Question

Which of the following statements is(are) correct?
[Note:$\left[x\right]$and $\left\{x\right\}$ denote the greatest integer less than or equal to $x$ and the fractional part of $x$ respectively.]

• A. $f\left(x\right)=\left[\ln x\right]+\sqrt{\left\{\ln x\right\}},x>1$ is continuous at $x=e$
• B. If $f$ is a one-one mapping from set $A$ to $A$ then $f$ is onto.
• C. $f:[-1,1]\to [-1,1],f\left(x\right)=x^2\text{sgn}\left(x\right)$ is a bijective function, where $\text{sgn}\left(x\right)$ denotes signum function of $x$.
• D. If $f$ is a onto mapping from set $A$ to $A$ then $f$ is one-one.

Answer 1

Answer: (A), (C)

The correct options are
A $f\left(x\right)=\left[\ln x\right]+\sqrt{\left\{\ln x\right\}},x>1$ is continuous at $x=e$
C $f:[-1,1]\to [-1,1],f\left(x\right)=x^2\text{sgn}\left(x\right)$ is a bijective function, where $\text{sgn}\left(x\right)$ denotes signum function of $x$.

$f$ is one-one from $A\to A$ $\Rightarrow f$ is onto from $A$-$A$ & vice-versa is only true if $A$ is a finite set.
Ex-$f:[0,4]\to [0,4],f\left(x\right)=\sqrt{x}$ is one -one but not onto.

Ex-$f:R\to R,f\left(x\right)=x(x-1)(x-2)$ is onto but not one-one.

Now
$f\left(x\right)=x^2\text{sgn}\left(x\right)f\left(x\right)=\{\begin{array}{c}{x}^2,x>0\\ 0,x=0\\ -x^2,x<0\end{array}$
Clearly, $f\left(x\right)$ is bijective.

$f\left(x\right)=\left[\ln x\right]+\sqrt{\left\{\ln x\right\}}$
$f\left(e^{+}\right)=f\left(e\right)=f\left(e^{-}\right)=1$
$f$ is continuous at $x=e.$