Question 12:
Diagram 12 shows a geometrical mural drawn by members of the Art Club. The mural consists of a rectangle QRST and a semicircle PQBAT with centre P. SR = 1.84 m and ∠ SPR = 72o.
The club has allocated RM50 to decorate the mural. Sectors TPA and QPB will be painted green. A tube of paint costing RM6 will paint an area of 0.25 m2. Paint is only sold in whole tubes.
The perimeter of the blue coloured region will be taped with yellow ribbon. A roll of ribbon measuring 2 m costs RM10. Ribbon is only sold in whole rolls.
(a) Using π = 3.142, determine ∠ APB, in radians.
(b) Hence, determine whether the allocation is sufficient or not. Show your calculation.
[8 marks]
Answer:
(a) $$ \begin{aligned} \angle A P B & =\frac{72^{\circ}}{180^{\circ}} \times 3.142 \\ & =1.2568 \mathrm{rad} \end{aligned} $$
(b) $$ \begin{aligned} \frac{P S}{\sin 54^{\circ}} & =\frac{1.84 \mathrm{~m}}{\sin 72^{\circ}} \\ P S & =\frac{1.84 \mathrm{~m}}{\sin 72^{\circ}} \times \sin 54^{\circ} \\ P S & =1.5652 \mathrm{~m} \end{aligned} $$
$$ \begin{aligned} B R=A S & =1.5652 \mathrm{~m}-0.92 \mathrm{~m} \\ A S & =0.6452 \mathrm{~m} \end{aligned} $$
$$ \begin{aligned} S_{A B} & =(1.2568)(0.92) \\ & =1.1563 \mathrm{~m} \end{aligned} $$
$$ \begin{aligned} &\text { Perimeter of blue-coloured region }\\ &\begin{aligned} & =0.6452 \mathrm{~m}+1.1563 \mathrm{~m}+0.6452 \mathrm{~m}+1.84 \mathrm{~m} \\ & =4.2867 \mathrm{~m} \end{aligned} \end{aligned} $$
$$ \begin{aligned} &\text { Number of ribbon needed }\\ &\begin{aligned} & =4.2867 \mathrm{~m} \div 2 \mathrm{~m} \\ & =2.14 \approx 3 \text { rolls } \end{aligned} \end{aligned} $$
$$ \begin{aligned} \angle T P A & =\frac{54^{\circ}}{180^{\circ}} \times 3.142 \\ & =0.9426 \mathrm{rad} \end{aligned} $$
$$ \begin{aligned} &\text { Total area of blue-coloured region }\\ &\begin{aligned} & =2\left(\frac{1}{2}\right)(0.92)^2(0.9426) \\ & =0.7978 \mathrm{~m}^2 \end{aligned} \end{aligned} $$
$$ \begin{aligned} &\text { Number of paint needed }\\ &\begin{aligned} & =0.7978 \mathrm{~m}^2 \div 0.25 \mathrm{~m}^2 \\ & =3.1912 \approx 4 \text { tubes } \end{aligned} \end{aligned} $$
$$ \begin{aligned} &\text { Total cost }\\ &\begin{aligned} & =(3 \text { rolls } \text { × } \text { RM10 })+(4 \text { tubes } \text { × } \text { RM6 }) \\ & =\text { RM30 }+ \text { RM24 } \\ & =\text { RM54 } \end{aligned} \end{aligned} $$
Total cost > RM50, so the allocation is insufficient.
Diagram 12 shows a geometrical mural drawn by members of the Art Club. The mural consists of a rectangle QRST and a semicircle PQBAT with centre P. SR = 1.84 m and ∠ SPR = 72o.
The club has allocated RM50 to decorate the mural. Sectors TPA and QPB will be painted green. A tube of paint costing RM6 will paint an area of 0.25 m2. Paint is only sold in whole tubes.
The perimeter of the blue coloured region will be taped with yellow ribbon. A roll of ribbon measuring 2 m costs RM10. Ribbon is only sold in whole rolls.
(a) Using π = 3.142, determine ∠ APB, in radians.
(b) Hence, determine whether the allocation is sufficient or not. Show your calculation.
