@Ronan  It seems that the problem is in using  DocumentTools:-Tabulate command. If you use only  DataFrame , then everything is OK:

restart;
QQFProj := proc(q12::algebraic, q23::algebraic,
                q34::algebraic, q14::algebraic,prnt::boolean:=true)
  description "Projective quadruple quad formula and intermediate 13 and 24 quads. Useful for cyclic quadrilaterals";
  local qqf,q13,q24, sub1,sub2,sub3, R,values,DF,lens,varp;
  uses   DocumentTools;
  sub1:= (q12 + q23 + q34 + q14);
  sub2:=-4*(q12*q23*q34);
  sub3:=64*q12*q23;
  qqf:=(sub1+sub2)^2-sub3;
  q13:=(q12-q23)^2;
  q24:=varp*(q23-q34)^2;
 varp:=alpha;
  if prnt then
  
   values:=<qqf,q13,q24>;
   DF:=DataFrame(<values>, columns=[`"Values Equations"`],rows=[`#1  QQF`,`#2  Q13`,`#3  Q24`(varp)]);
   lens := [4 +8* max(op(length~(RowLabels(DF)))),4+ min(max( 10*(length~(values))),1000)];#op(length~(ColumnLabels(DF)0)
print(DF);
#   Tabulate(DF,width=add(lens),widthmode = pixels,weights = lens);
  return qqf,q13,q24
  end if;
  return qqf,q13,q24
end proc:
 q12:=1/2:q23:=9/10:q34:=25/26:q41:=9/130:#Cyclic quadrilateral
q12:=sqrt(17+a)/2:q23:=(r^2+t^2)^2/10:q34:=((a+b+c)^4/26):q41:=sqrt(17+b)/130:
Q:=QQFProj(q12,q23,q34,q41,true):

@Carl Love  The widest range for the parameter  m  is easy to find. It is determined by the slopes  m  of two tangents to the circle c2 passing through  J .

Download latex_of_laplace_nov_21_2024_Maple_2018.2.mw

@EugeneKalentev  Yes, I accidentally missed  x  when I was typing the code in Maple. The correct answer was given by vv . 

@Blanc But you omitted thу  assuming  option in your solution. Just copy the code from my answer as text and paste it into your worksheet.

@Blanc  It's strange that it doesn't work in Maple 2023. Try adding the  real  option. Upload your worksheet here.

restart;
f := (x, n) -> 2^n/(1 + n*2^n*x^2) - 2*x^2*(2^n)^2*n/(1 + n*2^n*x^2)^2;
solve(f(x,n)>0, x, parametric=full) assuming real, n::integer;
Eq:=simplify(f(x,n));
solve(Eq>0, x) assuming  real, n>0;
solve(Eq>0, x) assuming  real, n=0;
solve(Eq>0, x) assuming  real, n<0;

Also try using the  useassumptions  option:

restart;
f := (x, n) -> 2^n/(1 + n*2^n*x^2) - 2*x^2*(2^n)^2*n/(1 + n*2^n*x^2)^2;
solve(f(x,n)>0, x, parametric=full, useassumptions) n::integer;
Eq:=simplify(f(x,n));
solve(Eq>0, x, useassumptions) assuming  n>0;
solve(Eq>0, x, useassumptions) assuming  n=0;
solve(Eq>0, x, useassumptions) assuming  n<0;

Here is another solution for my example. It is based on a parametric representation of the curve  11*x^3-9*x^2*y+27*x*y^2+7*y^3-18*x^2-24*x*y+18*y^2=0. Well-known formulas are used. The results coincide with the results obtained by @Rouben Rostamian  :

restart;
P:=(x,y)->11*x^3-9*x^2*y+27*x*y^2+7*y^3-18*x^2-24*x*y+18*y^2:
solve(P(x,y)=0, {x(t),y(t)});
assign(%);
DP:=simplify(diff([x,y], t));
D2P:=simplify(diff(DP, t));
T:=solve({x=0,y=0}, explicit);
t1:=eval(t,T[1]); t2:=eval(t,T[2]);
simplify(rationalize(eval(DP[2]/DP[1], t=t1)));
simplify(rationalize(eval(DP[2]/DP[1], t=t2)));
expand(rationalize(eval((DP[1]*D2P[2]-DP[2]*D2P[1])/DP[1]^3, t=t1)));
expand(rationalize(eval((DP[1]*D2P[2]-DP[2]*D2P[1])/DP[1]^3, t=t2)));

@Rouben Rostamian  Thank you very much for your solution! 

@Rouben Rostamian  

I tried to apply your method to the following example of an implicitly defined real-valued function. The graph shows that in the neighborhood of the singular point (0,0) there is no single-valued function y(x) , but there are 2 intersecting smooth curves. What do the results (2/3 and infinity) mean geometrically?

Download implicit.mw

@AHSAN  Your problems with integration are related to the fact that for some values ​​of the parameter  k  the denominator of the expression  Pre  is 0 and the function is not defined (its graph goes to infinity). Therefore, we can first find such values ​​of the parameter  k  and use the corresponding assumption on  k  when integrating:

restart; 
Pre := -(6*(x-1))*(k-1)*x/(((x-1)*k-x)^2*(k+1)); 
sol := solve(denom(Pre) = 0, x); 
solve({sol[1] >= 0, sol[1] <= 1}, k); 
int(Pre, x = 0 .. 1)  assuming k > 0;

@AHSAN  I don't understand why you integrate the expression  Pre ? What does it have to do with your problem  "Need to plot maximum relation by considering Pre_max alog vertical axis and k along x axis or horizontal axis" ?

@mmcdara 

`(Probability(Total)=50)` = (Probability(Total <=50)+Probability(Total >=50)) - 1;

@nm  The answer is not necessarily the greatest common divisor of the numbers 45 and 63. The answer can be any common divisor of this numbers.

@Aixleft math 

restart;
solutions := solve({v1^2 = 2*g*(h-h1), (1/2)*g*t^2 = h2, v1*t+(1/2)*g*t^2 = h1}, {h, t, v1}, explicit): 
selected_solutions := select(p->op(1,eval(t,p))<>-1, [solutions])[]; 

@Raafat  Maybe you want these points to be connected by line segments:

S:=seq(P(1,2,n), n=0..10):
A:=plots:-pointplot([S][1..4], color=red, symbol=solidcircle, symbolsize=15):
B:=plot([S][1..4], linestyle=3, color=blue):
plots:-display(A,B, view=[0..110,0..16]);