A) \[-\frac{53}{3}\]
B) \[\frac{53}{3}\]
C) \[-\frac{55}{3}\]
D) \[\frac{55}{3}\]
Correct Answer: B
Solution :
Let\[f'(x)=3{{x}^{10}}-7{{x}^{8}}+5{{x}^{6}}-21{{x}^{3}}+3{{x}^{2}}-7\] \[f'(x)=30{{x}^{9}}-56{{x}^{7}}+30{{x}^{5}}-63{{x}^{5}}-63{{x}^{2}}+6x\] \[f'(1)=30-56+30-63+6\] \[=66-63-56=-53\] Consider\[\underset{\alpha \to 0}{\mathop{\lim }}\,\frac{f(1-\alpha )-f(1)}{{{\alpha }^{3}}+3\alpha }\] \[=\underset{\alpha \to 0}{\mathop{\lim }}\,\frac{f'(1-\alpha )(-1)-0}{3{{\alpha }^{2}}+3}\](By using L?hospital rule) \[=\frac{f'(1-0)(-1)}{3{{(0)}^{2}}+3}=\frac{-f'(1)}{3}=\frac{53}{3}\]You need to login to perform this action.
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