Question

Which of the following statements is/are CORRECT?
(Here, $\left[x\right]$ and $\left\{x\right\}$ denote greatest integer less than or equal to $x$ and 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 $\underset{x\to -2}{lim}\left(\frac{3x^2+ax+a+1}{x^2+x-2}\right)$ exists and equals $l,$ then $a+\frac{1}{l}=10$
• C. $f:[-1,1]\to [-1,1]$ defined by $f\left(x\right)=x^2\text{sgn}\left(x\right)$ is a bijective function, where $\text{sgn}\left(x\right)$ denotes the signum function of $x$
• D. If $f$ is continuous on $[-1,1],$ $f(-1)=4$ and $f\left(1\right)=3,$ then there exists a number $r$ such that $|r|<1$ and $f\left(r\right)=\pi$

Answer 1

Answer: (A), (B), (C), (D)

If $f$ is continuous on $[-1,1],$ $f(-1)=4$ and $f\left(1\right)=3,$ then there exists a number $r$ such that $|r|<1$ and $f\left(r\right)=\pi$

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

$\underset{x\to -2}{lim}\left(\frac{3x^2+ax+a+1}{x^2+x-2}\right)\left(\frac{0}{0}\right)\text{form}$
For limit to exist,
$3(-2{)}^2+a(-2)+a+1=0$
$\Rightarrow a=13$
Now, $\underset{x\to -2}{lim}\left(\frac{3x^2+13x+14}{x^2+x-2}\right)$
$=\underset{x\to -2}{lim}\frac{(3x+7)(x+2)}{(x+2)(x-1)}=-\frac{1}{3}$
Here $a=13$ and $l=-\frac{1}{3}$
So, $a+\frac{1}{l}=13+\left(\frac{1}{-1/3}\right)=13-3=10$

$f\left(x\right)=x^2\text{sgn}\left(x\right)=\{\begin{array}{cc}-x^2,& x<0\\ 0,& x=0\\ x^2,& x>0\end{array}$


Clearly, we can conclude from the graph that $f$ is bijective.

As, $f$ is continuous on $[-1,1]$ and
$f(-1)=4,$ $f\left(1\right)=3$
by the intermediate value theorem, there exists a number $r$ such that $|r|<1$ and $f\left(r\right)=\pi$