A) \[4[{{\alpha }^{3}}+f({{\alpha }^{4}})-{{\beta }^{3}}f(4)]\]
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\[g(x)=\int\limits_{0}^{{{x}^{4}}}{f(d)}dt\Rightarrow g(0)=0\] \[LMVT\] 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) \[g'(\alpha )+g'(\beta )=g(2)-g(0);\]but\[g'(x)=f({{x}^{4}}).4{{x}^{3}}\] \[\therefore \]\[4\left[ {{\alpha }^{3}}f({{\alpha }^{4}})+{{\beta }^{3}}f({{\beta }^{4}}) \right]=\int\limits_{0}^{16}{f(t)}\,dt\]
You need to login to perform this action.
You will be redirected in
3 sec
