List of top Mathematics Questions on Relations and functions asked in TS EAMCET

The domain of the real valued function \[ f(x)=\cos^{-1}\!\left(\log_{5}\frac{x}{5}\right)+\log_{5}\!\left(\cos^{-1}\frac{x}{5}\right) \] is:

If \(f:\mathbb R\to\mathbb R\) is defined by \[ f(x)=|x| \] and \(A=(0,1)\), then \(f^{-1}(A)\) is:

If $f:\mathbb{R}-\left\{\frac{2a}{3}\right\}\rightarrow \mathbb{R}$ is a function defined by \[ f(x)=\frac{2x+a}{3x-2a} \] and $(f\circ f)(x)=x$ for all $x\in \mathbb{R}-\left\{\frac{2a}{3}\right\}$, then $f(3)=$}

If $f:\mathbb{R}\rightarrow\mathbb{R}$ and $g:\mathbb{R}\rightarrow\mathbb{R}$ are defined by \[ f(x)=|x+1| \] and \[ g(x)= \begin{cases} e^{-x}, & x\le 0\\ x-1, & x>0 \end{cases} \] then $(f\circ g)(-2)+(g\circ f)(2)=$}

The domain and range of $f(x) = \frac{1}{\sqrt{|x|-x^2}}$ are A and B respectively. Then $A \cup B = $ ?

If the range of the real valued function $f(x) = \frac{x^2 + x + k}{x^2 - x + k}$ is $[\frac{1}{3}, 3]$, then $k =$

Let $f: R \rightarrow R$ be defined by $f(x) = 5^{|x|} + \text{sgn}(5^{-x})$, where sgn x denotes signum function of x. Then f is

A function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \begin{cases} 2x+3, & x \le 4/3 \\ -3x^2+8x, & x>4/3 \end{cases}$ is

If the domain of the real valued function $f(x) = \frac{1}{\sqrt{\log_{\frac{1}{3}}\left(\frac{x-1}{2-x}\right)}}$ is $(a,b)$, then $2b =$

If $D \subset \mathbb{R}$ and $f : D \to \mathbb{R}$ defined by $f(x) = \frac{x^2+x+a}{x^2-x+a}$ is a surjection then '$a$' lies in the interval