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(a) Sketch the graph of $y = 3 ext{cos}^{-1}(2x)$ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2003 - Paper 1

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(a) Sketch the graph of $y = 3 ext{cos}^{-1}(2x)$. Your graph must clearly indicate the domain and the range. (b) Find $\frac{d}{dx} (x \tan^{-1} x)$. (c) Evaluate... show full transcript

Worked Solution & Example Answer:(a) Sketch the graph of $y = 3 ext{cos}^{-1}(2x)$ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2003 - Paper 1

Step 1

Sketch the graph of $y = 3\text{cos}^{-1}(2x)$

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Answer

To sketch the graph of y=3cos1(2x)y = 3\text{cos}^{-1}(2x), we first determine the domain and range:

  • Domain: The inverse cosine function is defined for 12x1-1 \leq 2x \leq 1, leading to 0.5x0.5-0.5 \leq x \leq 0.5.

  • Range: The range of cos1(x)\text{cos}^{-1}(x) is from 00 to π\pi. Thus, multiplying by 33, the range of yy becomes [0,3π][0, 3\pi].

Next, we plot the function, ensuring it reflects these characteristics.

Step 2

Find $\frac{d}{dx} (x \tan^{-1} x)$

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Answer

To differentiate xtan1xx \tan^{-1} x, we use the product rule:

Let u=xu = x and v=tan1(x)v = \tan^{-1}(x), where:

  • dudx=1\frac{du}{dx} = 1,
  • dvdx=11+x2\frac{dv}{dx} = \frac{1}{1 + x^2}.

Using the product rule: ddx(uv)=udvdx+vdudx\frac{d}{dx}(u v) = u \frac{dv}{dx} + v \frac{du}{dx} We get: ddx(xtan1x)=x11+x2+tan1(x)1.\frac{d}{dx} (x \tan^{-1} x) = x \cdot \frac{1}{1 + x^2} + \tan^{-1}(x) \cdot 1.

Thus, the derivative is: x1+x2+tan1(x).\frac{x}{1 + x^2} + \tan^{-1}(x).

Step 3

Evaluate \( \int_{0}^{1} \frac{1}{\sqrt{2 - x^2}} \: dx \)

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Answer

To evaluate the integral ( \int_{0}^{1} \frac{1}{\sqrt{2 - x^2}} : dx ), we use the substitution x=2sinθx = \sqrt{2} \sin \theta. Then, we find:

  • dx=2cosθdθdx = \sqrt{2} \cos \theta d\theta.
  • When x=0x = 0, θ=0\theta = 0, and when x=1x = 1, θ=π4\theta = \frac{\pi}{4}.

The integral becomes: 0π42cosθ22sin2θdθ=0π4dθ.\int_{0}^{\frac{\pi}{4}} \frac{\sqrt{2} \cos \theta}{\sqrt{2 - 2\sin^2 \theta}} d\theta = \int_{0}^{\frac{\pi}{4}} d\theta.

Thus, evaluating gives: [θ]0π4=π4.\left[ \theta \right]_{0}^{\frac{\pi}{4}} = \frac{\pi}{4}.

Step 4

Find the coefficient of $x^4$ in the expansion of $(2 + x^2)^5$

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Answer

Using the binomial theorem: (a+b)n=k=0n(nk)ankbk,(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k, we identify a=2a = 2, b=x2b = x^2, and n=5n = 5.

To find the term containing x4x^4, we set k=2k = 2 (since x2k=x4x^{2k} = x^4): Term=(52)(2)52(x2)2=(52)(2)3(x4).\text{Term} = \binom{5}{2} (2)^{5-2} (x^2)^2 = \binom{5}{2} (2)^3 (x^4).

Calculating:

  • (52)=10\binom{5}{2} = 10 and (2)3=8(2)^3 = 8 gives: 10×8=80.10 \times 8 = 80. Therefore, the coefficient of x4x^4 is 80.80.

Step 5

Express $\cos x - \sin x$ in the form $R \cos(x + \alpha)$

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Answer

To express cosxsinx\cos x - \sin x in the form Rcos(x+α)R \cos(x + \alpha), we start from: R=A2+B2R = \sqrt{A^2 + B^2} where A=1A = 1 and B=1B = -1. Thus: R=12+(1)2=2.R = \sqrt{1^2 + (-1)^2} = \sqrt{2}.

Next, to find α\alpha, we use: tanα=BA=11=1α=π4.\tan \alpha = \frac{B}{A} = \frac{-1}{1} = -1 \Rightarrow \alpha = -\frac{\pi}{4}.

Thus, we can write cosxsinx\cos x - \sin x as: 2cos(x+π4).\sqrt{2} \cos\left(x + \frac{\pi}{4}\right).

Step 6

Hence, or otherwise, sketch the graph of $y = \cos x - \sin x$ for $0 \leq x \leq 2\pi$

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Answer

To sketch the graph:

  1. Identify the amplitude, which is 2\sqrt{2}, indicating the graph oscillates between 2-\sqrt{2} and 2\sqrt{2}.
  2. The period is 2π2\pi.
  3. Start at y=cos(0)sin(0)=10=1y = \cos(0) - \sin(0) = 1 - 0 = 1 and mark critical points within [0,2π][0, 2\pi].
  4. Sketch the wave based on the transformations identified: the graph will reflect shifts and inverses from the standard cosine and sine functions.
  5. Label key points at 0,π2,π,3π2,2π0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi.

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