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Is there a better method to solve this system of inequalities regarding the binomial distribution? … 0 && n >= k >= 1] // FunctionExpand Reduce[{eq1a, eq2a, 1 > 1 - p > 0, 1 > p > 0, n >= k >= 0, n \[Element] Integers, k \[Element] Integers}, k] Is there a better method to solve this system of inequalities …

My personal attempt is as follows. inequalities = {-1 <= a + b <= 1, 1 <= a - b <= 3}; RegionPlot[Evaluate@inequalities, {a, -5, 5}, {b, -5, 5}, PlotStyle -> Directive[Opacity[0.5], LightBlue], FrameLabel … -> {"a", "b"}, PlotLegends -> "Expressions"] targetFunction = b == 3 a - t; solution = Reduce[Join[inequalities, {targetFunction}], {a, b, t}] The issue is that the feasible region is not shaded for …

Let the solution set of the following system of two inequations $$x^2 - 4 x + 3 < 0 \\ x^2 - 6 x + 8 < 0$$ be a subset of the solution set of $$2 x^2 - 9 x + a < 0.$$ How can I find the range of param …

Let the following inequality $$4^xa - 2^x + 1 > 0$$ hold for any $x$. How can I find the range of parameter $a$? The following method doesn't work: Clear["Global`*"] ForAll[x, x ∈ Reals, a*4^x - 2^x + …

Let the solution set of the following inequation $$6 k - 2 x + k x^2 < 0, \quad k \neq 0$$ be an empty set. How can I find find the range of parameter $k$? The following code gives an answer: Reduce[{ …

Given the following two sets of reals: -(1/4) < x <= 0 || 0 < x <= 1/2 || x > 1/2 x <= -1 || -1 < x < 0 how can I calculate their union? The expected answer is: 0 > x > -1/4

Given that the solution set of inequation $$ax ^ 2-bx-1>0$$ is $-1/2<x<-1/3$, I want to calculate the solution set of inequation $$x ^ 2-bx-a\geq 0.$$ I have used the following code, but is there a fa …

How can I accurately find the range of $t$? Reduce seems to be unable to find it: Clear["Global`*"] Reduce[{a > b > 0, a b == 1, (b/2^a)/Log[2, a + b] == t}, t, {a, b}, Reals] (* Reduce::nsmet: This s …

In[21]:= FunctionDomain[Log[10, 3 - 4 (Sin[x])^2], x] Out[21]= C[1] \[Element] Integers && (1/3 (-\[Pi] + 6 \[Pi] C[1]) < x < 1/3 (\[Pi] + 6 \[Pi] C[1]) || 1/3 (2 \[Pi] + 6 \[Pi] C[1]) < …

Log[a, x + 1] > Log[a, x - 1] the inequality above is equation holds good under all circumstances You can see the range of x In[52]:= ForAll[x, x > 1, Log[a, x + 1] > Log[a, x - 1]] Resolve[%, Reals] …

This inequality: -1 + a E^x > 0 In[6]:= Reduce[-1 + a E^x > 0, x] During evaluation of In[6]:= Reduce::nsmet: This system cannot be solved with the methods available to Reduce. Out[6]= Reduce[-1 + …