{2}(3 \cdot 9 + 7) = \frac{9}{2} \cdot (27 + 7) = \frac{9}{2} \cdot 34 = 153 $
Step 3: Calculate the desired sum
Now subtract the two sums:
$ \sum_{r=10}^{40} (3r + 2) = 2540 - 153 = 2387 $
Thus, the sum is:
$ \sum_{r=10}^{40} (3r + 2) = 2387 $
Question 2
$$ y=(\sin 2x)\sqrt{3+2x} $$ Show that $$ \frac{dy}{dx} = \frac{\sin 2x+(A+Bx)\cos 2x}{\sqrt{3+2x}} $$ where $A$ and $B$ are integers to be found.Click to view the solution
Solution:
Problem:
$ \begin{aligned} y &= (\sin 2x)\sqrt{3 + 2x} \\ \text{Show that } \frac{dy}{dx} &= \frac{\sin 2x + (A + Bx)\cos 2x}{\sqrt{3 + 2x}} \text{ where } A \text{ and } B \text{ are integers to be found.} \end{aligned} $
We start by applying the product rule to the expression for $y$:
$ y = (\sin 2x)\sqrt{3 + 2x} $
Step 1: Apply the product rule.
The product rule states that:
$ \frac{dy}{dx} = \frac{d}{dx}(\sin 2x)\cdot\sqrt{3 + 2x} + \sin 2x \cdot \frac{d}{dx}(\sqrt{3 + 2x}) $
Step 2: Differentiate each part.
First term: Differentiate $ \sin 2x $.
Using the chain rule:
$ \frac{d}{dx}(\sin 2x) = 2\cos 2x $
Second term: Differentiate $ \sqrt{3 + 2x} $.
Using the chain rule:
$ \frac{d}{dx}(\sqrt{3 + 2x}) = \frac{1}{2\sqrt{3 + 2x}} \cdot \frac{d}{dx}(3 + 2x) = \frac{1}{2\sqrt{3 + 2x}} \cdot 2 = \frac{1}{\sqrt{3 + 2x}} $
Step 3: Substitute the derivatives into the product rule.
$ \frac{dy}{dx} = (2\cos 2x)\cdot\sqrt{3 + 2x} + (\sin 2x)\cdot \frac{1}{\sqrt{3 + 2x}} $
This simplifies to:
$ \frac{dy}{dx} = 2\cos 2x \cdot \sqrt{3 + 2x} + \frac{\sin 2x}{\sqrt{3 + 2x}} $
Step 4: Factor out $ \frac{1}{\sqrt{3 + 2x}} $.
$ \frac{dy}{dx} = \frac{1}{\sqrt{3 + 2x}} \left( 2\cos 2x \cdot (3 + 2x) + \sin 2x \right) $
Step 5: Expand the expression inside the parentheses.
$ 2\cos 2x \cdot (3 + 2x) = 6\cos 2x + 4x\cos 2x $
Thus:
$ \frac{dy}{dx} = \frac{1}{\sqrt{3 + 2x}} \left( 6\cos 2x + 4x\cos 2x + \sin 2x \right) $
Step 6: Match the desired form.
Compare with:
$ \frac{\sin 2x + (A + Bx)\cos 2x}{\sqrt{3 + 2x}} $
We see that:
- $\sin 2x$ matches directly
- $(A + Bx)\cos 2x = 6\cos 2x + 4x\cos 2x$
Therefore:
$ A = 6 \quad \text{and} \quad B = 4 $
Final Answer:
$ \frac{dy}{dx} = \frac{\sin 2x + (6 + 4x)\cos 2x}{\sqrt{3 + 2x}} $
where $A = 6$ and $B = 4$.
Question 3
Click to view the solution
Solution:
The graph of the function $ y = \frac{x}{2} + \frac{4}{x^2} $ for $ 0.8 \lt x \lt 7 $ is given.
