Further Mathematics | Study Mode

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Question 1
A
\(\frac{-x}{1 - x}, x \neq 1\) 
B
\(\frac{1}{1 - x}, x \neq 1\) 
C
\(\frac{-1}{1 - x}, x \neq 1\) 
D
\(\frac{x}{1 - x}, x \neq 1\) 
Explanation.
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\(x * y = x + y - xy\)

Let \(x^{-1}\) be the inverse of x, so that

\(x * x^{-1} = x + x^{-1} - x(x^{-1}) = 0\)

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

\(x^{-1}(x - 1) = x \implies x^{-1} = \frac{x}{x - 1}\)

= \(\frac{x}{-(1 - x)} = \frac{-x}{1 - x}, x \neq 1\)


Correct Option:
A
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Question 2
A
200 
B
90 
C
60 
D
Explanation.
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\(\sin \theta = \tan \theta \implies \frac{\sin \theta}{1} = \frac{\sin \theta}{\cos \theta}\)

Equating, we have

\(\cos \theta = 1 \implies \theta = \cos^{-1} 1\)

= \(0\)


Correct Option:
D
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Question 3
A
-6 
B
-1.2 
C
0.83 
D
1.2 
Explanation.
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\(a^{\frac{5}{6}} \times a^{\frac{-1}{n}} = 1\)

\(\implies a^{\frac{5}{6} + \frac{-1}{n}} = a^{0}\)

Equating bases, we have

\(\frac{5}{6} - \frac{1}{n} = 0\)

\(\frac{5n - 6}{6n} = 0\)

\(5n - 6 = 0 \implies 5n = 6\)

\(n = \frac{6}{5} = 1.20\)


Correct Option:
D
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Question 4
A
3 log 2 
B
4 log 2 
C
-3 log 2 
D
-4 log 2 
Explanation.
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\(\log \frac{1}{8} + \log \frac{1}{2} = \log 8^{-1} + \log 2^{-1}\)

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

= \(-3 \log 2 - 1 \log 2 = -4 \log 2\)


Correct Option:
D
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Question 6
A
\(-15a^{4}b^{2}\) 
B
\(15a^{4}b^{2}\) 
C
\(-15a^{3}b^{3}\) 
D
\(15a^{3}b^{3}\) 
Explanation.
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\((a - b)^{6} = ^{6}C_{0}(a)^{6}(-b)^{0} + ^{6}C_{1}(a)^{5}(-b)^{1} + ^{6}C_{2}(a)^{4}(-b)^{2} + ...\)

Third term = \(^{6}C_{2}(a)^{4}(-b)^{2} = \frac{6!}{(6-2)! 2!}(a^4)(b^2)\)

= \(15a^{4}b^{2}\)


Correct Option:
B
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Question 7
A
1 and -1 
B
-1 and 2 
C
1 and 2 
D
0 and -1 
Explanation.
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\(\sqrt{x} + \sqrt{x + 1} = \sqrt{2x + 1}\)

Squaring both sides, we have

\((\sqrt{x} + \sqrt{x + 1})^{2} = (\sqrt{2x + 1})^{2}\)

\(x + 2\sqrt{x(x + 1)} + x + 1 = 2x + 1\)

\(2x + 1 + 2\sqrt{x(x+1)} - (2x + 1) = 0\)

\((2\sqrt{x(x + 1)})^{2}= 0^{2} \implies 4(x(x + 1)) = 0\)

\(\therefore x(x + 1) = 0\)

\(x = \text{0 or -1}\)


Correct Option:
D
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Question 8
A
\(\frac{24}{5}\) 
B
\(\frac{8}{5}\) 
C
\(\frac{5}{8}\) 
D
\(\frac{5}{24}\) 
Explanation.
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\(2x^{2} - 6x + 5 = 0 \implies a = 2, b = -6, c = 5\)

\(\alpha + \beta = \frac{-b}{a} = \frac{-(-6)}{2} = 3\)

\(\alpha\beta = \frac{c}{a} = \frac{5}{2} \)

\(\frac{\beta}{\alpha} + \frac{\alpha}{\beta} = \frac{\beta^{2} + \alpha^{2}}{\alpha\beta}\)

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

= \(\frac{4}{\frac{5}{2}} = \frac{8}{5}\)


Correct Option:
B
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Question 9
A
(x - 3)(x - 2)(2x + 2) 
B
(x + 3)(x - 2)(x - 1) 
C
(x - 3)(x + 2)(2x -1) 
D
(x + 3)(x - 2)(2x - 1) 
Explanation.
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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 Option:
C
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Question 10
A
y + 5x + 3 = 0 
B
2y - 5x - 9 = 0 
C
5y + 2x - 8 = 0 
D
5y - 2x - 12 = 0 
Explanation.
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\(Line:2y + 5x - 6 = 0\)

\(2y = 6 - 5x \implies y = 3 - \frac{5x}{2}\)

Gradient = \(\frac{-5}{2}\)

For the line perpendicular to the given line, Gradient = \(\frac{-1}{\frac{-5}{2}} = \frac{2}{5}\)

The midpoint of P(4, 3) and Q(-6, 1) = \((\frac{-6 + 4}{2}, \frac{3 + 1}{2})\)

= (-1, 2).

Therefore, the line = \(\frac{y - 2}{x + 1} = \frac{2}{5}\)

\(2(x + 1) = 5(y - 2) \implies 5y - 2x - 12 = 0\)


Correct Option:
D
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