A) 0
B) ?1
C) ? 2
D) ? 3
Correct Answer: C
Solution :
To determine 'c' in Rolle's theorem, \[f'(c)=0\]. Here\[f'(x)=({{x}^{2}}+3x){{e}^{-(1/2)x}}.\left( -\frac{1}{2} \right)+(2x+3){{e}^{-(1/2)x}}\] \[={{e}^{-(1/2)x}}\left\{ -\frac{1}{2}({{x}^{2}}+3x)+2x+3 \right\}\] \[=-\frac{1}{2}{{e}^{-(x/2)}}\{{{x}^{2}}-x-6\}\] \[\therefore f'(c)=0\Rightarrow {{c}^{2}}-c-6=0\Rightarrow c=3,\,-2,\] But \[c=3\notin [-3,\,0].\]You need to login to perform this action.
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