Further Mathematics | Study Mode

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Question 2
A
\(\frac{5\pi}{12}\) 
B
\(\frac{3\pi}{4}\) 
C
\(\frac{5\pi}{6}\) 
D
\(\frac{7\pi}{6}\) 
Explanation.
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\(180 = \pi rads\)

\(1 = \frac{\pi}{180}\)

\(\therefore 75 = \frac{\pi}{180} \times 75 \)

= \(\frac{5\pi}{12}\)


Correct Option:
A
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Question 3
A
\(\frac{-1}{2}\) 
B
\(\frac{-1}{4}\) 
C
\(\frac{1}{4}\) 
D
\(\frac{1}{2}\) 
Explanation.
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\(\log_{9} 3 = \log_{9} (9^{\frac{1}{2}}) = \frac{1}{2}\log_{9} 9 = \frac{1}{2}\)

\(\frac{1}{2} + 2x = 1 \implies 2x = \frac{1}{2}\)

\(x = \frac{1}{4}\)


Correct Option:
C
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Question 4
A
\(\frac{-2}{\sqrt{3}}\) 
B
\(\frac{-\sqrt{3}}{2}\) 
C
\(\frac{\sqrt{3}}{4}\) 
D
\(\frac{4}{\sqrt{3}}\) 
Explanation.
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\(\cos (x + y) = \cos x \cos y - \sin x \sin y\)

\(\cos (\frac{\pi}{2} + \frac{\pi}{3}) = \cos \frac{\pi}{2} \cos \frac{\pi}{3} - \sin \frac{\pi}{2} \sin \frac{\pi}{3}\)

= \((0 \times \frac{1}{2}) - (1 \times \frac{\sqrt{3}}{2})\)

= \(0 - \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{2}\)


Correct Option:
B
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Question 5
A
-51 
B
-23 
C
29 
D
49 
Explanation.
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Using remainder theorem, the remainder when \(5x^{3} + 2x^{2} - 7x -5\) is divided by (x - 2) = f(2)

\(f(2) = 5(2^{3}) + 2(2^{2}) - 7(2) -5 = 40 + 8 - 14 - 5\)

= 29


Correct Option:
C
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Question 6
A
\(-1\frac{1}{4}\) 
B
\(-1\) 
C
\(\frac{4}{5}\) 
D
\(1\) 
Explanation.
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\(f(x) = \frac{3x + 1}{x^{2} - 1}\)

\(f(-3) = \frac{3(-3) + 1}{(-3)^{2} - 1} = \frac{-8}{8} = -1\)


Correct Option:
B
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Question 7
A
\(\frac{-5}{6}\) 
B
\(-\frac{4}{27}\) 
C
\(0\) 
D
\(\frac{2}{9}\) 
Explanation.
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\(\sqrt[3]{\frac{8}{27}} - (\frac{4}{9})^{\frac{-1}{2}} \)

\(\frac{2}{3} - (\frac{9}{4})^{\frac{1}{2}}\)

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

= \(\frac{-5}{6}\)


Correct Option:
A
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Question 8
A
\(x < -1, x < -\frac{1}{3}\) 
B
\(x > -1, x > -\frac{1}{3}\) 
C
\(x > \frac{1}{3}, x < -1\) 
D
\(x < \frac{1}{3}, x > -1\) 
Explanation.
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\(3x^{2} + 4x + 1 > 0 \)

\(3x^{2} + 3x + x + 1 > 0\)

\(3x(x + 1) + 1(x + 1) > 0\)

\((3x + 1)(x + 1) > 0\)

\(3x + 1 > 0 \implies 3x > -1 \)

\(x > -\frac{1}{3}\)

\(x + 1 > 0 \implies x > -1\)

\(x > -1, x > -\frac{1}{3}\)


Correct Option:
B
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Question 9
A
B
\(\sqrt{3}\) 
C
\(\sqrt{11}\) 
D
\(\sqrt{6}\) 
Explanation.
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The equation of a circle is given as \((x - a)^{2} + (y - b)^{2} = r^{2}\)

Expanding this, we have \(x^{2} + y^{2} - 2ax - 2by + a^{2} + b^{2} = r^{2}\)

Comparing with the given equation, \(3x^{2} + 3y^{2} + 6x -12y + 6 = 0 \equiv x^{2} + y^{2} + 2x - 4y + 2 = 0\) (making the coefficients of \(x^{2}\) and \(y^{2}\) = 1, we get that

\(-2a = 2\implies a = -1\)

\(2b = 4\implies b = 2\)

\(r^{2} - a^{2} - b^{2} = -2\)

\(\therefore r^{2} - (-1)^{2} - (2)^{2} = -2\implies r^{2} = -2+ 1 + 4 = 3\)

\(r = \sqrt{3}\)


Correct Option:
B
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Question 10
A
13 
B
15 
C
17 
D
26 
Explanation.
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\(f(x) = p + qx\)

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

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

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

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


Correct Option:
A
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