Further Mathematics | Study Mode

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Question 1
A
-2 
B
-\(\frac{1}{2}\) 
C
\(\frac{1}{2}\) 
D
Explanation.
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\(log_{y}\frac{1}{8} = 3 \implies y^{3} = \frac{1}{8}\) (Laws of logarithm)

\(y^{3} = \frac{1}{2^{3}} = (\frac{1}{2})^{3}\)

Equating both sides, we have

\(y = \frac{1}{2}\)


Correct Option:
C
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Question 2
A
\(-4\sqrt{3}\) 
B
\(\frac{-4\sqrt{3}}{3}\) 
C
\(\frac{-3\sqrt{3}}{4}\) 
D
\(\frac{-3\sqrt{3}}{4}\) 
Explanation.
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\(a \Delta b\) = \(\frac{a+b}{\sqrt{ab}}\)

\(-3\Delta -1\) = \(\frac{-3 + -1}{\sqrt{-3\times -1}}\)

\(\frac{-4}{\sqrt{3}}\), rationalising, we have

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


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

\((1-\sqrt{3})^{2} = (1-\sqrt{3})(1-\sqrt{3})\)

\(1 - 2\sqrt{3} + 3 = 4 - 2\sqrt{3}\)

\(\frac{1}{4-2\sqrt{3}}\)

After rationalising (multiplying the denominator and numerator with \(4+2\sqrt{3}\), we have

\(\frac{4+2\sqrt{3}}{4} = 1 + \frac{1}{2}\sqrt{3}\)


Correct Option:
B
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Question 4
A
3, 4 
B
\(\pm3\) 
C
\(\pm5\) 
D
\(\pm6\) 
Explanation.
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For equal roots, we have that \(b^{2} = 4ac\), so, given a=1, b = -k and c = 9,

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

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


Correct Option:
D
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Question 5
A
\(x<2\) 
B
\(x \leq 2\) 
C
\(x = 2\) 
D
\(x > -2\) 
Explanation.
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\(f : x \to \sqrt{4 -2x}\) defined on the set of real numbers, R, which has range from \((-\infty, \infty)\) but because of the root sign, it is defined from \([0, \infty)\).

This is because the root of numbers only has real number values from 0 and upwards.

\(\sqrt{4-2x} \geq 0 \implies 4-2x \geq 0\)

\(-2x \geq -4; x \leq 2\)


Correct Option:
B
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Question 6
A
-5 
B
-3 
C
\(-\frac{1}{2}\) 
D
Explanation.
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Let \(f(x) = y\), then we have

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

Let \(f^{1}(x) = x; x = 2y-1\), replacing y with x,

\(f^{1}(x) = 2x - 1 \implies f^{1}(-2) = 2(-2) -1= -5\)


Correct Option:
A
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Question 7
A
20 
B
12 
C
-10 
D
-22 
Explanation.
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Taking the LCM of the right hand side of the equation, we have

\(\frac{4(2x-3) - 2(x+5)}{(x+5)(2x-3)} = \frac{6x+m}{2x^{2}+7x-15}\)

Comparing the numerators, we have

\(4(2x-3) - 2(x+5) = 6x+m\)

\(8x-12-2x-10 = 6x -22 = 6x + m\)

\(\implies m = -22\)


Correct Option:
D
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Question 9
A
(-2,4) 
B
(\(\frac{-2}{3}, \frac{4}{3}\)) 
C
(\(\frac{2}{3}, \frac{-4}{3}\)) 
D
(2, -4) 
Explanation.
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The equation for a circle with centre coordinates (a, b) and radius r is

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

Expanding the above equation, we have

\(x^{2} - 2ax +a^{2} + y^{2} - 2by + b^{2} - r^{2} = 0\) so that

\(x^{2} - 2ax + y^{2} - 2by = r^{2} - a^{2} - b^{2}\)

Taking the original equation given, \(3x^{2} + 3y^{2} - 4x + 8y = 2\) and making the coefficients of \(x^{2}\) and \(y^{2}\) = 1,

\(x^{2} + y^{2} - \frac{4x}{3} + \frac{8y}{3} = \frac{2}{3}\), comparing, we have

\(2a = \frac{4}{3}; 2b = \frac{-8}{3}\)

\(\implies a = \frac{2}{3}; b = \frac{-4}{3}\)


Correct Option:
C
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Question 10
A
43.5 
B
46 
C
48.5 
D
51 
Explanation.
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\(T_{n} = a + (n-1)d\)

\( d = T_{2} - T_{1} = T_{3} - T_{2} = -1.5 - (-4) = 2.5\)

\(T_{21} = -4 + (21 - 1) \times 2.5 = -4 + (20\times 2.5) \)

= \(-4 + 50 = 46\)


Correct Option:
B
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