Short Questions (Question 11 – 13)

Question 11:
Given that the graph of function

f (x) = h x^{3} + \frac{k}{x^{2}}

has a gradient function

f' (x) = 12 x^{2} - \frac{258}{x^{3}}

such that h and k are constants. Find the values of h and k.

Solution:

\begin{array}{l} f (x) = h x^{3} + \frac{k}{x^{2}} = h x^{3} + k x^{- 2} \\ f' (x) = 3 h x^{2} - 2 k x^{- 3} \\ f' (x) = 3 h x^{2} - \frac{2 k}{x^{3}} \\ But it is given that f' (x) = 12 x^{2} - \frac{258}{x^{3}} \\ Hence, by comparison, \\ 3 h = 12 and          2 k = 258 \\ h = 4 k = 129 \end{array}

Question 12:

\begin{array}{l} Given that y = \frac{x^{2}}{x + 3}, show that \frac{d y}{d x} = \frac{x^{2} + 6 x}{{(x + 3)}^{2}} \\ Find \frac{d^{2} y}{d x^{2}} in the simplest form . \end{array}

Solution:

\begin{array}{l} y = \frac{x^{2}}{x + 3} \\ \frac{d y}{d x} = \frac{(x + 3) (2 x) - x^{2} .1}{{(x + 3)}^{2}} = \frac{2 x^{2} + 6 x - x^{2}}{{(x + 3)}^{2}} \\ \frac{d y}{d x} = \frac{x^{2} + 6 x}{{(x + 3)}^{2}} (shown) \\ \frac{d^{2} y}{d x^{2}} = \frac{{(x + 3)}^{2} (2 x + 6) - (x^{2} + 6 x) .2 (x + 3)}{{(x + 3)}^{4}} \\ \frac{d^{2} y}{d x^{2}} = \frac{(x + 3) [(x + 3) (2 x + 6) - 2 (x^{2} + 6 x)]}{{(x + 3)}^{4}} \\ \frac{d^{2} y}{d x^{2}} = \frac{[2 x^{2} + 6 x + 6 x + 18 - 2 x^{2} - 12 x]}{{(x + 3)}^{3}} \\ \frac{d^{2} y}{d x^{2}} = \frac{18}{{(x + 3)}^{3}} \end{array}

Question 13:

If y = x^{2} + 4 x, show that x^{2} \frac{d^{2} y}{d x^{2}} - 2 x \frac{d y}{d x} + 2 y = 0.

Solution:

\begin{array}{l} y = x^{2} + 4 x \\ \frac{d y}{d x} = 2 x + 4 \\ \frac{d^{2} y}{d x^{2}} = 2 \\ x^{2} \frac{d^{2} y}{d x^{2}} - 2 x \frac{d y}{d x} + 2 y \\ = x^{2} (2) - 2 x (2 x + 4) + 2 (x^{2} + 4 x) \\ = 2 x^{2} - 4 x^{2} - 8 x + 2 x^{2} + 8 x \\ = 0 (Shown) \end{array}