JEE Main 2023 — Functions Question with Solution

$$f(x)=\frac{x^2+2x+1}{(x^2+1)}$$

$$\Rightarrow f^{\prime }(x)=\frac{(x^2+1)(2x+2)-(x^2+2x+1)(2x)}{{(x^2+1)}^2}$$

$$\Rightarrow f^{\prime }(x)=\frac{2-2x^2}{{(x^2+1)}^2}$$

$$=\frac{2\left(1+x\right)\left(1-x\right)}{{\left(x^2+1\right)}^2}$$

$$\therefore$$ $$f^{\prime }(x)=0$$ at x = 1 and x = -1. So, x = 1 and x = -1 are point of maxima/minima.

Option A : $$f(x)$$ is many-one in $$(-\infty ,-1)$$.

As x = -1 is a point of maxima/minima. So it is a boundary point in the range $$(-\infty ,-1)$$. So, $$f(x)$$ is one-one in $$(-\infty ,-1)$$.

$$\therefore$$ Option A is incorrect.

Option B : $$f(x)$$ is one-one in $$(-\infty ,\infty )$$.

As, x = 1 and x = -1 are point of maxima/minima which is present inside the range $$(-\infty ,\infty )$$. So, $$f(x)$$ is many-one function in $$(-\infty ,\infty )$$.

$$\therefore$$ Option B is incorrect.

Option C : $$f(x)$$ is one-one in $$[1,\infty )$$ but not in $$(-\infty ,\infty )$$.

As x = 1 is a point of maxima/minima. So it is a boundary point in the range $$[1,\infty )$$. So, $$f(x)$$ is one-one in $$[1,\infty )$$.

As, x = 1 and x = -1 are point of maxima/minima which is present inside the range $$(-\infty ,\infty )$$. So, $$f(x)$$ is many-one function in $$(-\infty ,\infty )$$ .

$$\therefore$$ Option C is correct.

Option D : $$f(x)$$ is many-one in $$(1,\infty )$$.

As x = 1 is a point of maxima/minima. So it is a boundary point in the range $$[1,\infty )$$. So, $$f(x)$$ is one-one in $$[1,\infty )$$.

$$\therefore$$ Option D is incorrect.

Note :

Methods to Check One-One Function :

(i) If a function is one-one, any line parallel to $x$-axis cuts the graph of the function maximum at one point.

(ii) Any continuous function which is entirely increasing or decreasing (no maxima/minima is present) in whole domain will always be one-one function.