If f(x)=∫1x 12+t4dt, then

f(2)=∫12 12+t4dt, 1 12+t4>118⇒118<12+t4<13⇒∫12 118dt<∫12 12+t4dt<∫12 13dt⇒118<f(2)<13

If f(x)=∫1x 12+t4dt, then

f(2)=∫12 12+t4dt, 1 12+t4>118⇒118<12+t4<13⇒∫12 118dt<∫12 12+t4dt<∫12 13dt⇒118<f(2)<13

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If $f (x) = \int_{1}^{x} \frac{1}{2 + t^{4}} d t$ , then

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$1 / 18 < f (2) < 1 / 3$

$f (2) < 1 / 2 (or) f (2) > 2$

$f (2) < 1 / 3$

$f (2) > 1 / 3$

answer is A.

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Detailed Solution

$\begin{array}{l} f (2) = \int_{1}^{2} \frac{1}{2 + t^{4}} d t, 1 < t < 2 \Rightarrow 1 < t^{4} < 16 \Rightarrow 3 < 2 + t^{4} < 18 \\ \Rightarrow \frac{1}{3} > \frac{1}{2 + t^{4}} > \frac{1}{18} \\ \Rightarrow \frac{1}{18} < \frac{1}{2 + t^{4}} < \frac{1}{3} \\ \Rightarrow \int_{1}^{2} \frac{1}{18} d t < \int_{1}^{2} \frac{1}{2 + t^{4}} d t < \int_{1}^{2} \frac{1}{3} d t \Rightarrow \frac{1}{18} < f (2) < \frac{1}{3} \end{array}$