Find the equation of a line perpendicular to the line 4y = 7x + 3 which passes through (-3, 1)
7y + 4x + 5 = 0
7y - 4x - 5 = 0
3y - 5x + 2 = 0
3y + 5x - 2 = 0
Correct answer is A
Equation: 4y = 7x + 3
\(\implies y = \frac{7}{4} x + \frac{3}{4}\)
Slope = coefficient of x = \(\frac{7}{4}\)
Slope of perpendicular line = \(\frac{-1}{\frac{7}{4}}\)
= \(\frac{-4}{7}\)
The perpendicular line passes (-3, 1)
\(\therefore\) Using the equation of line \(y = mx + b\)
m = slope and b = intercept.
\(y = \frac{-4}{7} x + b\)
To find the intercept, substitute y = 1 and x = -3 in the equation.
\(1 = \frac{-4}{7} (-3) + b\)
\(1 = \frac{12}{7} + b\)
\(b = \frac{-5}{7}\)
\(\therefore y = \frac{-4}{7} x - \frac{5}{7}\)
\(7y + 4x + 5 = 0\)
Find the distance between the points C(2, 2) and D(5, 6).
13 units
7 units
12 units
5 units
Correct answer is D
= \(\sqrt{(5 - 2)^2 + (6 - 2)^2}\)
= \(\sqrt{3^2 + 4^2}\)
= \(\sqrt{9 + 16}\)
= \(\sqrt{25}\)
= 5 units
Differentiate \(\frac{2x}{\sin x}\) with respect to x.
\(2 \cot x \sec x (1 + \tan x)\)
\(2 \csc x - x \cot x\)
\(2x \csc x + \tan x\)
\(2\csc x(1 - x\cot x)\)
Correct answer is D
No explanation has been provided for this answer.
If y = 8x\(^3\) - 3x\(^2\) + 7x - 1, find \(\frac{\mathrm d^2 y}{\mathrm d x^2}\).
48x - 6
11x\(^2\) + 6x - 7
32x + 7
24x\(^2\) - 6x + 7
Correct answer is A
y = 8x\(^3\) - 3x\(^2\) + 7x - 1
\(\frac{\mathrm d^2 y}{\mathrm d x^2} = \frac{\mathrm d}{\mathrm d x} (\frac{\mathrm d y}{\mathrm d x})\)
= \(\frac{\mathrm d}{\mathrm d x} (24x^2 - 6x + 7)\)
\(\frac{\mathrm d^2 y}{\mathrm d x^2} = 48x - 6\)
\(\begin{vmatrix} 6 & 2 \\ 1 & 6 \end{vmatrix}\)
\(\begin{vmatrix} \frac{2}{11} & \frac{1}{12}\\ \frac{3}{20} & \frac{1}{10} \end{vmatrix}\)
\(\begin{vmatrix} -3 & 2 \\ -1 & -6 \end{vmatrix}\)
\(\begin{vmatrix} \frac{3}{10} & \frac{1}{10} \\ \frac{-1}{20} & \frac{3}{20}\end{vmatrix}\)
Correct answer is D
\(A = \begin{vmatrix} 3 & -2 \\ 1 & 6 \end{vmatrix}\)
|A| = (3 x 6) - (-2 x 1)
= 18 + 2
= 20.
A\(^{-1}\) = \(\frac{1}{20} \begin{vmatrix} 6 & 2 \\ -1 & 3 \end{vmatrix}\)
= \(\begin{vmatrix} \frac{6}{20} & \frac{2}{20} \\ \frac{-1}{20} & \frac{3}{20} \end{vmatrix}\)
= \(\begin{vmatrix} \frac{3}{10} & \frac{1}{10} \\ \frac{-1}{20} & \frac{3}{20} \end{vmatrix}\)