Let $$f:\left( 1,3 \right) \rightarrow R$$ be a function defined by $$f(x)=\cfrac { x\left[ x \right] }{ 1+{ x }^{ 2 } } $$, where $$[x]$$ denotes the greatest integer $$\le x$$. Then the range of $$f$$ is:
$$f(x) = \dfrac{x [x]}{x^2 + 1} \, : \, (1, 3) \rightarrow R$$ $$f(x) = \begin{cases} \dfrac{x}{x^2 + 1} & x \, \in (1, 2)\\ \dfrac{2x}{x^2 + 1} & x \, \in (2, 3)\end{cases}$$ $$f'(x) = \begin{cases} \dfrac{(x^2 + 1). 1- x.2x}{(x^2 + 1)^2} = \dfrac{1 - x^2}{(x^2 + 1)^2} = - \dfrac{(x^2 - 1)}{(x^2 + 1)^2} \,(-ve) \, \downarrow \\ \dfrac{(x^2 + 1) . 2 - 2x . 2x}{(x^2 + 1)^2} = \dfrac{2 - 2x^2}{(x^2 + 1)^2} = \dfrac{2 - 2x^2}{(x^2 + 1)^2} = \dfrac{-(2x^2 - 2)}{(x^2 + 1)^2} \downarrow \end{cases}$$ $$f(x) $$ is decreasing $$f^n$$. $$f(1) = \dfrac{1}{2}, f(2^-) = \dfrac{2}{5}$$ $$f(2^+) = \dfrac{4}{5} , f(3) = \dfrac{6}{10}$$ $$y \in \left(\dfrac{2}{5} , \dfrac{1}{2} \right) \cup \left(\dfrac{6}{10} , \dfrac{4}{5} \right]$$
Let $$f$$ be an injective function with domain $$\{x, y, z\}$$ and range $$\{1, 2, 3\}$$ such that exactly one of the following statements is correct and the remaining are false $$f(x) =1, f(y) \neq 1, f(z) \neq$$ 2. The value of $$f^{-1}(1)$$ is:
Left $$f$$ be an odd function defined on the set of real numbers such that for $$x\geq 0, f(x)=3\sin x+4\cos x$$. Then $$f(x)$$ at $$x= \displaystyle -\frac{11\pi}{6}$$ is equal to :
Let $$ f_{k}(x)=\displaystyle \frac{1}{k}(\sin ^{k}x+\cos ^{k}x)$$ where $$ x\in R$$ and $$ k\geq 1$$. Then $$ f_{4}(x)-f_{6}(x)$$ equals
For $$x\in R, x\neq 0, x\neq 1$$, let $$f_{0}(x) = \dfrac {1}{1 - x}$$ and $$f_{n + 1} (x) = f_{0} (f_{n}(x)), n = 0, 1, 2, ....$$ Then the value of $$f_{100}(3) + f_{1}\left (\dfrac {2}{3}\right ) + f_{2} \left (\dfrac {3}{2}\right )$$ is equal to:
If $$f(x) = \dfrac {2^{x} + 2^{-x}}{2}$$, then $$f (x + y)\cdot f (x - y)$$ is
Let $$A = \left \{x_{1}, x_{2}, x_{3}, ...., x_{7}\right \}, B = \left \{y_{1}, y_{2}, y_{3}\right \}$$. The total number of functions $$f : A\rightarrow B$$ that are on to and there are exactly three element $$x$$ in $$A$$ such that $$f(x) = y_{2}$$ is equal to