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Question 1: If $x = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}$ and $y = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$, find $x^2 + y^2$.
Solution:
$x = \frac{(\sqrt{3}+\sqrt{2})^2}{3-2} = 3+2+2\sqrt{6} = 5+2\sqrt{6}$.
Similarly, $y = 5-2\sqrt{6}$.
$x+y = 10$ and $xy = 1$.
$x^2+y^2 = (x+y)^2 - 2xy = (10)^2 - 2(1) = 100 - 2 = 98$. -
Question 2: Simplify: $\frac{1}{1+x^{a-b}} + \frac{1}{1+x^{b-a}}$.
Solution:
$= \frac{1}{1+\frac{x^a}{x^b}} + \frac{1}{1+\frac{x^b}{x^a}}$
$= \frac{x^b}{x^b+x^a} + \frac{x^a}{x^a+x^b} = \frac{x^b+x^a}{x^b+x^a} = 1$. -
Question 3: If $5^{x-3} \times 3^{2x-8} = 225$, find $x$.
Solution:
$225 = 25 \times 9 = 5^2 \times 3^2$.
Comparing powers of 5: $x-3 = 2 \Rightarrow x=5$.
Comparing powers of 3: $2x-8 = 2 \Rightarrow 2(5)-8 = 2$ (Verified).
So, $x=5$. -
Question 4: Evaluate: $(1^3 + 2^3 + 3^3)^{1/2}$.
Solution:
$= (1 + 8 + 27)^{1/2} = (36)^{1/2} = \sqrt{36} = 6$. -
Question 5: If $x = 9 - 4\sqrt{5}$, find the value of $\sqrt{x} - \frac{1}{\sqrt{x}}$.
Solution:
$x = 5 + 4 - 2(2\sqrt{5}) = (\sqrt{5}-2)^2$.
$\sqrt{x} = \sqrt{5}-2$.
$\frac{1}{\sqrt{x}} = \frac{1}{\sqrt{5}-2} = \sqrt{5}+2$.
$\sqrt{x} - \frac{1}{\sqrt{x}} = (\sqrt{5}-2) - (\sqrt{5}+2) = -4$. -
Question 6: If $x = \frac{1}{2-\sqrt{3}}$, find the value of $x^3 - 2x^2 - 7x + 5$.
Solution:
$x = 2+\sqrt{3} \Rightarrow x-2 = \sqrt{3}$.
Squaring both sides: $(x-2)^2 = 3 \Rightarrow x^2-4x+4=3 \Rightarrow x^2-4x+1=0$.
Divide $x^3-2x^2-7x+5$ by $x^2-4x+1$:
$x^3-2x^2-7x+5 = x(x^2-4x+1) + 2x^2-8x+5 = 0 + 2(x^2-4x+1) + 3 = 3$. -
Question 7: Simplify: $\frac{1}{1+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + \dots + \frac{1}{\sqrt{8}+\sqrt{9}}$.
Solution:
Rationalizing each term: $(\sqrt{2}-1) + (\sqrt{3}-\sqrt{2}) + \dots + (\sqrt{9}-\sqrt{8})$.
All intermediate terms cancel out.
Result $= \sqrt{9} - 1 = 3 - 1 = 2$. -
Question 8: If $2^x = 3^y = 12^z$, prove that $\frac{1}{z} = \frac{1}{y} + \frac{2}{x}$.
Solution:
Let $2^x = 3^y = 12^z = k$. Then $2=k^{1/x}, 3=k^{1/y}, 12=k^{1/z}$.
Since $12 = 2^2 \times 3$, substitute values: $k^{1/z} = (k^{1/x})^2 \times k^{1/y}$.
$k^{1/z} = k^{2/x + 1/y}$. Equating powers: $\frac{1}{z} = \frac{2}{x} + \frac{1}{y}$. -
Question 9: Find the value of $x$ if $(\frac{3}{4})^6 \times (\frac{16}{9})^5 = (\frac{4}{3})^{x+2}$.
Solution:
LHS: $(\frac{3}{4})^6 \times ((\frac{4}{3})^2)^5 = (\frac{4}{3})^{-6} \times (\frac{4}{3})^{10} = (\frac{4}{3})^4$.
So, $(\frac{4}{3})^4 = (\frac{4}{3})^{x+2} \Rightarrow 4 = x+2 \Rightarrow x=2$. -
Question 10: If $x = \frac{\sqrt{p+2q} + \sqrt{p-2q}}{\sqrt{p+2q} - \sqrt{p-2q}}$, show that $qx^2 - px + q = 0$.
Solution:
Apply Componendo and Dividendo: $\frac{x+1}{x-1} = \frac{2\sqrt{p+2q}}{2\sqrt{p-2q}}$.
Square both sides: $\frac{(x+1)^2}{(x-1)^2} = \frac{p+2q}{p-2q}$.
Apply C&D again: $\frac{(x+1)^2+(x-1)^2}{(x+1)^2-(x-1)^2} = \frac{2p}{4q}$.
$\frac{2(x^2+1)}{4x} = \frac{p}{2q} \Rightarrow \frac{x^2+1}{2x} = \frac{p}{2q}$.
$q(x^2+1) = px \Rightarrow qx^2 - px + q = 0$.