JEE Mathematics 2026 Question with Solution

Given: $f : (0,\infty) \rightarrow \mathbb{R}$ with

$$f(x) = \left| \log_e x \right| - |x - 1|$$

Find: Which of the statements (I), (II) and (III) are true.

Since absolute value expressions are involved, split the domain at $x = 1$.

For $0 < x < 1$,

$$|\log x| = -\log x, \qquad |x-1| = 1-x$$

So,

$$f(x) = -\log x - (1-x)$$

Differentiating,

$$f'(x) = -\frac{1}{x} + 1$$

For $0 < x < 1$, we have $\frac{1}{x} > 1$, hence

$$f'(x) < 0$$

Therefore, $f$ is decreasing in $(0,1)$, so statement (II) is false.

For $x > 1$,

$$|\log x| = \log x, \qquad |x-1| = x-1$$

So,

$$f(x) = \log x - (x-1)$$

Differentiating,

$$f'(x) = \frac{1}{x} - 1$$

For $x > 1$, we have $\frac{1}{x} < 1$, hence

$$f'(x) < 0$$

Therefore, $f$ is decreasing in $(1,\infty)$, so statement (III) is true.

At $x = 1$, the solution states that the left-hand and right-hand derivatives of both terms exist and are finite. Hence, $f$ is differentiable for all $x > 0$, so statement (I) is true.

Therefore, only statements (I) and (III) are true. The correct option is C.

Given:

$$f(x) = \left| \log_e x \right| - |x - 1|, \qquad x > 0$$

Find: The correct combination of true statements.

Step 1: Check statement (I).

The expression contains $|\log x|$ and $|x-1|$, so the only possible point of non-differentiability is $x=1$. According to the provided solution, the left-hand and right-hand derivatives at $x=1$ exist and are finite. Hence statement (I) is taken as true.

Step 2: Check statement (II) on $(0,1)$.

For $0<x<1$,

$$\log x < 0$$

so

$$|\log x| = -\log x$$

and also

$$|x-1| = 1-x$$

Therefore,

$$f(x) = -\log x - (1-x)$$

Now differentiate:

$$f'(x) = -\frac{1}{x} + 1$$

Since $0<x<1$, we get $\frac{1}{x}>1$, therefore

$$-\frac{1}{x}+1<0$$

So $f$ is decreasing on $(0,1)$. Hence statement (II) is false.

Step 3: Check statement (III) on $(1,\infty)$.

For $x>1$,

$$\log x > 0$$

so

$$|\log x| = \log x$$

and

$$|x-1| = x-1$$

Therefore,

$$f(x) = \log x - (x-1)$$

Differentiate:

$$f'(x) = \frac{1}{x} - 1$$

Since $x>1$, we have $\frac{1}{x}<1$, therefore

$$\frac{1}{x}-1<0$$

So $f$ is decreasing on $(1,\infty)$. Hence statement (III) is true.

Conclusion: Statement (I) is true, statement (II) is false, and statement (III) is true. Therefore, the correct option is C.