The table shows the operation * on the set {x, y, z, w}.
| * | X | Y | Z | W |
| X | Y | Z | X | W |
| Y | Z | W | Y | X |
| Z | X | Y | Z | W |
| W | W | X | W | Z |
Find the identity of the element.
W
Y
Z
X
Correct answer is C
From the table, x * z = x, y * z = y, z * z = z and w * z = w
∴ z is the identity element
If\((\frac{1}{9})^{2x-1} = (\frac{1}{81})^{2-3x}\)find the value of x
\(-\frac{5}{8}\)
\(-\frac{3}{4}\)
\(\frac{3}{4}\)
\(-\frac{5}{8}\)
Correct answer is D
\((\frac{1}{9})^{2x-1} = (\frac{1}{81})^{2-3x}\)
\((\frac{1}{9})^{2x-1} = (\frac{1}{9})^{2(2-3x)}\)
\((\frac{1}{9})^{2x-1} = (\frac{1}{9})^{4-6x}\)
Since the bases are equal, powers can be equated
= 2x - 1 = 4 - 6x
= 2x + 6x = 4 + 1
= 8x = 5
\(\therefore x = \frac{5}{8}\)
\(\frac{63}{65}\)
\(\frac{48}{65}\)
\(\frac{56}{65}\)
\(\frac{16}{65}\)
Correct answer is A
\(sin x = \frac{4}{5}\) and \(cos y = \frac{12}{13}\)
x is obtuse i.e sin x = + ve while cos x = + ve
\(cos x=\frac{3}{5}==>cos x=-\frac {3}{5}(obtuse)\)
\(sin y= \frac{5}{13}\)
\(sin (x-y) = sin x\) \(cos y - cos x\) \(sin y\)
\(sin(x-y) = \frac{4}{5}\times\frac{12}{13}-(-\frac{3}{5})\times\frac{5}{13}\)
\(sin(x-y) = \frac{48}{65}-(-\frac{3}{13})\)
\(\therefore sin (x-y) = \frac{48}{65} + \frac{3}{13} = \frac{63}{65}\)
Evaluate \(\int^1_0 x(x^2-2)^2 dx\)
\(\frac{6}{7}\)
\(1\frac{1}{6}\)
\(\frac{1}{7}\)
\(3\frac{1}{6}\)
Correct answer is B
\(\int^1_0 x(x^2-2)^2 dx\)
\((x^2-2)^2=x^4-2x^2-2x^2+4\)
=\(x^4-4x^2+4\)
\(x(x^2-2)^2=x(x^4-4x^2+4)\)
=\(x^5-4x^3+4x\)
\(\int^1_0 x(x^2-2)^2 dx = \int^1_0 x^5 - 4x^3 + 4x dx\)
=\((\frac{x^6}{6} - x^4+2x^2)^1_0\)
= \((\frac{(1)^6}{6} - (1)^4 +2(1)^2)-(\frac{(0)^6}{6} - (0)^4+2(0)^2)\)
=\(\frac{7}{6} - 0 =\frac{7}{6}\)
\(\therefore 1\frac{1}{6}\)
19 \(ms ^{-2}\)
21 \(ms ^{-2}\)
41 \(ms ^{-2}\)
31 \(ms ^{-2}\)
Correct answer is C
\(S = 5t^3 - \frac{19}{2} t^2 + 6t - 4\)
\(v(t)=\frac{dS}{dt}=15t^2-19t+6\)
\(a(t)=\frac{dv}{dt}=30t-19\)
∴a(2)=30(2)-19=60-19=41 \(ms ^{-2}\)