Classification of Functions, Domain and Range and Even, Odd Functions — JEE Main Maths PYQ MCQs with Solutions

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: 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: 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: 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: $f(x)=\dfrac{\log_2(x+3)}{(x+1)(x+2)}$. 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: 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: 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: $\dfrac{2x-1}{2x^3+3x^2+x}>0\Rightarrow\dfrac{2x-1}{x(2x^2+3x+1)}>0\Rightarrow\dfrac{2x-1}{x(2x+1)(x+1)}>0$. 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 $S$.