JEE Mathematics 2025 Question with Solution

Given: $f(x) = \log x$ and $g(x) = \frac{x^4 - 2x^3 + 3x^2 - 2x + 2}{2x^2 - 2x + 1}$.

Find: The domain of $f \circ g$.

For $f(g(x))$ to be defined, we need

$$g(x) > 0$$

because $\log x$ is defined only for positive arguments.

Now examine the denominator:

$$2x^2 - 2x + 1$$

The quadratic equation

$$2x^2 - 2x + 1 = 0$$

has discriminant

$$\Delta = (-2)^2 - 4 \times 2 \times 1 = 4 - 8 = -4$$

Since the discriminant is negative, the denominator has no real roots, so it is never zero for any real $x$.

Thus, the expression $g(x)$ is defined for all real $x$. The solution further concludes that $g(x) > 0$ for all real $x$, so $f(g(x))$ is defined for every real number.

Therefore, the domain of $f \circ g$ is $\mathbb{R}$. The correct option is A.

Given: $f(x) = \log x$ and $g(x) = \frac{x^4 - 2x^3 + 3x^2 - 2x + 2}{2x^2 - 2x + 1}$.

Find: The set of all real $x$ for which $f(g(x))$ exists.

Since

$$f(x) = \log x$$

we must have

$$g(x) > 0$$

Now,

$$g(x) = \frac{x^4 - 2x^3 + 3x^2 - 2x + 2}{2x^2 - 2x + 1}$$

First check the denominator. For

$$2x^2 - 2x + 1 = 0$$

the discriminant is

$$\Delta = 4 - 8 = -4$$

So the denominator has no real zero and does not vanish for any real $x$.

The extracted solution then observes that the expression remains positive and concludes

$$g(x) > 0 \quad \text{for all } x \in \mathbb{R}$$

Hence every real number is allowed in the composition.

Therefore, the domain is $\mathbb{R}$.