$ \begin{aligned} 3x^3 - 12x^2 + 8 &= 0 \\ \text{divided by } x^2:\quad 3x - 12 + \frac{8}{x^2} &= 0 \\ \frac{8}{x^2} &= -3x + 12 \\ \frac{4}{x^2} &= -\frac{3}{2}x + 6 \\ \frac{x}{2} + \frac{4}{x^2} &= -x + 6 \\ y &= -x + 6 \end{aligned} $
Thus the equation $ 3x^3 - 12x^2 + 8 = 0 $ is equivalent to finding the intersection between the curve $ y = \frac{x}{2} + \frac{4}{x^2} $ and the straight line $ y = -x + 6 $.
Step 1: Add the straight line.
We plot the straight line $ y = -x + 6 $ together with the curve.
Step 2: Graphical estimation.
Step 3: Estimate the intersection points.
From the graph, the intersections occur approximately at:
- $ x \approx 0.9 $
- $ x \approx 3.8 $
Final Answer:
$ x \approx 0.9 \quad \text{and} \quad x \approx 3.8 $
Question 4
A particle $P$ is moving along the $x$-axis. At time $t$ seconds, $t \geqslant 0$, the velocity, $v \mathrm{~m}/\mathrm{s}$, of $P$ is given by $v=2t^2-16t+30$(a) Find the acceleration, in $\mathrm{m}/\mathrm{s}^2$, of $P$ when $t=5$
[2]
$P$ comes to instantaneous rest at the points $M$ and $N$ at times $t_1$ seconds and $t_2$ seconds where $t_2>t_1$
(b) Find the exact distance $MN$
[8]
Click to view the solution
Solution:
(a) Acceleration at $t = 5$:
The acceleration $a$ is:
$ a = \frac{dv}{dt} $
$ a = \frac{d}{dt}(2t^2 - 16t + 30) = 4t - 16 $
At $t = 5$:
$ a = 4(5) - 16 = 4 \, \mathrm{m/s}^2 $
Answer: $4 \, \mathrm{m/s}^2$
(b) Distance $MN$:
The particle is at rest when $v = 0$:
$ 2t^2 - 16t + 30 = 0 $
$ t^2 - 8t + 15 = 0 $
$ (t - 5)(t - 3) = 0 $
$ t_1 = 3, \quad t_2 = 5 $
Displacement:
$ x = \int_3^5 (2t^2 - 16t + 30)\,dt $
$ x = \left[\frac{2t^3}{3} - 8t^2 + 30t\right]_3^5 $
$ x(3) = 36 $
$ x(5) = \frac{100}{3} $
$ MN = \left| x(5) - x(3) \right| = \left| \frac{100}{3} - 36 \right| = \frac{8}{3} $
Final Answer: $ \frac{8}{3} \, \mathrm{m} $
Question 5
A solid cuboid has width $x \mathrm{~cm}$, length $4x \mathrm{~cm}$ and height $h \mathrm{~cm}$.The volume of the cuboid is $75 \mathrm{~cm}^3$ and the surface area of the cuboid is $S \mathrm{~cm}^2$
(a) Show that $$ S=8x^2+\frac{375}{2x} $$
[4]
Given that $x$ can vary, using calculus,
(b) (i) find to 3 significant figures, the value of $x$ for which $S$ is a minimum,
(ii) justify that this value of $x$ gives a minimum value of $S$
[5]
(c) Find, to 3 significant figures, the minimum value of $S$
[2]
Click to view the solution
Solution:
(a)
Volume:
$ 4x^2 h = 75 \Rightarrow h = \frac{75}{4x^2} $
Surface area:
$ S = 2(4x^2 + xh + 4xh) = 2(4x^2 + 5xh) $
Substitute $h$:
$ S = 2\left(4x^2 + \frac{375}{4x}\right) = 8x^2 + \frac{375}{2x} $
(b)(i)
$ \frac{dS}{dx} = 16x - \frac{375}{2x^2} $
Set to zero:
$ 16x - \frac{375}{2x^2} = 0 $
$ 32x^3 - 375 = 0 \Rightarrow x^3 = \frac{375}{32} $
$ x = \sqrt[3]{\frac{375}{32}} \approx 2.27 \, \mathrm{cm} $
(b)(ii)
$ \frac{d^2S}{dx^2} = 16 + \frac{375}{x^3} $
Since this is positive, $S$ is minimized.
