Further Mathematics | Study Mode

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Question 2
A
-1 
B
-2 
C
-3 
D
-4 
Explanation.
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\(x + 3 = 0 \implies x = -3\)

Using remainder theorem, if x + 3 is a factor, f(-3) = 0.

\(f(-3) = (-3)^{3} + 3(-3)^{2} + n(-3) - 12 = 0\)

\(-27 + 27 - 3n - 12 = 0 \implies -3n = 12\)

\(n = -4\)


Correct Option:
D
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Question 3
A

-4

 
B

-1

 
C

3

 
D

4

 
Explanation.
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\(y = 1 - 3x + 2x^{3}\)

\(\frac{\mathrm d y}{\mathrm d x} = -3 + 6x^{2}\)

At (1, 0), \(\frac{\mathrm d y}{\mathrm d x} = -3 + 6(1^{2}) = -3 + 6 = 3\)

\(y = mx - 3 \implies \frac{\mathrm d y}{\mathrm d x} = m = 3\) (Tangent with equal gradient)


Correct Option:
C
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Question 4
A
\(x^{2} + y^{2} - 4x - 6y - 12 = 0\) 
B
\(x^{2} + y^{2} - 4x + 6y - 12 = 0\) 
C
\(x^{2} + y^{2} + 4x + 6y - 12 = 0\) 
D
\(x^{2} + y^{2} + 4x - 6y - 12 = 0\) 
Explanation.
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Equation of a circle with centre coordinates (a, b) : \((x - a)^{2} + (y - b)^{2} = r^{2}\)

Area of circle = \(\pi r^{2} = 25\pi cm^{2} \implies r^{2} = 25 \)

\(\therefore r = 5cm\)

(a, b) = (-2, 3)

Equation: \((x - (-2))^{2} + (y - 3)^{2} = 5^{2}\)

\(x^{2} + 4x + 4 + y^{2} - 6y + 9 = 25 \implies x^{2} + y^{2} + 4x - 6y + 13 - 25 = 0\)

= \(x^{2} + y^{2} + 4x - 6y - 12 = 0\)


Correct Option:
D
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Question 5
A
\(- \sqrt{3}\) 
B
\(-\frac{\sqrt{3}}{2}\) 
C
\(\frac{\sqrt{3}}{2}\) 
D
\(\sqrt{3}\) 
Explanation.
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\(\sin \theta = \frac{\sqrt{3}}{2} \implies opp = \sqrt{3}; hyp= 2\)

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

\(\cos \theta = \frac{1}{2}\)

\(\sin 2\theta = \sin (180 - \theta) = \sin \theta = \frac{\sqrt{3}}{2}\)

\(\cos 2\theta = \cos (180 - \theta) = -\cos \theta = -\frac{1}{2}\)

\(\tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} = \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}\)

= \(- \sqrt{3}\)


Correct Option:
A
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Question 7
A
\(a * b = \frac{1}{a} + \frac{1}{b}\) 
B
\(a * b = a + b - ab\) 
C
\(a * b = 2a + 2b + ab\) 
D
\(a * b = a - b + ab\) 
Explanation.
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All other options given are commutative i.e. \(a * b = b * a\), except option D.

\(a * b = a - b + ab\)

\(b * a = b - a + ba\)

\(a - b = -(b - a) \neq b - a\)


Correct Option:
D
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Question 8
A
\(6 + \sqrt{7}\) 
B
\(3 + \sqrt{7}\) 
C
\(3 - \sqrt{7}\) 
D
\(6 - \sqrt{7}\) 
Explanation.
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Rationalizing \(\frac{2}{3 - \sqrt{7}}\) by multiplying through with \(3 + \sqrt{7}\),

\(\frac{2}{3 - \sqrt{7}} \frac{(3 + \sqrt{7})}{(3 + \sqrt{7})} = \frac{6 + 2\sqrt{7}}{9 - 7}\)

= \(\frac{6 + 2\sqrt{7}}{2} = 3 + \sqrt{7}\)


Correct Option:
B
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Question 9
A
\(\frac{25}{4} - m\) 
B
\(\frac{25}{4} - 2m\) 
C
\(\frac{25}{4} + m\) 
D
\(\frac{25}{4} + 2m\) 
Explanation.
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\(2x^{2} - 5x + m = 0\)

\(a = 2, b = -5, c = m\)

\(\alpha + \beta = \frac{-b}{a} = \frac{5}{2}\)

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

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

= \((\frac{5}{2})^{2} - 2(\frac{m}{2}) \)

= \(\frac{25}{4} - m\)


Correct Option:
A
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