MCQ
${f}(x) = \int\limits_{{x^2}}^{{x^2} + 1} {\,{e^{ - {t^2}}}} \,dt\,$ એ ...... આંતરલમાં વધતું વિધેય છે.
- A$(0, \infty )$
- ✓$(-\infty , 0)$
- C$[-1, 1]$
- D$[0, \infty )$
✓
Answer
Correct option: B.
$(-\infty , 0)$
b
અહી, ${f}(x)\,\, = \,\,\int\limits_{{x^2}}^{{x^2} + 1} {{e^{ - {t^2}}}\,dt} \,\,\,\,\,$
અહી, ${f}(x)\,\, = \,\,\int\limits_{{x^2}}^{{x^2} + 1} {{e^{ - {t^2}}}\,dt} \,\,\,\,\,$
$\therefore \,\,{f}'\,(x)\,\, = \,\,{e^{ - {{\left( {{x^2}\, + \,1} \right)}^2}}}\,\,\,2x\, - \,e{\,^{ - {{\left( {{x^2}} \right)}^2}}}\,\,2x\,\,\,\,$
$\,\therefore \,\,{f}'(x)\,\, = \,\,2x\,\left( {{e^{ - {x^4}}} - 2{x^2}\, - \,1\, - \,{e^{ - {x^4}}}} \right) $
$= \,\,2x{e^{ - {x^4} - 2{x^2} - 1}}\left( {1\, - \,{e^{2{x^2}}}\, + \,1} \right)$
જો $x\,\, \leqslant \,\,0\,\,$ તો ${e^{2{x^2} + 1}}\,\, > \,\,1\,\,\,\,\,$
$\therefore \,\,1\, - \,{e^{2{x^2}\, + \,1}}\,\, < \;0\,\,\,\,\,\,\,$
$\therefore \,\,x\,\left( {1\, - \,{e^{2{x^2} + 1}}} \right)\,\, \geqslant \,\,0\,\,\,\,\,$
$\therefore \,\,{f}'\,(x)\,\, \geqslant \,\,0\,\,\left[ {\because \,\,\frac{1}{{{e^{{{\left( {{x^2}\, + 1} \right)}^2}}}}}\, > \,0} \right]$
$(-\infty , 0) $ માં $ f$ વધતું વિધેય છે.
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