(c)
$ S = 8x^2 + \frac{375}{2x} $
Substitute $x \approx 2.27$:
$ S \approx 124 \, \mathrm{cm}^2 $
Final Answer: $124 \, \mathrm{cm}^2$
Question 6
Solve the equation $$ \log_2 x^3+\log_1 x^2-3\log_x 2=0 $$ giving your answers to 3 significant figures.[8]
Click to view the solution
Solution:
Step 1:
$ \log_2 x^3 = 3\log_2 x $
Step 2:
$ \log_4 x^2 = 2\log_4 x = 2 \cdot \frac{\log_2 x}{2} = \log_2 x $
Step 3:
$ \log_x 2 = \frac{1}{\log_2 x} \quad \Rightarrow \quad -3\log_x 2 = -\frac{3}{\log_2 x} $
Step 4:
$ 3\log_2 x + \log_2 x - \frac{3}{\log_2 x} = 0 $
$ 4\log_2 x - \frac{3}{\log_2 x} = 0 $
Step 5:
$ 4(\log_2 x)^2 - 3 = 0 $
$ (\log_2 x)^2 = \frac{3}{4} $
$ \log_2 x = \pm \frac{\sqrt{3}}{2} $
Step 6:
$ x = 2^{\frac{\sqrt{3}}{2}} \quad \text{or} \quad x = 2^{-\frac{\sqrt{3}}{2}} $
Step 7:
$ x \approx 1.82 \quad \text{or} \quad x \approx 0.549 $
Final Answer:
$ \boxed{x \approx 1.82 \quad \text{and} \quad x \approx 0.549} $
Question 7
The equation of a curve is $$ y=\sqrt{\frac{\mathrm{e}^{4x}}{2x-3}} $$ When $x$ is increased to $(x+\delta x)$, $y$ increases to $(y+\delta y)$ where $\delta x$ and $\delta y$ are small.(a) Show that $$ \delta y \approx \frac{\mathrm{e}^{2x}(4x-7)}{(2x-3)^{\frac{3}{2}}}\delta x $$
[7]
Given that $x=2.5$
(b) find an estimate, to 2 significant figures, of the value of $\delta y$ when the value of $x$ increases by $0.2\%$
[3]
Click to view the solution
Solution:
(a)
$ y = \left(\frac{\mathrm{e}^{4x}}{2x - 3}\right)^{\frac{1}{2}} $
$ \frac{dy}{dx} = \frac{1}{2} \left(\frac{\mathrm{e}^{4x}}{2x - 3}\right)^{-\frac{1}{2}} \cdot \frac{d}{dx}\left(\frac{\mathrm{e}^{4x}}{2x - 3}\right) $
$ \frac{d}{dx}\left(\frac{\mathrm{e}^{4x}}{2x - 3}\right) = \frac{(2x - 3)(4\mathrm{e}^{4x}) - \mathrm{e}^{4x}(2)}{(2x - 3)^2} = \frac{\mathrm{e}^{4x}(8x - 14)}{(2x - 3)^2} $
$ \left(\frac{\mathrm{e}^{4x}}{2x - 3}\right)^{-\frac{1}{2}} = \frac{(2x - 3)^{\frac{1}{2}}}{\mathrm{e}^{2x}} $
$ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{(2x - 3)^{\frac{1}{2}}}{\mathrm{e}^{2x}} \cdot \frac{\mathrm{e}^{4x}(8x - 14)}{(2x - 3)^2} $
$ \frac{dy}{dx} = \frac{\mathrm{e}^{2x}(4x - 7)}{(2x - 3)^{\frac{3}{2}}} $
Hence,
$ \delta y \approx \frac{\mathrm{e}^{2x}(4x - 7)}{(2x - 3)^{\frac{3}{2}}}\delta x $
(b)
$ \delta x = 0.2\% \text{ of } 2.5 = 0.002 \times 2.5 = 0.005 $
$ \delta y \approx \frac{\mathrm{e}^{5}(3)}{2^{\frac{3}{2}}} \cdot 0.005 $
$ \mathrm{e}^5 \approx 148.413 $
$ \delta y \approx \frac{148.413 \times 3}{2.828} \times 0.005 \approx 0.79 $
Final Answer: $0.79$
Question 8
$$ f'(x)=18x^2-2x+13 $$ Given that $(2x-1)$ is a factor of $f(x)$ show that the curve with equation $y=f(x)$ has only one intersection with the $x$-axis.Click to view the solution
Solution:
Step 1: Expression for $f(x)$ by integration:
$ f(x) = \int (18x^2 - 2x + 13)\,dx $
$ f(x) = 6x^3 - x^2 + 13x + C $
where $C$ is a constant.
