Statement-1: The function $$x^2(e^x + e^{-x})$$ is increasing for all $$x > 0$$.
Statement-2: The functions $$x^2 e^x$$ and $$x^2 e^{-x}$$ are increasing for all $$x > 0$$ and the sum of two increasing functions in any interval $$(a, b)$$ is an increasing function in $$(a, b)$$.

To solve this problem, we need to evaluate the truth of Statement-1 and Statement-2 and determine which option correctly describes their relationship.

Statement-1 claims that the function $$ f(x) = x^2 (e^x + e^{-x}) $$ is increasing for all $$ x > 0 $$. A function is increasing in an interval if its derivative is non-negative in that interval. Therefore, we must compute the derivative of $$ f(x) $$ and check if it is non-negative for all $$ x > 0 $$.

First, rewrite $$ f(x) $$ as $$ f(x) = x^2 e^x + x^2 e^{-x} $$. Now, compute the derivative $$ f'(x) $$ using the product rule for each term. For the first term, $$ g(x) = x^2 e^x $$, the derivative is:

$$ g'(x) = \frac{d}{dx}(x^2) \cdot e^x + x^2 \cdot \frac{d}{dx}(e^x) = 2x e^x + x^2 e^x = e^x x (x + 2) $$

For the second term, $$ h(x) = x^2 e^{-x} $$, the derivative is:

$$ h'(x) = \frac{d}{dx}(x^2) \cdot e^{-x} + x^2 \cdot \frac{d}{dx}(e^{-x}) = 2x e^{-x} + x^2 (-e^{-x}) = e^{-x} (2x - x^2) = e^{-x} x (2 - x) $$

Thus, the derivative of $$ f(x) $$ is:

$$ f'(x) = g'(x) + h'(x) = e^x x (x + 2) + e^{-x} x (2 - x) $$

Factor out $$ x $$ (since $$ x > 0 $$, $$ x \neq 0 $$):

$$ f'(x) = x \left[ e^x (x + 2) + e^{-x} (2 - x) \right] $$

Define $$ k(x) = e^x (x + 2) + e^{-x} (2 - x) $$. Since $$ x > 0 $$, the sign of $$ f'(x) $$ depends on $$ k(x) $$. We need to check if $$ k(x) \geq 0 $$ for all $$ x > 0 $$.

Evaluate $$ k(x) $$ at some points:

• At $$ x = 0 $$: $$ k(0) = e^0 (0 + 2) + e^0 (2 - 0) = 1 \cdot 2 + 1 \cdot 2 = 4 > 0 $$
• At $$ x = 1 $$: $$ k(1) = e^1 (1 + 2) + e^{-1} (2 - 1) = 3e + \frac{1}{e} \approx 3 \cdot 2.718 + 0.3678 \approx 8.154 + 0.3678 = 8.5218 > 0 $$
• At $$ x = 2 $$: $$ k(2) = e^2 (2 + 2) + e^{-2} (2 - 2) = 4e^2 + 0 \approx 4 \cdot 7.389 = 29.556 > 0 $$
• At $$ x = 3 $$: $$ k(3) = e^3 (3 + 2) + e^{-3} (2 - 3) = 5e^3 - e^{-3} \approx 5 \cdot 20.0855 - 0.0498 \approx 100.4275 - 0.0498 = 100.3777 > 0 $$
• At $$ x = 4 $$: $$ k(4) = e^4 (4 + 2) + e^{-4} (2 - 4) = 6e^4 - 2e^{-4} \approx 6 \cdot 54.598 - 2 \cdot 0.0183 \approx 327.588 - 0.0366 = 327.5514 > 0 $$
• At $$ x = 10 $$: $$ k(10) = e^{10} (10 + 2) + e^{-10} (2 - 10) = 12e^{10} - 8e^{-10} \approx 12 \cdot 22026.46579 - 8 \cdot 4.54 \times 10^{-5} \approx 264317.589 - 0.0003632 \approx 264317.5886 > 0 $$

