If \(f(x) = 3x^{3} + 8x^{2} + 6x + k\) and \(f(2) = 1\), find the value of k.
-67
-61
61
67
Correct answer is A
\(f(x) = 3x^{3} + 8x^{2} + 6x + k\)
\(f(2) = 3(2^{3}) + 8(2^{2}) + 6(2) + k = 1\)
\(\implies 24 + 32 + 12 + k = 1\)
\(68 + k = 1 \therefore k = 1 - 68 = -67\)
8
6
5
4
Correct answer is C
\((x * y) = \frac{x+y}{2}\)
\((3 * b) = \frac{3+b}{2}\)
\(x \circ y = \frac{x^{2}}{y}\)
\((\frac{3+b}{2}) \circ 48 = \frac{(\frac{3+b}{2})^{2}}{48} = \frac{1}{3}\)
\(\frac{(3+b)^{2}}{48 \times 4} = \frac{1}{3}\)
\((3 + b)^{2} = \frac{48 \times 4}{3} = 64\)
\(b^{2} + 6b + 9 = 64 \implies b^{2} + 6b + 9 - 64 = 0\)
\(b^{2} + 6b - 55 = 0 \implies b^{2} - 5b + 11b - 55 = 0\)
\(b(b - 5) + 11(b - 5) = 0 \implies (b - 5) = \text{0 or (} b + 11) = 0\)
Since b > 0, b - 5 = 0
b = 5.
Given that \(3x + 4y + 6 = 0\) and \(4x - by + 3 = 0\) are perpendicular, find the value of b.
4
3
\(\frac{1}{3}\)
\(\frac{1}{4}\)
Correct answer is B
When you have two lines, \(y_{1}, y_{2}\), perpendicular to each other, the product of their slopes = -1.
\(3x + 4y + 6 = 0 \implies 4y = -6 - 3x\)
\(\therefore y = \frac{-6}{4} - \frac{3}{4}x\)
\(\frac{\mathrm d y}{\mathrm d x} = \frac{-3}{4}\)
Also, \(4x - by + 3 = 0 \implies by = 4x + 3\)
\(y = \frac{4}{b}x + \frac{3}{b}\)
\(\frac{\mathrm d y}{\mathrm d x} = \frac{4}{b}\)
\(\frac{-3}{4} \times \frac{4}{b} = -1 \implies \frac{4}{b} = \frac{4}{3}\)
\(b = 3\)
Simplify: \((1 - \sin \theta)(1 + \sin \theta)\)
\(\sin^{2} \theta\)
\(\sec^{2} \theta\)
\(\tan^{2} \theta\)
\(\cos^{2} \theta\)
Correct answer is D
\((1 + \sin \theta)(1 - \sin \theta) = 1 - \sin \theta + \sin \theta - \sin^{2} \theta\)
\(= 1 - \sin^{2} \theta\)
Recall, \(\cos^{2} \theta + \sin^{2} \theta = 1\)
\(\therefore 1 - \sin^{2} \theta = \cos^{2} \theta\).
If \(\frac{1}{5^{-y}} = 25(5^{4-2y})\), find the value of y.
4
2
-4
-5
Correct answer is B
\(\frac{1}{5^{-y}} = 25(5^{4-2y})\)
\(\implies 5^{y} = (5^{2})(5^{4-2y})\)
\(5^{y} = 5^{2+4-2y}\)
Comparing bases, we have
\(y = 6 - 2y\)
\(3y = 6 \implies y = 2\)