Equations and Inequalities with a Parameter: Difficult Problems with Solutions

Solution:
Whenever the square root exists, [tex]1+\sqrt{x^{2}+mx+5}\ge 1\gt 0[/tex], so the only domain condition is [tex]x^{2}+mx+5\ge 0[/tex].
Case [tex]0\lt m\lt 1[/tex]. The function [tex]\log_{m}[/tex] is decreasing, so the inequality is equivalent to
[tex]2\le 1+\sqrt{x^{2}+mx+5}\iff x^{2}+mx+5\ge 1\iff x^{2}+mx+4\ge 0[/tex]
Its discriminant is [tex]m^{2}-16\lt 0[/tex], so it holds for every real [tex]x[/tex] (and then the domain condition holds too): infinitely many solutions, so this case is rejected.
Case [tex]m\gt 1[/tex]. The function [tex]\log_{m}[/tex] is increasing, so the inequality is equivalent to
[tex]\sqrt{x^{2}+mx+5}\le 1\iff 0\le x^{2}+mx+5\le 1\iff -1\le x^{2}+mx+4\le 0[/tex]
If [tex]1\lt m\lt 4[/tex], the discriminant [tex]m^{2}-16[/tex] is negative, so [tex]x^{2}+mx+4\gt 0[/tex] for every [tex]x[/tex]: no solutions.
If [tex]m=4[/tex], we need [tex](x+2)^{2}\le 0[/tex], i.e. [tex]x=-2[/tex], and indeed [tex]x^{2}+4x+5=1\ge 0[/tex] there: exactly one solution.
If [tex]m\gt 4[/tex], the trinomial [tex]x^{2}+mx+4[/tex] has two distinct roots; it equals [tex]0[/tex] at the smaller root and is negative just to the right of it, so by continuity it takes values in [tex]\left[-1 , 0\right][/tex] on a whole interval: infinitely many solutions.
Hence [tex]m=4[/tex].

Solution:
For [tex]x\ge 0[/tex] we have [tex]2|x|-2x=0[/tex], so the equation becomes [tex]0=a-x[/tex], i.e. [tex]x=a[/tex]. Hence there is a positive root if and only if [tex]a\gt 0[/tex], and then it is the only non-negative root. So [tex]a\gt 0[/tex] is necessary.
For [tex]x\lt 0[/tex] we have [tex]2|x|-2x=-4x[/tex] and the equation is [tex]\sqrt{-4x}=a-x[/tex]. Since [tex]a\gt 0[/tex] and [tex]x\lt 0[/tex], the right-hand side is positive, so squaring gives an equivalent equation:
[tex]-4x=(a-x)^2\iff x^2+(4-2a)x+a^2=0[/tex]
It must have two distinct negative roots. The discriminant condition is
[tex]\frac{D}{4}=(2-a)^2-a^2=4-4a\gt 0\iff a\lt 1[/tex]
By Vieta's formulas [tex]x_1+x_2=2a-4\lt 0[/tex] and [tex]x_1x_2=a^2\gt 0[/tex] for [tex]0\lt a\lt 1[/tex], so both roots are negative. For [tex]a=1[/tex] there is only the double root [tex]x=-1[/tex], i.e. just one negative root. Therefore [tex]a\in\left(0 , 1\right)[/tex].

