Find the coefficient of \(x^{4}\) in the expansion of \((1-2x)^{6}\)
-320
-240
240
320
Correct answer is C
\(^{6}C_{4}(1)^{6-4}(-2x)^{4}\) = \(15\times1\times16x^{4} = 240x^{4}\)
The coefficient of \(x^{4}\)= 240
Given that \(\frac{6x+m}{2x^{2}+7x-15} \equiv \frac{4}{x+5} - \frac{2}{2x-3}\), find the value of m
20
12
-10
-22
Correct answer is D
Taking the LCM of the right hand side of the equation, we have
\(\frac{4(2x-3) - 2(x+5)}{(x+5)(2x-3)} = \frac{6x+m}{2x^{2}+7x-15}\)
Comparing the numerators, we have
\(4(2x-3) - 2(x+5) = 6x+m\)
\(8x-12-2x-10 = 6x -22 = 6x + m\)
\(\implies m = -22\)
Given that \(f(x) = \frac{x+1}{2}\), find \(f^{1}(-2)\).
-5
-3
\(-\frac{1}{2}\)
5
Correct answer is A
Let \(f(x) = y\), then we have
\(y = \frac{x+1}{2} \implies 2y = x+1; x = 2y-1\)
Let \(f^{1}(x) = x; x = 2y-1\), replacing y with x,
\(f^{1}(x) = 2x - 1 \implies f^{1}(-2) = 2(-2) -1= -5\)
\(x<2\)
\(x \leq 2\)
\(x = 2\)
\(x > -2\)
Correct answer is B
\(f : x \to \sqrt{4 -2x}\) defined on the set of real numbers, R, which has range from \((-\infty, \infty)\) but because of the root sign, it is defined from \([0, \infty)\).
This is because the root of numbers only has real number values from 0 and upwards.
\(\sqrt{4-2x} \geq 0 \implies 4-2x \geq 0\)
\(-2x \geq -4; x \leq 2\)
Find the coordinates of the centre of the circle \(3x^{2}+3y^{2} - 4x + 8y -2=0\)
(-2,4)
(\(\frac{-2}{3}, \frac{4}{3}\))
(\(\frac{2}{3}, \frac{-4}{3}\))
(2, -4)
Correct answer is C
The equation for a circle with centre coordinates (a, b) and radius r is
\((x-a)^{2} + (y-b)^{2} = r^{2}\)
Expanding the above equation, we have
\(x^{2} - 2ax +a^{2} + y^{2} - 2by + b^{2} - r^{2} = 0\) so that
\(x^{2} - 2ax + y^{2} - 2by = r^{2} - a^{2} - b^{2}\)
Taking the original equation given, \(3x^{2} + 3y^{2} - 4x + 8y = 2\) and making the coefficients of \(x^{2}\) and \(y^{2}\) = 1,
\(x^{2} + y^{2} - \frac{4x}{3} + \frac{8y}{3} = \frac{2}{3}\), comparing, we have
\(2a = \frac{4}{3}; 2b = \frac{-8}{3}\)
\(\implies a = \frac{2}{3}; b = \frac{-4}{3}\)