Let f: R arrow R and g: R arrow R be defined as f(x)= beginarrayll\log _e x, & x\gt 0 e^-x, & x \leq 0\endarray. and g(x)= beginarrayllx, & x \geqslant 0 e^x, & x\lt 0\endarray. . Then, gof : R arrow R is :

Functions PYQ — JEE Main 2024 · 1 Feb · Shift 1 · Q41 with Solution

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Correct answer: B

Given functions

f(x)= \begin{cases} \ln x, & x>0 \\ e^{-x}, & x\le 0 \end{cases} \qquad g(x)= \begin{cases} x, & x\ge 0 \\ e^x, & x<0 \end{cases}

We need to study the composite function

(g\circ f)(x)=g(f(x)).

Find $g(f(x))$ for different values of $x$

We split according to the definition of $f$ .

Case 1: $x>0$

Then

f(x)=\ln x.

Now apply $g$ to $\ln x$ .

If $\ln x\ge 0$ , i.e. $x\ge 1$ , then

g(f(x))=g(\ln x)=\ln x.

- If $\ln x<0$, i.e. $0<x<1$, then

g(f(x))=g(\ln x)=e^{\ln x}=x.

So for $x>0$,

(g\circ f)(x)= \begin{cases} x, & 0<x<1,\ \ln x, & x\ge 1. \end{cases}

### Case 2: $x\le 0$ Then

f(x)=e^{-x}.

Since $e^{-x}>0$, we use the branch of $g$ for nonnegative input:

g(f(x))=g(e^{-x})=e^{-x}.

Thus,

(g\circ f)(x)= \begin{cases} e^{-x}, & x\le 0,\ x, & 0<x<1,\ \ln x, & x\ge 1. \end{cases}

--- 3. **Check whether $g\circ f$ is one-one** A function is one-one if different inputs give different outputs. Observe:

(g\circ f)(0)=e^{0}=1.

Also,

(g\circ f)(1)=\ln 1=0.

These are different, so this alone does not show failure. Let us test if one output can come from two different inputs. Take output $1$: - For $x=0$, $$ (g\circ f)(0)=1. $$ - For $x=e$, since $e\ge 1$,

(g\circ f)(e)=\ln e=1.

Thus,

(g\circ f)(0)=(g\circ f)(e)=1, \quad 0\ne e.

So $g\circ f$ is **not one-one**. --- 4. **Check whether $g\circ f$ is onto** We find the range of each branch. - For $x\le 0$:

y=e^{-x} \ge 1.

Range from this branch is $$[1,\infty).$$ - For $0<x<1$:

y=x,

so range is $$ (0,1). $$ - For $x\ge 1$:

y=\ln x,

so range is $$ [0,\infty). $$ Combining all these,

\text{Range}(g\circ f)=[0,\infty).

But codomain is $\mathbb{R}$. Negative real numbers are not attained. Hence $g\circ f$ is **not onto**. --- 5. **Conclusion** The function $g\circ f$ is: - **not one-one**, - **not onto**. Therefore the correct option is

\boxed{\text{B: neither one-one nor onto}}

More from Functions

View written solutionFree

Correct answer: B

Given functions

f(x)= \begin{cases} \ln x, & x>0 \\ e^{-x}, & x\le 0 \end{cases} \qquad g(x)= \begin{cases} x, & x\ge 0 \\ e^x, & x<0 \end{cases}

We need to study the composite function

(g\circ f)(x)=g(f(x)).

Find $g(f(x))$ for different values of $x$

We split according to the definition of $f$ .

Case 1: $x>0$

Then

f(x)=\ln x.

Now apply $g$ to $\ln x$ .

If $\ln x\ge 0$ , i.e. $x\ge 1$ , then

g(f(x))=g(\ln x)=\ln x.

- If $\ln x<0$, i.e. $0<x<1$, then

g(f(x))=g(\ln x)=e^{\ln x}=x.

So for $x>0$,

(g\circ f)(x)= \begin{cases} x, & 0<x<1,\ \ln x, & x\ge 1. \end{cases}

### Case 2: $x\le 0$ Then

f(x)=e^{-x}.

Since $e^{-x}>0$, we use the branch of $g$ for nonnegative input:

g(f(x))=g(e^{-x})=e^{-x}.

Thus,

(g\circ f)(x)= \begin{cases} e^{-x}, & x\le 0,\ x, & 0<x<1,\ \ln x, & x\ge 1. \end{cases}

--- 3. **Check whether $g\circ f$ is one-one** A function is one-one if different inputs give different outputs. Observe:

(g\circ f)(0)=e^{0}=1.

Also,

(g\circ f)(1)=\ln 1=0.

These are different, so this alone does not show failure. Let us test if one output can come from two different inputs. Take output $1$: - For $x=0$, $$ (g\circ f)(0)=1. $$ - For $x=e$, since $e\ge 1$,

(g\circ f)(e)=\ln e=1.

Thus,

(g\circ f)(0)=(g\circ f)(e)=1, \quad 0\ne e.

So $g\circ f$ is **not one-one**. --- 4. **Check whether $g\circ f$ is onto** We find the range of each branch. - For $x\le 0$:

y=e^{-x} \ge 1.

Range from this branch is $$[1,\infty).$$ - For $0<x<1$:

y=x,

so range is $$ (0,1). $$ - For $x\ge 1$:

y=\ln x,

so range is $$ [0,\infty). $$ Combining all these,

\text{Range}(g\circ f)=[0,\infty).

But codomain is $\mathbb{R}$. Negative real numbers are not attained. Hence $g\circ f$ is **not onto**. --- 5. **Conclusion** The function $g\circ f$ is: - **not one-one**, - **not onto**. Therefore the correct option is

\boxed{\text{B: neither one-one nor onto}}

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