A) \[\left[ \frac{1}{\sqrt{2}},\,\,1 \right]\]
B) \[[0,\,\,1]\]
C) \[\left[ \frac{1}{2},\,\,2 \right]\]
D) \[\left[ \frac{3}{4},\,\,1 \right]\]
Correct Answer: A
Solution :
Let \[I=\int_{0}^{1}{\frac{dx}{\sqrt{1+{{x}^{4}}}}}\] Here, \[0\le x\le 1\Rightarrow 1\le (1+{{x}^{4}})\le 2\] Þ \[1\le \sqrt{1+{{x}^{4}}}\le \sqrt{2}\Rightarrow \frac{1}{\sqrt{2}}\le \frac{1}{\sqrt{1+{{x}^{4}}}}\le 1\] Þ \[\frac{1}{\sqrt{2}}\le \int_{0}^{1}{\frac{dx}{\sqrt{1+{{x}^{4}}}}\le 1}\] Hence \[\left[ \frac{1}{\sqrt{2}},\,1 \right]\] is the smallest interval, such that \[I\in \left[ \frac{1}{\sqrt{2}},\,\,1 \right]\]. Note: If \[m=\]least value of \[f(x)\]and M= greatest value of \[f(x)\]in [a, b], then \[m(b-a)\le \int_{a}^{b}{f(x)dx\le M(b-a)}\].
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