All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

5.
(i)
\frac{dy}{dx} = 3x^{2} + 2kx + 7

When x = 1, ~ \frac{dy}{dx} = 0

\Rightarrow 3 + 2k + 7 = 0

k = -5

\Rightarrow 1^{3} - 5 (1)^{2} + 7 + c = 2

\therefore, c = -1

(ii)
\frac{dy}{dx} = 3x^{2} - 10 x + 7 = 0

x= \frac{7}{3} \mathrm{~or~} x = 1

When x = \frac{7}{3}, y = \frac{22}{27}

\therefore, (\frac{7}{3}, \frac{22}{27})

(iii)

Graph of 5
Graph of 5(iii)

(iv)
Area = \int_1^2 x^{3} - 5 x^{2} + 7x - 1 dx

= \frac{x^{4}}{4} - \frac{5x^{3}}{3} + \frac{7x^{2}}{2} - x \biggl|_1^2

= \frac{19}{12}

KS Comments

Students must be careful to leave answers in fractions and not just decimals. They should also check their answers with the Graphing Calculator.

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