Solution:
Put [tex]t=2^{x}\gt 0[/tex]. Since [tex]x\mapsto 2^{x}[/tex] maps [tex]\mathbb{R}[/tex] one-to-one onto [tex]\left(0 , +\infty\right)[/tex], the number of roots [tex]x[/tex] equals the number of distinct positive roots of
[tex]f(t)=t^{2}-kt-k+3=0[/tex]
Its discriminant and Vieta's formulas give
[tex]D=k^{2}+4k-12=(k+6)(k-2),\qquad t_{1}+t_{2}=k,\qquad t_{1}t_{2}=3-k[/tex]
If [tex]k\gt 3[/tex], then [tex]t_{1}t_{2}\lt 0[/tex], the roots have opposite signs and exactly one is positive: one root [tex]x[/tex].
If [tex]k=3[/tex], the equation is [tex]t^{2}-3t=0[/tex] with roots [tex]0[/tex] (rejected, since [tex]t\gt 0[/tex]) and [tex]3[/tex]: one root [tex]x=\log_{2}3[/tex].
If [tex]D=0[/tex], the double root is [tex]t=\frac{k}{2}[/tex]: for [tex]k=2[/tex] it is [tex]t=1\gt 0[/tex], giving exactly one root [tex]x=0[/tex]; for [tex]k=-6[/tex] it is [tex]t=-3\lt 0[/tex], giving no roots.
If [tex]2\lt k\lt 3[/tex], then [tex]D\gt 0[/tex], [tex]t_{1}t_{2}\gt 0[/tex] and [tex]t_{1}+t_{2}\gt 0[/tex], so there are two distinct positive roots [tex]t[/tex] and hence two roots [tex]x[/tex].
If [tex]k\lt -6[/tex], then [tex]t_{1}t_{2}\gt 0[/tex] and [tex]t_{1}+t_{2}=k\lt 0[/tex], so both roots [tex]t[/tex] are negative; if [tex]-6\lt k\lt 2[/tex], then [tex]D\lt 0[/tex] and there are no real roots. In both cases there are no roots [tex]x[/tex].
Hence [tex]k\in\left\{2\right\}\cup\left[3 , +\infty\right)[/tex].

Solution:
Put [tex]t=2^x\gt 0[/tex]. Then [tex]2^{-x}=\frac{1}{t}[/tex], and multiplying by [tex]t[/tex] gives
[tex]\left(a^2+a-6\right)t^2-2at+1=0[/tex]
Since [tex]x=\log_2 t[/tex], different roots [tex]t[/tex] give different roots [tex]x[/tex], and [tex]x\gt 0\iff t\gt 1[/tex], [tex]x\lt 0\iff 0\lt t\lt 1[/tex]. So the equation in [tex]t[/tex] must have one root in [tex]\left(0 , 1\right)[/tex] and one root in [tex]\left(1 , +\infty\right)[/tex].
If [tex]a^2+a-6=0[/tex], i.e. [tex]a=2[/tex] or [tex]a=-3[/tex], the equation is linear and has at most one root, so these values are rejected.
Let [tex]c=a^2+a-6\ne 0[/tex] and [tex]f(t)=ct^2-2at+1[/tex]. The number [tex]1[/tex] lies strictly between two real roots exactly when [tex]c\,f(1)\lt 0[/tex], and both roots are then positive exactly when their product [tex]t_1t_2=\frac{1}{c}[/tex] is positive. So the conditions are
[tex]c\gt 0,\qquad f(1)\lt 0[/tex]
First,
[tex]c\gt 0\iff (a+3)(a-2)\gt 0\iff a\lt -3\text{ or }a\gt 2[/tex]
Second,
[tex]f(1)=a^2+a-6-2a+1=a^2-a-5\lt 0\iff \frac{1-\sqrt{21}}{2}\lt a\lt \frac{1+\sqrt{21}}{2}[/tex]
Since [tex]\frac{1-\sqrt{21}}{2}\gt -3[/tex] and [tex]\frac{1+\sqrt{21}}{2}\gt 2[/tex] (because [tex]3\lt\sqrt{21}\lt 7[/tex]), the intersection is [tex]2\lt a\lt\frac{1+\sqrt{21}}{2}[/tex], i.e. [tex]a\in\left(2 , \frac{1+\sqrt{21}}{2}\right)[/tex].

