JC1 H2 Mathematics: Inequalities

A repeated factor gives an isolated solution

Solve $\dfrac{(x+1)^2(x-1)}{x+2}\geq0$.

The critical points are $-2,-1,1$. The denominator excludes $-2$; the numerator is zero at $-1$ and 1.

Include the numerator zeros because the inequality is non-strict. Hence $x<-2$, or $x=-1$, or $x\geq1$. The sign stays negative on both sides of $-1$, but equality at the point still satisfies the question. For $F(x)>0$, the answer is only $x<-2$ or $x>1$.

Complete the square before drawing signs

For $k>1$, solve $\dfrac{kx^2+1}{x^2+(k+1)x+k}>-1$.

Both $k+1$ and the completed square expression are positive. The sign is therefore the sign of the denominator. Since $-k<-1$,

The endpoints are excluded because the original denominator is zero. Check $x=0$: the original fraction is $1/k>-1$, consistent with the right interval.

Compare two moduli by squaring

Solve $|x-6|<|9-4x|$.

Both sides are non-negative, so squaring preserves the inequality.

Hence $x<1$ or $x>3$. Equality at 1 and 3 is excluded. Check $x=2$: $4<1$ is false, as expected.

If only one side has a modulus, do not square until the other side's sign is known. For example, $|x-1|>x-3$ is automatically true when $x<3$, since the right side is negative.

A reciprocal substitution can lose a valid input

First solve $\dfrac{t+3}{t^2+t-2}>-1$, then hence solve $\dfrac{x+3x^2}{1+x-2x^2}>-1$.

The numerator is positive except at $t=-1$. The denominator is positive outside its roots, so $t<-2$ or $t>1$.

For $x\ne0$, put $t=1/x$. Dividing numerator and denominator by $x^2$ matches the first inequality.

Now check $x=0$ in the original second inequality: its left side is 0, and $0>-1$ is true. Hence the complete answer is $-\frac12<x<1$. At both endpoints the denominator is zero.

Use the cosine graph on the stated interval

It is known that $\dfrac{2t-1}{2t^2-1}\leq1$ holds for $t<-1/\sqrt2$, or $0\leq t<1/\sqrt2$, or $t\geq1$. Hence solve $\dfrac{2\cos x-1}{\cos2x}\leq1$ for $0\leq x\leq\pi$.

Use $\cos2x=2\cos^2x-1$ and replace $t$ by $\cos x$. On the stated interval cosine decreases from 1 to −1.

Thus $x=0$, or $\frac\pi4<x\leq\frac\pi2$, or $\frac{3\pi}4<x\leq\pi$. The two quarter-turn boundary values excluded above make $\cos2x=0$.

Discard a negative replacement interval

Solve $\dfrac{3x^2}{1-2|x|}>1$.

Put $t=|x|\geq0$, so $x^2=t^2$.

The sign diagram gives $t<-1$ or $1/3<t<1/2$. Since $t\geq0$, retain only the second interval.

The inner boundaries give equality; the outer boundaries make the denominator zero. All four are excluded.