CDC Model Question 2083 – Grade 10 Optional Mathematics

4. Matrices \(A = \begin{pmatrix}2 & 1\\1 & 3\end{pmatrix}\) and \(B = \begin{pmatrix}a\\b\end{pmatrix}\) are given.

(a)Find \(A^T\).[1]
(b)Find \(A^{-1}\).[1]
(c)The constraints are \(2x + y \leq 8\), \(x + 3y \leq 9\), \(x \geq 0\), \(y \geq 0\). Show the feasible region on a graph.[2]

5. Given \(\sin(A + B) = \sin A\cos B + \cos A\sin B\).

(a)How can you convert this into \(\sin 2A = 2\sin A\cos A\)?[1]
(b)If \(\sin 2A = 2\sin A\cos A\), show that \(\sin 2A = \dfrac{2\tan A}{1 + \tan^2 A}\).[1]

6. From the top of a cliff of height 100 m, two points on the same straight line on the ground are observed with angles of depression 60° and 45°.

7. Given a conditional identity:

\[\cos 2A + \cos 2B + \cos 2C = -1 - 4\cos A\cos B\cos C\]

(a)Write \(2\cos A\cos B\) in terms of sum or difference.[1]
(b)Verify the identity for \(A = B = C = 60^\circ\).[2]
(c)Why can the identity not be verified for \(A = 90^\circ, B = 60^\circ, C = 45^\circ\)?[1]

8. A line \(L_1\) passes through \(A(1, 2)\) and \(B(5, 6)\).

(a)Write the matrix that transforms \(A(1, 2)\) into \(A'(-2, -1)\).[1]
(b)\(L_1\) makes an acute angle \(\theta\) with \(L_2 : 2x - y + 3 = 0\). Find \(\tan\theta\).[2]

9. \(x^2 + y^2 - 6x - 4y - 12 = 0\) is a circle.

(a)Write the equation of a circle with centre \((h, k)\) and radius \(r\).[1]
(b)Find the centre and radius of the given circle.[2]
(c)Use the two-point formula to find the equation of the radius through point \(P(8, 2)\) on the circle.[2]

10. Points \(A(2, 1)\) and \(B(6, 3)\).

(a)Apply the combined transformation \(T \circ R\) where \(R\) is reflection in the \(Y\)-axis and \(T\) is translation by \(\begin{pmatrix}3\\2\end{pmatrix}\). Find the images of \(A\) and \(B\).[2]
(b)Plot the object and final image on the same graph paper.[1]

11. Triangle \(ABC\) has vertices \(A(2, 3)\), \(B(4, 1)\), \(C(6, 5)\). \(D\) and \(E\) are midpoints of \(AB\) and \(AC\).

(a)Write the position vector of \(D\) in terms of \(A\) and \(B\).[1]
(b)Find the position vector of \(D\) in \(\vec{i},\,\vec{j}\) form.[1]
(c)Prove by vector method: \(\overrightarrow{DE} = \dfrac{1}{2}\overrightarrow{BC}\).[1]

12. The box plots below show battery life of Battery \(X\) and Battery \(Y\).

(a)Which battery has higher median battery life?[1]
(b)The coefficient of quartile deviation of Battery \(Y\) is 0.098. Calculate that of Battery \(X\).[1]
(c)Which battery would you suggest for a remote signal tower and why?[1]

13. A function \(p(x)\) is defined as:

\[p(x) = \begin{cases} 2x + 1, & 1 \leq x \leq 3 \\ x + 4, & x > 3 \end{cases}\]

14. Triangle \(OAB\) with \(O(0,0)\), \(A(1,0)\), \(B(1,1)\) and transformation matrix \(M = \begin{pmatrix}0 & 1\\1 & 2\end{pmatrix}\).

(a)Find the image \(O'A'B'\) under \(M\).[2]
(b)Find the slope of the altitude from \(O\) to \(AB\).[1]
(c)Is \(M\) singular? If not, how can you make it singular?[1]
(d)Dolma claims that \(M\) and \(M^T\) give the same transformation. Do you agree? Justify with an example.[2]

15. Shreya cuts a wooden cone to form a conic section.

(a)Name the shape when the cut is inclined more than the semi-vertical angle but less than 90°.[1]
(b)For a semicircle \(ABC\) with diameter \(AC\), prove using vectors that \(\angle ABC = 90^\circ\) for any \(B\) on the arc.[2]
(c)Would the claim hold if \(B\) lies inside the semicircle? Explain.[1]
(d)If \((AB)^2 = \sqrt{x+9}\), \((BC)^2 = \sqrt{x}\) and \(AC = 3\) units, find \(x\).[2]

16. A hiker climbs a hill with height \(h = f(x) = mx + 50\) (metres). The slope \(m\) is given by

\[m = \frac{\cos\theta - \cos 3\theta}{\sin 3\theta - \sin\theta}\]

(a)Prove that \(m = \tan 2\theta\).[2]
(b)If \(\tan\theta = \dfrac{1}{3}\), find \(m\).[2]
(c)Using \(f^{-1}(x)\), find the horizontal distance when the hiker is at height 150 m.[2]
(d)Bina claims: “If the angle with the \(X\)-axis doubles, the slope doubles.” Verify with a specific \(\theta\).[1]

17. A bus stop is to be placed at position \(k\) metres along a road. The table gives population in each section:

The total squared walking distance is \(s(k) = 10k^2 - 76k + 178\).

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