Let f (t) be a real-valued function whose second derivative is positive for $-\mathrm{\infty }<t<\mathrm{\infty }.$ Which of the following is/are always true ?

The correct option is f (t) cannot have two distinct local minima.

Let us analyze the properties of the function $f\left(t\right)$ given the condition on its second derivative.
We are given that the second derivative of the real-valued function $f\left(t\right)$ is strictly positive for all real numbers $t$:
$$f^{″}\left(t\right)>0\text{for all}-\infty <t<\infty$$

A function whose second derivative is strictly positive on an interval is strictly convex on that interval.
One of the fundamental properties of a strictly convex function is that any local minimum is also a unique global minimum. Consequently, a strictly convex function cannot have more than one local minimum.

To prove this mathematically, suppose for contradiction that $f\left(t\right)$ has two distinct local minima at $t=a$ and $t=b$ (where $a<b$).
Since these are local minima of a differentiable function, the first derivative must vanish at these points:
$$f^{\prime }\left(a\right)=0\text{and}{f}^{\prime }\left(b\right)=0$$

However, since $f^{″}\left(t\right)>0$ for all $t$, the first derivative $f^{\prime }\left(t\right)$ must be a strictly increasing function on the entire real line.
If $f^{\prime }\left(t\right)$ is strictly increasing and $a<b$, we must have:
$$f^{\prime }\left(a\right)<f^{\prime }\left(b\right)$$ This contradicts our assumption that $f^{\prime }\left(a\right)=f^{\prime }\left(b\right)=0$.
Therefore, $f\left(t\right)$ cannot have two distinct local minima.

Let us briefly check why the other options are not always true:
1. f (t) has at least one local minimum: This is not always true. For example, the function $f\left(t\right)=e^t$ has $f^{″}\left(t\right)=e^t>0$ everywhere, but it has no local minimum anywhere on the real line.
2. f (t) has at least one local maximum: A strictly convex function cannot have any local maximum since $f^{″}\left(t\right)>0$ rule out the second-order condition for a maximum.
3. The minimum value of f (t) cannot be negative: This is false. Consider the function $f\left(t\right)=t^2-5$, which has $f^{″}\left(t\right)=2>0$ but has a negative minimum value of $-5$ at $t=0$.