Equations and Inequalities with a Parameter: Very Difficult Problems with Solutions

Solution:
On [tex]\left[0 , \pi\right][/tex] the function [tex]\cos x[/tex] is strictly decreasing and takes every value in [tex]\left[-1 , 1\right][/tex] exactly once, and [tex]3^{c}[/tex] is strictly increasing. Hence, with [tex]t=3^{\cos x}[/tex], the solutions [tex]x\in\left[0 , \pi\right][/tex] correspond one-to-one to the solutions [tex]t\in\left[\frac{1}{3} , 3\right][/tex] of the equation
[tex]t+\frac{3}{t}=a\iff t^2-at+3=0[/tex]
For [tex]t\gt 0[/tex] we have [tex]t+\frac{3}{t}\ge 2\sqrt{3}[/tex], so for [tex]a\lt 2\sqrt{3}[/tex] there are no solutions. Let [tex]a\ge 2\sqrt{3}[/tex]. By Vieta's formulas the product of the roots is [tex]3[/tex], so the roots are [tex]\tau[/tex] and [tex]\frac{3}{\tau}[/tex], where [tex]\tau\in\left(0 , \sqrt{3}\right][/tex] is the smaller one and
[tex]a=h(\tau)=\tau+\frac{3}{\tau}[/tex]
If [tex]\tau\lt \frac{1}{3}[/tex], then [tex]\frac{3}{\tau}\gt 9[/tex], so neither root lies in [tex]\left[\frac{1}{3} , 3\right][/tex]: no solutions.
If [tex]\frac{1}{3}\le \tau\lt 1[/tex], then [tex]\frac{3}{\tau}\in\left(3 , 9\right][/tex] lies outside, so there is exactly one solution.
If [tex]1\le \tau\lt \sqrt{3}[/tex], then also [tex]\frac{3}{\tau}\in\left(\sqrt{3} , 3\right][/tex], so there are two solutions.
If [tex]\tau=\sqrt{3}[/tex] (a double root, [tex]a=2\sqrt{3}[/tex]), there is exactly one solution.
The function [tex]h[/tex] is strictly decreasing on [tex]\left(0 , \sqrt{3}\right][/tex], because for [tex]0\lt t_1\lt t_2\le \sqrt{3}[/tex]
[tex]h(t_1)-h(t_2)=\left(t_2-t_1\right)\left(\frac{3}{t_1t_2}-1\right)\gt 0[/tex]
Therefore [tex]\tau\in\left[\frac{1}{3} , 1\right)[/tex] corresponds exactly to [tex]a\in\left(h(1) , h\left(\frac{1}{3}\right)\right]=\left(4 , \frac{28}{3}\right][/tex]. The answer is [tex]a\in\left\{2\sqrt{3}\right\}\cup\left(4 , \frac{28}{3}\right][/tex].

Solution:
Since [tex]x^2-2x-3=(x-3)(x+1)[/tex], the domain is [tex]x\ne 3[/tex], [tex]x\ne -1[/tex], and on it the equation becomes
[tex]\frac{|x-2|}{x+1}=a[/tex]
Case [tex]1[/tex]: [tex]x\ge 2[/tex], [tex]x\ne 3[/tex]. Then [tex]x-2=a(x+1)[/tex], i.e. [tex](1-a)x=a+2[/tex]. For [tex]a=1[/tex] this gives [tex]0=3[/tex], no root. Otherwise [tex]x=\frac{a+2}{1-a}[/tex], and
[tex]x\ge 2\iff \frac{a+2}{1-a}-2=\frac{3a}{1-a}\ge 0\iff 0\le a\lt 1[/tex]
Moreover, [tex]x=3[/tex] exactly when [tex]a+2=3(1-a)[/tex], i.e. [tex]a=\frac{1}{4}[/tex], and then this root is excluded. So Case [tex]1[/tex] gives one root for [tex]a\in\left[0 , \frac{1}{4}\right)\cup\left(\frac{1}{4} , 1\right)[/tex] and none for the other values of [tex]a[/tex].
Case [tex]2[/tex]: [tex]x\lt 2[/tex], [tex]x\ne -1[/tex]. Then [tex]2-x=a(x+1)[/tex], i.e. [tex](1+a)x=2-a[/tex]. For [tex]a=-1[/tex] this gives [tex]0=3[/tex], no root. Otherwise [tex]x=\frac{2-a}{1+a}[/tex], which is never [tex]-1[/tex] (that would mean [tex]2-a=-1-a[/tex]), and
[tex]x\lt 2\iff \frac{2-a}{1+a}-2=\frac{-3a}{1+a}\lt 0\iff a\lt -1\text{ or }a\gt 0[/tex]
Counting: for [tex]a\lt -1[/tex] there is one root (Case [tex]2[/tex]); for [tex]-1\le a\lt 0[/tex] there are none; for [tex]a=0[/tex] there is one root [tex]x=2[/tex] (from Case [tex]1[/tex] only).
For [tex]0\lt a\lt 1[/tex], [tex]a\ne \frac{1}{4}[/tex], there are two roots; for [tex]a=\frac{1}{4}[/tex] there is one root [tex]x=\frac{7}{5}[/tex], because the Case [tex]1[/tex] candidate [tex]x=3[/tex] is outside the domain; for [tex]a\ge 1[/tex] there is one root (Case [tex]2[/tex]).
Hence the solution is unique for [tex]a\in\left(-\infty , -1\right)\cup\left\{0 , \frac{1}{4}\right\}\cup\left[1 , +\infty\right)[/tex].

