WAEC Further Mathematics Past Questions & Answers - Page 123

\(x^{2}\sqrt{x} \equiv x^{2}. x^{\frac{1}{2}} = x^{\frac{5}{2}}\)

\(\implies \frac{\mathrm d y}{\mathrm d x} = x^{\frac{5}{2}}\)

\(y = \int x^{\frac{5}{2}} \mathrm d x\)

= \(\frac{x^{\frac{5}{2} + 1}}{\frac{5}{2} + 1} + c\)

= \(\frac{2x^{\frac{7}{2}}}{7} + c\)

With radius = 7cm, \(Area = \pi r^{2} = \frac{22}{7} \times 7^{2}\)

= \(154cm^{2}\)

The next second, radius = 7.5cm, \(Area = \pi r^{2} = \frac{22}{7} \times 7.5^{2}\)

= \(176cm^{2}\)

Change in area = \((176 - 154)cm^{2} = 22cm^{2}\)

\(\therefore\) The rate of increase = \(22cm^{2}s^{-1}\)

OR

\(Area (A) = \pi r^{2} \implies \frac{\mathrm d A}{\mathrm d r} = 2\pi r\)

Given \(\frac{\mathrm d r}{\mathrm d t} = 0.5\)

\(\frac{\mathrm d A}{\mathrm d r} \times \frac{\mathrm d r}{\mathrm d t} = \frac{\mathrm d A}{\mathrm d t}\)

\(\frac{\mathrm d A}{\mathrm d t} = 2\pi r \times 0.5 = 2 \times \frac{22}{7} \times 7 \times 0.5\)

= \(22cm^{2}s^{-1}\)

\(3x^{2} - x - 6 = y\)

\(a = 3, b = -1, c = -6\)

Minimum value of x = \(\frac{-b}{2a} = \frac{-(-1)}{6} = \frac{1}{6}\)

\(y(\frac{1}{6}) = 3(\frac{1}{6}^{2}) - \frac{1}{6} - 6 = -6\frac{1}{12}\)

Given : \(y = x^{3} - x^{2}\)

\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 2x\)

\(\therefore  \text{The gradient of the tangent at point (x = 2)} = 3(2^{2}) - 2(2) \)

= \(12 - 4 = 8\)

Recall, the tangent and the normal are perpendicular to each other and the product of the gradients of perpendicular lines = -1.

\(\implies \text{the gradient of the normal} = \frac{-1}{8}\)

\(\lim\limits_{x \to 3} \frac{2x^{2} + x - 21}{x - 3}\)

\(2x^{2} + x - 21 = 2x^{2} - 6x + 7x - 21 \) (by factorizing)

= \((2x + 7)(x - 3)\)

\(\therefore \lim\limits_{x \to 3} \frac{2x^{2} + x - 21}{x - 3} \equiv \lim\limits_{x \to 3} \frac{(2x+7)(x-3)}{x-3}\)

\(\lim\limits_{x \to 3} (2x + 7)  = 2(3) + 7 = 13\)