Correct Answer: A. 13

Explanation

\(f(x) = p + qx\)

\(f(1) = p + q(1) \implies p + q = 7 .... (1)\)

\(f(5) = p + 5q = 19 .....(2)\)

Solving for p and q using simultaneous equation, p = 4, q = 3

\(f(3) = 4 + 3(3) = 13\)

Explanation

(ai) \(p=\frac{4}{16}=\frac{1}{4}\)

\(∴q=1-\frac{1}{4}=\frac{3}{4}\)

P(The probability that he did not pick agreenball) = \(^nC_r p^rq^{n - r}\)

Where n = 5 and r = 0

\(=^5C_0(\frac{1}{4})^o(\frac{3}{4})^{5 - 0}\)

\(=^5C_0(\frac{1}{4})^o(\frac{3}{4})^5\)

\(=1\times1\times\frac{243}{1024}\)

\(=\frac{243}{1024}\)

(aii) Pr(at least three) = Pr(for 3 green balls) + Pr (for 4 green balls) + Pr (for 5 green balls)

\(^5C_3(\frac{1}{4})^3(\frac{3}{4})^{5 - 3}+^5C_4(\frac{1}{4})^4(\frac{3}{4})^{5 - 4}+^5C_5(\frac{1}{4})^5(\frac{3}{4})^{5 - 5}\)

\(=^5C_3(\frac{1}{4})^3(\frac{3}{4})^2+^5C_4(\frac{1}{4})^4(\frac{3}{4})^1+^5C_5(\frac{1}{4})^5(\frac{3}{4})^0\)

\(=10\times\frac{1}{6}4\times\frac{9}{16}+5\times\frac{1}{256}\times\frac{3}{4}+1\times\frac{1}{1024}\times1\)

\(=\frac{45}{512}+\frac{15}{1024}+\frac{1}{1024}\)

\(=\frac{53}{512}\)

(b) The sum of deviations from the mean is always equal to 0. This is a fundamental property of deviations and the definition of the mean.

\(= -2 + (m - 1) + (m^2 + 1) + (-1) + 2 + (2m - 1) + (-2) = 0\)

\(= -2 + m - 1 + m^2 + 1 - 1 + 2 + 2m - 1 - 2 = 0\)

\(= m^2 + 3m - 4 = 0\)

\(= m^2 + 4m - m - 4 = 0\)

= m(m + 4) - 1(m + 4) = 0

= (m + 4)(m - 1) = 0

This gives us two possible values for m: -4 and 1

So, the possible values of m are -4 and 1.

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. \((15i + 6j)ms^{-1}\)

Explanation

Recall, \(F = mass \times acceleration \implies acceleration = \frac{force}{mass}\)

= \(\frac{10i + 4j}{2} = (5i + 2j) ms^{-2}\)

= \(v = u + at \implies v \text{at 3 seconds} = 0 + (5i + 2j \times 3)\)

= \((15i + 6j) ms^{-1}\)

Correct Answer: C. (1, -1)

Explanation

\(y = 4x^{3} + kx^{2} - 6x + 4\)

\(\frac{\mathrm d y}{\mathrm d x} = 12x^{2} + 2kx - 6\)

At P(1, m)

\(\frac{\mathrm d y}{\mathrm d x} = 12 + 2k - 6 = 0\) (parallel to the x- axis)

\(6 + 2k = 0 \implies k = -3\)

\(P(1, m) \implies m = 4(1^{3}) - 3(1^{2}) - 6(1) + 4)

= -1

P = (1, -1)

Explanation

\(\tan 2A = \frac{2 \tan A}{1 - \tan^{2} A}\)

\(\tan 30 = \tan 2(15°)\)

\(\frac{\sqrt{3}}{3} = \frac{2 \tan 15}{1 - \tan^{2} 15°}\)

\(\frac{\sqrt{3}}{6} = \frac{\tan 15}{1 - tan^{2} 15°}\)

Let \(\tan 15° = x\)

\(\frac{\sqrt{3}}{6} = \frac{x}{1 - x^{2}}\)

\(\sqrt{3} (1 - x^{2}) = 6x \implies \sqrt{3} - \sqrt{3} x^{2} = 6x \)

\(\sqrt{3} x^{2} + 6x - \sqrt{3} = 0\)

Divide through by the coefficient of x\(^{2}\).

\(x^{2} + \frac{6}{\sqrt{3}} x - 1 = 0\)

\(\implies x^{2} + 2\sqrt{3} x - 1 = 0\)

\(x = \frac{-(2\sqrt{3}) \pm \sqrt{(2\sqrt{3})^{2} - 4(1)(-1)}}{2(1)}\)

\(x = \frac{-2\sqrt{3} \pm \sqrt{16}}{2} = \frac{-2\sqrt{3} \pm 4}{2}\)

\(x = -\sqrt{3} + 2 ; x = -\sqrt{3} - 2\)

Since 15° is in the first quadrant, \(\tan 15° > 0\)

\(\therefore \tan 15° = -\sqrt{3} + 2 \equiv 2 - \sqrt{3}\).

Correct Answer: D. (13, 157.38°)

Explanation

\(p = (5i - 12j); |p| = \sqrt{5^{2} + (-12)^{2}}\)

= \(\sqrt{169} = 13\)

\(\tan\theta = \frac{-12}{5} = -2.4 \implies \theta = -67.38°\)

Direction = \(90° - (-67.38°) = 157.38°\)

Correct Answer: D. 2.82

Explanation

f(20\(^o\)) = cos 20\(^o\) = 0.9397

g(0.9397) = 3(0.9397)

= 2.82

Explanation

Frequency Distribution Table

Correct Answer: B. \(\frac{-7}{8}\)

Explanation

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

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

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

From the equation, a = 2, b = -3, c =4

\(\alpha + \beta = \frac{-(-3)}{2} = \frac{3}{2}\)

\(\alpha\beta = \frac{4}{2} = 2\)

\(\alpha^{2} + \beta^{2} = (\frac{3}{2})^{2} - 2(2)) = \frac{9}{4} - 4 = \frac{-7}{4}\)

\(\implies \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\frac{-7}{4}}{2} = \frac{-7}{8}\)