(D)

   A = {1, 3, 7, 9, 11}, B = {2, 4, 5, 7, 8, 10, 12}

   $f:A\to B$ is one-one such that

   $f\left(1\right)+f\left(3\right)=14$

   $\left(f\right(1),f(3\left)\right)=\left\{\right(2,12),(4,10),(12,2),(10,4\left)\right\}$

   Since $f$ is one-one so $f\left(1\right)\ne f\left(3\right)$ so we cannot take (7, 7).

   $\therefore$  So, for $f\left(1\right)$ we have 4 choices and for $f\left(3\right)$ we have 4 choices and remaining 3 elements have 5! choices for mapping to be one-one.

   Total number of ways = $4\times 5!/2$

   $=\frac{480}{2}=240$                                  [$\because$ Pair (2, 12) and (12, 2) will be considered same]

(B)

  $\mathrm{We} \mathrm{have},f\left(x\right)=\frac{x^2+2x-15}{x^2-4x+9}$

  $f\left(x\right)=\frac{(x+5)(x-3)}{x^2-4x+9}$

  $f\left(x\right) \mathrm{has} \mathrm{two} \mathrm{roots} \mathrm{as} x=-5,3$

  $\mathrm{So} f(-5)=0 \mathrm{and} f\left(3\right)=0$

  $\mathrm{So} f \mathrm{cannot} \mathrm{be} \mathrm{one}-\mathrm{one}.$

  $\mathrm{Now} \mathrm{consider}, x^2-4x+9$

  $D=16-36=-20<0$

  $\mathrm{Now}, \mathrm{let} y=\frac{x^2+2x-15}{x^2-4x+9}$

  $\Rightarrow y{x}^2-4xy+9y=x^2+2x-15$

  $\Rightarrow x^2(y-1)-2x(2y+1)+(9y+15)=0$

       $D=4{(2y+1)}^2-4(y-1)(9y+15)\ge 0$          $[\because x\in R]$

 $\Rightarrow 4(4y^2+1+4y)-(36y^2-36y+60y-60)\ge 0$

 $\Rightarrow 16y^2+4+16y-36y^2-24y+60\ge 0$

 $\Rightarrow -20y^2-8y+64\ge 0$

 $\Rightarrow 5y^2+2y-16\le 0$

 $\Rightarrow (5y-8)(y+2)\le 0$

 $\Rightarrow y\in [-2,\frac{8}{5}] \text{which is the range of function }f$

 $\mathrm{So}, f \mathrm{is} \mathrm{not} \mathrm{onto} \mathrm{if} f:R\to R.$

 $\mathrm{Here} \mathrm{in} \mathrm{given} \mathrm{question} \mathrm{co}-\mathrm{domain} \mathrm{is} \mathrm{not} \mathrm{defined}.$

(C)

   Since, $-1\le \mathrm{sin\theta }\le 1$ $\forall \mathrm{\theta }$, so we have

   $-1\le \sin$ $5x\le 1$

   $\Rightarrow -1\le -\sin 5x\le 1$

   $\Rightarrow -1+7\le 7-\sin 5x\le 1+7$

   $\Rightarrow \frac{1}{8}\le \frac{1}{7-\sin 5x}\le \frac{1}{6}$

   $\Rightarrow \text{Range }f\left(x\right)=[\frac{1}{8},\frac{1}{6}]$

(D)

   $2310=2\times 3\times 5\times 7\times 11$

   $\therefore A=\{2, 3, 5, 7, 11\}$

   $f:A\to Z\text{ is a function such that}$

   $f\left(x\right)=\left[{\log }_2\right(x^2+\left[\frac{x^3}{5}\right]\left)\right]$

   $f\left(2\right)=\left[{\log }_2\right(4+[1.6]\left)\right]=\left[{\log }_2\right(4+1\left)\right]=\left[{\log }_{2}5\right]=\left[{\log }_2\right(2^2+1\left)\right]=2$

  $f\left(3\right)=\left[{\log }_2\right(9+5\left)\right]=\left[{\log }_2\right(2^3+6\left)\right]=3$

  $f\left(5\right)=\left[{\log }_2\right(25+25\left)\right]=\left[{\log }_2\right(2^5+18\left)\right]=5$

  $f\left(7\right)=\left[{\log }_2\right(49+68\left)\right]=\left[{\log }_2\right(2^6+53\left)\right]=6$

  $f\left(11\right)=\left[{\log }_2\right(121+266\left)\right]=\left[{\log }_2\right(2^8+131\left)\right]=8$

  $\text{Range of }f=\{2,3,5,6,8\}$

  $\text{Number of one-one functions}=5!=120$

(D)

   $y=f\left(x\right)=\{\begin{array}{cc}-a,& -a\le x\le 0\\ x+a,& 0<x\le a\end{array}$

   $y=f\left|x\right|=\{\begin{array}{cc}-a,& -a\le \left|x\right|\le 0\\ \left|x\right|+a,& 0<\left|x\right|\le a\end{array}$

   but $\left|x\right|<0$ is not possible.