Solution:
Let [tex]P(x)=x^2-2mx-4(m^2+1)[/tex] and [tex]Q(x)=x^2-4x-2m(m^2+1)[/tex]. The roots of the equation are the roots of [tex]P[/tex] together with the roots of [tex]Q[/tex]. For the discriminants we have
[tex]\frac{D_P}{4}=m^2+4(m^2+1)=5m^2+4\gt 0,\qquad \frac{D_Q}{4}=4+2m(m^2+1)=2(m+1)(m^2-m+2)[/tex]
So [tex]P[/tex] always has two distinct real roots, and since [tex]m^2-m+2\gt 0[/tex], [tex]Q[/tex] has real roots if and only if [tex]m\ge -1[/tex]. For [tex]m\lt -1[/tex] the equation has only the two roots of [tex]P[/tex], so [tex]m\ge -1[/tex]. Exactly three distinct roots are possible in two ways: (a) [tex]Q[/tex] has a double root which is not a root of [tex]P[/tex]; (b) [tex]Q[/tex] has two distinct roots and exactly one of them is a root of [tex]P[/tex].
(a) [tex]D_Q=0[/tex] gives [tex]m=-1[/tex]. Then [tex]Q(x)=(x-2)^2[/tex] and [tex]P(x)=x^2+2x-8=(x-2)(x+4)[/tex], so the only roots are [tex]2[/tex] and [tex]-4[/tex]; rejected.
(b) Let [tex]m\gt -1[/tex] and let [tex]x_0[/tex] be a common root. Then [tex]P(x_0)-Q(x_0)=0[/tex], i.e.
[tex](4-2m)x_0+(2m-4)(m^2+1)=0\iff (m-2)\left(m^2+1-x_0\right)=0[/tex]
If [tex]m=2[/tex], then [tex]P(x)=Q(x)=x^2-4x-20[/tex] and the equation has only two roots; rejected. Otherwise [tex]P\ne Q[/tex], so they have at most one common root, [tex]x_0=m^2+1[/tex]. Substituting it in [tex]P[/tex]:
[tex](m^2+1)^2-2m(m^2+1)-4(m^2+1)=(m^2+1)(m^2-2m-3)=(m^2+1)(m-3)(m+1)=0[/tex]
Since [tex]m\gt -1[/tex], we get [tex]m=3[/tex]. Check:
[tex]P(x)=x^2-6x-40=(x-10)(x+4),\qquad Q(x)=x^2-4x-60=(x-10)(x+6)[/tex]
The roots are [tex]10[/tex], [tex]-4[/tex] and [tex]-6[/tex], exactly three distinct roots. The answer is [tex]m=3[/tex].

