JEE Main Functions Practice Questions With Solutions

If $f(x)=\frac{4x+3}{6x-4},x\ne \frac{2}{3}$f \left(\right. x \left.\right) = \frac{4 x + 3}{6 x - 4} , x \neq \frac{2}{3} and $(f\circ f)(x)=g(x)$\left(\right. f \circ f \left.\right) \left(\right. x \left.\right) = g \left(\right. x \left.\right), where $g:\mathbb{R}-\left\{\frac{2}{3}\right\}\to \mathbb{R}-\left\{\frac{2}{3}\right\}$g : \mathbb{R} - \left{\right. \frac{2}{3} \left.\right} \rightarrow \mathbb{R} - \left{\right. \frac{2}{3} \left.\right}, then $(gogog)(4)$\left(\right. g o g o g \left.\right) \left(\right. 4 \left.\right) is equal to

If the domain of the function $f(x)={\log }_e\left(\frac{2x+3}{4x^2+x-3}\right)+{\cos }^{-1}\left(\frac{2x-1}{x+2}\right)$f \left(\right. x \left.\right) = log_{e} ⁡ \left(\right. \frac{2 x + 3}{4 x^{2} + x - 3} \left.\right) + cos^{- 1} ⁡ \left(\right. \frac{2 x - 1}{x + 2} \left.\right) is $(\alpha ,\beta ]$\left(\right. \alpha , \beta \left]\right., then the value of $5\beta -4\alpha$5 \beta - 4 \alpha is equal to

If the domain of the function $f(x)={\cos }^{-1}\left(\frac{2-|x|}{4}\right)+{\{{\log }_e(3-x)\}}^{-1}$f \left(\right. x \left.\right) = cos^{- 1} ⁡ \left(\right. \frac{2 - \left|\right. x \left|\right.}{4} \left.\right) + \left(\left{\right. log_{e} ⁡ \left(\right. 3 - x \left.\right) \left.\right}\right)^{- 1} is $[-\alpha ,\beta )-\{\gamma \}$\left[\right. - \alpha , \beta \left.\right) - \left{\right. \gamma \left.\right}, then $\alpha +\beta +\gamma$\alpha + \beta + \gamma is equal to :

If $f(x)=\{\begin{array}{cc}2+2x,& -1\le x<0\\ 1-\frac{x}{3},& 0\le x\le 3\end{array};g(x)=\{\begin{array}{cc}-x,& -3\le x\le 0\\ x,& 0<x\le 1\end{array}$f \left(\right. x \left.\right) = \left{\right. 2 + 2 x , & - 1 \leq x < 0 \\ 1 - \frac{x}{3} , & 0 \leq x \leq 3 ; g \left(\right. x \left.\right) = \left{\right. - x , & - 3 \leq x \leq 0 \\ x , & 0 < x \leq 1, then range of $(fog)(x)$\left(\right. f o g \left.\right) \left(\right. x \left.\right) is

Let $f:\mathbf{R}-\left\{\frac{-1}{2}\right\}\to \mathbf{R}$f : \mathbf{R} - \left{\right. \frac{- 1}{2} \left.\right} \rightarrow \mathbf{R} and $g:\mathbf{R}-\left\{\frac{-5}{2}\right\}\to \mathbf{R}$g : \mathbf{R} - \left{\right. \frac{- 5}{2} \left.\right} \rightarrow \mathbf{R} be defined as $f(x)=\frac{2x+3}{2x+1}$f \left(\right. x \left.\right) = \frac{2 x + 3}{2 x + 1} and $g(x)=\frac{|x|+1}{2x+5}$g \left(\right. x \left.\right) = \frac{\left|\right. x \left|\right. + 1}{2 x + 5}. Then, the domain of the function fog is :

The range of $f(x)=4{\sin }^{-1}\left(\frac{x^2}{x^2+1}\right)$f \left(\right. x \left.\right) = 4 sin^{- 1} ⁡ \left(\right. \frac{x^{2}}{x^{2} + 1} \left.\right) is

For $x\in \mathbb{R}$x \in \mathbb{R}, two real valued functions $f(x)$f \left(\right. x \left.\right) and $g(x)$g \left(\right. x \left.\right) are such that, $g(x)=\sqrt{x}+1$g \left(\right. x \left.\right) = \sqrt{x} + 1 and $f\circ g(x)=x+3-\sqrt{x}$f \circ g \left(\right. x \left.\right) = x + 3 - \sqrt{x}. Then $f(0)$f \left(\right. 0 \left.\right) is equal to

Let $\mathrm{D}$D be the domain of the function $f(x)={\sin }^{-1}({\log }_{3x}\left(\frac{6+2{\log }_{3}x}{-5x}\right))$f \left(\right. x \left.\right) = sin^{- 1} ⁡ \left(\right. log_{3 x} ⁡ \left(\right. \frac{6 + 2 log_{3} ⁡ x}{- 5 x} \left.\right) \left.\right). If the range of the function $\mathrm{g}:\mathrm{D}\to \mathbb{R}$g : D \rightarrow \mathbb{R} defined by $\mathrm{g}(x)=x-[x],([x]$g \left(\right. x \left.\right) = x - \left[\right. x \left]\right. , \left(\right. \left[\right. x \left]\right. is the greatest integer function), is $(\alpha ,\beta )$\left(\right. \alpha , \beta \left.\right), then ${\alpha }^2+\frac{5}{\beta }$\alpha^{2} + \frac{5}{\beta} is equal to

The domain of the function $f(x)=\frac{1}{\sqrt{[x{]}^2-3[x]-10}}$f \left(\right. x \left.\right) = \frac{1}{\sqrt{\left[\right. x \left]\right.^{2} - 3 \left[\right. x \left]\right. - 10}} is : ( where $[\mathrm{x}]$\left[\right. x \left]\right. denotes the greatest integer less than or equal to $x$x )