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a) Find the inverse of the function $y=x^3 - 2$ - HSC - SSCE Mathematics Extension 1 - Question 11 - 2016 - Paper 1

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a) Find the inverse of the function $y=x^3 - 2$. b) Use the substitution $u=x-4$ to find $\int \sqrt{x-4} \, dx$. c) Differentiate $3\tan^{-1}(2x)$. d) Evaluat... show full transcript

Worked Solution & Example Answer:a) Find the inverse of the function $y=x^3 - 2$ - HSC - SSCE Mathematics Extension 1 - Question 11 - 2016 - Paper 1

Step 1

Find the inverse of the function $y=x^3 - 2$

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Answer

To find the inverse, we first interchange xx and yy.

  1. Start with: y=x3−2y = x^3 - 2
  2. Interchange xx and yy: x=y3−2x = y^3 - 2
  3. Solve for yy: y3=x+2y^3 = x + 2 y=x+23y = \sqrt[3]{x + 2} Thus, the inverse function is: f−1(x)=x+23f^{-1}(x) = \sqrt[3]{x + 2}

Step 2

Use the substitution $u=x-4$ to find $\int \sqrt{x-4} \, dx$

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Answer

  1. Let u=x−4u = x - 4, then dx=dudx = du and x=u+4x = u + 4.
  2. Rewrite the integral: ∫x−4 dx=∫u du\int \sqrt{x-4} \, dx = \int \sqrt{u} \, du
  3. Now, compute the integral: =∫u1/2 du=u3/23/2+C=23u3/2+C= \int u^{1/2} \, du = \frac{u^{3/2}}{3/2} + C = \frac{2}{3} u^{3/2} + C
  4. Substitute back to get: =23(x−4)3/2+C= \frac{2}{3} (x-4)^{3/2} + C

Step 3

Differentiate $3\tan^{-1}(2x)$

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Answer

  1. Use the chain rule for differentiation: ddx[3tan⁡−1(2x)]=3⋅11+(2x)2⋅2\frac{d}{dx}[3\tan^{-1}(2x)] = 3 \cdot \frac{1}{1+(2x)^2} \cdot 2
  2. This simplifies to: =61+4x2= \frac{6}{1 + 4x^2}

Step 4

Evaluate $\lim_{x \to 0} \frac{2\sin x \cos x}{3x}$

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Answer

  1. Recognize that lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1.
  2. Rewrite the limit: lim⁡x→02sin⁡xcos⁡x3x=23⋅lim⁡x→0sin⁡xx⋅cos⁡x\lim_{x \to 0} \frac{2\sin x \cos x}{3x} = \frac{2}{3} \cdot \lim_{x \to 0} \frac{\sin x}{x} \cdot \cos x
  3. Now substituting: =23⋅1⋅1=23= \frac{2}{3} \cdot 1 \cdot 1 = \frac{2}{3}

Step 5

Solve $\frac{3}{2x+5} > 0$

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Answer

  1. The fraction is positive when its numerator and denominator have the same sign.
  2. Since the numerator 33 is always positive, we need to solve: 2x+5>02x + 5 > 0
  3. This simplifies to: x > -\frac{5}{2}$$

Step 6

Find the probability that she hits the bullseye with exactly one of her first three throws

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Answer

  1. We use the binomial probability formula: P(X=k)=(nk)pk(1−p)n−kP(X=k) = {n \choose k} p^k (1-p)^{n-k}
  2. In this case, n=3n=3, k=1k=1, and p=25p=\frac{2}{5}: P(X=1)=(31)(25)1(35)2P(X=1) = {3 \choose 1}\left(\frac{2}{5}\right)^1 \left(\frac{3}{5}\right)^{2} =3⋅25⋅925=54125= 3 \cdot \frac{2}{5} \cdot \frac{9}{25} = \frac{54}{125}

Step 7

Find the probability that she hits the bullseye with at least two of her first six throws

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Answer

  1. We find P(X≥2)P(X \geq 2): P(X≥2)=1−P(X=0)−P(X=1)P(X \geq 2) = 1 - P(X=0) - P(X=1)
  2. Calculate:
    • For X=0X=0: P(X=0)=(60)(25)0(35)6=72915625P(X=0) = {6 \choose 0}\left(\frac{2}{5}\right)^{0}\left(\frac{3}{5}\right)^{6} = \frac{729}{15625}
    • For X=1X=1: P(X=1)=(61)(25)1(35)5=6⋅25⋅2433125=291615625P(X=1) = {6 \choose 1}\left(\frac{2}{5}\right)^{1}\left(\frac{3}{5}\right)^{5} = 6 \cdot \frac{2}{5} \cdot \frac{243}{3125} = \frac{2916}{15625}
  3. Thus: P(X≥2)=1−729+291615625=1−364515625=1198015625P(X \geq 2) = 1 - \frac{729 + 2916}{15625} = 1 - \frac{3645}{15625} = \frac{11980}{15625}

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