(a) For each sketch, put a ring around the correct type of function shown.
$\mathbf{(i)}$ On the grid, sketch the curve $y= \sin x$ for $0^\circ\leqslant x\leqslant360^\circ.$
(ii) Solve the equation $\sin x+ 0.4= 0$ for $0^{\circ}\leqslant x\leqslant360^{\circ}$
▶️ Answer/Explanation
(a) Function Identification:
Graph (i): The curve shows an S-shape with two turning points, characteristic of a cubic function (e.g., y = x³).
Graph (ii): The curve approaches both axes but never touches them, showing vertical and horizontal asymptotes – this is typical of a reciprocal function (e.g., y = 1/x).
(b)(i) Sine Graph Sketch:
The sine curve has:
- Amplitude of 1 (peaks at y=1, troughs at y=-1)
- Period of 360° (completes one full cycle)
- Passes through (0°,0), (90°,1), (180°,0), (270°,-1), and (360°,0)
- Smooth, continuous wave pattern
(b)(ii) Trigonometric Equation Solution:
1. Rearrange the equation: \[ \sin x = -0.4 \] (We subtract 0.4 from both sides)
2. Find reference angle: \[ x_{\text{ref}} = \sin^{-1}(0.4) \approx 23.58° \] (Always use the positive value to find reference angle)
3. Locate solutions in proper quadrants:
Since sine is negative, solutions lie in:
- Quadrant III: 180° + reference angle \[ x = 180° + 23.58° = 203.58° \approx 203.6° \]
- Quadrant IV: 360° – reference angle \[ x = 360° – 23.58° = 336.42° \approx 336.4° \]
Final solutions: \(x = \boxed{203.6°}\) and \(x = \boxed{336.4°}\) (rounded to 1 decimal place).
By drawing a suitable tangent, estimate the gradient of the curve at the point P.
▶️ Answer/Explanation
Ans: −0.5 (accept −0.7 to −0.3)
Detailed Solution:
- Drawing the Tangent:
- Locate point P at x=2 on the curve
- Draw a straight line that just touches the curve at P and follows the curve’s slope
- Ensure the tangent is neither too steep nor too shallow
- Selecting Points:
- Choose two points on the tangent line that are easy to read from the grid
- Suggested points: (1.5, 4.2) and (2.5, 3.7)
- These points should be about 0.5 units either side of P for best accuracy
- Calculating Gradient:
Using the formula:
\[ \text{Gradient} = \frac{\Delta y}{\Delta x} = \frac{y_2 – y_1}{x_2 – x_1} \]Substitute the coordinates:
\[ \frac{3.7 – 4.2}{2.5 – 1.5} = \frac{-0.5}{1} = -0.5 \]
Important Notes:
- The negative gradient indicates the curve is decreasing at point P
- The acceptable range accounts for slight variations in tangent placement
- For best results:
- Use a ruler for the tangent line
- Choose points that are easy to read from the grid
- Take measurements as precisely as possible
Why This Works:
The tangent line represents the instantaneous rate of change of the curve at point P. By calculating its gradient, we estimate the derivative of the function at that point.