Correct Answer: C. 462

Explanation

\(A = \begin{pmatrix} -2 & 5 \\ 4 & 3 \end{pmatrix} ; B = \begin{pmatrix} 3 & 1 \\ 2 & 3 \end{pmatrix}\)

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

= \(\begin{pmatrix} -6 + 4 & 15 + 3 \\ -4 + 12 & 10 + 9 \end{pmatrix}\)

= \(\begin{pmatrix} -2 & 18 \\ 8 & 19 \end{pmatrix}\).

\(BA = 2 \begin{pmatrix} 3 & 7 \\ -2 & x \end{pmatrix} + \begin{pmatrix} y & 4 \\ 12 & -3 \end{pmatrix}\)

= \(\begin{pmatrix} 6 + y & 18 \\ 8 & 2x - 3 \end{pmatrix}\)

\(\therefore 6 + y = -2 \implies y = -8\)

\(\therefore 2x - 3 = 19 \implies x = 11\)

(b) \(f(x) = \frac{1}{2}x + 1 ; g(x) = \frac{5x - 1}{3}\)

(i) \(g^{-1} (x) \)

Let \(y = g(x)\)

\(y = \frac{5x - 1}{3} \implies 5x = 3y + 1\)

\(x = \frac{3y + 1}{5}\)

\(\therefore g^{-1} (x) = \frac{3x + 1}{5}\)

(ii) \(g^{-1} \circ f = g^{-1} [f(x)]\)

= \(\frac{3(\frac{1}{2}x + 1) + 1}{5}\)

= \(\frac{\frac{3}{2}x + 3 + 1}{5}\)

= \((\frac{3}{2}x + 4) \times \frac{1}{5}\)

= \(\frac{3x + 8}{10}\)

Correct Answer: A. 70.6°

Explanation

\(\tan \theta = \frac{m_{1} - m_{2}}{1 - m_{1}m_{2}}\)

\(m_{1} = \text{slope of 1st line } 4y = 3x + 5 \implies y = \frac{3}{4}x + \frac{5}{4}\)

\(m_{1} = \frac{3}{4}\)

\(m_{2} = \text{slope of 2nd line} 3y = 1 - 2x \implies y = \frac{1}{3} - \frac{2}{3}x\)

\(m_{2} = -\frac{2}{3}\)

\(\tan \theta = \frac{\frac{3}{4} - (-\frac{2}{3})}{1 - ((\frac{3}{4})(-\frac{2}{3}))} = \frac{\frac{17}{12}}{\frac{1}{2}}\)

\(\tan \theta = \frac{17}{6}\)

\(\theta \approxeq 70.6°\)

Correct Answer: B. \(\sin x^{2}\)

Explanation

\(f(x) = x^{2}, g(x) = \sin x\)

\(g \circ f = g(x^{2}) = \sin x^{2}\)

Explanation

(a) From y = (x\(^2\) - x + 1)(x - 2) - x + 1)(x - 2)

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

dy/dx = 3x\(^2\) - 6x + 3

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

(x - 1) = 0 twice ; x = 1

y = (1\(^2\) - 1 + 1) (1-2) ; y = -1

Gradient m\(_1\) of tangent = y/x = -1/1 = -1

Using m\(_1\)m\(_2\) = -1

Gradient, m\(_2\) of normal =1

Using y-y\(_1\) = m\(_2\)(x - x\(_1\))

y + 1 = x - 1; x - y - 2 = 0

x\(_1\),y\(_1\) = ( -1, 2), , x\(_2\),y\(_2\), =(3, -4)

Mid point of PR =\(\frac{3-1,-4+2}{2,2}\)

= (1-1)

Using \(\frac{y_2 - y_1}{x_2 - x_1}\) = \(\frac{y - y_1}{x - x_1}\)

\(\frac{1-{-1}}{5-1}\) = \(\frac{y-1}{x-5}\);

\(\frac{2}{4}\) = \(\frac{y-1}{x-5}\)

2(y-1) = x - 5; 2y - 2 = x - 5

2y - x = -3

x - 2y -3 =0

Correct Answer: D. \(\frac{167}{576}\)

Explanation

\(P(Jide) = \frac{1}{12}; P(\text{not Jide}) = \frac{11}{12}\)

\(P(Atu) = \frac{1}{6}; P(\text{not Atu}) = \frac{5}{6}\)

\(P(Obu) = \frac{1}{8}; P(\text{not Obu}) = \frac{7}{8}\)

\(P(\text{only one of them}) = P(\text{Jide not Atu not Obu}) + P(\text{Atu not Jide not Obu}) + P(\text{Obu not Jide not Atu})\)

= \((\frac{1}{12} \times \frac{5}{6} \times \frac{7}{8}) + (\frac{1}{6} \times \frac{11}{12} \times \frac{7}{8}) + (\frac{1}{8} \times \frac{11}{12} \times \frac{5}{6})\)

= \(\frac{35}{576} + \frac{77}{576} + \frac{55}{576}\)

= \(\frac{167}{576}\)

Correct Answer: D. 10

Explanation

\(p(blue) = \frac{\text{no of blue balls}}{\text{total no of balls}}\)

= \(\frac{2}{3} = \frac{k}{k + 5}\)

\(3k = 2(k + 5) \implies 3k = 2k + 10\)

\(3k - 2k = k = 10\)

Correct Answer: C. -1 ≤ x ≤ 1

Explanation

1 - x\(^2\) ≥ 0

-x\(^2\) ≥ -1

x\(^2\) ≤ 1

√x\(^2\) ≤ 1

|x| ≤ 1

-1 ≤ x ≤ 1

Correct Answer: A. P

Explanation

P \(\cap\) (Q U Q`)

P \(\cap\) U = P

Correct Answer: D. \(\pm6\)

Explanation

For equal roots, we have that \(b^{2} = 4ac\), so, given a=1, b = -k and c = 9,

\((-k)^{2} = 4\times1\times9 \implies k^{2} = 36\)

\(k = \sqrt{36} = \pm6\)