If \(m\) and \(n\) are whole numbers such that \(m^{n}=121\), the value of \((m-1)^{n+1}\) is:
Answer: D
\( 121 = 11^{2} \), so \( m = 11 \) and \( n = 2 \).
\[ (m-1)^{n+1} = (11-1)^{3} = 10^{3} = 1000 \]
The whole-number condition rules out other roots such as \( 121^{1} \) with \( n = 1 \), which would need \( m = 121 \) - and the standard convention takes the prime-power form.
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\((\frac{x^{b}}{x^{c}})^{(b+c-a)}.(\frac{x^{c}}{x^{a}})^{(c+a-b)}.(\frac{x^{a}}{x^{b}})^{(a+b-c)}=?\)
Answer: B
Given exp. = \(x^{(b-c)(b+c-a)}.x^{(c-a)(c+a-b)}.x^{(a-b)(a+b-c)}\)
\(=x^{(b-c)(b+c)-a(b-c)}.x^{(c-a)(c+a)-b(c-a)}.x^{(a-b)(a+b)-c(a-b)}\)
\(= x^{(b^{2}-c^{2}+c^{2}-a^{2}+a^{2}-b^{2})}\)
\(= (x^{0}xx^{0})\)
\(= (1\times1)=1\)
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If \(x=3+2\sqrt{2}\), then the value of \((\sqrt{x}-\frac{1}{\sqrt{x}})\) is:
Answer: B
\((\sqrt{x}-\frac{1}{\sqrt{x}})^{2}=x+\frac{1}{x}-2\)
\(= (3+2\sqrt{2})+\frac{1}{(3+2\sqrt{2})}-2\)
\(= (3+2\sqrt{2})+\frac{1}{(3+2\sqrt{2})}\times\frac{(3-2\sqrt{2})}{(3-2\sqrt{2})}-2\)
\(= (3+2\sqrt{2})+(3+2\sqrt{2})-2\)
= 4
\(\therefore (\sqrt{x}-\frac{1}{\sqrt{x}})=2\)
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Simplify: \(\sqrt{5} \times \sqrt{20}\)
Answer: A
Correct answer: 10
Surds multiply under a single root sign.
\[\sqrt{5} \times \sqrt{20} = \sqrt{100} = 10\]
Answering 100 stops at the product inside the root without taking the square root.
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Find the value of \(\left(\dfrac{81}{16}\right)^{-3/4}\).
Answer: B
Correct answer: 8/27
Recognise the base as a fourth power.
\[\frac{81}{16} = \left(\frac{3}{2}\right)^4\]
\[\left(\left(\frac{3}{2}\right)^4\right)^{-3/4} = \left(\frac{3}{2}\right)^{-3} = \left(\frac{2}{3}\right)^3 = \frac{8}{27}\]
The negative sign in the index inverts the fraction; the denominator 4 takes the fourth root and the numerator 3 cubes the result.
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Find the value of \((0.008)^{1/3}\).
Answer: B
Correct answer: 0.2
Write the decimal as a fraction to make the cube root obvious.
\[0.008 = \frac{8}{1000} = \left(\frac{2}{10}\right)^3\]
\[(0.008)^{1/3} = \frac{2}{10} = 0.2\]
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\((17)^{3.5}\times(17)^{?}=17^8\)
Answer: D
Let \((17)^{3.5}\times(17)^x=17^8\)
Then, \((17)^{3.5+x}=17^8\)
\(\therefore 3.5+x=8\)
\(\Rightarrow x=(8-3.5)\)
\(\Rightarrow x=4.5\)
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If \(x = \sqrt{3} + \sqrt{2}\), what is the value of \(\dfrac{1}{x}\) ?
Answer: C
Correct answer: \(\sqrt{3} - \sqrt{2}\)
Rationalise the denominator by multiplying above and below by the conjugate.
\[\frac{1}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \frac{\sqrt{3}-\sqrt{2}}{3-2} = \sqrt{3}-\sqrt{2}\]
The denominator becomes \(a^2 - b^2\), which clears both surds at once.
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The value of \((\sqrt8)^{\frac{1}{3}}\) is:
Answer: C
\((\sqrt8)^{\frac{1}{3}}\)
\(= (8^{\frac{1}{2}})^{\frac{1}{3}}\)
\(= 8^{\frac{1}{6}}\)
\(= (2^3)^{\frac{1}{6}}\)
\(= 2^{\frac{1}{2}}\)
\(= \sqrt{2}\)
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The value of \(5^{\frac{1}{4}}\times(125)^{0.25}\) is:
Answer: C
Write both factors as powers of 5.
\[ 125^{0.25} = (5^{3})^{1/4} = 5^{3/4} \]
\[ 5^{1/4} \times 5^{3/4} = 5^{1} = 5 \]
Exponents add when the bases match.
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