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Compare f(2) with 1/3 for f(x)=∫dt/(2+t⁴) | JEE

JEE Maths question with a full step-by-step solution.

Question
If f(x)=∫1xdt2+t4f(x)=\displaystyle\int_{1}^{x}\dfrac{dt}{2+t^{4}}, then
Af(2)<13f(2)<\dfrac13correct
Bf(2)>13f(2)>\dfrac13
Cf(2)=13f(2)=\dfrac13
Df(2)>1f(2)>1
Solution
Step 1: Differentiate by the Fundamental Theorem of Calculus: f′(x)=12+x4f'(x)=\dfrac{1}{2+x^{4}}, and note f(1)=0f(1)=0. Step 2: Apply the Lagrange Mean Value Theorem to ff on [1,2][1,2]. There exists c∈(1,2)c\in(1,2) with
f′(c)=f(2)−f(1)2−1=f(2)⇒f(2)=12+c4.f'(c)=\dfrac{f(2)-f(1)}{2-1}=f(2)\quad\Rightarrow\quad f(2)=\dfrac{1}{2+c^{4}}.
 Step 3: Bound c4c^{4}. Since 1<c<21<c<2, we have 1<c4<161<c^{4}<16, hence 3<2+c4<183<2+c^{4}<18, giving
118<12+c4<13.\dfrac{1}{18}<\dfrac{1}{2+c^{4}}<\dfrac{1}{3}.
 Step 4: Therefore
f(2)=12+c4<13.f(2)=\dfrac{1}{2+c^{4}}<\dfrac13.
 Correct answer: (1)
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