2
4
6
8
Correct answer is D
The gradient of the equation = \(\frac{y_{1} - y_{0}}{x_{1} - x_{0}} = \frac{\frac{5}{2} - 0}{\frac{5}{4} - 0}\)
= \(\frac{\frac{5}{2}}{\frac{5}{4}} = 2\)
The line passes through the origin, therefore the equation of the line is y = mx
\(y = 2x\)
When x = 4, y = 2 x 4 = 8.
1
2
3
4
Correct answer is B
\(Gradient = \frac{y_{1} - y_{0}}{x_{1} - x_{0}}\)
Line passing through \((0, 0)\) & \((1\frac{1}{4}, 2\frac{1}{2})\),
\(Gradient = \frac{2\frac{1}{2} - 0}{1\frac{1}{4} - 0} = \frac{\frac{5}{2}}{\frac{5}{4}} \)
Find the minimum value of \(y = x^{2} + 6x - 12\).
-21
-12
-6
-3
Correct answer is A
\(y = x^{2} + 6x - 12\)
\(\frac{\mathrm d y}{\mathrm d x} = 2x + 6 = 0\)
\(2x = -6 \implies x = -3\)
\(y(-3) = (-3)^{2} + 6(-3) - 12 = 9 - 18 - 12 = -21\)
\(\frac{9}{16}\)
\(\frac{81}{16}\)
\(9\)
\(9\frac{9}{16}\)
Correct answer is D
\(T_{n} = ar^{n-1}\) (for an exponential sequence)
\(T_{1} = 36 = a\)
\(T_{2} = ar = 36r = p\)
\(T_{3} = ar^{2} = 36r^{2} = \frac{9}{4}\)
\(T_{4} = ar^{3} = 36r^{3} = q\)
\(36r^{2} = \frac{9}{4} \implies r^{2} = \frac{\frac{9}{4}}{36} = \frac{1}{16}\)
\(r = \sqrt{\frac{1}{16}} = \frac{1}{4}\)
\( p = 36 \times \frac{1}{4} = 9 ; q = \frac{9}{4} \times \frac{1}{4} = \frac{9}{16}\)
\(p + q = 9 + \frac{9}{16} = 9\frac{9}{16}\)
Given that \(f : x \to \frac{2x - 1}{x + 2}, x \neq -2\), find \(f^{-1}\), the inverse of f
\(f^{-1} : x \to \frac{1+2x}{2-x}, x \neq 2\)
\(f^{-1} : x \to \frac{1-2x}{x+2}, x \neq -2\)
\(f^{-1} : x \to \frac{1-2x}{x-2}, x \neq 2\)
\(f^{-1} : x \to \frac{1+2x}{x+2}, x \neq -2\)
Correct answer is A
\(f(x) = \frac{2x - 1}{x + 2}\)
\(y = \frac{2x - 1}{x + 2}\)
\(x = \frac{2y - 1}{y + 2} \implies x(y + 2) = 2y - 1\)
\(xy - 2y = -1 - 2x \implies y = \frac{-1 - 2x}{x - 2}\)
\(f^{-1} : x \to \frac{1 + 2x}{2 - x} ; x \neq 2\)