A) \[-\frac{1}{24}\]
B) \[\frac{1}{12}\]
C) \[\frac{\sqrt{3}}{7}\]
D) \[-\frac{1}{12}\]
Correct Answer: B
Solution :
[b] For application of Roll?s theorem\[f(3)=f(4)\] \[\frac{9+\alpha }{21}=\frac{16+\alpha }{28}\Rightarrow 36+4\alpha =48+3\alpha \Rightarrow \alpha =12\] \[\Rightarrow \frac{2c}{{{c}^{2}}+12}=\frac{1}{c}\Rightarrow 2{{c}^{2}}={{c}^{2}}+12\Rightarrow {{c}^{2}}=12\] \[f''(x)=\frac{({{x}^{2}}+12)2-2x(2x)}{{{({{x}^{2}}+12)}^{2}}}+\frac{1}{{{x}^{2}}}\] \[f''(c)=\frac{(24)(2)-4(12)}{{{(12+12)}^{2}}}+\frac{1}{12}=\frac{1}{12}\]You need to login to perform this action.
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