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STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
MCQ
${d \over {dx}}\log (\sqrt {x - a} + \sqrt {x - b} ) = $
  • A
    ${1 \over {2[\sqrt {(x - a)} + \sqrt {(x - b)} ]}}$
  • ✓
    ${1 \over {2\sqrt {(x - a)(x - b)} }}$
  • C
    ${1 \over {\sqrt {(x - a)(x - b)} }}$
  • D
    None of these
✓

Answer

Correct option: B.
${1 \over {2\sqrt {(x - a)(x - b)} }}$
b
(b) $\frac{d}{{dx}}\log (\sqrt {x - a} + \sqrt {x - b} )$

$ = \left( {\frac{1}{{\sqrt {x - a} + \sqrt {x - b} }}} \right)\frac{1}{2}\left[ {\frac{1}{{\sqrt {x - a} }} + \frac{1}{{\sqrt {x - b} }}} \right]$

$ = \left[ {\frac{{\sqrt {x - a} + \sqrt {x - b} }}{{\sqrt {x - a} + \sqrt {x - b} }}} \right]\frac{1}{{2\sqrt {(x - a)(x - b)} }}$$ = \frac{1}{{2\sqrt {(x - a)(x - b)} }}$.

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