3y - x -14 = 0
3x + y + 1 = 0
3y + x + 1 = 0
3y + x + 14 = 0
Correct answer is D
\(3x - y + 11 = 0 \implies y = 3x + 11\)
\(Gradient = 3\)
Gradient of perpendicular line = \(\frac{-1}{3}\)
\(\therefore \frac{y - (-5)}{x - 1} = \frac{-1}{3}\)
\(3(y + 5) = 1 - x\)
\(3y + x + 15 - 1 = 3y + x + 14 = 0\)
If \(y^{2} + xy - x = 0\), find \(\frac{\mathrm d y}{\mathrm d x}\).
\(\frac{1 - y}{2y}\)
\(\frac{1 - 2y}{x}\)
\(\frac{1 - y}{x + 2y}\)
\(\frac{1}{x + 2y}\)
Correct answer is C
Given \(y^{2} + xy - x = 0\)
Using the method of implicit differentiation, we have
\(2y\frac{\mathrm d y}{\mathrm d x} + x\frac{\mathrm d y}{\mathrm d x} + y - 1 = 0\)
\(\frac{\mathrm d y}{\mathrm d x}(2y + x) = 1 - y\)
\(\frac{\mathrm d y}{\mathrm d x} = \frac{1 - y}{x + 2y}\)
Find \(\lim \limits_{x \to 3} \frac{x + 3}{x^{2} - x - 12}\)
-1
\(\frac{-1}{7}\)
\(\frac{1}{7}\)
1
Correct answer is A
\(\lim \limits_{x \to 3} \frac{x + 3}{x^{2} - x - 12} = \frac{3 + 3}{3^{2} - 3 - 12}\)
= \(\frac{6}{-6} = -1\)
4.0
6.5
5.0
10.6
Correct answer is B
\(f(x) = (x^{2} + 3)^{2}\)
Using the chain rule, \(\frac{\mathrm d y}{\mathrm d x} = \frac{\mathrm d y}{\mathrm d u} \times \frac{\mathrm d u}{\mathrm d x}\)
Let \(u = x^{2} + 3\) so that \(y = u^{2}\)
\(\frac{\mathrm d y}{\mathrm d u} = 2u\)
\(\frac{\mathrm d u}{\mathrm d x} = 2x\)
\(\therefore \frac{\mathrm d y}{\mathrm d x} = 2u(2x) = 4xu\)
But \(u = x^{2} + 3\),
\(\frac{\mathrm d y}{\mathrm d x} = 4x(x^{2} + 3)\)
At \(x = \frac{1}{2}, \frac{\mathrm d y}{\mathrm d x} = 4(\frac{1}{2})((\frac{1}{2})^{2} + 3)\)
= \(2 \times \frac{13}{4} = \frac{13}{2} = 6.5\)
\(7x^{2} - 4x - 5 = 0\)
\(7x^{2} - 4x + 5 = 0\)
\(7x^{2} + 4x - 5 = 0\)
\(7x^{2} + 4x + 5 = 0\)
Correct answer is B
Let the roots of the equation be \(\alpha\) and \(\beta\).
\(\alpha + \beta = \frac{-b}{a} = \frac{4}{7}\)
\(\alpha \beta = \frac{c}{a} = \frac{5}{7}\)
\(\implies a = 7, b = -4, c = 5\)
Equation: \(ax^{2} + bx + c = 0 \)
= \(7x^{2} - 4x + 5 = 0\)