Correct Answer: D. \(\frac{3x + 2}{x - 1},x ≠ 1\)

Explanation

\(f :x→\frac{x + 2}{x - 3},x ≠ 3, f = ?\)

Let \(f :x=y\)

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

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

\(=x-xy=-3y-2\)

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

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

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

\(∴f ^{-1} : x=\frac{3x + 2}{x - 1},x ≠ 1\)

Correct Answer: C. 59.5°

Explanation

\(\overrightarrow{r} . \overrightarrow{t} = |\overrightarrow{r}||\overrightarrow{t}|\cos \theta\)

\(\overrightarrow{r} . \overrightarrow{t} = (3i + 4j) . (-5i + 12j) = -15 + 48 = 33\)

\(|\overrightarrow{r}| = \sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5\)

\(|\overrightarrow{t}| = \sqrt{(-5)^{2} + 12^{2}| = \sqrt{169} = 13\)

\(\cos \theta = \frac{\overrightarrow{r} . \overrightarrow{t}}{|\overrightarrow{r}||\overrightarrow{t}|}\)

\(\cos \theta = \frac{33}{5 \times 13} = \frac{33}{65}\)

\(\theta = \cos^{-1} {\frac{33}{65}} \approxeq 59.5°\)

Correct Answer: C. (x - 3)(x + 2)(2x -1)

Explanation

Since f(3) = 0, then (x - 3) is a factor of f(x).

Dividing f(x) by (x - 3), we get \(2x^{2} + 3x - 2\).

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

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

Therefore, \(f(x) = (x - 3)(x + 2)(2x -1)\)

Correct Answer: D. \(\frac{13}{54}\)

Explanation

\(P(A)=\frac{1}{6},P(T)=\frac{1}{9}\)

Probability that only one of them will hit the target = \(P(A)\times P( \bar T ) + P( \bar A )\times P(T)\)

Where \(P( \bar T )\) is the probability that Tunde will not hit the target and \(P( \bar A )\) is the probability that Atta will not hit the target

\(P( \bar T )=1-\frac{1}{9}=\frac{8}{9}\)

\(P( \bar A )=1-\frac{1}{6}=\frac{5}{6}\)

Pr(only one) =\((\frac{1}{6}\times\frac{8}{9}) + (\frac{5}{6} \times \frac{1}{9}) =\frac{4}{27} + \frac{5}{54}\)

\(\therefore\) pr (only one) = \(\frac{13}{54}\)

Explanation

(a) \(\frac{2\sqrt{2}}{\sqrt{48} - \sqrt{8} - \sqrt{27}}\)

= \(\frac{2\sqrt{2}}{\sqrt{16 \times 3} - \sqrt{4 \times 2} - \sqrt{9 \times 3}}\)

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

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

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

= \(\frac{2\sqrt{6} + 4(2)}{3 + 2\sqrt{6} - 2\sqrt{6} - 4(2)}\)

= \(\frac{2\sqrt{6} + 8}{3 - 8}\)

= \(\frac{8 + 2\sqrt{6}}{-5}\)

= \(-\frac{8}{5} - \frac{2\sqrt{6}}{5}\)

= \(p = -\frac{8}{5}; q = -\frac{2}{5} ; r = 6\)

(b) \(V = A\log_{10} (M + N)\)

\(\log_{10} (M + N) = \frac{V}{A}\)

\(10^{\frac{V}{A}} = M + N \)

\(N = 10^{\frac{V}{A}} - M\)

Correct Answer: C. 14\(ms^{-2}\)

Explanation

\(accl = \frac{\mathrm d V}{\mathrm d t}\)

\(V = 3t^{2} + 2t - 1 \therefore a = \frac{\mathrm d V}{\mathrm d t} = 6t + 2\)

\(a \text{(after 2 seconds)} = (6\times 2) + 2 = 12+2 = 14ms^{-2}\)

Explanation

(a) \(\frac{\frac{3}{2} \log 27 - 3 \log 5\sqrt{5}}{\log 0.6}\)

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

= \(\frac{\frac{2}{3} \times 3 \log ^3 - 3 \times \frac{3}{2} \log 5}{\log 3 - log 5}\)

= \(\frac{\frac{9}{2}(\log 3 - \log 5)}{\log 3 - \log 5}\)

= \(\frac{9}{2} = 4\frac{1}{2}\)

(b)(i) \(A = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix}, B = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}\)

(ii) Transformation of A followed by B is BA.

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

= \(\begin{pmatrix} -1 & 7 \\ -1 & 4 \end{pmatrix}\)

Image of (-2, 2) is given by

\(\begin{pmatrix} -1 & 7 \\ -1 & 4 \end{pmatrix} \begin{pmatrix} -2 \\ 2 \end{pmatrix}\)

= \(\begin{pmatrix} 16 \\ 10 \end{pmatrix}\)

Correct Answer: D. (2x + 5)(x + 2)

Explanation

Divide \(2x^{3} + 3x^{2} - 17x - 30\) by \((x - 3)\). You get \(2x^{2} + 9x + 10\).

Factorizing, we have \(2x^{2} + 9x + 10 = 2x^{2} + 4x + 5x + 10\)

\(2x(x + 2) + 5(x + 2)\)

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

Correct Answer: A. 100N

Explanation

\(F = ma = m(\frac{v - u}{t})\)

\(F = 25(\frac{35 - 15}{5}) = 25 \times 4 = 100N\)

Correct Answer: C. 15

Explanation

\(=^6C_4\)

\(=\frac{6!}{(4!\times2!)}\)

\(=\frac{6\times5}{2\times1}\)

= 15