\(\frac{-7}{8}\)
\(\frac{-3}{8}\)
\(\frac{1}{8}\)
\(\frac{5}{8}\)
Correct answer is C
The remainder theorem states that if f(x) is divided by (x - a), the remainder is f(a).
\(f(x) = x^{3} - 2x + m\) divided by (x - 1), so that a = 1.
Remainder = \(f(1) = 1^3 - 2(1) + m = -1 + m\)
\(f(x) = 2x^{3} + x - m\) divided by (2x + 1), so that a = \(\frac{-1}{2}\)
\(f(\frac{-1}{2}) = 2(\frac{-1}{2}^{3}) + (\frac{-1}{2}) - m = \frac{-3}{4} - m\)
\(\implies m - 1 = \frac{-3}{4} - m\), collecting like terms,
\(2m = \frac{1}{4} \therefore m = \frac{1}{8}\)
If the solution set of \(x^{2} + kx - 5 = 0\) is (-1, 5), find the value of k.
-6
-4
4
5
Correct answer is B
Given x = (-1, 5) for the equation \(x^{2} + kx - 5 = 0\)
\(x = -1 \implies x + 1 = 0\); \(x = 5 \implies x - 5 = 0\)
\((x + 1)(x - 5) = 0\), expanding,
\(x^{2} - 5x + x - 5 = 0 \therefore x^{2} - 4x - 5 = 0\)
\(\therefore\) k = -4.
Factorize completely: \(x^{2} + x^{2}y + 3x - 10y + 3xy - 10\).
(x + 2)(x + 5)(y + 1)
(x + 2)(x - 5)(y + 1)
(x - 2)(x + 5)(y + 1)
(x - 2)(x - 5)(y + 1)
Correct answer is C
\(x^{2} + x^{2}y + 3x - 10y + 3xy -10\)
= \(x^{2} + x^{2}y + 3x + 3xy - 10y - 10 = x^{2}(1 + y) + 3x(1 + y) - 10(y + 1)\)
= \((x^{2} + 3x - 10)(y + 1)\)
= \((x^{2} + 3x - 10) = x^{2} - 2x + 5x - 10\)
= \(x(x - 2) + 5(x - 2) = (x - 2)(x +5)\)
\(\therefore x^{2} + x^{2}y + 3x - 10y + 3xy -10 = (x - 2)(x + 5)(y + 1)\).
If \(y = 4x - 1\), list the range of the domain \({-2 \leq x \leq 2}\), where x is an integer.
{-9, -1, 2,3, 4}
{-9, -2, 0, 1, 7}
{-5, -4, -3, -2}
{-9, -5, -1, 3, 7}
Correct answer is D
The elements of x are {-2, -1, 0, 1, 2}
\(y = 4x - 1\) = 4(-2) - 1 = -9; 4(-1) - 1 = -5; 4(0) - 1 = -1; 4(1) - 1 = 3; 4(2) - 1 = 7.
The range of x is {-9, -5, -1, 3 7}.
If \(f(x) = \frac{4}{x} - 1, x \neq 0\), find \(f^{-1}(7)\).
\(\frac{-3}{7}\)
\(0\)
\(\frac{1}{2}\)
\(4\)
Correct answer is C
\(f(x) = \frac{4}{x} - 1\). Let y = f(x)
\(y = \frac{4 - x}{x} \implies xy + x = 4\)
\(x(y + 1) = 4 \therefore x = \frac{4}{y + 1}\)
\(f^{-1}(7) = \frac{4}{7 + 1} = \frac{1}{2}\)