Question 1
Topic: SL 3.4
Given :
- A circle with centre O and radius 4cm.
The points P, Q and R lie on the circumference of the circle and PÔR = θ, where θ is measured in radians with length of arc PQR is 10cm
Find
(a) perimeter of the shaded sector.
(b) Angle θ and
(c) area of the shaded sector.
▶️Answer/Explanation
Answer:
(a) attempts to find perimeter
arc+2 × radius OR 10 + 4 + 4
= 18 (cm)
Perimeter of the shaded sector: The perimeter of the shaded sector is the sum of the lengths of the arc PQR and the radii OP and OR.
(b) 10 = 4θ
θ = \(\frac{10}{4}\) \(=\frac{5}{2}\), 2.5)
\(\theta =\frac{10}{4}\Rightarrow 2.5\)
(c) area = \(\frac{1}{2}(\frac{10}{4})(4^2)\) (=1.25 × 16)
= 20 (\(cm^2\))
The area of the shaded sector is calculated using the formula:
Area of sector \(=\frac{1}{2}r^{2}\theta \Rightarrow \frac{1}{2}4^{2}\left ( 2.5 \right )\)
Area of sector\(=\frac{1}{2}\times 16\times 2.5\Rightarrow 20 cm^{2}\)
Question 2
Topic: SL 2.8
Given:
- f(x) = 1 – \(\frac{1}{x-2}\), where \(x\epsilon \mathbb{R}\), x ≠ 2.
(a) If graph of y = f (x) has a vertical asymptote and a horizontal asymptote.
Then Find:
(i) the vertical asymptote;
(ii) the horizontal asymptote.
(b) What will be coordinates if graph of y = f (x) intersects at
(i) the y-axis;
(ii) the x-axis.
(c) sketch the graph of y = f (x), showing all the features found in parts (a) and (b).
▶️Answer/Explanation
Answer:
(a) (i) x = 2
(ii) y = 1
(i) The vertical asymptote occurs when the denominator of the fraction becomes zero. Since the denominator is x-2, the vertical asymptote is at x=2.
(ii) To find the horizontal asymptote, we look at the behavior of the function as x
x approaches positive or negative infinity, the term \(\frac{1}{x-2}\) becomes very close to zero. Thus, the horizontal asymptote is the horizontal line y = 1, as x approaches positive or negative infinity.
(b) (i) (0, \(\frac{3}{2}\))
To find the point where the graph intersects the y-axis, set x=0 in the function:
\(f(0) = 1 – \frac{1}{0-2}\Rightarrow 1+\frac{1}{2}\Rightarrow \frac{3}{2}\)
So, the coordinates of the point where the graph intersects the y-axis are \(\left (0,\frac{3}{2} \right )\)
(ii) (3, 0)
To find the point where the graph intersects the x-axis, set y=0 in the function:
\(0 = 1 – \frac{1}{x-2}\Rightarrow \frac{1}{x-2}=1\)
\(x-2=1\Rightarrow x=3\)
So, the coordinates of the point where the graph intersects the x-axis are (3, 0).
(c) 
two correct branches with correct asymptotic behaviour and intercepts clearly shown
To sketch the graph of y=f(x), we incorporate the features found in parts (a) and (b). We have:
- A vertical asymptote at x=2
- A horizontal asymptote at y = 1
- Intersecting the y-axis at \(\left (0,\frac{3}{2} \right )\).
- Intersecting the x-axis at (3, 0).