Solution:
Put [tex]u=3^{x^2}[/tex]. Since [tex]x^2\ge 0[/tex], we have [tex]u\ge 1[/tex]. The value [tex]u=1[/tex] corresponds to the single solution [tex]x=0[/tex], every value [tex]u\gt 1[/tex] corresponds to exactly two solutions [tex]x=\pm\sqrt{\log_3 u}[/tex], and values [tex]u\lt 1[/tex] give no solutions. The equation becomes
[tex]u^2-(2a+1)u+a^2-3=0[/tex]
It has at most two roots. Each root [tex]u\gt 1[/tex] contributes two solutions and the root [tex]u=1[/tex] contributes one, so three solutions are possible only if [tex]u=1[/tex] is a root and the other root is greater than [tex]1[/tex]. The value [tex]u=1[/tex] is a root when
[tex]1-(2a+1)+a^2-3=a^2-2a-3=0\iff a=3\text{ or }a=-1[/tex]
For [tex]a=3[/tex]: [tex]u^2-7u+6=0[/tex], so [tex]u=1[/tex] or [tex]u=6\gt 1[/tex], which gives [tex]x=0[/tex] and [tex]x=\pm\sqrt{\log_3 6}[/tex], exactly three solutions.
For [tex]a=-1[/tex]: [tex]u^2+u-2=0[/tex], so [tex]u=1[/tex] or [tex]u=-2\lt 1[/tex], which gives only [tex]x=0[/tex], one solution; rejected. Hence [tex]a=3[/tex].

Solution:
Domain: [tex]kx\gt 0[/tex], [tex]x+1\gt 0[/tex] and [tex]\lg(x+1)\ne 0[/tex], i.e. [tex]x\gt -1[/tex], [tex]x\ne 0[/tex], [tex]kx\gt 0[/tex]; in particular [tex]k\ne 0[/tex]. On the domain the equation is equivalent to
[tex]\lg(kx)=2\lg(x+1)=\lg\left((x+1)^2\right)\iff kx=(x+1)^2\iff x^2+(2-k)x+1=0[/tex]
For [tex]k\ne 0[/tex], [tex]x=-1[/tex] is not a root (substituting gives [tex]k=0[/tex]), so every root satisfies [tex]kx=(x+1)^2\gt 0[/tex] automatically; also [tex]x=0[/tex] is never a root. Hence a root of the quadratic equation is a root of the given equation if and only if [tex]x\gt -1[/tex].
The discriminant and Vieta's formulas give
[tex]D=(2-k)^2-4=k(k-4),\qquad x_1x_2=1,\qquad x_1+x_2=k-2[/tex]
If [tex]0\lt k\lt 4[/tex], then [tex]D\lt 0[/tex]: no roots. If [tex]k=4[/tex], the double root is [tex]x=1\gt -1[/tex]: exactly one root.
If [tex]k\gt 4[/tex], the two distinct roots have product [tex]1[/tex] and sum [tex]k-2\gt 0[/tex], so both are positive and both are valid: two roots.
If [tex]k\lt 0[/tex], then [tex]D\gt 0[/tex] and the two distinct roots have product [tex]1[/tex] and sum [tex]k-2\lt 0[/tex], so both are negative; since their product is [tex]1[/tex], one lies in [tex]\left(-1 , 0\right)[/tex] and the other is less than [tex]-1[/tex]. Only the first one is valid: exactly one root.
Therefore [tex]k\in\left(-\infty , 0\right)\cup\left\{4\right\}[/tex].

