Correct Answer: B. 11.4 N

Explanation

\(m=4.56kg;t=16s;v_1=(10ms^{-1},060^o);v_2=(50ms^{-1},060^o)\)

Notice that there is no change in the direction of the velocity

So,we don't need to write it in a vector form

\(a=\frac{?v}{t}=\frac{50 - 10}{16}\)

\(a=\frac{40}{16}=2.5ms^{-2}\)

F = ma = 4.56 x 2.5

∴ F = 11.4 N

Explanation

(a) Let \(y = \frac{x^{2} + 1}{(x + 1)^{2}}\)

Let \(u = x^{2} + 1 ; v = (x + 1)^{2}\)

\(\frac{\mathrm d u}{\mathrm d x} = 2x ; \frac{\mathrm d v}{\mathrm d x} = 2(x + 1)\)

Using the quotient rule,

\(\frac{\mathrm d y}{\mathrm d x} = \frac{v \frac{\mathrm d u}{\mathrm d x} - u \frac{\mathrm d v}{\mathrm d x}}{v^{2}}\)

= \(\frac{(x + 1)^{2} (2x) - (x^{2} + 1)(2(x + 1))}{((x + 1)^{2})^{2}}\)

= \(\frac{(x + 1)[(x + 1)(2x) - (x^{2} + 1)(2)}{(x + 1)^{4}}\)

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

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

(b)(i) \(\begin{vmatrix} 1 & 2 & -1 \\ 2 & 3 & -1 \\ -1 & 1 & 3 \end{vmatrix}\)

= \(1(9 + 1) - 2(6 - 1) - 1(2 + 3)\)

= \(10 - 10 - 5 = -5\)

(ii) \(x + 2y - z = 4\)

\(2x + 3y - z = 2\)

\(-x + y + 3z = -1\)

In matrix form, this is

\(\begin{pmatrix} 1 & 2 & -1 \\ 2 & 3 & -1 \\ -1 & 1 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \\ -1 \end{pmatrix}\)

Cofactors :

\(A_{11} = +(3 \times 3 + 1) = 10\)

\(A_{12} = -(2 \times 3 - 1) = -5\)

\(A_{13} = +(2 \times 1 + 3) = 6\)

\(A_{21} = -(2 \times 3 + 1) = -7\)

\(A_{22} = +(1 \times 3 - 1) = 2\)

\(A_{23} = -(1 \times 1 + 1 \times 2) = -3\)

\(A_{31} = +(2 \times -1 + 3) = 1\)

\(A_{32} = -(-1 + 2) = -1\)

\(A_{33} = +(1 \times 3 - 2 \times 2) = -1\)

\(C = \begin{pmatrix} 10 & -5 & 5 \\ -7 & 2 & -3 \\ 1 & -1 & -1 \end{pmatrix}\)

adj A = \(C^{T} = \begin{pmatrix} 10 & -7 & 1 \\ -5 & 2 & -1 \\ 5 & -3 & -1 \end{pmatrix}\)

= \(\frac{1}{-5} \begin{pmatrix} 10 & -7 & 1 \\ -5 & 2 & -1 \\ 5 & -3 & -1 \end{pmatrix}\)

= \(\begin{pmatrix} -2 & \frac{7}{5} & \frac{-1}{5} \\ 1 & \frac{-2}{5} & \frac{1}{5} \\ -1 & \frac{3}{5} & \frac{1}{5} \end{pmatrix}\)

\(\therefore x = A^{-1} . b \)

= \(\begin{pmatrix} -2 & \frac{7}{5} & \frac{-1}{5} \\ 1 & \frac{-2}{5} & \frac{1}{5} \\ -1 & \frac{3}{5} & \frac{1}{5} \end{pmatrix} \begin{pmatrix} 4 \\ 2 \\ -1 \end{pmatrix}\)

= \(\begin{pmatrix} - 8 + \frac{14}{5} + \frac{1}{5} \\ 4 - \frac{4}{5} + \frac{1}{5} \\ -4 + \frac{6}{5} - \frac{1}{5} \end{pmatrix}\)

= \(\begin{pmatrix} -5 \\ 3 \\ -3 \end{pmatrix}\)

x = -5 ; y = 3 ; z = -3.

