```latex \text{If } f : (0, \infty) \to \mathbb{R} \text{ be given by}

$$f(x) = \int_{1/x}^{x} e^{-\left(t + \frac{1}{t}\right)} \frac{dt}{t}.$$

\text{Then,} \begin{itemize} \item[(a)] f(x) \text{ is monotonically increasing on } [1, \infty) \item[(b)] f(x) \text{ is monotonically decreasing on } (0, 1] \item[(c)] f(x) + f\left(\frac{1}{x}\right) = 0, \forall x \in (0, \infty) \item[(d)] f(2^x) \text{ is an odd function of } x \text{ on } \mathbb{R} \end{itemize} ```

Key Ideas

1. Integral with variable limits
   For a function $f(x) = \int_{a(x)}^{b(x)} g(t)\,dt$ the derivative is $f'(x) = g\bigl(b(x)\bigr)\,b'(x) \; - \; g\bigl(a(x)\bigr)\,a'(x).$

2. Here

   • (a(x) = 1/x) and (b(x) = x).
   • (g(t) = \dfrac{e^{-(t + 1/t)}}{t}.)

3. Monotonicity
   If (f'(x) > 0) for the interval of interest, (f) is strictly increasing there.

4. Symmetry from swapping limits
   Reversing the order of an integral changes its sign: $\int_{p}^{q} g(t)\,dt = -\int_{q}^{p} g(t)\,dt.$ With limits (1/x) and (x), flipping them simply gives (-f(x)).

5. Oddness through an exponential substitution
   Put (x = 2^u). Then (1/x = 2^{-u}). If (f(2^u) + f(2^{-u}) = 0), the new function (F(u) = f(2^u)) satisfies (F(-u) = -F(u)) — the very definition of an odd function.

Imagine a see-saw of arrows

1. You pick a positive number x.
2. From 1/x you draw an arrow all the way up to x.
3. Every short piece of that arrow is weighted by a tiny number (e^{-(t + 1/t)}) (this makes pieces near (t = 1) heavier and far-away pieces lighter).
4. Add all those tiny weighted pieces – that total is f(x).

Because the same rule is used forward ((1/x \to x)) and backward ((x \to 1/x)), the function behaves like a mirror:

• Push the right end of the see-saw (make x bigger) and the total weight always goes up – it never dips.
• Flip the arrow (swap start and end) and you simply get the negative of what you had.

So:

• From 1 onward the weight only rises.
• From 1 down to tiny numbers it still rises (just in the opposite direction).
• The mirror rule gives (f(x) + f(1/x) = 0).
• Re-label x as (2^x) and the same mirror rule says the graph is an odd curve (symmetric through the origin).

Step-by-Step Solution

1. Set up the ingredients
   $g(t) = \frac{e^{-(t + 1/t)}}{t}, \quad a(x) = \frac{1}{x}, \quad b(x) = x$

2. Differentiate using Leibniz rule
   $f'(x) = g\bigl(b(x)\bigr)\,b'(x) \; - \; g\bigl(a(x)\bigr)\,a'(x)$ $\Rightarrow f'(x) = g(x) \cdot 1 - g\!\left(\frac{1}{x}\right) \!\left(-\frac{1}{x^{2}}\right)$ $\phantom{f'(x)} = g(x) + \frac{g(1/x)}{x^{2}}$

3. Simplify each term
   $g(x) = \frac{e^{-(x + 1/x)}}{x}$ $g\!\left(\frac{1}{x}\right) = \frac{e^{-(1/x + x)}}{1/x} = x\,e^{-(x + 1/x)}$ Hence $\frac{g(1/x)}{x^{2}} = \frac{x\,e^{-(x + 1/x)}}{x^{2}} = \frac{e^{-(x + 1/x)}}{x}$

4. Add them up
   $f'(x) = \frac{e^{-(x + 1/x)}}{x} + \frac{e^{-(x + 1/x)}}{x} = \frac{2\,e^{-(x + 1/x)}}{x}$

5. Sign of the derivative
   For every (x>0): (e^{-(x + 1/x)} > 0) and (x > 0). Therefore $f'(x) > 0 \quad \forall\,x>0.$ So (f) is strictly increasing everywhere on ((0,\infty)).

   • (a) True on ([1,\infty)).
   • (b) False (because it increases, not decreases, on ((0,1])).

6. Reciprocal relation
   (f(1/x) = \int_{x}^{1/x} g(t),dt = -\int_{1/x}^{x} g(t),dt = -f(x)).
   So $f(x) + f\!\left(\frac{1}{x}\right) = 0$ making statement (c) True.

7. Oddness after substitution
   Let (F(u) = f(2^{u})). Then $F(-u) = f(2^{-u}) = f\!\left(\frac{1}{2^{u}}\right) = -f(2^{u}) = -F(u).$ Therefore (F) is an odd function of (u), verifying statement (d) True.

[\boxed{\text{Correct options: (a), (c), (d)}}]