-4
-3
-1
2
Correct answer is B
\(P = begin{pmatrix} y - 2 & y - 1 \\ y - 4 & y + 2 \end{pmatrix}\)
\(|P| = (y - 2)(y + 2) - (y - 1)(y - 4) = (y^{2} - 4) - (y^{2} - 5y + 4) = -23\)
\(5y - 8 = -23 \implies 5y = -23 + 8 = -15\)
\(y = \frac{-15}{5} = -3\)
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
\(\frac{\mathrm d}{\mathrm d x}(x^2 + xy - 5) = \frac{\mathrm d (x^{2})}{\mathrm d x} + \frac{\mathrm d (xy)}{\mathrm d x} - \frac{\mathrm d (5)}{\mathrm d x} = 0\)
= \(2x + x\frac{\mathrm d y}{\mathrm d x} + y = 0\)
\(\implies x\frac{\mathrm d y}{\mathrm d x} = -(2x + y)\)
\(\frac{\mathrm d y}{\mathrm d x} = \frac{-(2x + y)}{x}\)