Solution:
Let [tex]f(x)=(2a-1)x^{2}+2(a+2)x+2[/tex]. If [tex]a=\frac{1}{2}[/tex], the inequality is [tex]5x+2\gt 0[/tex], i.e. [tex]x\gt -\frac{2}{5}[/tex], which contains infinitely many integers. If [tex]a\gt \frac{1}{2}[/tex], the leading coefficient is positive and [tex]f(x)\gt 0[/tex] for all sufficiently large [tex]x[/tex], again infinitely many integers. Hence [tex]a\lt \frac{1}{2}[/tex] and the parabola opens downwards. Its discriminant is
[tex]\frac{D}{4}=(a+2)^{2}-2(2a-1)=a^{2}+6\gt 0[/tex]
so there are two roots [tex]x_1\lt x_2[/tex] and the solution set is the interval [tex]\left(x_1 , x_2\right)[/tex]. Moreover
[tex]f(-1)=(2a-1)-2(a+2)+2=-3\lt 0,\qquad f(0)=2\gt 0[/tex]
hence [tex]-1\lt x_1\lt 0\lt x_2[/tex]. The interval contains no negative integers, and its integers are [tex]0, 1, 2, \ldots[/tex] up to [tex]x_2[/tex]. There are exactly three of them if and only if [tex]2[/tex] is a solution and [tex]3[/tex] is not, i.e. [tex]f(2)\gt 0[/tex] and [tex]f(3)\le 0[/tex]:
[tex]f(2)=12a+6\gt 0\iff a\gt -\frac{1}{2},\qquad f(3)=24a+5\le 0\iff a\le -\frac{5}{24}[/tex]
Both conditions agree with [tex]a\lt \frac{1}{2}[/tex]. For [tex]a=-\frac{5}{24}[/tex] the number [tex]3[/tex] is a root, so it does not satisfy the strict inequality and this endpoint is included; for [tex]a=-\frac{1}{2}[/tex] the number [tex]2[/tex] is a root and only [tex]0[/tex] and [tex]1[/tex] remain. The answer is [tex]a\in\left(-\frac{1}{2} , -\frac{5}{24}\right][/tex].

Solution:
For [tex]x=0[/tex] we have [tex]f(0)=8\gt 0[/tex] for every [tex]m[/tex]. For [tex]x=k[/tex] and [tex]x=-k[/tex], where [tex]k\in\mathbb{N}[/tex], the conditions are
[tex]k^2-2mk+8\ge 0\iff m\le \frac{k}{2}+\frac{4}{k},\qquad k^2+2mk+8\ge 0\iff m\ge -\left(\frac{k}{2}+\frac{4}{k}\right)[/tex]
So we need [tex]|m|\le h(k)=\frac{k}{2}+\frac{4}{k}[/tex] for every [tex]k\in\mathbb{N}[/tex], i.e. [tex]|m|[/tex] must not exceed the smallest value of [tex]h(k)[/tex] over the natural numbers. We have [tex]h(3)=\frac{3}{2}+\frac{4}{3}=\frac{17}{6}[/tex] and for every natural [tex]k[/tex]
[tex]h(k)-\frac{17}{6}=\frac{3k^2-17k+24}{6k}=\frac{(3k-8)(k-3)}{6k}\ge 0[/tex]
because no integer lies strictly between [tex]\frac{8}{3}[/tex] and [tex]3[/tex]. Hence the smallest value is [tex]\frac{17}{6}[/tex] (at [tex]k=3[/tex]) and the condition is [tex]-\frac{17}{6}\le m\le \frac{17}{6}[/tex].
This set is wider than [tex]\left[-2\sqrt{2} , 2\sqrt{2}\right][/tex], which comes from [tex]D\le 0[/tex] (the inequality for all real [tex]x[/tex]): for example, for [tex]m=\frac{17}{6}[/tex] the roots of [tex]f[/tex] are [tex]\frac{8}{3}[/tex] and [tex]3[/tex], and [tex]f[/tex] is negative only strictly between them, where there is no integer. The answer is [tex]m\in\left[-\frac{17}{6} , \frac{17}{6}\right][/tex].

