Test Index

Functions Part 1

Section: Mathematics

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Question : 25 of 100

Marks: +1, -0

Let

f: \mathbb{R} \rightarrow \mathbb{R}

f(x)=x-1

g: \mathbb{R} \setminus \{1,-1\} \rightarrow \mathbb{R}

g(x)=\frac{x^2}{x^2-1}

. Then the function fog is :

[26-Jun-2022-Shift-2]

one-one but not onto
onto but not one-one
both one-one and onto
neither one-one nor onto

Solution: 👈: Video Solution

f: R \rightarrow R

defined as

f(x)=x-1

and

g: R \rightarrow \{1,-1\} \rightarrow R, g(x)=\frac{x^2}{x^2-1}

Now

f \circ g(x)=\frac{x^2}{x^2-1}-1=\frac{1}{x^2-1}

\therefore

Domain of

f \circ g(x)=R \setminus \{-1,1\}

And range of

f \circ g(x)=(-\infty,-1] \cup (0, \infty)

Now,

\frac{d}{dx}(f \circ g(x))=\frac{-1}{x^2-1} \cdot 2x=\frac{2x}{1-x^2}

\therefore \frac{d}{dx}(f \circ g(x))>0

for

\frac{2x}{(1-x)(1+x)}>0

\Rightarrow \frac{x}{(x-1)(x+1)}<0

\therefore x \in (-\infty,-1) \cup (0,1)

and

\frac{d}{dx}(f \circ g(x))<0

for

x \in (-1,0) \cup (1, \infty)

\therefore f \circ g(x)

is neither one-one nor onto.

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