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Question 3 (15 marks) Use a SEPARATE writing booklet. - HSC - SSCE Mathematics Extension 2 - Question 3 - 2002 - Paper 1

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Question 3 (15 marks) Use a SEPARATE writing booklet.. (a) The diagram shows the graph of $y = f(x)$. Draw separate one-third page sketches of the graphs of the f... show full transcript

Worked Solution & Example Answer:Question 3 (15 marks) Use a SEPARATE writing booklet. - HSC - SSCE Mathematics Extension 2 - Question 3 - 2002 - Paper 1

Step 1

Draw separate one-third page sketches of the graphs of the following: (i) $y = -\frac{1}{\sqrt{x}}$

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Answer

To sketch the graph of y=1xy = -\frac{1}{\sqrt{x}}, note that:

  1. The domain of the function is x>0x > 0, as square roots are defined only for non-negative values.
  2. As xx approaches 00 from the right, yy approaches - rac{1}{\sqrt{0}} which tends to infty-\\infty.
  3. As xx increases, yy approaches 00 (but remains negative).

The graph will lie entirely below the x-axis, approaching the x-axis asymptotically as xx increases.

Step 2

Draw separate one-third page sketches of the graphs of the following: (ii) $y^2 = f(x)$

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Answer

The graph of y2=f(x)y^2 = f(x) utilizes the original graph of y=f(x)y = f(x), but reflects it about the x-axis. It is essential to:

  1. Identify the range of f(x)f(x) which may be negative.
  2. Draw the positive values of yy based on f(x)f(x) and reflect any negative values to the negative yy-axis, resulting in two branches.

Step 3

Draw separate one-third page sketches of the graphs of the following: (iii) $y = |f(x)|$

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Answer

To sketch y=f(x)y = |f(x)|, perform the following steps:

  1. Plot the graph of f(x)f(x) as given.
  2. Identify portions of the graph that are below the x-axis. These segments should be reflected above the x-axis.
  3. Retain the segments above the x-axis as they are. The resulting graph will be entirely non-negative.

Step 4

Draw separate one-third page sketches of the graphs of the following: (iv) $y = \ln(f(x))$

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Answer

For the graph of y=ln(f(x))y = \ln(f(x)), consider these important points:

  1. The function is defined only where f(x)>0f(x) > 0.
  2. Identify the x-values for which f(x)=1f(x) = 1; these will correspond to y=0y = 0.
  3. As f(x)f(x) increases, ln(f(x))\\ln(f(x)) will also increase.
  4. As f(x)f(x) approaches zero, the logarithmic function will tend to -\infty. Ensure to mark these critical points clearly on the graph.

Step 5

Show that the equation of the tangent at P is $x + p y = 2c p$. (i)

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Answer

To find the equation of the tangent line at point P(cp,cp)P\left(c_{p}, \frac{c}{p}\right) on the hyperbola xy=c2xy = c^2:

  1. The derivative of y=c2xy = \frac{c^2}{x} gives the slope, m=c2x2m = -\frac{c^2}{x^2}.
  2. Evaluate the slope at PP: mP=c2(cp)2m_P = -\frac{c^2}{(c_p)^2}.
  3. Using point-slope form, with point PP and slope mPm_P, formulate the equation: ycp=c2(cp)2(xcp)y - \frac{c}{p} = -\frac{c^2}{(c_p)^2}(x - c_p)
  4. Rearranging yields the equation of the tangent: x+py=2cpx + p y = 2c p.

Step 6

Show that T is the point \left( \frac{2c p q}{p + q}, \frac{2c}{p + q} \right). (ii)

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Answer

To find the coordinates of point TT, the intersection of the tangents at points PP and QQ:

  1. Write the equations of the tangents at points PP and QQ using the forms derived previously.
  2. Set these two equations equal to solve for xx and yy.
  3. After substituting the necessary expressions of pp and qq, determine the coordinates at TT as (2cpqp+q,2cp+q)\left( \frac{2c p q}{p + q}, \frac{2c}{p + q} \right).

Step 7

Show that the locus of T is a hyperbola, and state its eccentricity. (iii)

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Answer

To prove that the locus of TT is a hyperbola:

  1. Substitute the coordinates found for point TT into the relationship derived from the hyperbola's equation.
  2. Manipulate the terms to bring it into the standard hyperbola form: x2a2y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 where aa and bb are expressions derived from your coordinates.
  3. The eccentricity ee of the hyperbola can be derived from the relationship e=1+b2a2e = \sqrt{1 + \frac{b^2}{a^2}}, where terms for aa and bb are referenced from your manipulations.

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