If α and β are the roots of \(7x2 +12x - 4 = 0\),find the value of \(\frac{αβ}{(α + β)^2}\)
\( \frac{7}{36}\)
-\( \frac{36}{7}\)
\(\frac{36}{7}\)
-\( \frac{7}{36}\)
Correct answer is D
The general form of a quadratic equation is:
\(x^2\) -(sum of roots)\(x\) +(product of roots) = 0
\(7x^2+12x-4=0\)
Divide through by 7
=\(x^2+\frac{12}{7}x-\frac{4}{7}=0\)
=\(x^2-(-\frac{12}{7})x+(-\frac{4}{7})=0\)
\(\therefore\) sum of roots = \(-\frac{12}{7}\), and products of roots =\(-\frac{4}{7}\)
α + β = \(-\frac{12}{7}, αβ = -\frac{4}{7}\)
\(\frac{αβ}{(α + β)^2} = \frac{\frac{-4}{7}}{(\frac{-12}{7})^2}\)
=\(\frac{\frac{-4}{7}}{\frac{144}{49}}=-\frac{4}{7}\times\frac{49}{144}\)
\(\therefore - \frac{7}{36}\)
>Evaluate: \(\int(2x + 1)^3 dx\)
\(8(2x + 1)^2 + k\)
\(6(2x + 1)^2 + k\)
\(\frac{1}{8} (2x + 1)^4 + k\)
\(\frac{1}{6} (2x + 1)^4 + k\)
Correct answer is C
Using substitution method, Let \(u = 2x + 1\)
\(\frac{du}{dx}=2==>du=2dx==>dx=\frac{du}{2}\)
=\(\int\frac{u^3}{2} du = \frac{1}{2}\int u^3 du\)
=\(\frac{1}{2}(\frac{u^4}{4})=\frac{u^4}{8}\)
\(\therefore\frac{1}{8} (2x + 1)^4 + k\)
In how many ways can a committee of 3 women and 2 men be chosen from a group of 7 men and 5 women?
500
350
720
210
Correct answer is D
For choosing, its different 'combinations'
Options - 7 men, 5 women
To pick - 2 men, 3 women
∴ The number of ways to choose a committee of 3 women and 2 men from a group of 7 men and 5 women is:
=\(^5C_3 \times ^7C_2\)
=\(\frac{5\times4\times3}{3\times2}\times\frac{7\times6}{2\times1}\)
=\(10\times21\)
=210
2.0
1.4
1.8
1.6
Correct answer is B
n = 5
x̄ = \(\frac{∑x}{n} = \frac{85 + 84 + 83 + 86 + 87}{5} = \frac{425}{5} = 85\)
| \(x\) | \(x - x̄\) | \((x - x̄)^2\) |
| 85 | 0 | 0 |
| 84 | -1 | 1 |
| 83 | -2 | 4 |
| 86 | 1 | 1 |
| 87 | 2 | 4 |
| \(\Sigma(x - x̄)^2 = 10\) |
\(S.D = √\frac{∑(x - x̄ )}{n} = √\frac{10}{5}\)
∴ S.D = √2 =1.4
\(5\frac{2}{5}\)
\(4\frac{1}{5}\)
\(13\frac{1}{2}\)
\(6\frac{3}{4}\)
Correct answer is A
Sum to infinity of a G.P when /r/ < 1 = \(\frac{a}{1 - r}\)
a = \(\frac{9}{2},r = \frac{T_2}{T_1} = \frac{3}{4} + \frac{9}{2}\)
r = \(\frac{3}{4} \times \frac{2}{9} = \frac{1}{6}\)
\(S_∞ = \frac{\frac{9}{2}}{1-\frac{1}{6}} = \frac{\frac{9}{2}}{\frac{5}{6}} = \frac{27}{5}\)
\(\therefore S_∞ = 5\frac{2}{5}\)