Types of Functions — JEE Main Maths PYQ MCQs with Solutions

Solution: Let $y=f(x)=\dfrac{x^2}{1-x^2}$. Then $x^2=y(1-x^2)\Rightarrow x^2(1+y)=y\Rightarrow x^2=\dfrac{y}{1+y}$ (provided $y\neq -1$). Since $x^2\ge 0$, we need $\dfrac{y}{1+y}\ge 0$, which gives $y\in(-\infty,-1)\cup[0,\infty)=\mathbf{R}-[-1,0)$. For $f$ to be surjective, the range must equal the codomain, so $A=\mathbf{R}-[-1,0)$.

Solution: As $x\to 0^{+}$, $f(x)\to\infty$; at $x=1$, $f(x)=0$; and as $x\to\infty$, $f(x)\to 1$. So $f$ first decreases from $\infty$ to $0$ on $(0,1]$ and then increases from $0$ towards $1$ on $[1,\infty)$, so it is many-one (not injective). Also the actual range of $f$ works out to $[0,\infty)$, but the stated codomain is $(0,\infty)$, so range $\neq$ codomain and $f$ is not surjective either. Hence $f$ is neither injective nor surjective.

Solution: The multiples of $4$ in $\{1,\ldots,20\}$ are $\{4,8,12,16,20\}$ (5 elements), and the multiples of $3$ in the codomain are $\{3,6,9,12,15,18\}$ (6 elements). Since $f$ is an onto function between two sets of equal (finite) size, it must be a bijection. So the 5 elements $\{4,8,12,16,20\}$ must map injectively into the 6 available multiples of 3, which can be done in $^{6}C_{5}\times 5!=6!$ ways. The remaining 15 elements of the domain then map bijectively onto the remaining 15 elements of the codomain in $15!$ ways. Total number of onto functions $=15!\times 6!$.

Solution: Let $y=\dfrac{x}{1+x^2}$. Then $yx^2-x+y=0$; since $x\in\mathbf{R}$, the discriminant must be non-negative: $1-4y^2\ge 0\Rightarrow -\dfrac{1}{2}\le y\le \dfrac{1}{2}$. So the range of $f$ is $\left[-\dfrac{1}{2},\dfrac{1}{2}\right]$ (consistent with $f$ being an odd function attaining maximum $\dfrac12$ at $x=1$ and minimum $-\dfrac12$ at $x=-1$).

Solution: $g(n)=n+1$ if $n$ is odd and $g(n)=n-1$ if $n$ is even. Computing $f(g(n))$: if $n$ is odd, $g(n)=n+1$ is even and $f(n+1)=\dfrac{n+1}{2}$; if $n$ is even, $g(n)=n-1$ is odd and $f(n-1)=\dfrac{(n-1)+1}{2}=\dfrac{n}{2}$. In both cases $f(g(n))=f(n)$, i.e. $fog=f$. Since $f(2)=f(1)=1$, $fog$ is not one-one. But $f$ is onto: for any $m\in N$, if $m$ is even then $f(2m)=m$, and if $m$ is odd then $f(2m-1)=m$. Hence $fog$ is onto but not one-one.

Solution: One-one: if $f(x_1)=f(x_2)$ then $\dfrac{2x_1}{x_1-1}=\dfrac{2x_2}{x_2-1}\Rightarrow 2x_1x_2-2x_1=2x_1x_2-2x_2\Rightarrow x_1=x_2$, so $f$ is injective. Onto: solving $f(x)=4$ gives $x=2$, but $2$ is a positive integer and hence excluded from the domain $A$, so $f(x)$ can never equal $4$ (similarly other values are missed). Hence $f$ is into, not surjective. So $f$ is injective but not surjective.

Solution: Since $f\left(\dfrac{1}{x}\right)=f(x)$ for $x\neq 0$ (e.g. $f(1/2)=f(2)$), two different inputs give the same output, so $f$ is many-one (not injective). Solving $y=\dfrac{x}{1+x^2}$ gives $yx^2-x+y=0$; as $x\in\mathbf{R}$, the discriminant $1-4y^2\ge 0\Rightarrow y\in\left[-\dfrac12,\dfrac12\right]$, so the range equals the codomain and $f$ is surjective. Hence $f$ is surjective but not injective.

Solution: $f'(x)=6x^2-30x+36=6(x-2)(x-3)$, which is positive on $[0,2)$, negative on $(2,3)$ and positive again just after $3$ — so $f$ increases, then decreases, then increases on $[0,3]$, making it many-one. Evaluating $f(0)=1$, $f(2)=29$, $f(3)=28$, the range of $f$ over $[0,3]$ is $[1,29]$, which equals the codomain. Hence $f$ is onto but not one-one.

Solution: $(f-g)(x)=x$ if $x$ is rational and $-x$ if $x$ is irrational. For any real $y$: if $y$ is rational, take $x=y$; if $y$ is irrational, take $x=-y$. Each real value is attained by exactly one $x$, so $f-g$ is one-one and onto.

Solution: $f'(x)=\dfrac{1}{(1+x)^2}>0$ for all $x\ge 0$, so $f$ is strictly increasing, hence one-one. Setting $y=\dfrac{x}{1+x}$ gives $x=\dfrac{y}{1-y}\ge 0\Rightarrow 0\le y<1$, so the range is $[0,1)$, a proper subset of the codomain $[0,\infty)$. Hence $f$ is one-one but not onto.

Solution: $f'(x)=2+\cos x\ge 1>0$ for all $x\in R$, so $f$ is strictly increasing, hence one-one. Being continuous and strictly increasing with $f(x)\to\pm\infty$ as $x\to\pm\infty$, $f$ attains every real value, so it is also onto. Hence $f$ is one-to-one and onto.

Solution: Total number of functions from $E$ to $F$ is $2^4=16$. Of these, the two constant functions ($f(x)=1$ for all $x\in E$, and $f(x)=2$ for all $x\in E$) are not onto. So the number of onto functions $=16-2=14$.