Correct Answer: C. 150° and 330°

Explanation

√5 cosx + √15sinx = 0;

√5 (cosx + √3sinx) = 0

cosx = -sinx√3; \(\frac{sinx}{cos}\) = \(\frac{-1}{√3}\)

tanx = \(\frac{-1}{√3}\); x = 150° or 330°

Explanation

(ai) Let the quadratic equation be \(ax^2+bx+c\)

\(g(x)=ax^2+bx+c\)

Since \((2x+1)\) is a factor,\(x=-\frac{1}{2}\)is a root

\(∴g(-\frac{1}{2})=a(-\frac{1}{2})^2+b(-\frac{1}{2})+c=0\)

\(=a(\frac{1}{4})-b(\frac{1}{2})+c=0\)

Multiply through by 4

=a-2b+4c=0-----(i)

For(x-1)and(x-2),x=1 and x = 2 respectively

So,

\(g(1)=a(1)^2+b(1)+c=-6\)

=a+b+c=-6-----(ii)

and

\(g(2)=a(2)^2+b(2)+c=-5\)

= 4a+2b+c=-5

Adding equations (i) and (iii) gives;

=5a+5b=-5

Divide through by 5

=a+b=-1......(iv)

Adding equation (i) to 2 times equation (iv) gives:

3a+6c=-12.....(v)

Equation (v) minus three times equation (iv) gives:

=3c=-12-(-3)

=3c=-9

=c\(\frac{-9}{3}=-3\)

Substitute (-3) for c in equation (v)

=3a+6(-3)=-12

=3a-18=-12

=3a=-12+18

=3a=6

=a=\(\frac{6}{3}=2\)

Substitute 2 for a and (-3) for c in equation (ii)

=2+6-3=-6

=b-1=-6

=b=-6+1

=b=-5

\(\therefore g(x)=2x^2-5x-3\)

(aii) \(2\times2 - 5x - 3\)

\(= 2\times2 - 6x + x - 3\)

\(= 2x(x - 3) + 1(x - 3)\)

\(= (x - 3)(2x + 1)\)

zeros of \(g(x)\) are \(2x + 1 = 0\) and \(x - 3 = 0\)

∴ zeros of \(g(x)\) are -\frac{1}{2}\) and \(3\)

(b) \((\frac{x}{2}-1)^8\)

rth term is given as \(^nC_r-1 a^n-(r-1)b^r-1\)

\(a=\frac{x}{2},b=-1,n=8,r=3,r-1=2\)

3rd term = \(^8C_2(\frac{x}{2})^8-2(-1)^2\)

= \(28×(\frac{x}{2})^6\times1\)

∴3rd term = \(28\times \frac{x^6}{64}=\frac{7x^6}{16}\)

Correct Answer: D. 0.63

Explanation

(a)

Momentum before collision = Momentum after collision.

\(((3 \times 8) + 5m) = (3 + m) \times 6\)

\(24 + 5m = 18 + 6m\)

\(24 - 18 = 6m - 5m \implies m = 6kg\)

(b)

Momentum before collision = Momentum after collision

\((3 \times 8) - (5 \times 6) = (3 + 6)v\)

\(-6 =9v\)

\(\implies v = -0.67ms^{-1}\)

The bodies move in the direction of the second body.

Explanation

(a) \(\frac{5 + \sqrt{2}}{3 - \sqrt{2}} - \frac{5 - \sqrt{2}}{3 + \sqrt{2}}\)

= \(\frac{(5 + \sqrt{2})(3 + \sqrt{2}) - (5 - \sqrt{2})(3 - \sqrt{2})}{3^{2} - (\sqrt{2})^{2}}\)

= \(\frac{(15 + 5\sqrt{2} + 3\sqrt{2} + 2) - (15 - 5\sqrt{2} - 3\sqrt{2} + 2)}{9 - 2}\)

= \(\frac{17 + 8\sqrt{2} - 17 + 8\sqrt{2}}{7}\)

= \(\frac{0}{7} + \frac{16\sqrt{2}}{7}\)

\(\therefore a = 0 ; b = \frac{16}{7}\)

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

\(x + 5y + 2z = 5\)

\(2x + 3y + z = 0\)

We write the set of equations in matrix form.

\(\begin{pmatrix} 3 & -1 & -1 \\ 1 & 5 & 2 \\ 2 & 3 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -2 \\ 5 \\ 0 \end{pmatrix}\)

We find the inverse of the matrix, say A,

\(|A| = \begin{vmatrix} 3 & -1 & -1 \\ 1 & 5 & 2 \\ 2 & 3 & 1 \end{vmatrix} \)

= \(3(5 - 6) + 1(1 - 4) - 1(3 - 10) = 1\)

Cofactors, C

\(C = \begin{pmatrix} -1 & 3 & -7 \\ -2 & 5 & -11 \\ 3 & 7 & 16 \end{pmatrix}\)

\(C^{T} = \begin{pmatrix} -1 & -2 & 3 \\ 3 & 5 & 7 \\ -7 & -11 & 16 \end{pmatrix}\)

Inverse \(\frac{C^{T}}{|A|} = \begin{pmatrix} -1 & -2 & -3 \\ 3 & 5 & 7 \\ -7 & -11 & -16 \end{pmatrix}\)

\(x = \begin{pmatrix} -1 & -2 & -3 \\ 3 & 5 & 7 \\ -7 & -11 & -16 \end{pmatrix} \begin{pmatrix} -2 \\ 5 \\ 0 \end{pmatrix}\)

= \(\begin{pmatrix} 2 - 10 \\ -6 + 25 \\14 - 55 \end{pmatrix}\)

= \(\begin{pmatrix} -8 \\ 19 \\ -41 \end{pmatrix}\)

\(\therefore x = -8 ; y = 19 ; z = -41\)

Correct Answer: A. 1.7N

Explanation

\(F = F \cos \theta + F \sin \theta\)

where \(F \cos \theta = \text{horizontal component}\)

\(F \sin \theta = \text{vertical component}\)

Horizontal component of resultant = sum of horizontal compoents of individual forces

= \(8 \cos 30 + 10 \cos 150 = 6.928 - 8.66 \approxeq - 1.7N\)

Correct Answer: B. k = 5, r = 2

Explanation

\(x^{2} + 4x + k = (x + r)^{2} + 1\)

\(x^{2} + 4x + k = x^{2} + 2rx + r^{2} + 1\)

Comparing the LHS and RHS equations, we have

\(2r = 4 \implies r = 2\)

\(k = r^{2} + 1 = 2^{2} + 1 = 5\)

Correct Answer: D. \(\begin{pmatrix} -23 \\ 17 \end{pmatrix}\)

Explanation

\(\overrightarrow{OY} \equiv -\overrightarrow{YO}\)

Also, \(\overrightarrow{YO} + \overrightarrow{OX} = \overrightarrow{YX}\)

\(\therefore \overrightarrow{YO} = -\overrightarrow{OY} = - \begin{pmatrix} 16 \\ -11 \end{pmatrix} = \begin{pmatrix} -16 \\ 11 \end{pmatrix}\)

\(\overrightarrow{YX} = \begin{pmatrix} -16 \\ 11 \end{pmatrix} + \begin{pmatrix} -7 \\ 6 \end{pmatrix}\)

= \(\begin{pmatrix} -23 \\ 17 \end{pmatrix}\)

Correct Answer: C. {y : y \(\in\) R, y \(\neq\) 5}

Correct Answer: A. 120