If $f(x) = \log_{e}\left(\frac{1-x}{1+x}\right)$,$|x| < 1$,then $f\left(\frac{2x}{1+x^2}\right)$ is equal to

  • A
    $2f(x^2)$
  • B
    $(f(x))^2$
  • C
    $-2f(x)$
  • D
    $2f(x)$

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Similar Questions

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 $f: R \rightarrow R$ and $g: R \rightarrow R$ be functions defined by
$f(x)=\begin{cases} x|x| \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0 \end{cases}$ and $g(x)=\begin{cases} 1-2x, & 0 \leq x \leq \frac{1}{2} \\ 0, & \text{otherwise} \end{cases}$
Let $a, b, c, d \in R$. Define the function $h: R \rightarrow R$ by
$h(x)=a f(x)+b\left(g(x)+g\left(\frac{1}{2}-x\right)\right)+c(x-g(x))+d g(x), x \in R$
Match each entry in $List-I$ to the correct entry in $List-II$.
$List-I$$List-II$
$(P)$ If $a=0, b=1, c=0$ and $d=0$,then$(1)$ $h$ is one-one
$(Q)$ If $a=1, b=0, c=0$ and $d=0$,then$(2)$ $h$ is onto
$(R)$ If $a=0, b=0, c=1$ and $d=0$,then$(3)$ $h$ is differentiable on $R$
$(S)$ If $a=0, b=0, c=0$ and $d=1$,then$(4)$ the range of $h$ is $[0,1]$
$(5)$ the range of $h$ is $\{0,1\}$

The correct option is

Let $A = \{2, 3, 4, 5, 6\}$. Let $R$ be a relation on the set $A \times A$ defined by $(x, y) R (z, w)$ if and only if $x$ divides $z$ and $y \le w$. Then the number of elements in $R$ is . . . . . . .

Which of the following statements is false?

The function $f(x) = \text{sgn}(x) \cdot \sin x$ is

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