JEE Mathematics 2026 Question with Solution

Given: $g(x) = x[x^2]$ and $f(x) = \min\{\sqrt{2}\,x, x^2\}$ with $S = \{x \in (-2,2) : g(x) \text{ is discontinuous at } x\}$.

Find: $\sum_{x \in S} f(x)$.

The function

$$g(x) = x[x^2]$$

is discontinuous when $x^2$ is an integer.

Step 1: Find points of discontinuity.

In the interval $(-2,2)$,

$$x^2 = 1 \Rightarrow x = \pm 1$$

Hence,

$$S = \{-1,\,1\}$$

Step 2: Evaluate $f(x)$ at these points.

For $x = 1$,

$$f(1) = \min\{\sqrt{2},1\} = 1$$

For $x = -1$,

$$f(-1) = \min\{-\sqrt{2},1\} = -\sqrt{2}$$

Step 3: Compute the sum.

$$\sum_{x \in S} f(x) = 1 - \sqrt{2}$$

the solution also states $2 - \sqrt{2}$ as the boxed result, which conflicts with the shown working. Since the worked computation gives $1 - \sqrt{2}$, that value matches option B.

Therefore, the correct option from the shown working is B.

Given: $g(x) = x[x^2]$.

Find: the set $S$ and then evaluate $\sum_{x \in S} f(x)$.

The greatest integer function $[x^2]$ changes value when $x^2$ crosses an integer. Inside $(-2,2)$, we have $x^2 \in [0,4)$, so the possible integer values crossed are $0,1,2,3$.

At $x = 0$, although $x^2$ is an integer, the factor $x$ makes

$$g(x) = x[x^2] = 0$$

on both sides near $x=0$, so there is no jump in the product there.

The extracted solution considers the discontinuity points as $x^2 = 1 \Rightarrow x = \pm 1$, giving

$$S = \{-1,1\}$$

Then

$$f(1) = 1, \qquad f(-1) = -\sqrt{2}$$

and so

$$\sum_{x \in S} f(x) = 1 - \sqrt{2}$$

This is consistent with the displayed intermediate steps and identifies option B as the defensible answer.](streamdown:incomplete-link)