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1. (a) Find the coordinates of the point P that divides the interval joining (−3, 4) and (5, 6) internally in the ratio 1 : 3 - HSC - SSCE Mathematics Extension 1 - Question 1 - 2003 - Paper 1
Question 1
1. (a) Find the coordinates of the point P that divides the interval joining (−3, 4) and (5, 6) internally in the ratio 1 : 3. (b) Solve \( \frac{3}{x-2} \leq 1 \).... show full transcript
Worked Solution & Example Answer:1. (a) Find the coordinates of the point P that divides the interval joining (−3, 4) and (5, 6) internally in the ratio 1 : 3 - HSC - SSCE Mathematics Extension 1 - Question 1 - 2003 - Paper 1
Step 1
Find the coordinates of the point P that divides the interval joining (−3, 4) and (5, 6) internally in the ratio 1 : 3.
Answer
To find the coordinates of point P, use the section formula for internal division. The formula states that if point P divides the line segment joining points A(x1, y1) and B(x2, y2) in the ratio m:n, the coordinates of P are given by:
Here, A(−3, 4), B(5, 6), m = 1, and n = 3. Substituting the values:
Thus, the coordinates of point P are (−1, 4.5).
Step 2
Solve \( \frac{3}{x-2} \leq 1 \).
Answer
To solve the inequality ( \frac{3}{x-2} \leq 1 ), first isolate the term involving x:
- Rearranging gives: ( \frac{3}{x-2} - 1 \leq 0 )
- Finding a common denominator: ( \frac{3 - (x - 2)}{x-2} \leq 0 ) ( \frac{5 - x}{x-2} \leq 0 )
- The critical points are x = 2 and x = 5. Test the intervals ( (-\infty, 2), (2, 5), (5, \infty) ):
- For ( x < 2 ), the expression is positive.
- For ( 2 < x < 5 ), the expression is negative (solution interval).
- For ( x > 5 ), the expression is positive.
- Including x = 5 as zero satisfies the inequality: ( x \in [2, 5] ).
Step 3
Evaluate \( \lim_{x \to 0} \frac{3x}{\sin 2x} \).
Answer
To evaluate the limit ( \lim_{x \to 0} \frac{3x}{\sin 2x} ), apply L'Hôpital's Rule since both the numerator and denominator approach 0 when x approaches 0:
- Differentiate the numerator: ( 3 )
- Differentiate the denominator: ( 2\cos(2x) )
- Thus, the limit becomes: ( \lim_{x \to 0} \frac{3}{2 \cos(2x)} = \frac{3}{2 \times 1} = \frac{3}{2} ) Therefore, the limit is ( \frac{3}{2} ).
Step 4
A curve has parametric equations \( x = \frac{1}{2}, y = 3t^2 \). Find the Cartesian equation for this curve.
Answer
To find the Cartesian equation, express y in terms of x or vice versa. Given that x is constant:
- From the parametric equation, since x = ( \frac{1}{2} ), substitute this into the parametric equation for y: ( y = 3t^2 )
- As x does not depend on t, the relationship remains constant: Thus, the parametric form indicates that y depends on the value of t, implying: The Cartesian equation is not dependent on x as it is a vertical line at x = ( \frac{1}{2} ).
Step 5
Use the substitution \( u = x^2 + 1 \) to evaluate \( \int_{0}^{2} \frac{x}{x^2 + 1} dx \).
Answer
To evaluate the integral, follow these steps:
- Perform the substitution: ( u = x^2 + 1 ), thus ( du = 2x , dx ) or ( dx = \frac{du}{2x} ).
- Change the limits of integration:
- When x = 0, u = 1.
- When x = 2, u = 5.
- Substitute into the integral: ( \int_{0}^{2} \frac{x}{u} \cdot \frac{du}{2x} = \frac{1}{2} \int_{1}^{5} \frac{1}{u} du ).
- This evaluates to: ( \frac{1}{2} [\ln |u|]_{1}^{5} = \frac{1}{2} (\ln 5 - \ln 1) = \frac{1}{2} \ln 5 ). Thus, the evaluated integral is ( \frac{1}{2} \ln 5 ).