Solution:
The domain is [tex]x\ne a[/tex], [tex]x\ne b[/tex], [tex]x\ne c[/tex]. On it, multiplying by [tex](x-a)(x-b)(x-c)\ne 0[/tex] gives the equivalent equation [tex]g(x)=0[/tex], where
[tex]g(x)=(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=3x^2-2(a+b+c)x+ab+bc+ca[/tex]
For the quadratic trinomial [tex]g[/tex] we have
[tex]\frac{D}{4}=(a+b+c)^2-3(ab+bc+ca)=a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right][/tex]
Since [tex]a[/tex], [tex]b[/tex], [tex]c[/tex] are distinct, every square is positive, so [tex]\frac{D}{4}\gt 0[/tex] and [tex]g[/tex] has two distinct real roots. None of them is excluded by the domain, because
[tex]g(a)=(a-b)(a-c)\ne 0,\qquad g(b)=(b-a)(b-c)\ne 0,\qquad g(c)=(c-a)(c-b)\ne 0[/tex]
Hence both roots of [tex]g[/tex] are roots of the given equation, and since on the domain the equation is equivalent to [tex]g(x)=0[/tex], it has exactly two distinct real roots.

Solution:
Let [tex]x_{0}[/tex] be a real root of the equation for some real value of [tex]m[/tex]. Arranging the equality in powers of [tex]m[/tex], we get
[tex]x_{0}^{2}-2mx_{0}+2m^{2}-2m-3=0\iff 2m^{2}-2(x_{0}+1)m+x_{0}^{2}-3=0[/tex]
This is a quadratic equation in [tex]m[/tex] with real coefficients that has the real root [tex]m[/tex]; hence its discriminant is non-negative:
[tex]\frac{D}{4}=(x_{0}+1)^{2}-2(x_{0}^{2}-3)=-x_{0}^{2}+2x_{0}+7\ge 0\iff x_{0}^{2}-2x_{0}-7\le 0[/tex]
The roots of [tex]x^{2}-2x-7=0[/tex] are [tex]1\pm 2\sqrt{2}[/tex], therefore
[tex]1-2\sqrt{2}\le x_{0}\le 1+2\sqrt{2}[/tex]
This is what we had to prove. The bounds are attained: for [tex]m=1+\sqrt{2}[/tex] the roots are [tex]1[/tex] and [tex]1+2\sqrt{2}[/tex], and for [tex]m=1-\sqrt{2}[/tex] they are [tex]1[/tex] and [tex]1-2\sqrt{2}[/tex].

Solution:
Write the equation as [tex]a=f(x)[/tex], where [tex]f(x)=x-\left|2|x|-1\right|[/tex]. The number of solutions equals the number of common points of the graph of [tex]f[/tex] and the horizontal line [tex]y=a[/tex]. Opening the absolute values, we get
[tex]f(x)=\left\{\begin{array}{ll} 3x+1, & x\le -\frac{1}{2} \\ -x-1, & -\frac{1}{2}\le x\le 0 \\ 3x-1, & 0\le x\le \frac{1}{2} \\ 1-x, & x\ge \frac{1}{2} \end{array}\right.[/tex]
Hence [tex]f[/tex] increases from [tex]-\infty[/tex] to [tex]f\left(-\frac{1}{2}\right)=-\frac{1}{2}[/tex], decreases to [tex]f(0)=-1[/tex], increases to [tex]f\left(\frac{1}{2}\right)=\frac{1}{2}[/tex] and then decreases to [tex]-\infty[/tex]. We count the common points with the line [tex]y=a[/tex].
For [tex]a\gt \frac{1}{2}[/tex] there are no solutions, for [tex]a=\frac{1}{2}[/tex] there is one, and for [tex]-\frac{1}{2}\lt a\lt \frac{1}{2}[/tex] there are two.
For [tex]a=-\frac{1}{2}[/tex] there are three: [tex]x=-\frac{1}{2}[/tex], [tex]x=\frac{1}{6}[/tex], [tex]x=\frac{3}{2}[/tex].
For [tex]-1\lt a\lt -\frac{1}{2}[/tex] there are four (one on each interval of monotonicity).
For [tex]a=-1[/tex] there are three: [tex]x=-\frac{2}{3}[/tex], [tex]x=0[/tex], [tex]x=2[/tex]; for [tex]a\lt -1[/tex] there are two.
Hence the equation has exactly three solutions for [tex]a=-1[/tex] and for [tex]a=-\frac{1}{2}[/tex].

