4
2
\(\frac{1}{4}\)
-4
Correct answer is D
\(a \Delta b = a + b + 4\)
\(a \Delta e = a + e + 4 = a\)
\(e = a - a - 4 = -4\)
\(\begin{pmatrix} -6 & 17 \\ 3 & 1 \end{pmatrix}\)
\(\begin{pmatrix} -2 & 9 \\ 4 & 1 \end{pmatrix}\)
\(\begin{pmatrix} 0 & -6 \\ 9 & -8 \end{pmatrix}\)
\(\begin{pmatrix} 0 & 10 \\ 9 & -4 \end{pmatrix}\)
Correct answer is D
\(P = \begin{pmatrix} 2 & 1 \\ 5 & -3 \end{pmatrix} ; Q = \begin{pmatrix} 4 & -8 \\ 1 & -2 \end{pmatrix}\)
\(2P = \begin{pmatrix} 4 & 2 \\ 10 & -6 \end{pmatrix}\)
\(2P - Q = \begin{pmatrix} 4 & 2 \\ 10 & -6 \end{pmatrix} - \begin{pmatrix} 4 & -8 \\ 1 & -2 \end{pmatrix}\)
= \(\begin{pmatrix} 0 & 10 \\ 9 & -4 \end{pmatrix}\)
If \(y = x^{3} - x^{2} - x + 6\), find the values of x at the turning point.
\(\frac{1}{2}, 3\)
\(\frac{1}{3}, -\frac{1}{2}\)
\(1, -\frac{1}{3}\)
\(1, \frac{1}{3}\)
Correct answer is C
At turning point, \(\frac{\mathrm d y}{\mathrm d x} = 0\).
Given \(x^{3} - x^{2} - x + 6 \)
\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 2x - 1 = 0 \)
\(3x^{2} - 3x + x - 1 = 0 \implies (3x + 1)(x - 1) = 0\)
\(x = \frac{-1}{3}, 1\)
Evaluate \(\int_{-2}^{3} (3x^{2} - 2x - 12) \mathrm {d} x\)
-30
-18
-6
6
Correct answer is A
\(\int (3x^{2} - 2x - 12) \mathrm {d} x = \frac{3x^{2 + 1}}{2 + 1} - \frac{2x^{1 + 1}}{2} - 12x\)
= \(x^{3} - x^{2} - 12x\)
\((x^{3} - x^{2} - 12x)|_{-2}^{3} = ((3^{3}) - (3^{2}) - 12(3)) - ((-2^{3}) - (-2^{2}) - 12(-2))\)
= \((27 - 9 - 36) - (-8 - 4 + 24) = -18 - 12 = -30\)
If the midpoint of the line joining (1 - k, -4) and (2, k + 1) is (-k, k), find the value of k.
-4
-3
-2
-1
Correct answer is D
Midpoint between \((x_{1}, y_{1})\) and \((x_{2}, y_{2})\) = \((\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})\).
\((-k, k) = (\frac{2 + (1 + k)}{2}, \frac{-4 + (k + 1)}{2})\)
\(-k = \frac{k + 3}{2} \implies -2k = k + 3\)
\(-3k = 3 \implies k = -1\)