Further Mathematics | Study Mode

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Question 1
A
\(14(2\sqrt{2} + 6\sqrt{5} - 4\sqrt{10})\) 
B
\(\frac{1}{14}(2 - 3\sqrt{2} - 4\sqrt{5} - 6\sqrt{10})\) 
C
\(\frac{1}{14}(3\sqrt{2} + 4\sqrt{5} - 6\sqrt{10} - 2)\) 
D
\(14(2 + 3\sqrt{2} - 6\sqrt{5} + 4\sqrt{10})\) 
Explanation.
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\(\frac{1 - 2\sqrt{5}}{2 + 3\sqrt{2}} = (\frac{1 - 2\sqrt{5}}{2 + 3\sqrt{2}})(\frac{2 - 3\sqrt{2}}{2 - 3\sqrt{2}})\)

= \(\frac{2 - 3\sqrt{2} - 4\sqrt{5} + 6\sqrt{10}}{4 - 6\sqrt{2} + 6\sqrt{2} - 18}\)

= \(\frac{2 - 3\sqrt{2} - 4\sqrt{5} + 6\sqrt{10}}{-14}\)

= \(\frac{1}{14}(3\sqrt{2} + 4\sqrt{5} - 2 - 6\sqrt{10})\) (dividing through with the minus sign)


Correct Option:
C
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Question 2
A
\((\frac{2\pi}{3}, \frac{4\pi}{3})\) 
B
\((\frac{\pi}{6}, \frac{5\pi}{6})\) 
C
\((\frac{\pi}{5}, \frac{2\pi}{5})\) 
D
\((\frac{\pi}{3}, \frac{5\pi}{3})\) 
Explanation.
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\(2\cos x - 1 = 0 \implies 2\cos x = 1\)

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

\(x = \cos^{-1} (\frac{1}{2})\)

= \(\frac{\pi}{3}\) = \(\frac{5\pi}{3}\)


Correct Option:
D
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Question 3
A
(1, 2) 
B
(1, -2) 
C
(-1, 2) 
D
(-1, -2) 
Explanation.
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\(4(2^{x^2}) = 8^{x} \equiv (2^{2})(2^{x^2}) = (2^{3})^{x}\)

\(\implies 2^{2 + x^{2}} = 2^{3x}\)

Comparing bases, we have

\(2 + x^{2} = 3x \implies x^{2} - 3x + 2 = 0\)

\(x^{2} - 2x - x + 2 = 0 \)

\(x(x - 2) - 1(x - 2) = 0\)

\((x - 1) = 0\) or \((x - 2) = 0\)

\(x = \text{1 or 2}\)


Correct Option:
A
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Question 4
A
\(3^{2}\) 
B
\(3^{\frac{1}{2}}\) 
C
\(3^{\frac{1}{3}}\) 
D
\(2^{13}\) 
Explanation.
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\(\log_{3} x = \log_{9} 3 \implies \log_{3} x = \log_{9} 9^{\frac{1}{2}} = \frac{1}{2}\log_{9} 9\)

\(\log_{3} x = \frac{1}{2} \)

\(\therefore x = 3^{\frac{1}{2}}\)


Correct Option:
B
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Question 5
A
\(\frac{x^{7}}{8}\) 
B
\(\frac{7x^{6}}{16}\) 
C
\(\frac{7x^{5}}{4}\) 
D
\(\frac{35x^{4}}{8}\) 
Explanation.
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\((\frac{x}{2} - 1)^{8} = ^{8}C_{8}(\frac{x}{2})^{8}(-1)^{0} + ^{8}C_{7}(\frac{x}{2})^{7}(-1)^{1} + ^{8}C_{6}(\frac{x}{2})^{6}(-1)^{2} + ...\)

\(\text{The third term in the expansion =} ^{8}C_{6}(\frac{x}{2})^{6}(-1)^{2}\)

= \(\frac{8!}{6!2!}(\frac{x^{6}}{64})(1) \)

= \(28 \times \frac{x^{6}}{64} = \frac{7x^{6}}{16}\)


Correct Option:
B
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Question 7
A
P = 4 and Q = 2 
B
P = 2 and Q = 4 
C
P = 4 and Q = -2 
D
P = -2 and Q = 4 
Explanation.
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\(\frac{2x}{(x + 6)(x + 3)} = \frac{P}{x + 6} + \frac{Q}{x + 3}\)

\(\frac{2x}{(x + 6)(x + 3)} = \frac{P(x + 3) + Q(x + 6)}{(x + 6)(x + 3)}\)

Comparing equations, we have

\(2x = Px + 3P + Qx + 6Q\)

\(\implies 3P + 6Q = 0 ... (1) ; P + Q = 2 .... (2)\)

From equation (1), \(3P = -6Q \implies P = -2Q\)

\(\therefore -2Q + Q = -Q = 2 \)

\(Q = -2\)

\(P = -2Q = -2(-2) = 4\)

\(P = 4, Q = -2\)


Correct Option:
C
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Question 8
A
\(\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\) 
B
\(\begin{pmatrix} 27 & 12 \\ 16 & -15 \end{pmatrix}\) 
C
\(\begin{pmatrix} -20 & -6 \\ 12 & -8 \end{pmatrix}\) 
D
\(\begin{pmatrix} 11 & 12 \\ 30 & -11 \end{pmatrix}\) 
Explanation.
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\(P = \begin{pmatrix} -2& 1\\ 3& 4\end{pmatrix}; Q = \begin{pmatrix} 5& -3\\ 2& -1\end{pmatrix}\)

= \(PQ = \begin{pmatrix} -10+2 & 6-1 \\ 15+8 & -9-4 \end{pmatrix}\)

= \(\begin{pmatrix} -8 & 5 \\ 23 & -13 \end{pmatrix}\)

\(QP = \begin{pmatrix} -10-9 & 5-12 \\ -4-3 & 2-4 \end{pmatrix}\)

= \(\begin{pmatrix} -19 & -7 \\ -7 & -2 \end{pmatrix}\)

\(PQ - QP = \begin{pmatrix} -8 & 5 \\ 23 & -13 \end{pmatrix} - \begin{pmatrix} -19 & -7 \\ -7 & -2 \end{pmatrix}\)

= \(\begin{pmatrix} 11 & 12 \\ 30 & -11 \end{pmatrix}\)


Correct Option:
D
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Question 9
A
3x + 1 
B
x + 1 
C
2x + 1 
D
x + 2 
Explanation.
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Using the remainder theorem, if (x - a) is a factor of f(x), then f(a) = 0.

Check the options and get the answer.


Correct Option:
A
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Question 10
A
\(f^{-1} : x \to \frac{1+2x}{2-x}, x \neq 2\) 
B
\(f^{-1} : x \to \frac{1-2x}{x+2}, x \neq -2\) 
C
\(f^{-1} : x \to \frac{1-2x}{x-2}, x \neq 2\) 
D
\(f^{-1} : x \to \frac{1+2x}{x+2}, x \neq -2\) 
Explanation.
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\(f(x) = \frac{2x - 1}{x + 2}\)

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

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

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

\(f^{-1} : x \to \frac{1 + 2x}{2 - x} ; x \neq 2\)


Correct Option:
A
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