Given that \(\frac{\mathrm d y}{\mathrm d x} = \sqrt{x}\), find y.
\(2x^{\frac{3}{2}} + c\)
\(\frac{2}{3}x^{\frac{3}{2}} + c\)
\(\frac{3}{2}x^{\frac{3}{2}} + c\)
\(\frac{2}{3}x^{2} + c\)
Correct answer is B
\(\frac{\mathrm d y}{\mathrm d x} = \sqrt{x} = x^{\frac{1}{2}}\)
\(y = \int x^{\frac{1}{2}} \mathrm {d} x\)
= \(\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + c = \frac{x^{\frac{3}{2}}}{\frac{3}{2}} + c \)
= \(\frac{2}{3}x^{\frac{3}{2}} + c\)
Find the range of values of x for which \(x^{2} + 4x + 5\) is less than \(3x^{2} - x + 2\)
\(x > \frac{-1}{2}, x > 3\)
\(x < \frac{-1}{2}, x > 3\)
\(\frac{-1}{2} \leq x \leq 3\)
\(\frac{-1}{2} < x < 3\)
Correct answer is B
No explanation has been provided for this answer.
\(\frac{1}{3}\)
\(\frac{1}{2}\)
\(2\)
\(3\)
Correct answer is B
\(T_{n} = ar^{n - 1}\)
\(T_{4} = ar^{4 - 1} = ar^{3} = 192\)
\(T_{9} = ar^{9 - 1} = ar^{8} = 6\)
Dividing \(T_{9}\) by \(T_{4}\),
\(r^{8 - 3} = \frac{6}{192}\)
\(r^{5} = \frac{1}{32} = (\frac{1}{2})^{5}\)
\(r = \frac{1}{2}\)
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 = \(\frac{y - 2}{x + 1} = \frac{2}{5}\)
\(2(x + 1) = 5(y - 2) \implies 5y - 2x - 12 = 0\)