Correct Answer: D. \(\frac{19}{66}\)

Explanation

\(P(\text{two same color balls}) = P(\text{2 red}) + P(\text{2 white}) + P(\text{2 black})\)

\(P(\text{2 red}) = \frac{3}{12} \times \frac{2}{11} = \frac{1}{22}\)

\(P(\text{2 white}) = \frac{4}{12} \times \frac{3}{11} = \frac{1}{11}\)

\(P(\text{2 black}) = \frac{5}{12} \times \frac{4}{11} = \frac{5}{33}\)

\(P(\text{2 same color balls}) = \frac{1}{22} + \frac{1}{11} + \frac{5}{33} = \frac{19}{66}\)

Correct Answer: C. \(35q^3x^4\)

Explanation

rth term of a binomial expansion = \(^nC_r - _1a^{n - (r - 1)} b^{r - 1}\)

n = 7, r = 5 ∴ r - 1 = 4

5th term = \(^7C_4 q^{7 - 4} x^4\)

= \(^7C_4 q^3x^4\)

\(\therefore 35q^3x^4\)

Correct Answer: C. 150 m

Explanation

The acceleration is between the times 3 secs and 8 secs.

Distance = \(velocity \times time\)

= \(30 \times (8 - 3) = 30 \times 5\)

= \(150 m\)

Correct Answer: A. 1 sec and 7 sec

Explanation

\(s = ut + \frac{1}{2}at^{2}\)

\(s = ut - \frac{1}{2}gt^{2}\) (Upward movement against gravity)

\(35 = 40t - \frac{1}{2}10t^{2}\)

\(35 = 40t - 5t^{2}\)

\(5t^{2} - 40t + 35 = 0\)

\(t^{2} - 8t + 7 = 0\)

\((t - 1)(t - 7) = 0 \implies t = \text{1 sec and 7 sec}\)

Correct Answer: A. (x - 2)

Explanation

Divide \((2x^{3} - 5x^{2} - x + 6)\) by \((2x^{2} - x - 3)\) to get the other factor. Be careful of sign conventions.

Correct Answer: B. 63.167

Explanation

Put 1 + (0.998) = 1 + x;

x = (0.998)

Hence; 1 + 6(0.998) + 15(0.998)(0.998)\(^2\) + 20(0.998)\(^3\) + 15(0.998)\(^4\) + 6(0.998)\(^5\) + (0.998)\(^6\)

= 1 + 5.988 + 14.790 + 19.880 + 14.880 + 5.940 + 0.990

≈ 63.167

Correct Answer: A. 2

Explanation

\(Variance (\sigma^{2}) = \frac{\sum (x - \mu)^2}{n}\)

The mean \((\mu)\) of the data = \(\frac{11 + 12 + 13 + 14 + 15}{5} = \frac{65}{5} = 13\)

\(\sigma^{2} = \frac{10}{5} = 2\)

Explanation

Line : \(2y = x + 3\)

\(\therefore y = \frac{1}{2} (x + 3)\)

\(x^{2} + y^{2} - 2x + 6y - 15 = 0\)

\(x^{2} + (\frac{1}{2} (x + 3))^{2} - 2x + 6(\frac{1}{2} (x + 3)) - 15 = 0\)

\(x^{2} + \frac{x^{2}}{4} + \frac{6x}{4} + \frac{9}{4} + 3x + 9 - 15 = 0\)

\(\frac{5x^{2}}{4} + \frac{10x}{4} - \frac{15}{4} = 0\)

\(\equiv 5x^{2} + 10x - 15 = 0\)

\(5x^{2} - 5x + 15x - 15 = 0 \implies 5x(x - 1) + 15(x - 1) = 0\)

\((5x + 15)(x - 1) = 0 \implies \text{x = -3 or 1}\)

When \(x = -3\), \(y = \frac{1}{2} (-3 + 3) = 0\)

When \(x = 1\), \(y = \frac{1}{2} (1 + 3) = 2\)

The coordinates are M(-3, 0) and N(1, 2).