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The point P divides the interval joining A(-1,-2) to B(9,3) internally in the ratio 4 : 1 - HSC - SSCE Mathematics Extension 1 - Question 1 - 2011 - Paper 1
Question 1
The point P divides the interval joining A(-1,-2) to B(9,3) internally in the ratio 4 : 1. Find the coordinates of P. (b) Differentiate \( \frac{\sin^2 x}{x} \) wit... show full transcript
Worked Solution & Example Answer:The point P divides the interval joining A(-1,-2) to B(9,3) internally in the ratio 4 : 1 - HSC - SSCE Mathematics Extension 1 - Question 1 - 2011 - Paper 1
Step 1
Find the coordinates of P.
Answer
To find point P that divides the line segment joining points A(-1, -2) and B(9, 3) in the ratio 4:1, we can use the section formula:
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:
[ P\left( \frac{mx2 + nx1}{m+n}, \frac{my2 + ny1}{m+n} \right) ]
Substituting in the known values:
- m = 4, n = 1,
- A(-1, -2): (x1, y1) = (-1, -2)
- B(9, 3): (x2, y2) = (9, 3)
Thus, the coordinates of P are: [ P\left( \frac{4 \cdot 9 + 1 \cdot (-1)}{4 + 1}, \frac{4 \cdot 3 + 1 \cdot (-2)}{4 + 1} \right) = P\left( \frac{36 - 1}{5}, \frac{12 - 2}{5} \right) = P\left( \frac{35}{5}, \frac{10}{5} \right) = P(7, 2) ]
Step 2
Differentiate \( \frac{\sin^2 x}{x} \) with respect to x.
Answer
To differentiate ( \frac{\sin^2 x}{x} ) with respect to x, we use the quotient rule. If we let u = ( \sin^2 x ) and v = ( x ), then:
[ \frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} ]
First, we find ( \frac{du}{dx} ) and ( \frac{dv}{dx} ):
- ( \frac{du}{dx} = 2\sin x \cdot \cos x = \sin 2x )
- ( \frac{dv}{dx} = 1 )
Now applying the quotient rule: [ \frac{d}{dx}\left( \frac{\sin^2 x}{x} \right) = \frac{x \cdot \sin(2x) - \sin^2 x \cdot 1}{x^2} = \frac{x \sin(2x) - \sin^2 x}{x^2} ]
Step 3
Solve \( \frac{4 - x}{x} < 1 \).
Answer
To solve the inequality ( \frac{4 - x}{x} < 1 ), first rearrange it:
[ \frac{4 - x}{x} - 1 < 0 \implies \frac{4 - x - x}{x} < 0 \implies \frac{4 - 2x}{x} < 0 ]
Now, find the critical points by setting the numerator and denominator to zero:
- Numerator: ( 4 - 2x = 0 \implies x = 2 )
- Denominator: ( x = 0 )
The critical points are x = 0 and x = 2. To determine the intervals, test values in the intervals created:
- For ( x < 0 ): Choose x = -1, ( \frac{4 - 2(-1)}{-1} = \frac{6}{-1} < 0 ) (true)
- For ( 0 < x < 2 ): Choose x = 1, ( \frac{4 - 2(1)}{1} = 2 > 0 ) (false)
- For ( x > 2 ): Choose x = 3, ( \frac{4 - 2(3)}{3} = \frac{-2}{3} < 0 ) (true)
Thus, the solution is ( (-\infty, 0) \cup (2, \infty) ).
Step 4
Step 5
Step 6
What is the range of the function \( f(x) = \ln(x^2 + e) \ ?
Answer
To find the range of the function ( f(x) = \ln(x^2 + e) ), observe that:
- The term ( x^2 + e ) is always positive for all real x (since ( e > 0 )).
- The minimum value of ( x^2 ) is 0, so the minimum value of ( x^2 + e ) is ( e ).
- As x approaches infinity or negative infinity, ( x^2 ) approaches infinity, leading ( f(x) ) to approach infinity.
Thus, the range of the function is: [ [\ln(e), \infty) = [1, \infty) ]