\(\begin{vmatrix} 6 & 2 \\ 1 & 6 \end{vmatrix}\)
\(\begin{vmatrix} \frac{2}{11} & \frac{1}{12}\\ \frac{3}{20} & \frac{1}{10} \end{vmatrix}\)
\(\begin{vmatrix} -3 & 2 \\ -1 & -6 \end{vmatrix}\)
\(\begin{vmatrix} \frac{3}{10} & \frac{1}{10} \\ \frac{-1}{20} & \frac{3}{20}\end{vmatrix}\)
Correct answer is D
\(A = \begin{vmatrix} 3 & -2 \\ 1 & 6 \end{vmatrix}\)
|A| = (3 x 6) - (-2 x 1)
= 18 + 2
= 20.
A\(^{-1}\) = \(\frac{1}{20} \begin{vmatrix} 6 & 2 \\ -1 & 3 \end{vmatrix}\)
= \(\begin{vmatrix} \frac{6}{20} & \frac{2}{20} \\ \frac{-1}{20} & \frac{3}{20} \end{vmatrix}\)
= \(\begin{vmatrix} \frac{3}{10} & \frac{1}{10} \\ \frac{-1}{20} & \frac{3}{20} \end{vmatrix}\)
5
8
6
3
Correct answer is B
\(\begin{vmatrix} 2 & -5 & 3 \\ x & 1 & 4 \\ 0 & 3 & 2 \end{vmatrix} = 132\)
\(\implies 2 \begin{vmatrix} 1 & 4 \\ 3 & 2 \end{vmatrix} - (-5) \begin{vmatrix} x & 4 \\ 0 & 2 \end{vmatrix} + 3 \begin{vmatrix} x & 1 \\ 0 & 3 \end{vmatrix} = 132\)
\(2(2 - 12) + 5(2x) + 3(3x) = 132\)
\(-20 + 10x + 9x = 132\)
\(19x = 152\)
\(x = 8\)
If 2x\(^2\) + x - 3 divides x - 2, find the remainder.
7
3
5
6
Correct answer is A
When you divide a polynomial p(x) by (x - a), the remainder = p(a)
i.e. In the case of 2x\(^2\) + x - 3 \(\div\) (x - 2), the remainder = p(2).
= 2(2)\(^2\) + 2 - 3
= 8 + 2 - 3
= 7.
Find the polynomial if given q(x) = x\(^2\) - x - 5, d(x) = 3x - 1 and r(x) = 7.
3x\(^3\) - 4x\(^2\) - 14x + 12
3x\(^2\) + 3x - 7
3x\(^3\) + 4x\(^2\) + 14x - 12
3x\(^2\) - 3x + 4
Correct answer is A
Given q(x) [quotient], d(x) [divisor] and r(x) [remainder], the polynomial is gotten by multiplying the quotient and the divisor and adding the remainder.
i.e In this case, the polynomial = (x\(^2\) - x - 5)(3x - 1) + 7.
= (3x\(^3\) - x\(^2\) - 3x\(^2\) + x - 15x + 5) + 7
= (3x\(^3\) - 4x\(^2\) - 14x + 5) + 7
= 3x\(^3\) - 4x\(^2\) - 14x + 12
Determine the values for which \(x^2 - 7x + 10 \leq 0\)
2 \(\leq\) x \(\geq\) 5
-2 \(\leq\) x \(\leq\) 3
-2 \(\leq\) x \(\geq\) 3
2 \(\leq\) x \(\leq\) 5
Correct answer is D
\(x^2 - 7x + 10 \leq 0\)
Solve for \(x^2 - 7x + 10 = 0\)
We have, (x - 5)(x - 2) \(\leq\) 0.
Conditions:
Case 1: (x - 5) \(\leq\) 0, (x - 2) \(\geq\) 0.
\(\implies\) x \(\leq\) 5; x \(\geq\) 2.
2 \(\leq\) x \(\leq\) 5.
Choosing x = 3,
3\(^2\) - 7(3) + 10 = 9 - 21 + 10
= -2 \(\leq\) 0.
\(\therefore\) 2 \(\leq\) x \(\leq\) 5.