Solution:
A real number [tex]u[/tex] is a root of the equation for some real [tex]a[/tex] exactly when the equation, viewed as a quadratic in [tex]a[/tex],
[tex]a^2+6(1-u)a+3u^2-u-3=0[/tex]
has a real solution [tex]a[/tex], i.e. when its discriminant is non-negative:
[tex]\frac{D}{4}=9(1-u)^2-\left(3u^2-u-3\right)=6u^2-17u+12=(2u-3)(3u-4)\ge 0[/tex]
So the possible real roots are exactly the numbers [tex]u\in\left(-\infty , \frac{4}{3}\right]\cup\left[\frac{3}{2} , +\infty\right)[/tex]. Since [tex]\frac{4}{3}\lt \sqrt{2}\lt \frac{3}{2}[/tex], the possible values closest to [tex]\sqrt{2}[/tex] are [tex]\frac{4}{3}[/tex] and [tex]\frac{3}{2}[/tex]. We compare the distances:
[tex]\sqrt{2}-\frac{4}{3}\lt \frac{3}{2}-\sqrt{2}\iff 2\sqrt{2}\lt \frac{17}{6}\iff 12\sqrt{2}\lt 17\iff 288\lt 289[/tex]
which is true. So the minimum of [tex]\left|u-\sqrt{2}\right|[/tex] is attained for [tex]u=\frac{4}{3}[/tex]. For it the discriminant in [tex]a[/tex] is zero, so [tex]a=-3(1-u)=3u-3=1[/tex]. Check: for [tex]a=1[/tex] the equation is [tex]3x^2-7x+4=0[/tex] with roots [tex]1[/tex] and [tex]\frac{4}{3}[/tex]. The answer is [tex]a=1[/tex].

Solution:
The left side is defined only for [tex]a\ne 0[/tex] and [tex]\sin x\ne 0[/tex]. Put [tex]u=\left|a\sin x\right|\gt 0[/tex]. Since [tex]a^{2}\sin ^{2}x+9\gt 0[/tex], the left side equals
[tex]\frac{u^{2}+9}{u}=u+\frac{9}{u}=6+\frac{(u-3)^{2}}{u}\ge 6[/tex]
with equality only for [tex]u=3[/tex]. For the right side
[tex]-25\cos^{2}x+30\cos x-3=6-(5\cos x-3)^{2}\le 6[/tex]
with equality only for [tex]\cos x=\frac{3}{5}[/tex]. Therefore the inequality can hold only if both sides are equal to [tex]6[/tex], i.e. [tex]\cos x=\frac{3}{5}[/tex] and [tex]\left|a\sin x\right|=3[/tex]. From [tex]\cos x=\frac{3}{5}[/tex] we get [tex]\left|\sin x\right|=\frac{4}{5}[/tex], so [tex]\frac{4}{5}|a|=3[/tex] and [tex]|a|=\frac{15}{4}[/tex].
Conversely, for [tex]a=\pm\frac{15}{4}[/tex] the number [tex]x=\arccos\frac{3}{5}[/tex] gives [tex]\left|a\sin x\right|=3[/tex] and both sides equal [tex]6[/tex], so the inequality has a solution. The answer is [tex]a=\pm\frac{15}{4}[/tex].

Solution:
For every [tex]x[/tex] we have [tex]1+\sin ^2(ax)\ge 1[/tex] and [tex]\cos x\le 1[/tex]. Hence equality holds exactly when [tex]\cos x=1[/tex] and [tex]\sin (ax)=0[/tex]. The first condition gives [tex]x=2k\pi[/tex], [tex]k\in\mathbb{Z}[/tex], and then the second one becomes
[tex]\sin (2k\pi a)=0\iff 2ka\in\mathbb{Z}[/tex]
For [tex]k=0[/tex] this is always true, so [tex]x=0[/tex] is a solution for every [tex]a[/tex]. The solution is unique exactly when [tex]2ka\notin\mathbb{Z}[/tex] for every integer [tex]k\ne 0[/tex].
If [tex]a[/tex] is rational, [tex]a=\frac{p}{q}[/tex] with [tex]p\in\mathbb{Z}[/tex], [tex]q\in\mathbb{N}[/tex], then for [tex]k=q[/tex] we get [tex]2qa=2p\in\mathbb{Z}[/tex], so [tex]x=2q\pi\ne 0[/tex] is a second solution. If [tex]a[/tex] is irrational, then [tex]2ka[/tex] is irrational for every integer [tex]k\ne 0[/tex], so [tex]x=0[/tex] is the only solution.
Therefore the equation has a unique solution exactly when [tex]a[/tex] is irrational, i.e. [tex]a\in\mathbb{R}\setminus\mathbb{Q}[/tex].

