If \(\begin{vmatrix} m-2 & m+1 \\ m+4 & m-2 \end{vmatrix} = -27\), find the value of m.
\(3\frac{8}{9}\)
\(3\)
\(2\frac{1}{2}\)
\(2\)
Correct answer is B
\(\begin{vmatrix} m-2 & m+1 \\ m+4 & m-2 \end{vmatrix} = -27\)
\((m^{2} - 4m + 4) - (m^{2} + 5m + 4) = -27\)
\(-9m = -27 \implies m = 3\)
20
21
22
23
Correct answer is D
\(T_{n} = a + (n - 1)d\) (for a linear or arithmetic progression)
Given: \(T_{n} = 71, a = -6, d = -2\frac{1}{2} - (-6) = 3\frac{1}{2}\)
\(\implies 71 = -6 + (n - 1)\times 3\frac{1}{2}\)
\(71 = -6 + 3\frac{1}{2}n - 3\frac{1}{2} = -9\frac{1}{2} + 3\frac{1}{2}n\)
\(71 + 9\frac{1}{2} = 3\frac{1}{2}n \implies n = \frac{80\frac{1}{2}}{3\frac{1}{2}}\)
\(= 23\)
\(\frac{1}{3}\)
\(\frac{2}{3}\)
\(\frac{3}{4}\)
\(\frac{4}{3}\)
Correct answer is B
\(T_{n} = ar^{n-1}\) (for a geometric progression)
\(T_{3} = ar^{3-1} = ar^{2} = \frac{8}{3}\)
\(T_{6} = ar^{6-1} = ar^{5} = \frac{64}{81}\)
Dividing \(T_{6}\) by \(T_{3}\),
\(\frac{ar^{5}}{ar^{2}} = \frac{\frac{64}{81}}{\frac{8}{3}} \implies r^{3} = \frac{8}{27}\)
\(\therefore r = \sqrt[3]{\frac{8}{27}} = \frac{2}{3}\)
Find the fourth term in the expansion of \((3x - y)^{6}\).
\(-540x^{3}y^{3}\)
\(-540x^{4}y^{2}\)
\(-27x^{3}y^{3}\)
\(540x^{4}y^{2}\)
Correct answer is A
Listing out in order, the terms of this expansion have coefficients: \(^{6}C_{6}, ^{6}C_{5}, ^{6}C_{4}, ^{6}C_{3}, ^{6}C_{2}, ^{6}C_{1}, ^{6}C_{0}\)
The 4th term = \(^{6}C_{3}(3x)^{3}(-y)^{3} = 20 \times 27x^{3} \times -y^{3}\)
= \(-540x^{3}y^{3}\).
Find the coefficient of \(x^{3}\) in the expansion of \([\frac{1}{3}(2 + x)]^{6}\)
\(\frac{135}{729}\)
\(\frac{149}{729}\)
\(\frac{152}{729}\)
\(\frac{160}{729}\)
Correct answer is D
\([\frac{1}{3}(2 + x)]^{6} = (\frac{2}{3} + \frac{x}{3})^{6}\)
The coefficient of \(x^{3}\) is
\(^{6}C_{3}(\frac{2}{3})^{3}(\frac{1}{3})^{3}x^{3} = (\frac{6!}{3!3!})(\frac{8}{27})(\frac{1}{27})x^{3}\)
= \(\frac{160}{729}\)