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STD 12 - 10. vector algebra — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 10. vector algebra1 Mark
MCQ
If $|a|\, = a$ and $|b|\, = b,$ then ${\left( {\frac{a}{{{a^2}}} - \frac{b}{{{b^2}}}} \right)^2} = $
  • A
    ${\left( {\frac{{a + b}}{{ab}}} \right)^2}$
  • B
    $\frac{{{{(a - b)}^2}}}{{ab}}$
  • ✓
    ${\left( {\frac{{a - b}}{{ab}}} \right)^2}$
  • D
    $\frac{{{{(a + b)}^2}}}{{ab}}$
✓

Answer

Correct option: C.
${\left( {\frac{{a - b}}{{ab}}} \right)^2}$
c
(c) ${\left( {\frac{a}{{{a^2}}} - \frac{b}{{{b^2}}}} \right)^2} = \frac{{{a^2}}}{{{a^4}}} + \frac{{{b^2}}}{{{b^4}}} - \frac{{2a\,.\,b}}{{{a^2}{b^2}}}$,{$\because $ $a^2 = a^2$ etc.}

$ = \frac{{{a^2} + {b^2} - 2a\,.\,b}}{{{a^2}{b^2}}} = {\left( {\frac{{a - b}}{{ab}}} \right)^2}.$

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