Solution:
Let [tex]f(x)=8x^2+bx+c[/tex]. Then [tex]f(0)=c[/tex], [tex]f\left(\frac{1}{2}\right)=2+\frac{b}{2}+c[/tex], [tex]f(1)=8+b+c[/tex], and therefore, for all [tex]b[/tex] and [tex]c[/tex],
[tex]f(0)-2f\left(\frac{1}{2}\right)+f(1)=c-4-b-2c+8+b+c=4[/tex]
Suppose [tex]\left|f(x)\right|\le 1[/tex] on [tex]\left[0 , 1\right][/tex]. Then [tex]f(0)\le 1[/tex], [tex]-2f\left(\frac{1}{2}\right)\le 2[/tex] and [tex]f(1)\le 1[/tex], so the left side of the identity is at most [tex]4[/tex]. Since it equals [tex]4[/tex], all three estimates are equalities: [tex]f(0)=1[/tex], [tex]f\left(\frac{1}{2}\right)=-1[/tex], [tex]f(1)=1[/tex]. Hence [tex]c=1[/tex] and [tex]b=-8[/tex] (and indeed [tex]f\left(\frac{1}{2}\right)=2-4+1=-1[/tex]), so no other pair is possible.
Conversely, for [tex]b=-8[/tex], [tex]c=1[/tex]:
[tex]f(x)=8x^2-8x+1=2(2x-1)^2-1[/tex]
For [tex]x\in\left[0 , 1\right][/tex] we have [tex](2x-1)^2\in\left[0 , 1\right][/tex], so [tex]f(x)\in\left[-1 , 1\right][/tex], i.e. [tex]\left|f(x)\right|\le 1[/tex]. The only pair is [tex]b=-8[/tex], [tex]c=1[/tex].

Solution:
Let [tex]A=4x^2-(8a-1)x+5a^2[/tex] and [tex]B=x^2-2(a+1)x-a^2[/tex]; the domain is [tex]A\gt 0[/tex], [tex]B\gt 0[/tex]. Note that
[tex]A-B=3x^2-(6a-3)x+6a^2=3\left(x^2+(1-2a)x+2a^2\right)[/tex]
so the equation can be written as [tex]\frac{A-B}{3}+\lg A=\lg B[/tex], i.e.
[tex]\lg A+\frac{A}{3}=\lg B+\frac{B}{3}[/tex]
The function [tex]g(t)=\lg t+\frac{t}{3}[/tex] is strictly increasing on [tex]\left(0 , +\infty\right)[/tex], so [tex]g(A)=g(B)\iff A=B[/tex]. Hence the equation is equivalent to [tex]Q(x)=x^2+(1-2a)x+2a^2=0[/tex] together with [tex]B\gt 0[/tex] (then [tex]A=B\gt 0[/tex] automatically). For a root of [tex]Q[/tex] we have [tex]x^2=(2a-1)x-2a^2[/tex], so
[tex]B=(2a-1)x-2a^2-2(a+1)x-a^2=-3x-3a^2,\qquad B\gt 0\iff x\lt -a^2[/tex]
So we need exactly one root of [tex]Q[/tex] that is less than [tex]-a^2[/tex]. For the discriminant [tex]D[/tex] of [tex]Q[/tex], the abscissa [tex]x_v[/tex] of the vertex of its parabola and the value at [tex]-a^2[/tex] we have
[tex]D=(1-2a)^2-8a^2=1-4a-4a^2,\qquad x_v=a-\frac{1}{2},\qquad Q\left(-a^2\right)=a^4+2a^3+a^2=a^2(a+1)^2\ge 0[/tex]
If [tex]D\lt 0[/tex], there are no solutions. If [tex]D=0[/tex], i.e. [tex]a=\frac{-1\pm\sqrt{2}}{2}[/tex], then [tex]4a^2+4a=1[/tex] and [tex]x_v+a^2=\frac{2a^2+2a-1}{2}=-\frac{1}{4}\lt 0[/tex], so the double root satisfies [tex]x\lt -a^2[/tex]: exactly one solution.
If [tex]D\gt 0[/tex], i.e. [tex]\frac{-1-\sqrt{2}}{2}\lt a\lt \frac{-1+\sqrt{2}}{2}[/tex], then [tex]4a^2+4a\lt 1[/tex], so [tex]x_v+a^2=\frac{2a^2+2a-1}{2}\lt 0[/tex] and the vertex lies to the left of [tex]-a^2[/tex]. When [tex]Q\left(-a^2\right)\gt 0[/tex], both roots are less than [tex]-a^2[/tex]: two solutions. When [tex]Q\left(-a^2\right)=0[/tex], i.e. [tex]a=0[/tex] or [tex]a=-1[/tex] (both in this interval), the number [tex]-a^2[/tex] is itself a root (rejected, since then [tex]B=0[/tex]) and the other root is smaller: exactly one solution ([tex]x=-1[/tex] for [tex]a=0[/tex], [tex]x=-2[/tex] for [tex]a=-1[/tex]).
The answer is [tex]a\in\left\{\frac{-1-\sqrt{2}}{2} , -1 , 0 , \frac{-1+\sqrt{2}}{2}\right\}[/tex].

