Question

The minimum value of $x^{2} + \dfrac{1}{1 + x^{2}}$ is at}

• A. $x = 0$
• B. $x = 1$
• C. $x = 4$
• D. $x = 3$

Hint:

When a function involves $x^2$ throughout, substitute $t = x^2 \geq 0$ to convert it to a single-variable optimisation problem.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To find the minimum, we substitute $t = x^2 \ge 0$ and minimise $f(t) = t + \dfrac{1}{1+t}$.
Step 2: Detailed Explanation:
$f'(t) = 1 - \dfrac{1}{(1+t)^2} = 0 \Rightarrow (1+t)^2 = 1 \Rightarrow t = 0$.
Since $t = x^2$, $t = 0$ corresponds to $x = 0$. Checking $f''(t)>0$ confirms a minimum.
Step 3: Final Answer:
The minimum occurs at $x = 0$.