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જો $a = \sqrt {(21)} - \sqrt {(20)} $ અને $b = \sqrt {(18)} - \sqrt {(17),} $ તો
$a = b$
$a + b = 0$
$a > b$
$a < b$
Solution
(d) $a – b = \sqrt {21} – \sqrt {20} – \sqrt {18} + \sqrt {17} $
= $(\sqrt {21} – \sqrt {18} ) – (\sqrt {20} – \sqrt {17} )$
= ${{(\sqrt {21} – \sqrt {18} )(\sqrt {21} + \sqrt {18} )} \over {\sqrt {21} + \sqrt {18} }} – {{20 – 17} \over {\sqrt {20} + \sqrt {17} }}$
= $3\,\left[ {{1 \over {\sqrt {21} + \sqrt {18} }} – {1 \over {\sqrt {20} + \sqrt {17} }}} \right]$
= ${{3\,[\sqrt {20} + \sqrt {17} – \sqrt {21} – \sqrt {18} ]} \over {(\sqrt {21} + \sqrt {18} )\,(\sqrt {20} + \sqrt {17} )}}$
= ${{3\,[(\sqrt {20} – \sqrt {21} ) + (\sqrt {17} – \sqrt {18)} ]} \over {(\sqrt {21} + \sqrt {18} )\,(\sqrt {20} + \sqrt {17} )}}$
= ${{ – 3\,[(\sqrt {21} – \sqrt {20} ) + (\sqrt {18} – \sqrt {17} )} \over {(\sqrt {21} + \sqrt {18} )\,(\sqrt {20} + \sqrt {17} )}} < 0$,
$\therefore a < b$.