Step 2: Use $f\left(\frac{1}{2}\right) = 0$ to find $C$:
$ f\left(\frac{1}{2}\right) = 6\left(\frac{1}{2}\right)^3 - \left(\frac{1}{2}\right)^2 + 13\left(\frac{1}{2}\right) + C $
$ = \frac{3}{4} - \frac{1}{4} + \frac{13}{2} + C $
$ = \frac{2}{4} + \frac{26}{4} + C = 7 + C $
$ C = -7 $
$ f(x) = 6x^3 - x^2 + 13x - 7 $
Step 3: Perform synthetic division to confirm $(2x - 1)$ is a factor:
Rewrite $(2x - 1)$ as $x - \frac{1}{2}$. The coefficients are $6, -1, 13, -7$:
$ \begin{array}{r|rrrr} \frac{1}{2} & 6 & -1 & 13 & -7 \\ & & 3 & 1 & 7 \\ \hline & 6 & 2 & 14 & 0 \end{array} $
The remainder is $0$, so $(2x - 1)$ is a factor.
Step 4: Factorize $f(x)$:
$ f(x) = \left(x - \frac{1}{2}\right)(6x^2 + 2x + 14) = (2x - 1)(3x^2 + x + 7) $
Discriminant:
$ \Delta = 1^2 - 4(3)(7) = -83 $
Since $\Delta \lt 0$, there are no real roots from the quadratic.
Step 5: Conclusion:
$ x = \frac{1}{2} $
Therefore, the curve intersects the $x$-axis exactly once.
Question 9
(a) Using the formulae on page 2, show that- $$ \cos^2 A=\frac{\cos 2A+1}{2} $$
- $$ \sin^2 A=\frac{1-\cos 2A}{2} $$
(b) Show that $$ (2\sin x-\cos x)(\sin x-3\cos x) = \frac{1}{2}(\cos 2x-7\sin 2x-5) $$
$$ y=(2\sin x-\cos x)(\sin x-3\cos x) $$ (c) Solve, for $0^\circ \leq x \leq 180^\circ$, the equation, $$ \frac{dy}{dx}=0 $$ Give your answers to the nearest whole number.
Click to view the solution
Solution:
(a)(i)
$ \cos(A+B) = \cos A \cos B - \sin A \sin B $
Let $A = B$:
$ \cos 2A = \cos^2 A - \sin^2 A $
$ \cos 2A = \cos^2 A - (1 - \cos^2 A) $
$ \cos 2A = 2\cos^2 A - 1 $
$ \cos^2 A = \frac{\cos 2A + 1}{2} $
(a)(ii)
$ \cos 2A = 1 - 2\sin^2 A $
$ 2\sin^2 A = 1 - \cos 2A $
$ \sin^2 A = \frac{1 - \cos 2A}{2} $
(b)
$ \text{LHS} = (2 \sin x - \cos x)(\sin x - 3 \cos x) $
$ = 2\sin^2 x - 6\sin x \cos x - \sin x \cos x + 3\cos^2 x $
$ = 2\sin^2 x - 7\sin x \cos x + 3\cos^2 x $
$ \text{RHS} = \frac{1}{2}(\cos 2x - 7\sin 2x + 5) $
$ = \frac{1}{2}(\cos^2 x - \sin^2 x) - \frac{7}{2}(2\sin x \cos x) + \frac{5}{2} $
$ = \frac{1}{2}\cos^2 x - \frac{1}{2}\sin^2 x - 7\sin x \cos x + \frac{5}{2}(\sin^2 x + \cos^2 x) $
$ = 2\sin^2 x - 7\sin x \cos x + 3\cos^2 x $
Thus LHS = RHS.
