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Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 2 - 2003 - Paper 1
Question 2
Use a SEPARATE writing booklet. (a) Sketch the graph of $y = 3 ext{cos}^{-1}(2x)$. Your graph must clearly indicate the domain and the range. (b) Find \( \frac{d}... show full transcript
Worked Solution & Example Answer:Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 2 - 2003 - Paper 1
Step 1
Sketch the graph of $y = 3 \text{cos}^{-1}(2x)$
Answer
To sketch the graph of ( y = 3 \text{cos}^{-1}(2x) ), we first determine the domain and range.
Domain: The function ( \text{cos}^{-1}(u) ) is defined for ( -1 \leq u \leq 1 ). Therefore, set ( 2x ) such that: [-1 \leq 2x \leq 1 \rightarrow -\frac{1}{2} \leq x \leq \frac{1}{2}]. Thus, the domain is ( x \in [-\frac{1}{2}, \frac{1}{2}] ).
Range: The range of ( \text{cos}^{-1}(u) ) is ( [0, \pi] ). Multiplying by 3 gives us the range ( [0, 3\pi] ).
The graph should be plotted with ( x ) values from ( -\frac{1}{2} ) to ( \frac{1}{2} ), and the corresponding ( y ) values will vary from 0 to 3π.
Step 2
Step 3
Evaluate \( \int_0^1 \frac{1}{\sqrt{2 - x^2}} \, dx \)
Answer
This integral can be evaluated using a trigonometric substitution. Let ( x = \sqrt{2} \sin(t) ) where ( dx = \sqrt{2} \cos(t) , dt. )
The limits change as follows:
- When ( x = 0, t = 0 )
- When ( x = 1, t = \frac{\pi}{4} )
The integral becomes: [ \int_0^{\frac{\pi}{4}} \frac{\sqrt{2} \cos(t)}{\sqrt{2 - 2 \sin^2(t)}} , dt = \int_0^{\frac{\pi}{4}} dt = \frac{\pi}{4}. ]
Step 4
Find the coefficient of \( x^4 \) in the expansion of \( (2 + x^2)^5 \)
Answer
To find the coefficient of ( x^4 ), we use the binomial expansion: [ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k. ]
In this case:
- Let ( a = 2 ) and ( b = x^2 ), and ( n = 5 ).
- We want the term where ( k = 2 ) (since we need ( x^4 )).
The term is: [ \binom{5}{2} (2)^{5-2} (x^2)^2 = 10 \cdot 2^3 = 10 \cdot 8 = 80. ] Thus, the coefficient of ( x^4 ) is 80.
Step 5
Express \( \text{cos} \, x - \text{sin} \, x \) in the form \( R \text{cos}(x + \alpha) \)
Answer
We can express ( A \text{cos}(x) + B \text{sin}(x) ) terms into the format ( R \text{cos}(x + \alpha) ), where:
- ( R = \sqrt{A^2 + B^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2} )
- ( \tan(\alpha) = \frac{B}{A} = \frac{-1}{1} = -1 ), which gives ( \alpha = -\frac{\pi}{4} ).
Thus, we have: [ \text{cos} , x - \text{sin} , x = \sqrt{2} \text{cos}(x + (-\frac{\pi}{4})). ]
Step 6
Hence, or otherwise, sketch the graph of \( y = \text{cos} \, x - \text{sin} \, x \) for \( 0 \leq x \leq 2\pi \)
Answer
To sketch the graph of ( y = \text{cos} , x - \text{sin} , x ), use the expression obtained above: [ y = \sqrt{2} \text{cos}(x - \frac{\pi}{4}). ]
This graph oscillates between ( -\sqrt{2} ) and ( \sqrt{2} ). It will have zeros at points where ( x - \frac{\pi}{4} = \frac{\pi}{2} + n\pi ) leading to shifts in the sine and cosine values accordingly.