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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
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 \( \frac{3}{x-2} \leq 1 \). c) Ev... show full transcript
Worked Solution & Example Answer: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 that divides the line segment joining points A(−3, 4) and B(5, 6) in the ratio 1:3, we can use the section formula:
If a point P divides the line segment joining A(x1, y1) and B(x2, y2) in the ratio m:n, the coordinates (x, y) of P can be calculated as:
[ x = \frac{mx_2 + nx_1}{m+n}, \quad y = \frac{my_2 + ny_1}{m+n} ]
Here, m = 1, n = 3:
[ x = \frac{1 \cdot 5 + 3 \cdot (-3)}{1+3} = \frac{5 - 9}{4} = \frac{-4}{4} = -1 ]
[ y = \frac{1 \cdot 6 + 3 \cdot 4}{1+3} = \frac{6 + 12}{4} = \frac{18}{4} = \frac{9}{2} ]
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 ), we first set the inequality in a simpler form:
[ \frac{3}{x-2} - 1 \leq 0 ]
Rearranging gives us:
[ \frac{3 - (x-2)}{x-2} \leq 0 \quad \Rightarrow \quad \frac{5 - x}{x-2} \leq 0 ]
Next, we identify the critical points by solving for when the numerator and denominator are zero:
[ 5 - x = 0 \quad \Rightarrow \quad x = 5] [ x - 2 = 0 \quad \Rightarrow \quad x = 2]
Now, testing intervals around critical points 2 and 5:
- For ( x < 2 ): Positive
- For ( 2 < x < 5 ): Negative
- For ( x > 5 ): Positive
Thus, the solution is: [ 2 < x \leq 5 ]
Step 3
Evaluate \( \lim_{x \to 0} \frac{3x}{\sin 2x}. \)
Answer
To evaluate ( \lim_{x \to 0} \frac{3x}{\sin 2x} ), we use the fact that ( \sin x \sim x ) as ( x \to 0 ).
Thus, we rewrite: [ \lim_{x \to 0} \frac{3x}{\sin 2x} = \lim_{x \to 0} \frac{3x}{2x} \times 2 = \lim_{x \to 0} \frac{3}{2} = \frac{3}{2}. ]
So, ( \lim_{x \to 0} \frac{3x}{\sin 2x} = \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
The parametric equations are given by ( x = \frac{1}{2} ) and ( y = 3t^2 ). Since x is a constant, we can express the Cartesian equation simply as:
Given that x = ( \frac{1}{2} ), we can express t in terms of y from the equation for y:
[ y = 3t^2 \quad \Rightarrow \quad t^2 = \frac{y}{3}. ]
Since x does not depend on t, the Cartesian equation is simply: [ x = \frac{1}{2} \quad \text{and} \quad y = 3t^2 \text{ (where t is defined by y)}. ]
Step 5
Use the substitution \( u = x^2 + 1 \) to evaluate \( \int_{0}^{2} \frac{x}{x^2 + 1} \, dx \).
Answer
We start by making the substitution ( u = x^2 + 1 ). Therefore, ( du = 2x , dx ) or equivalently ( dx = \frac{du}{2x} ).
Now, changing the limits of integration:
When ( x = 0 ), ( u = 0^2 + 1 = 1 ), and when ( x = 2 ), ( u = 2^2 + 1 = 5 ).
Now we can rewrite the integral:
[ \int_{0}^{2} \frac{x}{x^2 + 1} , dx = \int_{1}^{5} \frac{x}{u} \frac{du}{2x} = \frac{1}{2} \int_{1}^{5} \frac{du}{u}. ]
This can be evaluated as: [ \frac{1}{2} \left[ \ln|u| \right]_{1}^{5} = \frac{1}{2} (\ln 5 - \ln 1) = \frac{1}{2} \ln 5. ]