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Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 11 - 2014 - Paper 1

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Use a SEPARATE writing booklet. (a) Solve $$ \left( x + \frac{2}{y} \right)^2 - 6 \left( x + \frac{2}{y} \right) + 9 = 0. $$ (b) The probability that it rains on ... show full transcript

Worked Solution & Example Answer:Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 11 - 2014 - Paper 1

Step 1

Solve \(\left( x + \frac{2}{y} \right)^2 - 6 \left( x + \frac{2}{y} \right) + 9 = 0\)

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Answer

Let ( z = x + \frac{2}{y} ). Then the equation becomes ( z^2 - 6z + 9 = 0 ), which factors to ( (z - 3)^2 = 0 ). Thus, ( z = 3 ), leading to ( x + \frac{2}{y} = 3 ). Solving for ( y ), under the assumption that ( y \neq 0 ), we find ( y = \frac{2}{3 - x} ). Lastly, we need to find when ( y = 0 ): setting ( \frac{2}{y} = 0 ) gives no solution. Therefore, the solutions are ( x = 1 ) and ( x = 2 ).

Step 2

Write an expression for the probability that it rains on fewer than 3 days in November.

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Answer

The number of days it rains can be modeled by a binomial distribution with parameters ( n = 30 ) and ( p = 0.1 ). The expression for the probability that it rains on fewer than 3 days is:

P(X<3)=P(X=0)+P(X=1)+P(X=2)P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

where ( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}. )

Step 3

Sketch the graph \(y = 6\tan^{-1}(x)\).

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Answer

To sketch the graph of ( y = 6\tan^{-1}(x) ):

  1. The function (\tan^{-1}(x)) approaches ( \frac{\pi}{2} ) as ( x ) increases. Therefore, ( y ) approaches ( 6 \cdot \frac{\pi}{2} = 3\pi \approx 9.42 )
  2. For ( x \to -\infty ), ( y \to -3\pi
  3. The range of the graph is ( (-3\pi, 3\pi) ).

Step 4

Evaluate \(\int_{2}^{5} \frac{x}{\sqrt{x-1}} \,dx\) using the substitution \(x = u^2 + 1\).

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Answer

Using the substitution ( x = u^2 + 1 ):

  1. Then ( dx = 2u , du ) and the limits change as follows:
    • For ( x = 2 ), ( u = 1 )
    • For ( x = 5 ), ( u = 2 ).
  2. The integral becomes:
12u2+1u2(2u)du=212(u+1u)du\int_{1}^{2} \frac{u^2 + 1}{\sqrt{u^2}} (2u) \, du = 2\int_{1}^{2} (u + \frac{1}{u}) \,du
  1. Evaluating gives:
= 2\left[ \frac{u^2}{2} + \ln|u| \right]_{1}^{2} = 2\left( 2 + \ln(2) - \frac{1}{2} - 0 \right) = 3 + 2\ln(2)\.

Step 5

Solve \(\sqrt{x^2 + 5} > 6\).

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Answer

Squaring both sides gives:

x2+5>6    x2+5>36    x2>31    x>31 or x<31.\sqrt{x^2 + 5} > 6 \\ \implies x^2 + 5 > 36 \\ \implies x^2 > 31 \\ \implies x > \sqrt{31} \, \text{ or } \, x < -\sqrt{31}.

Step 6

Differentiate \(e^{x} \ln x\).

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Answer

Using the product rule for differentiation,

ddx(exlnx)=exlnx+ex1x.\frac{d}{dx}\left( e^{x} \ln x \right) = e^{x} \ln x + e^{x} \cdot \frac{1}{x}.

Thus, the derivative is ( e^{x} \ln x + \frac{e^{x}}{x} ).

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