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five topics for early success - Mini Test 1

Submitted by roamingfree on Sat, 08/21/2021 - 19:22
skills
Apply the skills and knowledge of the introductory topics to exam-style questions
Question

Complete the following five questions

  1. (2 Marks) By completing the square, express $t^2 + 12t-8$ in the form $(t+a)^2 + b$, where $a$ and $b$ are integers
  2. (4 Marks) Express the following in the form $at^b + ct^d$ where $a,\;b,\;c,\;d$ are rational numbers. $$\frac{\sqrt{t^3} + 16t^2}{(2t)^5}$$
     
  3. (4 Marks) Sketch the quadratic curve defined by the equation $y = -3t^2 +30t + 15$
     
  4. (4 Marks) Determine the three angles of a triangle, with sides $8cm,\;11cm$ and $14cm$
     
  5. (4 Marks) The points $A(1,3)$, $B(6,5)$, $C(4,2)$ and $D(2,k)$ are such that the line segments AB and CD are perpendicular. Determine the value of $k$.
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All answers in full solution
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  1. Completing the square on this expression gives
    $$\begin{align}t^2 + 12t - 8 &= (t+6)^2 - 6^2 - 8\\  &= (t+6)^2 - 44\end{align}$$

  2. $$\begin{align}\frac{\sqrt{t^3} + 16t^2}{(2t)^5} &=\frac{t^\frac{3}{2}+ 16t^2}{32t^5}\\ &= \frac{1}{32}t^{-\frac{7}{2}} + \frac{1}{2}t^{-3}\end{align} $$

  3. solution curve
  4. Apply in sequence, the cosine rule, then the sine rule.solution triangle

    Recall $a^2 = b^2 + c^2 - 2bc\cos(A)$. Let $A = 14,\; B =11,\; C = 8$, then $$\begin{align}\cos(A) &= \frac{11^2 + 8^2 - 14^2}{2\times 11 \times 8} \\ &= \frac{-11}{2\times 11 \times 8} \\ &= -\frac{1}{16} \end{align}$$ this gives $A = 93.59^\circ$. We now have a side and the opposite angle and so can apply the sine rule. $$\begin{align}\frac{14}{\sin(93.59)} &= \frac{8}{\sin B} \\ \sin B &= \frac{8\sin(93.59)}{14} \\ B &= 34.77^\circ\end{align}$$ The third angle can just be calculated from $A+B+C = 180$, to give $C = 180 - (93.59 + 34.77) = 51.64.$ A final comment on this question is to observe that our answers are consistent with the appearance of our diagram, with $A$ being approximately $90^\circ$ and $B$ being the smallest angle.

  5. The gradient of $AB$ is given by $m_{AB} = \frac{5-3}{6-1}= \frac{2}{5}$. The gradient of $CD$ is given by $m_{CD} = \frac{k-2}{2-4} = \frac{1}{2}(2-k)$. As $AB$ and $CD$ are perpendicular, then $$\begin{align}m_{AB}\times m_{CD} &= -1 \\ \frac{2}{5}\times \frac{1}{2}(2-k) &= -1 \\ 2-k &= -5 \\ k &= 7\end{align}$$