Let f, g: R arrow R be defined as : f(x)=|x-1| and g(x)= \begincasese^x, & x \geq 0 x+1, & x \leq 0 .\endcases Then the function f(g(x)) is

Functions PYQ — JEE Main 2024 · 5 Apr · Shift 2 · Q32 with Solution

View written solutionFree

Correct answer: A

Given functions

$f(x)=|x-1|$

and

e^x, & x\ge 0,\\ x+1, & x\le 0. \end{cases}$$ We need to study the composite function $$h(x)=f(g(x))=|g(x)-1|.$$ --- 2. **Compute $f(g(x))$ piecewise** Since $g$ itself is piecewise, we evaluate separately. ### Case 1: $x\ge 0$ Then $$g(x)=e^x$$ so $$h(x)=f(g(x))=|e^x-1|.$$ Now for $x\ge 0$, we have $e^x\ge 1$, hence $$|e^x-1|=e^x-1.$$ So for $x\ge 0$, $$h(x)=e^x-1.$$ ### Case 2: $x\le 0$ Then $$g(x)=x+1$$ so $$h(x)=f(g(x))=|(x+1)-1|=|x|.$$ For $x\le 0$, $|x|=-x$. Thus $$h(x)=-x.$$ Hence the composite function is $$h(x)=f(g(x))=\begin{cases} -x, & x\le 0,\\ e^x-1, & x\ge 0. \end{cases}$$ At $x=0$, both give $0$, so this is consistent. --- 3. **Check whether $h$ is one-one** A function is one-one if different inputs always give different outputs. Let us test values from the two branches: - For $x=-1$, $$h(-1)=-(-1)=1.$$ - For $x=\ln 2$ (note that $\ln 2>0$), $$h(\ln 2)=e^{\ln 2}-1=2-1=1.$$ Thus, $$h(-1)=h(\ln 2)=1,$$ but $$-1\ne \ln 2.$$ Therefore, $h$ is **not one-one**. --- 4. **Check whether $h$ is onto** For onto $\mathbb R$, the range must be all real numbers. Now, - For $x\le 0$, $$h(x)=-x\ge 0.$$ - For $x\ge 0$, $$h(x)=e^x-1\ge 0.$$ So in every case, $$h(x)\ge 0.$$ Hence the range of $h$ is contained in $[0,\infty)$, not all of $\mathbb R$. In fact, - from the branch $-x$ for $x\le 0$, we get all values in $[0,\infty)$, - and from $e^x-1$ for $x\ge 0$, we also get $[0,\infty)$. So the range is exactly $$[0,\infty).$$ Therefore, $h$ is **not onto** as a function from $\mathbb R$ to $\mathbb R$. --- 5. **Conclusion** The function $f(g(x))$ is - **not one-one**, and - **not onto**. So the correct option is: $$\boxed{\text{A: neither one-one nor onto}}.$$ --- 6. **Comparison with stored correct answer** Stored correct answer: **A** Our derived answer: **A** They match.

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View written solutionFree

Correct answer: A

Given functions

$f(x)=|x-1|$

and

e^x, & x\ge 0,\\ x+1, & x\le 0. \end{cases}$$ We need to study the composite function $$h(x)=f(g(x))=|g(x)-1|.$$ --- 2. **Compute $f(g(x))$ piecewise** Since $g$ itself is piecewise, we evaluate separately. ### Case 1: $x\ge 0$ Then $$g(x)=e^x$$ so $$h(x)=f(g(x))=|e^x-1|.$$ Now for $x\ge 0$, we have $e^x\ge 1$, hence $$|e^x-1|=e^x-1.$$ So for $x\ge 0$, $$h(x)=e^x-1.$$ ### Case 2: $x\le 0$ Then $$g(x)=x+1$$ so $$h(x)=f(g(x))=|(x+1)-1|=|x|.$$ For $x\le 0$, $|x|=-x$. Thus $$h(x)=-x.$$ Hence the composite function is $$h(x)=f(g(x))=\begin{cases} -x, & x\le 0,\\ e^x-1, & x\ge 0. \end{cases}$$ At $x=0$, both give $0$, so this is consistent. --- 3. **Check whether $h$ is one-one** A function is one-one if different inputs always give different outputs. Let us test values from the two branches: - For $x=-1$, $$h(-1)=-(-1)=1.$$ - For $x=\ln 2$ (note that $\ln 2>0$), $$h(\ln 2)=e^{\ln 2}-1=2-1=1.$$ Thus, $$h(-1)=h(\ln 2)=1,$$ but $$-1\ne \ln 2.$$ Therefore, $h$ is **not one-one**. --- 4. **Check whether $h$ is onto** For onto $\mathbb R$, the range must be all real numbers. Now, - For $x\le 0$, $$h(x)=-x\ge 0.$$ - For $x\ge 0$, $$h(x)=e^x-1\ge 0.$$ So in every case, $$h(x)\ge 0.$$ Hence the range of $h$ is contained in $[0,\infty)$, not all of $\mathbb R$. In fact, - from the branch $-x$ for $x\le 0$, we get all values in $[0,\infty)$, - and from $e^x-1$ for $x\ge 0$, we also get $[0,\infty)$. So the range is exactly $$[0,\infty).$$ Therefore, $h$ is **not onto** as a function from $\mathbb R$ to $\mathbb R$. --- 5. **Conclusion** The function $f(g(x))$ is - **not one-one**, and - **not onto**. So the correct option is: $$\boxed{\text{A: neither one-one nor onto}}.$$ --- 6. **Comparison with stored correct answer** Stored correct answer: **A** Our derived answer: **A** They match.

Functions question

Functions question

More from Functions

Functions question

Functions question

More from Functions