(c)
$ y = \frac{1}{2}(\cos 2x - 7\sin 2x + 5) $
$ \frac{dy}{dx} = \frac{1}{2}(-2\sin 2x - 14\cos 2x) $
$ \frac{dy}{dx} = -\sin 2x - 7\cos 2x $
Set $ \frac{dy}{dx} = 0 $:
$ -\sin 2x - 7\cos 2x = 0 $
$ \sin 2x = -7\cos 2x $
$ \tan 2x = -7 $
$ 2x \approx -81.87^\circ + 180^\circ n $
For $0^\circ \lt x \lt 180^\circ$:
$ 2x \approx 98.13^\circ,\; 278.13^\circ $
$ x \approx 49.06^\circ,\; 139.06^\circ $
Final Answer:
$ x \approx 49^\circ,\; 139^\circ $
Question 10
$O$, $A$ and $B$ are fixed points such that $$ \overrightarrow{OA}=(b+1)\mathbf{i}+\mathbf{j} $$ $$ \overrightarrow{AB}=3\mathbf{i} $$ The unit vector parallel to $\overrightarrow{OB}$ is $$ \frac{\sqrt{17}}{34}(3a-2)\mathbf{i}-b\mathbf{j} $$ Given that $a$ and $b$ are constants where $a>0$ and $b>0$ find the exact value of- $a$
- $b$
Click to view the solution
Solution:
Given:Points $O$, $A$, and $B$ are fixed such that:
$$ \overrightarrow{OA} = (b+1) \mathbf{i} + b \mathbf{j} $$ $$ \overrightarrow{AB} = 3 \mathbf{i} $$ $$ \text{The unit vector parallel to } \overrightarrow{OB} \text{ is } \frac{\sqrt{17}}{34} \left[ (3a + 2) \mathbf{i} + b \mathbf{j} \right]. $$ We need to find the vector $ \overrightarrow{OB} $ and the relationship between $a$ and $b$.
Step 1: Expression for $ \overrightarrow{OB} $
First, note that: $$ \overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB}. $$ Substitute: $$ \overrightarrow{OB} = \left( (b+1) \mathbf{i} + b \mathbf{j} \right) + 3 \mathbf{i}. $$ Simplify: $$ \overrightarrow{OB} = (b+4) \mathbf{i} + b \mathbf{j}. $$ Thus: $$ \overrightarrow{OB} = (b+4) \mathbf{i} + b \mathbf{j}. $$Step 2: Unit Vector Parallel to $ \overrightarrow{OB} $
$$ \hat{u} = \frac{\overrightarrow{OB}}{|\overrightarrow{OB}|} $$ Magnitude: $$ |\overrightarrow{OB}| = \sqrt{(b+4)^2 + b^2} $$ $$ |\overrightarrow{OB}| = \sqrt{2b^2 + 8b + 16} $$ Unit vector: $$ \hat{u} = \frac{(b+4) \mathbf{i} + b \mathbf{j}}{\sqrt{2b^2 + 8b + 16}} $$Step 3: Given Unit Vector
$$ \frac{\sqrt{17}}{34} \left[ (3a + 2) \mathbf{i} + b \mathbf{j} \right] $$ Equate: $$ \frac{(b+4) \mathbf{i} + b \mathbf{j}}{\sqrt{2b^2 + 8b + 16}} = \frac{\sqrt{17}}{34} \left[ (3a + 2) \mathbf{i} + b \mathbf{j} \right] $$Step 4: $ \mathbf{i} $ components
$$ \frac{b+4}{\sqrt{2b^2 + 8b + 16}} = \frac{\sqrt{17}}{34}(3a + 2) $$Step 5: $ \mathbf{j} $ components
$$ \frac{b}{\sqrt{2b^2 + 8b + 16}} = \frac{\sqrt{17}}{34}b $$ Since $b \neq 0$: $$ \frac{1}{\sqrt{2b^2 + 8b + 16}} = \frac{\sqrt{17}}{34} $$ Square: $$ \frac{1}{2b^2 + 8b + 16} = \frac{17}{1156} $$ $$ 1156 = 17(2b^2 + 8b + 16) $$ $$ 34b^2 + 136b - 884 = 0 $$ $$ b^2 + 4b - 26 = 0 $$ $$ b = \frac{-4 \pm \sqrt{120}}{2} $$ $$ b = -2 \pm \sqrt{30} $$ Thus: $$ b_1 = -2 + \sqrt{30}, \quad b_2 = -2 - \sqrt{30} $$ Since $b>0$, $ b = -2 + \sqrt{30} $Step 6: Find $a$
