Let [f:R R] be a function such that [f(x) = x^2 + 2x + 1 x^2 + 1].

Question

Let $$f:R \to R$$ be a function such that $$f(x) = {{{x^2} + 2x + 1} \over {{x^2} + 1}}$$. Then

Accepted Answer

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

$$ \Rightarrow f'(x) = {{({x^2} + 1)(2x + 2) - ({x^2} + 2x + 1)(2x)} \over {{{({x^2} + 1)}^2}}}$$

$$ \Rightarrow f'(x) = {{2 - 2{x^2}} \over {{{({x^2} + 1)}^2}}}$$

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

$$ \therefore $$ $f'(x) = 0$ 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.

Solutions · 1