As $$ x \to \infty $$, $$ e^x (x + 2) $$ dominates and goes to $$ \infty $$, while $$ e^{-x} (2 - x) $$ goes to 0, so $$ k(x) \to \infty $$. To ensure $$ k(x) > 0 $$ for all $$ x > 0 $$, find the minimum by examining the derivative $$ k'(x) $$:

$$ k'(x) = \frac{d}{dx} \left[ e^x (x + 2) \right] + \frac{d}{dx} \left[ e^{-x} (2 - x) \right] $$

Using the product rule:

$$ \frac{d}{dx} \left[ e^x (x + 2) \right] = e^x (x + 2) + e^x \cdot 1 = e^x (x + 3) $$

$$ \frac{d}{dx} \left[ e^{-x} (2 - x) \right] = (-e^{-x}) (2 - x) + e^{-x} \cdot (-1) = e^{-x} \left[ -(2 - x) - 1 \right] = e^{-x} (x - 3) $$

So,

$$ k'(x) = e^x (x + 3) + e^{-x} (x - 3) $$

Set $$ k'(x) = 0 $$:

$$ e^x (x + 3) + e^{-x} (x - 3) = 0 $$

Multiply both sides by $$ e^x $$ (positive):

$$ e^{2x} (x + 3) + (x - 3) = 0 $$

$$ e^{2x} (x + 3) = 3 - x $$

At $$ x = 0 $$: $$ e^0 (0 + 3) = 3 $$ and $$ 3 - 0 = 3 $$, so $$ x = 0 $$ is a solution. Check the sign of $$ k'(x) $$:

• For $$ x < 0 $$ (e.g., $$ x = -1 $$): $$ k'(-1) = e^{-1} (-1 + 3) + e^{1} (-1 - 3) = \frac{2}{e} - 4e \approx \frac{2}{2.718} - 4 \cdot 2.718 \approx 0.735 - 10.872 = -10.137 < 0 $$
• For $$ x > 0 $$ (e.g., $$ x = 1 $$): $$ k'(1) = e^1 (1 + 3) + e^{-1} (1 - 3) = 4e - \frac{2}{e} \approx 4 \cdot 2.718 - \frac{2}{2.718} \approx 10.872 - 0.735 = 10.137 > 0 $$

Thus, $$ k'(x) < 0 $$ for $$ x < 0 $$ and $$ k'(x) > 0 $$ for $$ x > 0 $$, meaning $$ k(x) $$ has a minimum at $$ x = 0 $$. Since $$ k(0) = 4 > 0 $$ and $$ k(x) $$ increases for $$ x > 0 $$, $$ k(x) > 0 $$ for all $$ x > 0 $$. Therefore, $$ f'(x) = x k(x) > 0 $$ for all $$ x > 0 $$, so $$ f(x) $$ is strictly increasing for $$ x > 0 $$. Statement-1 is true.

Statement-2 claims that both $$ g(x) = x^2 e^x $$ and $$ h(x) = x^2 e^{-x} $$ are increasing for all $$ x > 0 $$, and that the sum of two increasing functions is increasing. We already computed $$ g'(x) = e^x x (x + 2) $$. For $$ x > 0 $$, $$ e^x > 0 $$, $$ x > 0 $$, and $$ x + 2 > 0 $$, so $$ g'(x) > 0 $$, meaning $$ g(x) $$ is strictly increasing for $$ x > 0 $$.

For $$ h(x) = x^2 e^{-x} $$, we have $$ h'(x) = e^{-x} x (2 - x) $$. For $$ x > 0 $$, $$ e^{-x} > 0 $$ and $$ x > 0 $$, so the sign depends on $$ (2 - x) $$:

• If $$ 0 < x < 2 $$, $$ 2 - x > 0 $$, so $$ h'(x) > 0 $$ (increasing)
• If $$ x > 2 $$, $$ 2 - x < 0 $$, so $$ h'(x) < 0 $$ (decreasing)

Thus, $$ h(x) $$ is not increasing for all $$ x > 0 $$ because it decreases for $$ x > 2 $$. Therefore, Statement-2 is false.

Since Statement-1 is true and Statement-2 is false, we select the option that matches this. The options are:

A. Statement-1 is false; Statement-2 is true.

B. Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

C. Statement-1 is true; Statement-2 is false.

D. Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is Option C.