JEE Mathematics 2026 Question with Solution

Given: Two statements about functions on $\mathbb{R}$.

Find: Which of the two statements is correct.

For Statement I: For $x \ge 0$,

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

which is strictly increasing.

For $x<0$,

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

which is also strictly increasing.

Also,

$$\lim_{x\to-\infty} f(x)=-1, \qquad \lim_{x\to\infty} f(x)=1$$

Hence the function is strictly increasing on $(-\infty,\infty)$, so it is one-one.

Therefore, Statement I is true.

For Statement II: The function

$$f(x)=\frac{x^2+4x-30}{x^2-8x+18}$$

is a rational function with quadratic numerator and denominator. From the given solution, distinct values of $x$ can give the same function value, so the function is many-one.

Therefore, Statement II is true.

So, both statements are true. The correct option is B.

Given:

1. $f(x)=\frac{x}{1+|x|}$
2. $f(x)=\frac{x^2+4x-30}{x^2-8x+18}$

Find: Truth values of Statement I and Statement II.

For the first function, split by the definition of modulus:

$$|x|=x \text{ for } x\ge 0, \qquad |x|=-x \text{ for } x<0$$

So,

$$f(x)=\frac{x}{1+x} \text{ for } x\ge 0$$

and

$$f(x)=\frac{x}{1-x} \text{ for } x<0$$

Both expressions are strictly increasing on their respective intervals. The end behavior is

$$\lim_{x\to-\infty} f(x)=-1, \qquad \lim_{x\to\infty} f(x)=1$$

Hence the function increases throughout its domain, so it is one-one.

Thus, Statement I is true.

For the second function, the provided solution concludes that the rational function is not one-one and takes the same value at more than one input. Hence it is many-one.

Thus, Statement II is true.

Therefore, the correct option is B.