Let f:(1,3) → mathrm{R} be a function defined by f(mathrm{x})=frac{mathrm{x}[mathrm{x}]}{1+mathrm{

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1.Relation and Function

hard

Let $f:(1,3) \rightarrow \mathrm{R}$ be a function defined by

$f(\mathrm{x})=\frac{\mathrm{x}[\mathrm{x}]}{1+\mathrm{x}^{2}},$ where $[\mathrm{x}]$ denotes the greatest

integer $\leq \mathrm{x} .$ Then the range of $f$ is

$\left(\frac{3}{5}, \frac{4}{5}\right)$

$\left(\frac{2}{5}, \frac{3}{5}\right] \cup\left(\frac{3}{4}, \frac{4}{5}\right)$

$\left(\frac{2}{5}, \frac{4}{5}\right]$

$\left(\frac{2}{5}, \frac{1}{2}\right) \cup\left(\frac{3}{5}, \frac{4}{5}\right]$

(JEE MAIN-2020)

Solution

$f(\mathrm{x})=\left\{\begin{array}{ll}{\frac{\mathrm{x}}{\mathrm{x}^{2}+1}} & {; \quad \mathrm{x} \in(1,2)} \\ {\frac{2 \mathrm{x}}{\mathrm{x}^{2}+1}} & {; \quad \mathrm{x} \in[2,3)}\end{array}\right.$

$f(\mathrm{x})$ is decreasing function

$f(x) \in\left(\frac{2}{5}, \frac{1}{2}\right) \cup\left(\frac{3}{5}, \frac{4}{5}\right]$

Standard 12

Mathematics

Let $A= \{1, 2, 3, 4\}$ and $R : A \to A$ be the relation defined by $R = \{ (1, 1), (2, 3), (3, 4), ( 4, 2) \}$. The correct statement is

hard

(JEE MAIN-2013)

If for the function $f(x) = \frac{1}{4}{x^2} + bx + 10$ ; $f\left( {12 – x} \right) = f\left( x \right)\,\forall \,x\, \in \,R$ , then the value of $'b'$ is

normal

If $f( x + y )=f( x ) f( y )$ and $\sum \limits_{ x =1}^{\infty} f( x )=2, x , y \in N$ where $N$ is the set of all natural numbers, then the value of $\frac{f(4)}{f(2)}$ is

hard

(JEE MAIN-2020)

Product of all the solution of the equation ${x^{1 + {{\log }_{10}}x}} = 100000x$ is

normal

If $y = 3[x] + 1 = 4[x -1] -10$, then $[x + 2y]$ is equal to (where $[.]$ is $G.I.F.$)

normal

Let $f:(1,3) \rightarrow \mathrm{R}$ be a function defined by $f(\mathrm{x})=\frac{\mathrm{x}[\mathrm{x}]}{1+\mathrm{x}^{2}},$ where $[\mathrm{x}]$ denotes the greatest integer $\leq \mathrm{x} .$ Then the range of $f$ is

Solution

Similar Questions

Let $A= \{1, 2, 3, 4\}$ and $R : A \to A$ be the relation defined by $R = \{ (1, 1), (2, 3), (3, 4), ( 4, 2) \}$. The correct statement is

If for the function $f(x) = \frac{1}{4}{x^2} + bx + 10$ ; $f\left( {12 – x} \right) = f\left( x \right)\,\forall \,x\, \in \,R$ , then the value of $'b'$ is

If $f( x + y )=f( x ) f( y )$ and $\sum \limits_{ x =1}^{\infty} f( x )=2, x , y \in N$ where $N$ is the set of all natural numbers, then the value of $\frac{f(4)}{f(2)}$ is

Product of all the solution of the equation ${x^{1 + {{\log }_{10}}x}} = 100000x$ is

If $y = 3[x] + 1 = 4[x -1] -10$, then $[x + 2y]$ is equal to (where $[.]$ is $G.I.F.$)

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Let $f:(1,3) \rightarrow \mathrm{R}$ be a function defined by

$f(\mathrm{x})=\frac{\mathrm{x}[\mathrm{x}]}{1+\mathrm{x}^{2}},$ where $[\mathrm{x}]$ denotes the greatest

integer $\leq \mathrm{x} .$ Then the range of $f$ is