Correct Answer: D. 8

Explanation

The gradient of the equation = \(\frac{y_{1} - y_{0}}{x_{1} - x_{0}} = \frac{\frac{5}{2} - 0}{\frac{5}{4} - 0}\)

= \(\frac{\frac{5}{2}}{\frac{5}{4}} = 2\)

The line passes through the origin, therefore the equation of the line is y = mx

\(y = 2x\)

When x = 4, y = 2 x 4 = 8.

Correct Answer: B. 3

Correct Answer: A. 20

Explanation

Since 2 students must be included, we have to arrange the remaining 3 students from the 6 students left

= \(^{6}C_{3} = \frac{6!}{(6-3)!3!}\)

= 20 ways

Correct Answer: D. \( p \wedge q\)

Explanation

Spearman's rank correlation coefficient = \(1 - \frac{6 \sum d^{2}}{n(n^{2} - 1)}\)

= \(1 - \frac{6(10)}{5(5^{2} - 1)}\)

= \(1 - \frac{60}{120}\)

= \(0.5\)

Correct Answer: C. 81°

Explanation

\(\overrightarrow{x} . \overrightarrow{y} = |\overrightarrow{x}||\overrightarrow{y}|\cos\theta\)

\(\overrightarrow{x} . \overrightarrow{y} = (i - 3j) . (6i + j) = 6 - 3 = 3\)

\(|\overrightarrow{x}| = \sqrt{1^{2} + (-3)^{2}} = \sqrt{10}\)

\(|\overrightarrow{y}| = \sqrt{6^{2} + 1^{2}} = \sqrt{37}\)

\(\therefore 3 = (\sqrt{10})(\sqrt{37})\cos \theta\)

\(\cos\theta = \frac{3}{\sqrt{370}} = 0.1559\)

\(\theta = \cos^{-1} 0.1559 \approxeq 81°\)

Explanation

(a)

Let O(x, y) be the centre of the circle. O is the midpoint of the two points with given coordinates.

\(x = \frac{2 + 6}{2} = \frac{8}{2} = 4\)

\(y = \frac{1 + 7}{2} = \frac{8}{2} = 4\)

The centre is (4, 4).

(b) Radius of circle = \(\sqrt{(4 - 1)^{2} + (4 - 2)^{2}}\)

= \(\sqrt{3^{2} + 2^{2}}\)

= \(\sqrt{13} units\).

Equation of the circle : \((x - 4)^{2} + (y - 4)^{2} = (\sqrt{13})^{2}\)

\((x^{2} - 8x + 16) + (y^{2} - 8y + 16) = 13\)

\(x^{2} + y^{2} - 8x - 8y + 32 - 13 = 0\)

\(x^{2} + y^{2} - 8x - 8y + 19 = 0\)

Correct Answer: A. \(\frac{5}{4}\)

Explanation

Given, \(x^{2} + x - 2 = 0\), a = 1, b = 1 and c = -2.

\(\alpha + \beta = \frac{-b}{a} = \frac{-1}{1} = -1\)

\(\alpha\beta = \frac{c}{a} = \frac{-2}{1} = -2\)

\(\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}} = \frac{\beta^{2} + \alpha^{2}}{(\alpha\beta)^{2}}\)

\(\beta^{2} + \alpha^{2} = (\alpha + \beta)^{2} - 2\alpha\beta = (-1)^{2} - 2(-2) = 1 + 4 = 5\)

\(\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}} = \frac{5}{(-2)^{2}} = \frac{5}{4}\).

Correct Answer: D. 22

Explanation

\(S_{n} = 2n^{2} - 5\)

\(T_{n} = S_{n} - S_{n - 1}\)

\(T_{6} = S_{6} - S_{5}\)

= \((2(6^{2} - 5) - (2(5^{2} - 5) = 62 - 40 = 22\)

Correct Answer: A. 3x + 1

Explanation

Using the remainder theorem, if (x - a) is a factor of f(x), then f(a) = 0.

Check the options and get the answer.