Function $f(x)$ is such that $f(x) = a\ln x + \dfrac{x^2}{2}$ where $a > 0$ is a parameter. If $\dfrac{f(x_1) - f(x_2)}{x_1 - x_2} \geq 2$ $\forall\, x_1, x_2 \in (0, \infty)$ and $x_1 \neq x_2$, then possible value of '$a$' can be:
Step-by-Step Solution
Key Concept: The key idea is to transform the given condition $\frac{f(x_1) - f(x_2)}{x_1 - x_2} \geq k$ into a condition on the derivative of a related function. Specifically, this implies that the function $g(x) = f(x) - kx$ is non-decreasing for all $x$, which means $g'(x) \geq 0$ (or equivalently $f'(x) \geq k$) for all $x$ in the domain.
Step 1: The condition $\frac{f(x_1)-f(x_2)}{x_1-x_2} \geq 2$ for all $x_1, x_2 \in (0,\infty)$ with $x_1 \neq x_2$ means that $f(x) - 2x$ is a non-decreasing function on $(0,\infty)$.
Step 2: Define $g(x) = f(x) - 2x = a\ln x + \frac{x^2}{2} - 2x$. For $g$ to be non-decreasing, we need $g'(x) \geq 0$ for all $x \in (0,\infty)$.
Step 3: Compute $g'(x)$:
$$g'(x) = \frac{a}{x} + x - 2$$
Step 4: Require $g'(x) \geq 0$ for all $x > 0$:
$$\frac{a}{x} + x - 2 \geq 0 \implies a + x^2 - 2x \geq 0 \implies a \geq 2x - x^2 = -(x^2 - 2x)$$
Step 5: Find the maximum of $h(x) = 2x - x^2$ on $(0,\infty)$. $h'(x) = 2 - 2x = 0 \implies x = 1$. $h(1) = 2 - 1 = 1$.
Step 6: Therefore, we need $a \geq 1$. Since $a > 0$, the condition is $a \geq 1$.
Step 7: Check the options: $a = 0$ (not valid, also $a > 0$ required), $a = 1/2 < 1$ (not valid), $a = 3/2 \geq 1$ (valid), $a = 1 \geq 1$ (valid). So possible values are $a = 3/2$ and $a = 1$, i.e., options (c) and (d).
Correct Answer: 3, 4