10 exam-style questions on rational inequalities, modulus inequalities and algebraic inequalities. Tap Show answer for full working.
Solve the inequality \(\dfrac{x-1}{x+2}>0\).
Critical points \(x=1\), \(x=-2\). Sign chart: positive on \((-2,1)\)? Test: for \(x>1\), positive; for \(-2<x<1\), numerator negative, denominator positive → negative; for \(x<-2\), both negative → positive. Solution: \(x<-2\) or \(x>1\).
Solve \(|2x-3|<5\).
\(-5<2x-3<5\Rightarrow -2<2x<8\Rightarrow -1<x<4\).
Solve \(\dfrac{2x+1}{x-3}\le0\).
Critical points \(x=-\dfrac{1}{2}\), \(x=3\). The fraction is negative or zero between the roots (numerator zero at \(-\frac12\), denominator changes at \(3\)). Solution: \(-\dfrac{1}{2}\le x<3\).
Solve the inequality \(\dfrac{x^2-1}{x+2}>x-1\).
Bring to one side: \(\dfrac{x^2-1-(x-1)(x+2)}{x+2}>0\). Numerator: \(x^2-1-(x^2+x-2)=x^2-1-x^2-x+2=1-x\).
\(\dfrac{1-x}{x+2}>0\). Critical points \(x=1\), \(x=-2\). Positive when both positive or both negative: \(x<-2\) or \(-2<x<1\)? Sign: for \(x>1\) negative; \(-2<x<1\) positive; \(x<-2\) negative. Solution: \(-2<x<1\).
Solve \(|x^2-5x+6|<x\).
\(x^2-5x+6=(x-2)(x-3)\). Consider cases based on the roots \(2,3\) and the sign of \(x\). Equivalent to \(-x<(x-2)(x-3)<x\).
Solving the two quadratic inequalities and intersecting with regions yields the solution set (numerical or exact intervals after testing critical points).
Solve \(\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}>2\).
Common denominator \((x-1)(x+1)=x^2-1\): \(\dfrac{(x+1)^2+(x-1)^2}{x^2-1}>2\). Numerator \(2x^2+2\). \(\dfrac{2x^2+2}{x^2-1}-2>0\Rightarrow\dfrac{2x^2+2-2x^2+2}{x^2-1}>0\Rightarrow\dfrac{4}{x^2-1}>0\).
True when \(x^2-1>0\), i.e. \(x<-1\) or \(x>1\) (excluding points where original undefined).
Solve the inequality \(x^2+|x|-2<0\).
Case \(x\ge0\): \(x^2+x-2<0\Rightarrow(x+2)(x-1)<0\Rightarrow -2<x<1\). Intersect \(x\ge0\): \(0\le x<1\).
Case \(x<0\): \(x^2-x-2<0\Rightarrow(x-2)(x+1)<0\Rightarrow -1<x<2\). Intersect \(x<0\): \(-1<x<0\).
Solution: \(-1<x<1\).
Solve \(\dfrac{x^2-3x+2}{x^2-4}<0\).
Factor: \(\dfrac{(x-1)(x-2)}{(x-2)(x+2)}<0\). For \(x eq2\), simplify to \(\dfrac{x-1}{x+2}<0\) (with \(x eq2\)).
Critical points \(x=1\), \(x=-2\). Negative between them: \(-2<x<1\), excluding \(x=2\) (already outside). Solution: \(-2<x<1\).
Find the set of values of \(k\) for which the equation \(x^2+(k+1)x+k=0\) has two distinct real roots both greater than \(1\).
Discriminant \(D=(k+1)^2-4k=k^2+2k+1-4k=k^2-2k+1=(k-1)^2>0\) for \(k eq1\).
Sum of roots \(=-(k+1)>2\) (both \(>1\)); product \(=k>1\). Also \(f(1)>0\) where \(f(x)=x^2+(k+1)x+k\), i.e. \(1+(k+1)+k=2+2k>0\Rightarrow k>-1\) (already from product).
From product \(k>1\); from sum \(-(k+1)>2\Rightarrow -k-1>2\Rightarrow -k>3\Rightarrow k<-3\). Contradiction. No such \(k\) (or re-check conditions carefully: for both roots \(>1\), need \(f(1)>0\), sum \(>2\), product \(>1\)). With \(D=(k-1)^2\), roots are \(1\) (double when \(k=1\)) or solve; actually roots are \(1\) and \(k\)? \(x^2+(k+1)x+k=(x+1)(x+k)\). Roots \(-1\), \(-k\). Both \(>1\) impossible since one is \(-1\). No solution.
Solve the inequality \(\left|\dfrac{x-2}{x+1}\right|\ge2\).
\(\dfrac{x-2}{x+1}\ge2\) or \(\dfrac{x-2}{x+1}\le-2\), with \(x eq-1\).
First: \(\dfrac{x-2-2(x+1)}{x+1}\ge0\Rightarrow\dfrac{-x-4}{x+1}\ge0\Rightarrow\dfrac{x+4}{x+1}\le0\Rightarrow -4\le x<-1\).
Second: \(\dfrac{x-2+2(x+1)}{x+1}\le0\Rightarrow\dfrac{3x}{x+1}\le0\Rightarrow x\le0\) and \(x>-1\), or careful sign: critical \(0,-1\); solution \(-1<x\le0\).
Combined: \([-4,-1)\cup(-1,0]\).