   $f\left|x\right|=\{\begin{array}{cc}-x+a,& -a\le x<0\\ x+a,& 0<x\le a\end{array}$

   $y=\left|f\right(x\left)\right|=\{\begin{array}{cc}a,& -a\le x\le 0\\ x+a,& 0<x\le a\end{array}$

   $g\left(x\right)=\frac{f\left|x\right|-\left|f\right(x\left)\right|}{2}=\{\begin{array}{c}\frac{-x+a-a}{2}=-\frac{x}{2} \mathrm{if} -a\le x\le 0\\ \frac{x+a-x-a}{2}=0 \mathrm{if} 0<x\le a\end{array}\begin{array}{}\end{array}$

   $g:[-a,a]\to [-a,a]$ is neither one-one nor onto as set $[0,a]$ has only one image i.e. 0.

(D)

   $\text{For domain, }4-x^2\ne 0\Rightarrow x\ne \pm 2$

   $x^2-25\ge 0; x^2\ge 25\Rightarrow x\in (-\infty ,-5]\cup [5,\infty )$

   $\text{Also, }{x}^2+2x-15>0\Rightarrow (x+5)(x-3)>0$

   $\Rightarrow x\in (-\infty ,-5)\cup (3,\infty )$ $\therefore D_f=(-\infty ,-5)\cup [5,\infty )$

   $\text{So, }\alpha =-5\text{ and }\beta =5$ $\therefore {\alpha }^2+{\beta }^3=25+125=150$

(D)

   There are two cases arise:

   Case I : Let $a$ is even $\therefore$ $f\left(a\right)=a-1$ (odd)

   $\Rightarrow f\left(f\right(a\left)\right)=f(a-1)=2a-2$ (even)

   $\Rightarrow f\left(f\right(f\left(a\right)\left)\right)=f(2a-2)=2a-2-1=2a-3$

   $\Rightarrow 21=2a-3\Rightarrow a=12$

   $\mathrm{So}, \underset{x\to a^{-}}{\lim }\{\frac{|x{|}^3}{a}-[\frac{x}{a}\left]\right\}=\underset{x\to {12}^{-}}{\lim }\{\frac{\left|x^3\right|}{12}-[\frac{x}{12}\left]\right\}$

   $=144$                                                 $(\because x<12)$

   Case II : Let $a$ is odd

   $\therefore f\left(a\right)=2a\text{ (even)}\Rightarrow f\left(f\right(a\left)\right)=f\left(2a\right)=2a-1\text{ (odd)}$

   $\Rightarrow f\left(f\right(f\left(a\right)\left)\right)=f(2a-1)=2(2a-1)$

   $\Rightarrow 21=4a-2\Rightarrow a=\frac{23}{4}\notin N$

(C)

   Given, $f:N-\left\{1\right\}\to N$

   $f\left(n\right)=$the highest prime factor of $n$

   For one-one: If $n=4,f\left(n\right)=2.$

   If $n=8,f\left(n\right)=2$  $\therefore$$f$ is not one-one.

   For onto : Range = All prime numbers

   Co-domain = Set of natural numbers

  $\Rightarrow$ Range $\ne$ Co-domain  $\therefore$$f$ is not onto.

(D)

(D)

Given, 

   $f\left(x\right)=\{\begin{array}{cc}2+2x,& -1\le x<0\\ 1-\frac{x}{3},& 0\le x\le 3\end{array}$ and $g\left(x\right)=\{\begin{array}{cc}-x,& -3\le x\le 0\\ x,& 0<x\le 1\end{array}$

   $fog(-3)=f\left(3\right)=0 ; fog(-2)=f\left(2\right)=\frac{1}{3} ; fog(-1)=f\left(1\right)=\frac{2}{3}$

   $fog\left(0\right)=f\left(0\right)=1 ; fog\left(1\right)=f\left(1\right)=\frac{2}{3}$

   $\therefore$  $\text{ Range of }fog\left(x\right)=[0,1]$