Question:

Step 1: Understand the function definition. We are given $f(x) = \int_{2}^{x} \frac{dt}{1+t^4}$. We need to find the value of $f(3)$ and compare it with $\frac{1}{17}$. Step 2: Express $f(3)$ as a definite integral. By substituting $x = 3$ into the function definition: $$f(3) = \int_{2}^{3} \frac{dt}{1+t^4}$$ Step 3: Establish bounds on $t^4$ for $t \in [2,3]$. For any $t$ in the interval $[2,3]$, we have: $$t^4 \geq 2^4 = 16$$ Therefore: $$1 + t^4 \geq 17$$ Step 4: Find an upper bound for the integrand. Since $1 + t^4 \geq 17$, we can write: $$\frac{1}{1+t^4} \leq \frac{1}{17}$$ Step 5: Apply the inequality to the integral. Using the property that if $g(t) \leq h(t)$ on $[a,b]$, then $\int_a^b g(t)\,dt \leq \int_a^b h(t)\,dt$: $$f(3) = \int_{2}^{3} \frac{dt}{1+t^4} \leq \int_{2}^{3} \frac{1}{17}\,dt = \frac{1}{17}(3-2) = \frac{1}{17}$$ Step 6: Determine if equality can hold. Equality in the inequality $\int_{2}^{3} \frac{dt}{1+t^4} \leq \int_{2}^{3} \frac{1}{17}\,dt$ would require: $$\frac{1}{1+t^4} = \frac{1}{17} \text{ for all } t \in [2,3]$$ This would mean $t^4 = 16$ for all $t \in [2,3]$. However, for $t > 2$, we have $t^4 > 16$, so this condition is never satisfied throughout the interval. Step 7: Conclude with the strict inequality. Since the integrand $\frac{1}{1+t^4}$ is strictly less than $\frac{1}{17}$ for all $t \in (2,3]$, we have: $$f(3) < \frac{1}{17}$$ **Final Answer:** The correct option is **Option 1: $f(3) < \dfrac{1}{17}$**