Differentiate \(x^{2} + xy - 5 = 0\)
\(\frac{-(2x + y)}{x}\)
\(\frac{(2x - y)}{x}\)
\(\frac{-x}{2x + y}\)
\(\frac{(2x + y)}{x}\)
Correct answer is A
\(\frac{\mathrm d}{\mathrm d x}(x^2 + xy - 5) = \frac{\mathrm d (x^{2})}{\mathrm d x} + \frac{\mathrm d (xy)}{\mathrm d x} - \frac{\mathrm d (5)}{\mathrm d x} = 0\)
= \(2x + x\frac{\mathrm d y}{\mathrm d x} + y = 0\)
\(\implies x\frac{\mathrm d y}{\mathrm d x} = -(2x + y)\)
\(\frac{\mathrm d y}{\mathrm d x} = \frac{-(2x + y)}{x}\)
y + 5x + 3 = 0
2y - 5x - 9 = 0
5y + 2x - 8 = 0
5y - 2x - 12 = 0
Correct answer is D
\(Line: 2y + 5x - 6 = 0\)
\(2y = 6 - 5x \implies y = 3 - \frac{5x}{2}\)
Gradient = \(\frac{-5}{2}\)
For the line perpendicular to the given line, Gradient = \(\frac{-1}{\frac{-5}{2}} = \frac{2}{5}\)
The midpoint of P(4, 3) and Q(-6, 1) = \((\frac{-6 + 4}{2}, \frac{3 + 1}{2})\)
= (-1, 2).
Therefore, the line = \(\frac{y - 2}{x + 1} = \frac{2}{5}\)
\(2(x + 1) = 5(y - 2) \implies 5y - 2x - 12 = 0\)
Given that \(f(x) = 2x^{3} - 3x^{2} - 11x + 6\) and \(f(3) = 0\), factorize f(x)
(x - 3)(x - 2)(2x + 2)
(x + 3)(x - 2)(x - 1)
(x - 3)(x + 2)(2x -1)
(x + 3)(x - 2)(2x - 1)
Correct answer is C
Since f(3) = 0, then (x - 3) is a factor of f(x).
Dividing f(x) by (x - 3), we get \(2x^{2} + 3x - 2\).
\(2x^{2} + 3x - 2 = 2x^{2} - x + 4x - 2\)
\(x(2x - 1) + 2(2x - 1) = (x + 2)(2x - 1)\)
Therefore, \(f(x) = (x - 3)(x + 2)(2x -1)\)
\(\frac{24}{5}\)
\(\frac{8}{5}\)
\(\frac{5}{8}\)
\(\frac{5}{24}\)
Correct answer is B
\(2x^{2} - 6x + 5 = 0 \implies a = 2, b = -6, c = 5\)
\(\alpha + \beta = \frac{-b}{a} = \frac{-(-6)}{2} = 3\)
\(\alpha\beta = \frac{c}{a} = \frac{5}{2} \)
\(\frac{\beta}{\alpha} + \frac{\alpha}{\beta} = \frac{\beta^{2} + \alpha^{2}}{\alpha\beta}\)
\(\frac{(\alpha + \beta)^{2} - 2\alpha\beta}{\alpha\beta} = \frac{3^{2} - 2(\frac{5}{2})}{\frac{5}{2}}\)
= \(\frac{4}{\frac{5}{2}} = \frac{8}{5}\)
If \(\sqrt{x} + \sqrt{x + 1} = \sqrt{2x + 1}\), find the possible values of x.
1 and -1
-1 and 2
1 and 2
0 and -1
Correct answer is D
\(\sqrt{x} + \sqrt{x + 1} = \sqrt{2x + 1}\)
Squaring both sides, we have
\((\sqrt{x} + \sqrt{x + 1})^{2} = (\sqrt{2x + 1})^{2}\)
\(x + 2\sqrt{x(x + 1)} + x + 1 = 2x + 1\)
\(2x + 1 + 2\sqrt{x(x+1)} - (2x + 1) = 0\)
\((2\sqrt{x(x + 1)})^{2}= 0^{2} \implies 4(x(x + 1)) = 0\)
\(\therefore x(x + 1) = 0\)
\(x = \text{0 or -1}\)