[8 marks]
Answer:
(a) $$ \begin{aligned} \angle A P B & =\frac{72^{\circ}}{180^{\circ}} \times 3.142 \\ & =1.2568 \mathrm{rad} \end{aligned} $$
(b) $$ \begin{aligned} \frac{P S}{\sin 54^{\circ}} & =\frac{1.84 \mathrm{~m}}{\sin 72^{\circ}} \\ P S & =\frac{1.84 \mathrm{~m}}{\sin 72^{\circ}} \times \sin 54^{\circ} \\ P S & =1.5652 \mathrm{~m} \end{aligned} $$
$$ \begin{aligned} B R=A S & =1.5652 \mathrm{~m}-0.92 \mathrm{~m} \\ A S & =0.6452 \mathrm{~m} \end{aligned} $$
$$ \begin{aligned} S_{A B} & =(1.2568)(0.92) \\ & =1.1563 \mathrm{~m} \end{aligned} $$
$$ \begin{aligned} &\text { Perimeter of blue-coloured region }\\ &\begin{aligned} & =0.6452 \mathrm{~m}+1.1563 \mathrm{~m}+0.6452 \mathrm{~m}+1.84 \mathrm{~m} \\ & =4.2867 \mathrm{~m} \end{aligned} \end{aligned} $$
$$ \begin{aligned} &\text { Number of ribbon needed }\\ &\begin{aligned} & =4.2867 \mathrm{~m} \div 2 \mathrm{~m} \\ & =2.14 \approx 3 \text { rolls } \end{aligned} \end{aligned} $$
$$ \begin{aligned} \angle T P A & =\frac{54^{\circ}}{180^{\circ}} \times 3.142 \\ & =0.9426 \mathrm{rad} \end{aligned} $$
$$ \begin{aligned} &\text { Total area of blue-coloured region }\\ &\begin{aligned} & =2\left(\frac{1}{2}\right)(0.92)^2(0.9426) \\ & =0.7978 \mathrm{~m}^2 \end{aligned} \end{aligned} $$
$$ \begin{aligned} &\text { Number of paint needed }\\ &\begin{aligned} & =0.7978 \mathrm{~m}^2 \div 0.25 \mathrm{~m}^2 \\ & =3.1912 \approx 4 \text { tubes } \end{aligned} \end{aligned} $$
$$ \begin{aligned} &\text { Total cost }\\ &\begin{aligned} & =(3 \text { rolls } \text { × } \text { RM10 })+(4 \text { tubes } \text { × } \text { RM6 }) \\ & =\text { RM30 }+ \text { RM24 } \\ & =\text { RM54 } \end{aligned} \end{aligned} $$
Total cost > RM50, so the allocation is insufficient.
Question 13:
It is given that the graph of the quadratic function f(x) = kx2 + 3kx + 10k2, such that k is a constant, has a maximum value of 49/4.
