-4
0
4
12
Correct answer is D
\(Gradient = \frac{y_{1} - y_{2}}{x_{1} - x_{2}} \)
\(\frac{1}{2} = \frac{5 - 9}{4 - x}\)
\(\frac{1}{2} = \frac{-4}{4 - x} \implies -8 = 4 - x\)
\(x = 4 + 8 = 12\)
12
11
\(\frac{11}{2}\)
\(\frac{9}{2}\)
Correct answer is C
\(f(x) = 2x - 1; g(x) = x^{2} + 1\)
\(f(x) = 2x - 1\)
\(f(y) = 2y - 1 \)
\(x = 2y - 1 \implies 2y = x + 1\)
\(y = \frac{x + 1}{2}\)
\(g(3) = 3^{2} + 1 = 10\)
\(f^{-1}(10) = \frac{10 + 1}{2} = \frac{11}{2}\)
130
192
210
260
Correct answer is B
\(S_{n} = \frac{n}{2}(2a + (n - 1)d)\) ( for an arithmetic progression)
\(S_{3} = 18 = \frac{3}{2}(2(4) + (3 - 1) d) \)
\(18 = \frac{3}{2} (8 + 2d)\)
\(18 = 12 + 3d \implies 3d = 6\)
\(d = 2\)
\(\therefore T_{1} = 4 \implies T_{2} = 4 + 2 = 6; T_{3} = 6 + 2 = 8\)
Their product = \(4 \times 6 \times 8 = 192\)
Which of the following is the same as \(\sin (270 + x)°\)?
\(\sin x\)
\(\tan x\)
\(- \sin x \)
\(- \cos x\)
Correct answer is D
No explanation has been provided for this answer.
\(x^{2} - 3x + 2 = 0\)
\(x^{2} - 2x + 3 = 0\)
\(x^{2} - 3x - 2 = 0\)
\(x^{2} - 2x - 3 = 0\)
Correct answer is A
Given \(ax^{2} + bx + c = 0 (\text{general form of a quadratic equation})\)
We have \(x^{2} + \frac{b}{a}x + \frac{c}{a} = 0\)
\( x^{2} - (-\frac{b}{a})x + \frac{c}{a} = 0\)
\(\implies x^{2} - (\alpha + \beta)x + (\alpha \beta) = 0\)
\(\therefore \text{The equation is} x^{2} - 3x + 2 = 0\)