Find the angle between \((5i + 3j)\) and \((3i - 5j)\)
180°
90°
45°
0°
Correct answer is B
\(a . b = |a||b|\cos \theta\)
\(\cos \theta = \frac{a . b}{|a||b|}\)
= \( \frac{(5i + 3j).(3i - 5j)}{(\sqrt{5^2 + 3^2})(\sqrt{3^{2} + (-5)^{2}})}\)
= \(\frac{0}{34} = 0\)
\(\theta = \cos^{-1} 0 = 90°\)
\(4\sqrt{2}\)
\(6\sqrt{2}\)
\(2\sqrt{10}\)
\(4\sqrt{10}\)
Correct answer is C
\(BC = BA + AC\)
Given, \(AB\), then \(BA = - AB\)
= \(AB = \begin{pmatrix} 4 \\ 3 \end{pmatrix} \implies BA = \begin{pmatrix} -4 \\ -3 \end{pmatrix}\)
\(\therefore BC = \begin{pmatrix} -4 \\ -3 \end{pmatrix} + \begin{pmatrix} 2 \\ -3 \end{pmatrix}\)
= \(\begin{pmatrix} -2 \\ -6 \end{pmatrix}\)
\(|BC| = \sqrt{(-2)^{2} + (-6)^{2}} = \sqrt{40} \)
= \(2\sqrt{10}\)
Integrate \((x - \frac{1}{x})^{2}\) with respect to x.
\(\frac{1}{3}(x - \frac{1}{x})^{3} + c\)
\(\frac{x^{3}}{3} - x\sqrt{\frac{1}{x^{3}}} + c\)
\(\frac{x^{3}}{3} - 2x + \frac{1}{x^{3}} + c\)
\(\frac{x^3}{3} - 2x - \frac{1}{x} + c\)
Correct answer is D
\((x - \frac{1}{x})^{2} = x^2 - 2 + \frac{1}{x^2}\)
\(\int (x^2 + \frac{1}{x^2} - 2) \mathrm {d} x\)
= \(\int (x^2 + x^{-2} - 2) \mathrm {d} x\)
= \(\frac{x^3}{3} - 2x - \frac{1}{x}\)
If \(Px^{2} + (P+1)x + P = 0\) has equal roots, find the values of P.
\(\text{-1 and }\frac{-1}{3}\)
\(\text{1 and }\frac{-1}{3}\)
\(\text{-1 and }\frac{1}{3}\)
\(\text{1 and }\frac{1}{3}\)
Correct answer is B
For equal roots, \(b^{2} - 4ac = 0\)
From the equation, \(a = P, b = (P+1), c = P\)
\((P+1)^{2} - 4(P)(P) = P^{2} + 2P + 1 - 4P^{2} = 0\)
\(-3P^{2} + 2P + 1 = 0 \implies 3P^{2} - 2P - 1 = 0\)
\(3P^{2} - 3P + P - 1 = 0\)
\(3P(P - 1) + 1(P - 1) = 0\)
\(P = \text{1 or }\frac{-1}{3}\)
Omission
Original entry
Principle
Commission
Correct answer is C
No explanation has been provided for this answer.