Tag: maths

Clearly, one of $4^{\frac{1}{5}}$ and $5^{\frac{1}{4}}$ must be the smallest.
$\frac{1}{5}$<$\frac{1}{4}$.
So, $\displaystyle 4^{\frac{4}{20}}$ < $5^{\frac{5}{20}}$

$12<13<15<17\ \Rightarrow \sqrt{12}<\sqrt{13}<\sqrt{15}<\sqrt{17}$

$\sqrt[4]{64}=\sqrt[4]{2^6}=2\sqrt[4]{2^2}=2\sqrt{2}=2\sqrt[6]{2^3}=2\sqrt[6]{8}$
$\sqrt[6]{128}=\sqrt[6]{2^7}=2\sqrt[6]{2}$
$\sqrt[4]{64}>\sqrt[6]{128}$

(B) $\sqrt[3]{4}, \sqrt[4]{5},  \sqrt[4]{6}, \sqrt[3]{8}$

$=4^{1/3}, 5^{1/4}, 6^{1/4}, 8^{1/3}$

L.C.M of 3 & 4 $=12$

So, the given surds can be written as,

$=4^{4/12}, 5^{3/12}, 6^{3/12}, 8^{4/12}$

$=(4^{4})^{1/12}, (5^{3})^{1/12}, (6^{3})^{1/12}, (8^{4})^{1/12}$

$=(256)^{1/12}, (125)^{1/12}, (216)^{1/12}, (4096)^{1/12}$

$\therefore $ The smallest one is $\sqrt[4]{5}$

We first consider $\sqrt { 11 } -\sqrt { 10 }$ as follows:

(B) $\frac{p}{q}=\frac{\sqrt{32}-\sqrt{24}}{\sqrt{50}-\sqrt{48}}\times \frac{\sqrt{50}+\sqrt{48}}{\sqrt{50}+\sqrt{48}}$

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

$=(2\sqrt{2}-\sqrt{6})(5\sqrt{2}+4\sqrt{3})> 1$

$\therefore p> q$

(A) Given, $x=\sqrt{2}+1$ and $y=\sqrt{17}-\sqrt{2}$

$\dfrac{x}{y}=\dfrac{\sqrt{2}+1}{\sqrt{17}-\sqrt{2}}\times \dfrac{\sqrt{17}+\sqrt{2}}{\sqrt{17}+\sqrt{2}}$

$=\dfrac{(\sqrt{2}+1)(\sqrt{17}+\sqrt{2})}{17-2}$

$=\dfrac{(\sqrt{2}+1)(\sqrt{17}+\sqrt{2})}{15}< 1$

$\therefore x< y$