Functions — JEE Main Maths PYQ MCQs with Solutions

Solution: $f(x)=\dfrac{\log_2(x+3)}{(x+1)(x+2)}$. For the numerator, $x+3>0\Rightarrow x>-3$ ...(i). For the denominator, $(x+1)(x+2)\ne0\Rightarrow x\ne-1,-2$ ...(ii). From (i) and (ii), domain is $(-3,\infty)/\{-1,-2\}$.

Solution: For $y$ to be defined: $1-x>0$, $1-x\ne1$ and $x+2\ge0$, i.e. $x<1$, $x\ne0$ and $x\ge-2$. So $-2\le x<1$, excluding $x=0$, i.e. $x\in(-2,1)$ excluding $\{0\}$.

Solution: $f(x^2)=|x^2-1|\neq (x-1)^2=\{f(x)\}^2$ in general, so (a) is false; $f(x+y)=|x+y-1|\neq f(x)+f(y)$ in general, so (b) is false; $f(|x|)=||x|-1|\neq |x-1|=|f(x)|$ in general, so (c) is false. Hence, the answer is none of the above.

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 $=16-2=14$.

Solution: Let $y = 3x - 5 \Rightarrow x = \dfrac{y + 5}{3}$. So $f^{-1}(x) = \dfrac{x + 5}{3}$.

Solution: $f(x) = x - [x] = \{x\}$, the fractional part function, satisfies $f(x+1) = f(x)$ for all $x$, so it is periodic with period $1$. The functions $\sin(1/x)$ and $x\cos x$ are not periodic.

Solution: Clearly $f_1(x)=\dfrac{a^x+a^{-x}}{2}$ and $f_2(x)=\dfrac{a^x-a^{-x}}{2}$. So $f_1(x+y)+f_1(x-y)=\dfrac12\left[a^{x+y}+a^{-(x+y)}+a^{x-y}+a^{-(x-y)}\right]=\dfrac12\left[a^xa^y+\dfrac{1}{a^xa^y}+\dfrac{a^x}{a^y}+\dfrac{a^y}{a^x}\right]=\dfrac12\left[a^x\left(a^y+\dfrac1{a^y}\right)+\dfrac{1}{a^x}\left(a^y+\dfrac1{a^y}\right)\right]=2f_1(x)f_1(y)$.

Solution: Let $y=\dfrac{x^2+x+2}{x^2+x+1}$. Then $y-1=\dfrac{1}{x^2+x+1}$ (so $y\ne1$), giving $x^2(y-1)+x(y-1)-(y-2)=0$. Since $x$ is real, discriminant $\ge0$: $(y-1)^2-4(y-1)(y-2)\ge0\Rightarrow(y-1)(-3y+7)\ge0\Rightarrow1\le y\le\dfrac73$. Excluding $y=1$, range $=\left(1,\dfrac73\right]$.

Solution: $f(x)=(1+b^2)\left(x+\dfrac{b}{1+b^2}\right)^2+1-\dfrac{b^2}{1+b^2}$, so $m(b)=1-\dfrac{b^2}{1+b^2}=\dfrac{1}{1+b^2}$. This is always positive and, as $b$ varies over all reals, $m(b)$ ranges from just above $0$ up to $1$ (attained at $b=0$). So range $=(0,1]$.

Solution: $\dfrac{2x-1}{2x^3+3x^2+x}>0\Rightarrow\dfrac{2x-1}{x(2x^2+3x+1)}>0\Rightarrow\dfrac{2x-1}{x(2x+1)(x+1)}>0$. Using the wavy curve method with critical points $-1,-\dfrac12,0,\dfrac12$, the solution set is $x\in(-\infty,-1)\cup\left(-\dfrac12,0\right)\cup\left(\dfrac12,\infty\right)$. Since $\left(-\infty,-\dfrac32\right)\subset(-\infty,-1)$ and $\left(\dfrac12,3\right)\subset\left(\dfrac12,\infty\right)$, options (a) and (d) are contained in $S$.

Solution: $g(S)=\{x: x^2\in[0,4]\}=[-2,2]$, so $f(g(S))=[0,4]=S$. Also $f(S)=[0,16]$, so $g(f(S))=\{x: x^2\in[0,16]\}=[-4,4]\neq g(S)=[-2,2]$. Hence $g(f(S))=g(S)$ is not true.

Solution: Since $f(x+y)=f(x)f(y)$ and $f(1)=2$, we get $f(x)=2^x$. So $\sum_{k=1}^{10}f(a+k)=2^a\sum_{k=1}^{10}2^k=2^a\cdot\dfrac{2(2^{10}-1)}{2-1}=2^{a+1}(2^{10}-1)$. Equating to $16(2^{10}-1)$ gives $2^{a+1}=16=2^4\Rightarrow a+1=4\Rightarrow a=3$.

Solution: Comparing $f(x+y)=f(x)+f(y)+xy$ with $f(x)=ax^2+bx+c$ gives $c=0$ and $2a=1\Rightarrow a=\dfrac12$; with $a+b+c=3$, $b=\dfrac52$. So $f(x)=\dfrac{x^2}{2}+\dfrac{5x}{2}$. $\sum_{n=1}^{10}f(n)=\dfrac12(385)+\dfrac52(55)=192.5+137.5=330$.

Solution: $(gof)(x)=\sin x^2$, so $(gogof)(x)=\sin(\sin x^2)$ and $(fogogof)(x)=(\sin(\sin x^2))^2$. Setting them equal: $\sin(\sin x^2)\{\sin(\sin x^2)-1\}=0\Rightarrow \sin(\sin x^2)=0$ (since $\sin(\sin x^2)=1$ is impossible as $|\sin\theta|\le1$ strictly less near required value) $\Rightarrow x^2=n\pi\Rightarrow x=\pm\sqrt{n\pi}$, $n\in\{0,1,2,\ldots\}$.

