MCQ
વિધેય $f(x) = \sin \left( {\log (x + \sqrt {{x^2} + 1} )} \right)$ એ . . ..
- Aયુગ્મ વિધેય
- ✓અયુગ્મ વિધેય
- Cયુગ્મ વિધેય કે અયુગ્મ વિધેય પૈકી એકપણ નહી.
- Dઆવર્તીય વિધેય
==> $f( - x) = \sin \,[\log \,( - x + \sqrt {1 + {x^2}} )]$
==> $f( - x) = \sin \,\log \left( {(\sqrt {1 + {x^2}} - x)\frac{{(\sqrt {1 + {x^2}} + x)}}{{(\sqrt {1 + {x^2}} + x)}}} \right)$
==> $f( - x) = \sin \,\log \,\left[ {\frac{1}{{(x + \sqrt {1 + {x^2}} )}}} \right]$
==> $f( - x) = \sin \left[ {\log {{(x + \sqrt {1 + {x^2}} )}^{ - 1}}} \right]$
==> $f( - x) = \sin \left[ { - \log (x + \sqrt {1 + {x^2}} )} \right]$
==> $f( - x) = - \sin \left[ {\log (x + \sqrt {1 + {x^2}} )} \right]$==> $f( - x) = - f(x)$
$f(x)$ is odd function.
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