(a) By using the method of completing the square, find the value of k. [5 marks]
(b) Hence, sketch the graph of f(x) for -6 ≤ x ≤ 3. [3 marks]
Answer:
(a)
$$ \begin{aligned} & f(x)=k x^2+3 k x+10 k^2 \\ & f(x)=k\left(x^2+3 x+10 k\right) \\ & f(x)=k\left[x^2+3 x+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+10 k\right] \\ & f(x)=k\left[\left(x+\frac{3}{2}\right)^2-\frac{9}{4}+10 k\right] \\ & f(x)=k\left(x+\frac{3}{2}\right)^2-\frac{9}{4} k+10 k^2 \end{aligned} $$
$$ \begin{aligned} &f(x) \text { maximum }, k<0\\ &\therefore k=-1 \end{aligned} $$
$$ \begin{aligned} &\ \text { Maximum } \text { value }=\frac{49}{4}\\ &\begin{aligned} -\frac{9}{4} k+10 k^2 & =\frac{49}{4} \\ -9 k+40 k^2 & =49 \\ 40 k^2-9 k-49 & =0 \\ (40 k-49)(k+1) & =0 \\ k=\frac{49}{40} \quad, \quad k & =-1 \end{aligned} \end{aligned} $$
(b)
$$ \begin{aligned} &\begin{aligned} f(x) & =(-1)\left(x+\frac{3}{2}\right)^2-\frac{9}{4}(-1)+10(-1)^2 \\ f(x) & =-\left(x+\frac{3}{2}\right)^2+\frac{49}{4} \\ f(-6) & =-\left(-6+\frac{3}{2}\right)^2+\frac{49}{4} \\ & =-8 \end{aligned}\\ &\begin{aligned} f(3) & =-\left(3+\frac{3}{2}\right)^2+\frac{49}{4} \\ & =-8 \end{aligned} \end{aligned} $$
$$ \begin{aligned} f(x) & =-\left(x+\frac{3}{2}\right)^2+\frac{49}{4} \\ -\left(x+\frac{3}{2}\right)^2+\frac{49}{4} & =0 \\ -\left(x+\frac{3}{2}\right)^2 & =-\frac{49}{4} \\ \left(x+\frac{3}{2}\right)^2 & =\frac{49}{4} \\ x+\frac{3}{2} & = \pm \frac{7}{2} \end{aligned} $$
$$ \begin{array}{lll} x=-\frac{3}{2}+\frac{7}{2} & , & x=-\frac{3}{2}-\frac{7}{2} \\ x=2 & , & x=-5 \end{array} $$
It is given that the graph of the quadratic function f(x) = kx2 + 3kx + 10k2, such that k is a constant, has a maximum value of 49/4.
(a) By using the method of completing the square, find the value of k. [5 marks]
(b) Hence, sketch the graph of f(x) for -6 ≤ x ≤ 3. [3 marks]
Answer:
(a)
$$ \begin{aligned} & f(x)=k x^2+3 k x+10 k^2 \\ & f(x)=k\left(x^2+3 x+10 k\right) \\ & f(x)=k\left[x^2+3 x+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+10 k\right] \\ & f(x)=k\left[\left(x+\frac{3}{2}\right)^2-\frac{9}{4}+10 k\right] \\ & f(x)=k\left(x+\frac{3}{2}\right)^2-\frac{9}{4} k+10 k^2 \end{aligned} $$
$$ \begin{aligned} &f(x) \text { maximum }, k<0\\ &\therefore k=-1 \end{aligned} $$
$$ \begin{aligned} &\ \text { Maximum } \text { value }=\frac{49}{4}\\ &\begin{aligned} -\frac{9}{4} k+10 k^2 & =\frac{49}{4} \\ -9 k+40 k^2 & =49 \\ 40 k^2-9 k-49 & =0 \\ (40 k-49)(k+1) & =0 \\ k=\frac{49}{40} \quad, \quad k & =-1 \end{aligned} \end{aligned} $$
(b)
$$ \begin{aligned} &\begin{aligned} f(x) & =(-1)\left(x+\frac{3}{2}\right)^2-\frac{9}{4}(-1)+10(-1)^2 \\ f(x) & =-\left(x+\frac{3}{2}\right)^2+\frac{49}{4} \\ f(-6) & =-\left(-6+\frac{3}{2}\right)^2+\frac{49}{4} \\ & =-8 \end{aligned}\\ &\begin{aligned} f(3) & =-\left(3+\frac{3}{2}\right)^2+\frac{49}{4} \\ & =-8 \end{aligned} \end{aligned} $$
$$ \begin{aligned} f(x) & =-\left(x+\frac{3}{2}\right)^2+\frac{49}{4} \\ -\left(x+\frac{3}{2}\right)^2+\frac{49}{4} & =0 \\ -\left(x+\frac{3}{2}\right)^2 & =-\frac{49}{4} \\ \left(x+\frac{3}{2}\right)^2 & =\frac{49}{4} \\ x+\frac{3}{2} & = \pm \frac{7}{2} \end{aligned} $$
$$ \begin{array}{lll} x=-\frac{3}{2}+\frac{7}{2} & , & x=-\frac{3}{2}-\frac{7}{2} \\ x=2 & , & x=-5 \end{array} $$