Solution: Each of the pairs $f(x)=\sin^2x,\,g(x)=\sqrt x$; $f(x)=\sin x,\,g(x)=|x|$; and $f(x)=x^2,\,g(x)=\sin\sqrt x$ satisfies both given relations, so $f$ and $g$ are not uniquely determined.

Solution: $f(x)f(y)-\dfrac12\left[f(x/y)+f(xy)\right]=\cos(\log x)\cos(\log y)-\dfrac12\left[2\cos(\log x)\cos(\log y)\right]=0$, which is none of the given values.

Solution: Since $\dfrac{\pi}{2}\sin x\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$, $f(x)\in\left[-\dfrac12,\dfrac12\right]$, so (a) is true. $fog(x)=f(t)$ with $t=g(x)\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$, which also gives range $\left[-\dfrac12,\dfrac12\right]$, so (b) is true. $\lim_{x\to0}\dfrac{f(x)}{g(x)}=\dfrac{\pi}{6}$ using $\lim_{\theta\to0}\sin\theta/\theta=1$, so (c) is true. $(gof)(x)=1\Rightarrow \sin\{f(x)\}=\dfrac2\pi$, but $f(x)\in\left[-\dfrac12,\dfrac12\right]$ makes $\sin\{f(x)\}\in\left[-\dfrac12,\dfrac12\right]$ approximately, which never equals $\dfrac2\pi\approx0.637$, so (d) is false.

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: $g(n)=n+1$ 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: $f'(x)=6x^2-30x+36=6(x-2)(x-3)$, 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: By the given definitions, $f\{f^{-1}(B)\} = B$ holds only when $B \subseteq f(X)$ (the range of $f$); in general $f\{f^{-1}(B)\} \subseteq B$. So option (c) is the correct statement.

Solution: Let $y = 2^{x(x-1)}$. Taking $\log_2$: $\log_2 y = x^2 - x \Rightarrow x^2 - x - \log_2 y = 0 \Rightarrow x = \dfrac{1 + \sqrt{1 + 4\log_2 y}}{2}$ (rejecting the negative-sign root since $x \ge 1$). So $f^{-1}(x) = \dfrac{1}{2}\left(1 + \sqrt{1 + 4\log_2 x}\right)$.

Solution: $f(-x) = [\log(\sec x - \tan x)]^3 = [-\log(\sec x + \tan x)]^3 = -f(x)$, so $f$ is odd. Also $f'(x) = 3\sec x\,[\log(\sec x + \tan x)]^2 > 0$ on $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$, so $f$ is strictly increasing, hence one-one. As $x \to \pm\dfrac{\pi}{2}$, $\sec x + \tan x \to \infty$ or $0^+$, so $\log(\sec x+\tan x) \to \pm\infty$ and $f(x) \to \pm\infty$, giving range $R$, so $f$ is onto. It is not even.

Solution: For the domain: $4-x^2\ne0\Rightarrow x\ne\pm2$ ...(i). Also $\log_{10}(x^3-x)$ needs $x^3-x>0\Rightarrow x(x-1)(x+1)>0$. By the wavy curve method, $x\in(-1,0)\cup(1,\infty)$ ...(ii). Combining (i) and (ii), domain $=(-1,0)\cup(1,2)\cup(2,\infty)$.

Solution: For the domain we need $\sin^{-1}(2x)+\dfrac{\pi}{6}\ge0$, and $-\dfrac{\pi}{2}\le\sin^{-1}(2x)\le\dfrac{\pi}{2}$. So $\sin^{-1}(2x)\ge-\dfrac{\pi}{6}\Rightarrow2x\ge\sin\left(-\dfrac{\pi}{6}\right)=-\dfrac12\Rightarrow x\ge-\dfrac14$. Also $-1\le2x\le1\Rightarrow-\dfrac12\le x\le\dfrac12$. Combining, $x\in\left[-\dfrac14,\dfrac12\right]$.

Solution: Given $2^x+2^y=2$, $\forall x,y\in R$. Since $2^x,2^y>0$ for all real $x,y$, we get $2^x<2^x+2^y=2$, i.e. $0<2^x<2$. Taking $\log_2$ throughout, $-\infty<x<1$.

Solution: $f(\theta)=\sin\theta(\sin\theta+3\sin\theta-4\sin^3\theta)=\sin\theta(4\sin\theta-4\sin^3\theta)=4\sin^2\theta(1-\sin^2\theta)=4\sin^2\theta\cos^2\theta=(2\sin\theta\cos\theta)^2=(\sin2\theta)^2\ge0$, which is true for all real $\theta$.

Solution: $\varphi(x)=h(f(g(x)))=h(\sqrt{\tan x})=\dfrac{1-\tan x}{1+\tan x}=\tan\left(\dfrac{\pi}{4}-x\right)$. So $\varphi\left(\dfrac{\pi}{3}\right)=\tan\left(\dfrac{\pi}{4}-\dfrac{\pi}{3}\right)=\tan\left(-\dfrac{\pi}{12}\right)=-\tan\dfrac{\pi}{12}=\tan\dfrac{11\pi}{12}$.

Solution: $f\left(\dfrac{2x}{1+x^2}\right)=\log_e\left(\dfrac{1-\frac{2x}{1+x^2}}{1+\frac{2x}{1+x^2}}\right)=\log_e\left(\dfrac{(1-x)^2}{(1+x)^2}\right)=2\log_e\left(\dfrac{1-x}{1+x}\right)=2f(x)$.