Solution:
Each solution is determined by its [tex]x[/tex], because [tex]y=\left|x^2-kx\right|-k[/tex]. Substituting this [tex]y[/tex] into the first equation, we see that the number of solutions equals the number of real roots of
[tex]x^2+\left(\left|x^2-kx\right|-k\right)^2=k^2\iff x^2\left(1+(x-k)^2\right)=2k\left|x\right|\left|x-k\right|\qquad (1)[/tex]
The number [tex]x=0[/tex] is always a root (it gives the solution [tex](0 , -k)[/tex]).
Case [tex]k\le 0[/tex]. The right side of [tex](1)[/tex] is non-positive, while the left side is non-negative and vanishes only for [tex]x=0[/tex]. So [tex]x=0[/tex] is the only root and the system has exactly one solution.
Case [tex]0\lt k\lt 6[/tex]. For [tex]x\ne 0[/tex] we divide [tex](1)[/tex] by [tex]\left|x\right|[/tex]:
[tex]\left|x\right|\left(1+(x-k)^2\right)=2k\left|x-k\right|\qquad (2)[/tex]
The number [tex]x=k[/tex] is not a root, since then the left side is [tex]k\ne 0[/tex] and the right side is [tex]0[/tex].
(i) For [tex]x\gt k[/tex] put [tex]u=x-k\gt 0[/tex]. The difference between the left and the right side of [tex](2)[/tex] is
[tex](u+k)\left(1+u^2\right)-2ku=u^3+u+k(u-1)^2\gt 0[/tex]
so there are no roots with [tex]x\gt k[/tex].
(ii) For [tex]0\lt x\lt k[/tex], equation [tex](2)[/tex] reads [tex]h(x)=x\left(1+(x-k)^2\right)-2k(k-x)=0[/tex]. Here [tex]h(0)=-2k^2\lt 0[/tex], [tex]h(k)=k\gt 0[/tex] and
[tex]h'(x)=1+2k+(x-k)(3x-k)\ge 1+2k-\frac{k^2}{3}=\frac{12-(k-3)^2}{3}\gt 0[/tex]
because the quadratic [tex](x-k)(3x-k)=3x^2-4kx+k^2[/tex] has minimum value [tex]-\frac{k^2}{3}[/tex] (at [tex]x=\frac{2k}{3}[/tex]), and [tex]0\lt k\lt 6[/tex] gives [tex](k-3)^2\lt 9[/tex]. Thus [tex]h[/tex] is continuous and strictly increasing on [tex]\left[0 , k\right][/tex], so it has exactly one root in [tex]\left(0 , k\right)[/tex].
(iii) For [tex]x\lt 0[/tex], equation [tex](2)[/tex] reads [tex]g(x)=-x\left(1+(x-k)^2\right)-2k(k-x)=0[/tex]. For [tex]x\lt 0[/tex] we have [tex](x-k)(3x-k)=3x^2-4kx+k^2\gt k^2[/tex], hence
[tex]g'(x)=2k-1-(x-k)(3x-k)\lt 2k-1-k^2=-(k-1)^2\le 0[/tex]
so [tex]g[/tex] is strictly decreasing on [tex]\left(-\infty , 0\right][/tex]. Since [tex]g(x)\to +\infty[/tex] as [tex]x\to -\infty[/tex] (the leading term is [tex]-x^3[/tex]) and [tex]g(0)=-2k^2\lt 0[/tex], [tex]g[/tex] has exactly one root in [tex]\left(-\infty , 0\right)[/tex].
Hence for [tex]0\lt k\lt 6[/tex] the system has exactly [tex]3[/tex] solutions. So for every [tex]k\lt 6[/tex] the number of solutions is [tex]1[/tex] or [tex]3[/tex], which is odd.