Substitute into: $$ \frac{b+4}{\sqrt{2b^2 + 8b + 16}} = \frac{\sqrt{17}}{34}(3a + 2) $$ For $b = -2 + \sqrt{30}$: $$ b+4 = 2 + \sqrt{30} $$ $$ b^2 = 34 - 4\sqrt{30} $$ $$ 2b^2 + 8b + 16 = 68 $$ $$ \sqrt{68} = 2\sqrt{17} $$ So: $$ \frac{2 + \sqrt{30}}{2\sqrt{17}} = \frac{\sqrt{17}}{34}(3a + 2) $$ Multiply: $$ 34 + 17\sqrt{30} = 17(3a + 2) $$ $$ 17\sqrt{30} = 51a $$ $$ a = \frac{\sqrt{30}}{3} $$ Final Answer: $$ a = \frac{\sqrt{30}}{3} $$Question 11
Given that $$ f(x)=10+6x-x^2 $$ can be written in the form $$ A(x+B)^2+C $$ where $A$, $B$ and $C$ are constants,(a) find the value of $A$, the value of $B$ and the value of $C$
[4]
(b) Hence, or otherwise, find
(i) the value of $x$ for which $f(x)$ has its greatest value
(ii) the greatest value of $f(x)$
[2]
The curve $C$ has equation $y=f(x)$.
The curve $S$ with equation $$ y=x^3-x+13 $$ intersects curve $C$ at two points.
(c) Find the $x$-coordinate of each of these two points.
[3]
(d) Use algebraic integration to find the exact area of the finite region bounded by the curve $C$ and the curve $S$.
[5]
Click to view the solution
Solution:
(a) Writing $ f(x) = 10 + 6x - x^2 $ in the form $ A(x + B)^2 + C $
We start with: $$ f(x) = -x^2 + 6x + 10 $$ Factor out $-1$: $$ f(x) = -(x^2 - 6x) + 10 $$ Complete the square: $$ x^2 - 6x = (x - 3)^2 - 9 $$ Substitute: $$ f(x) = -((x - 3)^2 - 9) + 10 $$ Simplify: $$ f(x) = -(x - 3)^2 + 19 $$ Thus, $ A = -1 $, $ B = -3 $, and $ C = 19 $.(b) (i) Value of $ x $ for maximum $ f(x) $
Since: $$ f(x) = -(x - 3)^2 + 19 $$ Maximum occurs at: $$ x = 3 $$(b) (ii) Greatest value of $ f(x) $
$$ f(3) = 19 $$(c) Intersection points of $ f(x) $ and $ S(x) $
$$ f(x) = 10 + 6x - x^2, \quad S(x) = x^2 - x + 13 $$ Set equal: $$ 10 + 6x - x^2 = x^2 - x + 13 $$ Simplify: $$ -2x^2 + 7x - 3 = 0 $$ $$ 2x^2 - 7x + 3 = 0 $$ Factorize: $$ (2x - 1)(x - 3) = 0 $$ Solutions: $$ x = \frac{1}{2}, \quad x = 3 $$(d) Area between $ f(x) $ and $ S(x) $
$$ \text{Area} = \int_{\frac{1}{2}}^3 [f(x) - S(x)] \, dx $$ $$ f(x) - S(x) = -2x^2 + 7x - 3 $$ So: $$ \text{Area} = \int_{\frac{1}{2}}^3 (-2x^2 + 7x - 3) \, dx $$ Integrate: $$ \left[-\frac{2x^3}{3} + \frac{7x^2}{2} - 3x \right]_{\frac{1}{2}}^3 $$ At $x = 3$: $$ -18 + \frac{63}{2} - 9 = 4.5 $$ At $x = \frac{1}{2}$: $$ -\frac{1}{12} + \frac{7}{8} - \frac{3}{2} = -\frac{17}{24} $$ Final area:Exact area = $\boxed{\frac{125}{24}}$
Post a Comment