Column-1	Column-2
A) let F:$R\to R$ be given by $f\left(x\right)=\left\{\begin{array}{l}5x\\ x^2+6\end{array}\right.$ if $x\in Q$ if $x\in R-Q$ then $f$  is continuous at x=p,and  x=q  then $|p-q|$ is	p)1
B) The function $\text{f}\left(\text{x}\right)={\left[\text{x}\right]}^2-\left[x^2\right]$(where[y] is the largest integer$\le y$)is discontinuous at all integers except K then 4K is	q)2
C)The number of points of discontinuity of h(x)= $\frac{1}{1-e^{-\left(\frac{x-1}{x-2}\right)}}$is	r)3
D)   $f\left(x\right)=x\left[3-\log \left(\frac{\sin x}{x}\right)\right]-2$  to be Continuous at x=0, the f(0)=-k then $\frac{6}{k}$ is	s)4

 

a) F is continuous when $5x=x^2+6$

$\Rightarrow x^2-5x+6=0$

$\Rightarrow x=2,3$

b) Clearly $f\left(x\right)=0$ for each integral value of x

            Also, if $0<x<1$ then $0<x^2<1$

            $\left[x\right]=0$ and $\left[x^2\right]=0$

            $\therefore f\left(x\right)=0$for $0<x<1$

            Again if $1\le x<\sqrt{2}$then $1\le x^2<2$

            $\left[x\right]=1,\left[x^2\right]=1$

            $\therefore f\left(x\right)=0$ is continuous at $x=1$

            However at points x other than integer and not lying between 0 and $\sqrt{2}$, $f\left(x\right)\ne 0$

            c) here $h\left(x\right)=f\left(g\left(x\right)\right)=\frac{x-1}{x-2}$and

            $f\left(x\right)=\frac{1}{1-e^{-x}}$

            Therefore, $f\left(g\left(x\right)\right)$ is discontinuous when $1-e^{-x}=0$ or $x=0$ and $g\left(x\right)$ is discontinuous

            d) $f\left(0\right)=\underset{x\to 0}{\lim }x[3-\log \left(\frac{\sin x}{x}\right)]-2$

            $=0-2=-2$