If the function $f:(-\infty,-1] \rightarrow(a, b]$ defined by $f(x)=e^{x^3-3 x+1}$ is one-one and onto,then the distance of the point $P(2 b+4, a+2)$ from the line $x+e^{-3} y=4$ is:

Let $A=\{-1, 0, 1, 2\}$ and $B=\{-4, -2, 0, 2\}$. Let $f, g: A \rightarrow B$ be functions defined by $f(x)=x^{2}-x$ for $x \in A$ and $g(x)=2\left|x-\frac{1}{2}\right|-1$ for $x \in A$. Are $f$ and $g$ equal? Justify your answer.

Let $f, g: R \rightarrow R$ be functions defined by $f(x) = \begin{cases} [x] & x < 0 \\ |1-x| & x \geq 0 \end{cases}$ and $g(x) = \begin{cases} e^x - x & x < 0 \\ (x-1)^2 - 1 & x \geq 0 \end{cases}$ where $[x]$ denotes the greatest integer less than or equal to $x$. Then,the function $(f \circ g)(x)$ is discontinuous at exactly

If $f : R \rightarrow R$ is defined by $f(x) = \begin{cases} x + 4, & x < -4 \\ 3x + 2, & -4 \leq x < 4 \\ x - 4, & x \geq 4 \end{cases}$ then the correct matching of List-$I$ from List-$II$ is :
List-$I$
$(A) f(-5) + f(-4)$
$(B) f(|f(-8)|)$
$(C) f(f(-7) + f(3))$
$(D) f(f(f(f(0)))) + 1$
List-$II$
$(i) 14$
$(ii) 4$
$(iii) -11$
$(iv) -1$
$(v) 1$
$(vi) 0$

Let $f : R \to R$ be a function defined by $f(x) = - \frac{|x|^3 + |x|}{1 + x^2}$; then the graph of $f(x)$ lies in the :-

Let the function $f:[0,1] \rightarrow \mathbb{R}$ be defined by $f(x) = \frac{4^x}{4^x+2}$. Then the value of $f\left(\frac{1}{40}\right) + f\left(\frac{2}{40}\right) + f\left(\frac{3}{40}\right) + \dots + f\left(\frac{39}{40}\right) - f\left(\frac{1}{2}\right)$ is: