Correct Answer: B. 2

Explanation

f (x) = x\(^2\) + px + q

\(\frac{dy}{dx}\) = 2x + p

2x + p = 0

2(-3) + p = 0

p = +6

f(-2) = -2\(^2\) + p(-2) + q

= -6 2p + q = -10

-2(-16) + q = -10

-12 + q = -10

q = -10 + 12

= 2

Explanation

\(p = \begin{pmatrix} 5 \\ 3 \end{pmatrix}; q = \begin{pmatrix} -1 \\ 2 \end{pmatrix}; r = \begin{pmatrix} 17 \\ 5 \end{pmatrix}\)

\(r = \alpha p + \beta q\)

\(\begin{pmatrix} 17 \\ 5 \end{pmatrix} = \alpha (\begin{pmatrix} 5 \\ 3 \end{pmatrix}) + \beta (\begin{pmatrix} -1 \\ 2 \end{pmatrix}\)

\(17 = 5\alpha - \beta ... (1); 5 = 3\alpha + 2\beta .... (2)\)

\((1) \times 2 : 34 = 10\alpha - 2\beta ... (3)\)

\((3) + (2) : 39 = 13\alpha \implies \alpha = 3\)

\(5 = 3(3) + 2\beta \implies 2 \beta = 5 - 9 = -4\)

\(\beta = -2\)

\(r = 3p - 2q \implies 2q = 3p - r\)

\(q = \frac{1}{2} (3p - r)\)

Correct Answer: D. - 5

Explanation

2x + 3y = 10 .......x2

3x - 2y = -11 ........x3

4x + 6y = 20

9x - 6y = -33

\(\overline{\frac{13x}{13} = \frac{-13}{13}}\), x = 1

from

2x + 3y = 10

2(-1) + 3y = 10

-2 + 3y = 10

3y = 10 + 2

\(\frac{3y}{3} = \frac{12}{3}\), y = 4

x - y = -1 - 4

= -5

Correct Answer: C. -\(\frac{5}{12}\)

Explanation

P * q = \(\frac{q^2 - p^2}{2pq}\).

3 * 2 (where p = 3, q = 2)

i.e 3 *2 = \(\frac{3^2 - 2^2}{2 *3 * 2}\)

= \(\frac{4 - 9}{12}\)

= \(\frac{-5}{12}\)

Correct Answer: B. 79.7°

Explanation

\(m . n = |m||n|\cos \theta\)

\((3i + 4j) . (2i - j) = 6 - 4 = 2\)

\(2 = |(3i + 4j)||(2i - j)| \cos \theta\)

\(|3i + 4j| = \sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5\)

\(|2i - j| = \sqrt{2^{2} + (-1)^{2}} = \sqrt{5}\)

\(2 = 5(\sqrt{5})(\cos \theta)\)

\(\cos \theta = \frac{2}{5\sqrt{5}} = 0.08\sqrt{5}\)

\(\theta = \cos^{-1} 0.1788 = 79.7°\)

Explanation

(a) \(\log_{10} p = a ; \log_{10} q = b ; \log_{10} s = c\)

\(\log_{10} (\frac{p^{\frac{1}{3}}q^{4}}{s^{2}} = \log_{10} p^{\frac{1}{3}} + \log_{10} q^{4} - \log_{10} s^{2}\)

= \(\frac{1}{3}\log_{10} p + 4\log_{10} q - 2\log_{10} s\)

= \(\frac{1}{3}a + 4b - 2c\)

(b) Area of circle = \(\pi r^{2}\)

Given r = 6cm.

\(\frac{\mathrm d A}{\mathrm d r} = 2\pi r\)

\(\frac{\mathrm d A}{\mathrm d t} = \frac{\mathrm d A}{\mathrm d r} \times \frac{\mathrm d r}{\mathrm d t}\)

\(20 = 2\pi (6) \times \frac{\mathrm d r}{\mathrm d t}\)

\(\frac{\mathrm d r}{\mathrm d t} = \frac{20}{12\pi}\)

= \(\frac{5}{3\pi}\)

Correct Answer: A. {x: -1 ? x ? 3}

Explanation

Let x\(^2\) - 2x - 3 = 0;

(x+1)(x-3) = 0

The zeros of the function are x = -1 and 3

The solution set is { x: -1 ≤ x ≤ 3}

Correct Answer: C. (8 - 3n)y + 1

Explanation

Tn = a + (n - 1)d

Tn = 5y + 1 (n - 1) -3y

Tn = 5y + 1 - 3ny + 3y

Tn = 8y - 3ny + 1

Tn = (8 - 3n) y + 1

Correct Answer: D. 135.5\(^o\) , 224.5\(^o\)

Explanation

cos x = -0.7133

x = cos \(^{-1}\)(0.7133)

= 44.496\(^o\)

x = 180 - 44.495\(^o\)

x = 135.5\(^o\)

and x = 180\(^o\) + 44.495\(^o\)

= 224.5\(^o\)

Correct Answer: D. 5

Explanation

The equation of a circle is given as: \((x - a)^{2} + (y - b)^{2} = r^{2}\)

Expanding, we have: \(x^{2} + y^{2} - 2ax - 2by + a^{2} + b^{2} = r^{2}\)

\(\implies x^{2} + y^{2} - 2ax - 2by = r^{2} - a^{2} - b^{2}\)

Comparing with the given equation: \(3x^{2} + 3y^{2} + 24x - 12y = 15\)

Making the coefficients of \(x^{2}\) and \(y^{2}\) = 1, we have

\(x^{2} + y^{2} + 8x - 4y = 5\)

\(2a = -8 \implies a = -4\)

\(2b = 4 \implies b = 2\)

\(r^{2} - a^{2} - b^{2} = 5 \implies r^{2} = 5 + (-4)^{2} + (2)^{2} = 5 + 16 + 4 = 25\)

\(\therefore r = 5\)