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The functions of $f(x) = -\tan^{-1}\left(\frac{1}{2}x\right)$ and $g(x) = \cos(x + 90^\circ)$ for $-180^\circ \leq x \leq 180^\circ$ are given - NSC Mathematics - Question 6 - 2016 - Paper 2

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The-functions-of-$f(x)-=--\tan^{-1}\left(\frac{1}{2}x\right)$-and-$g(x)-=-\cos(x-+-90^\circ)$-for-$-180^\circ-\leq-x-\leq-180^\circ$-are-given-NSC Mathematics-Question 6-2016-Paper 2.png

The functions of $f(x) = -\tan^{-1}\left(\frac{1}{2}x\right)$ and $g(x) = \cos(x + 90^\circ)$ for $-180^\circ \leq x \leq 180^\circ$ are given. 6.1 Make a neat sket... show full transcript

Worked Solution & Example Answer:The functions of $f(x) = -\tan^{-1}\left(\frac{1}{2}x\right)$ and $g(x) = \cos(x + 90^\circ)$ for $-180^\circ \leq x \leq 180^\circ$ are given - NSC Mathematics - Question 6 - 2016 - Paper 2

Step 1

Make a neat sketch, on the same system of axes of both graphs on the grid provided in the SPECIAL ANSWER BOOK. Indicate all intercepts with the axes and coordinates of the turning points.

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Answer

To create the sketch of the functions:

  1. Identify Characteristics of Each Function:

    • For f(x)=tan1(12x)f(x) = -\tan^{-1}\left(\frac{1}{2}x\right):
      • As xx \to -\infty, f(x)π2f(x) \to -\frac{\pi}{2}
      • As xx \to \infty, f(x)π2f(x) \to \frac{\pi}{2}
      • The function is sigmoid shaped.
    • For g(x)=cos(x+90)=sin(x)g(x) = \cos(x + 90^\circ) = -\sin(x):
      • The amplitude is 1 and the period is 360360^\circ.
      • It oscillates between -1 and 1.
  2. Calculate Intercepts:

    • For f(x)f(x):
      • The function crosses the y-axis at f(0)=0f(0) = 0 (intercept).
    • For g(x)g(x):
      • The function crosses the y-axis at g(0)=0g(0) = 0 (intercept).
  3. Plot the Graphs:

    • Draw both functions on the same axes using a grid.
    • Clearly indicate the intercept points and turning points:
      • The turning points of g(x)g(x) occur at multiples of 9090^\circ.
      • Mark the intercept points on the axes clearly.
  4. Label Everything Clearly:

    • Title your graph, label axes, and indicate turning points and intercepts clearly.

Step 2

Give the value(s) of $x$ for which: $\cos(x + 90^\circ) \leq -\tan^{-1}\left(\frac{1}{2}x\right)$

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Answer

To find the values of xx:

  1. Rearrange the Inequality:

    • We have: sin(x)tan1(12x)-\sin(x) \leq -\tan^{-1}\left(\frac{1}{2}x\right)
    • This simplifies to: sin(x)tan1(12x)\sin(x) \geq \tan^{-1}\left(\frac{1}{2}x\right).
  2. Analyze Each Function:

    • an1(12x) an^{-1}\left(\frac{1}{2}x\right) is an increasing function, while sin(x)\\sin(x) oscillates between -1 and 1.
  3. Graph the Functions:

    • Sketch the graphs of sin(x)\sin(x) and tan1(12x)\tan^{-1}\left(\frac{1}{2}x\right) to find the intersection points within the given domain ([-180^\circ, 180^\circ]).
  4. Solve for Intersection Points:

    • Use numerical methods or graphing techniques to find the intersection points, which satisfy the inequality.
    • Identify and list the valid xx values within the specified range.

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