1. k = -2
2. sqrt(26)/26
3. 1.75
4. -3/x – 2/x^2 + 4/(x-2)
5(i) v = 0.263
5(ii) around v = 0.231
5(iii) around f = 0.120
6. (i) Remember that when there is a mixture of trigo functions which cannot fit into a single trigo identity, convert all the given trigo functions into sine and cosine. Start with the LHS, the more complicated expression. You will reach the key stage where
LHS = 1 / [(1 – sin^2@)/(sin@cos@)] = 1/ (cos^2@/sin@cos@) = 1/(cos@/sin@) = tan@
6(ii) acute angle, so @ = 1.05.
7(i) A = (h, 2h); B = (7h/4, 2h); C = (h, h/2)
7(ii) Area of trapezium = 21 sqr units.
8. Integrate f'(x) to get f(x) with a +c. Use f(pi/2) = 0 to get c = 1/4. Differentiate f'(x) to get f”(x)= 4cos4x + 2sin2x. Proceed to add f”(x) to 4f(x) to get the RHS.
9(i) Pure inequality question. Solve 2x^2 + 5x – 12 > 0 by sketching a quadratic graph. Ans: {x : x < -4 U x > 3/2}
9(ii) Show that the (b^2 – 4ac) of (4x^2 + 4x + 1 = 0) is zero.
9(iii) a = -1.
To be continued ….
