There are 7 boys in a class of 20. Find the number of ways of selecting 3 girls and 2 boys
1638
2730
6006
7520
Correct answer is C
No of boys in the class = 7; Girls = 20-7 = 13
No of selection = \(^{13}C_{3} \times ^{7}C_{2} = \frac{13!}{(13-3)!3!} \times \frac{7!}{(7-2)!2!}\)
= \(286\times21 = 6006\)
Simplify \(\frac{\sqrt{128}}{\sqrt{32} - 2\sqrt{2}}\)
\(2\sqrt{2}\)
\(3\sqrt{2}\)
3
4
Correct answer is D
\(\sqrt{128} = \sqrt{64\times2} = 8\sqrt{2}\)
\(\sqrt{32} = \sqrt{16\times2} = 4\sqrt{2}\)
Simplifying, we have \(\frac{8\sqrt{2}}{4\sqrt{2} - 2\sqrt{2}} = \frac{8\sqrt{2}}{2\sqrt{2}}\)
= 4
Evaluate \(\int_{-1}^{0} (x+1)(x-2) \mathrm{d}x\)
\(\frac{7}{6}\)
\(\frac{5}{6}\)
\(\frac{-5}{6}\)
\(\frac{-7}{6}\)
Correct answer is D
Expanding \((x+1)(x-2) = x^{2} - 2x + x - 2 = x^{2} - x - 2\)
\(\int_{-1}^{0} (x^{2} - x - 2) \mathrm{d}x = [\frac{x^{3}}{3} - \frac{x^{2}}{2} - 2x]_{-1}^{0}\)
= \([\frac{0}{3} - \frac{0}{2} - 2\times0 - (\frac{-1^{3}}{3} - \frac{-1^{2}}{2} - 2\times-1)]\)
= \(0 + \frac{1}{3} + \frac{1}{2} - 2 = \frac{-7}{6}\)
Note: This can also be solved using integration by parts.
\(\int uv \mathrm{d}x = u\int v \mathrm{d}x - \int u'(\int v \mathrm{d}x)\mathrm{d}x\).
If \(y = \frac{1+x}{1-x}\), find \(\frac{dy}{dx}\).
\(\frac{2}{(1-x)^{2}}\)
\(\frac{-2}{(1-x)^{2}}\)
\(\frac{-1}{\sqrt{1-x}}\)
\(\frac{1}{\sqrt{1-x}}\)
Correct answer is A
\(y = \frac{1+x}{1-x}\)
Using quotient rule, \(\frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^{2}}\), we have
\(\frac{dy}{dx} = \frac{(1-x)(1) - (1+x)(-1)}{(1-x)^{2}} = \frac{(1 - x +1 +x)}{(1-x)^{2}}\)
= \(\frac{2}{(1-x)^{2}}\).
Given that \(f(x) = 5x^{2} - 4x + 3\), find the coordinates of the point where the gradient is 6.
(4,1)
(4,-2)
(1,4)
(1,-2)
Correct answer is C
\(f(x) = 5x^{2} - 4x + 3\)
\(f'(x) = 10x - 4 = 6 \implies 10x = 10; x=1\)
When x = 1, f(x) = y = \(5(1^{2}) - 4(1) + 3 = 4\)
The coordinates are (1,4)