A) \[4[{{\alpha }^{3}}f({{\alpha }^{4}})-{{\beta }^{3}}f({{\beta }^{4}})]\]
B) \[4[{{\alpha }^{3}}f({{\alpha }^{4}})+{{\beta }^{3}}f({{\beta }^{4}})]\]
C) \[4[{{\alpha }^{4}}f({{\alpha }^{3}})+{{\beta }^{4}}f({{\beta }^{3}})]\]
D) \[4[{{\alpha }^{2}}f({{\alpha }^{2}})+{{\beta }^{2}}f({{\beta }^{2}})]\]
Correct Answer: B
Solution :
\[I=\int\limits_{0}^{16}{f(t)dt}\]Consider, \[g(x)=\int\limits_{0}^{{{x}^{4}}}{f(t)dt\Rightarrow g(0)=0}\] LMVT for g in [0, 1] gives, some\[\alpha \in (0,1)\]such that \[\frac{g(1)-g(0)}{1-0}-g'(\alpha )\] ?..(1) Similarly LMVT in [1, 2] gives some \[\beta \in (1,2)\]such that \[\frac{g(2)-g(1)}{2-1}=g'(\beta )\] ?(2) Eq. (1) + Eq. (2) \[g'(\alpha )+g'(\beta )=g(2)-\underbrace{g(0)}_{zero};\] but \[g'(x)=f({{x}^{4}}).4{{x}^{3}}\] \[\therefore \]\[4\left[ {{\alpha }^{3}}f({{\alpha }^{4}})+{{\beta }^{3}}f({{\beta }^{4}}) \right]=\int\limits_{0}^{16}{f(t)dt}\]
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