Solution:
Fix [tex]a[/tex]. The expression [tex]A\cos x+B\sin x[/tex] takes exactly the values of the interval [tex]\left[-\sqrt{A^2+B^2} , \sqrt{A^2+B^2}\right][/tex]. Here [tex]A=a^2-9[/tex], [tex]B=6a[/tex] and
[tex]A^2+B^2=(a^2-9)^2+36a^2=a^4+18a^2+81=(a^2+9)^2[/tex]
Therefore the least value of the left side is [tex]-(a^2+9)[/tex]. The inequality has a solution [tex]x[/tex] if and only if this least value does not exceed the right side:
[tex]-(a^2+9)\le ab\iff a^2+ab+9\ge 0[/tex]
This must hold for every real [tex]a[/tex]. A quadratic trinomial in [tex]a[/tex] with a positive leading coefficient is non-negative for all [tex]a[/tex] if and only if its discriminant is non-positive:
[tex]D=b^2-36\le 0\iff -6\le b\le 6[/tex]
For [tex]b=\pm 6[/tex] we get [tex](a\pm 3)^2\ge 0[/tex], which is true, so the endpoints are included. The answer is [tex]b\in\left[-6 , 6\right][/tex].

Solution:
For [tex]x\in\left(0 , \frac{\pi}{2}\right)[/tex] both [tex]\sin x[/tex] and [tex]\cos x[/tex] are positive, and the equation reads [tex]\frac{\sin x+\cos x}{\sin x\cos x}=\lambda[/tex]. Put
[tex]t=\sin x+\cos x=\sqrt{2}\sin \left(x+\frac{\pi}{4}\right)[/tex]
Since [tex]x+\frac{\pi}{4}\in\left(\frac{\pi}{4} , \frac{3\pi}{4}\right)[/tex], the variable [tex]t[/tex] takes exactly the values in [tex]\left(1 , \sqrt{2}\right][/tex], and [tex]\sin x\cos x=\frac{t^2-1}{2}[/tex]. The equation becomes
[tex]\lambda=\frac{2t}{t^2-1}=\frac{2}{t-\frac{1}{t}}[/tex]
On [tex]\left(1 , \sqrt{2}\right][/tex] the denominator [tex]t-\frac{1}{t}[/tex] is positive and strictly increasing, from values arbitrarily close to [tex]0[/tex] up to [tex]\sqrt{2}-\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}[/tex]. Hence the right side is continuous and strictly decreasing, grows without bound as [tex]t\to 1^{+}[/tex], and at [tex]t=\sqrt{2}[/tex] (i.e. at [tex]x=\frac{\pi}{4}[/tex]) it equals
[tex]\frac{2}{\frac{\sqrt{2}}{2}}=2\sqrt{2}[/tex]
Its set of values is [tex]\left[2\sqrt{2} , +\infty\right)[/tex], so the equation has a root in [tex]\left(0 , \frac{\pi}{2}\right)[/tex] exactly when [tex]\lambda\ge 2\sqrt{2}[/tex].

Solution:
Put [tex]t=x^2-2x+2=(x-1)^2+1[/tex], so [tex]t\ge 1[/tex]. For [tex]t=1[/tex] we get only [tex]x=1[/tex], and for [tex]t\gt 1[/tex] two distinct values
[tex]x=1\pm\sqrt{t-1}[/tex]
Both are positive if and only if [tex]1-\sqrt{t-1}\gt 0[/tex], i.e. [tex]t\lt 2[/tex] (for [tex]t=2[/tex] one of them is [tex]0[/tex], and for [tex]t\gt 2[/tex] one of them is negative). Different values of [tex]t[/tex] give different values of [tex]x[/tex]. The equation becomes
[tex]g(t)=t^2-mt+3=0[/tex]
It has at most two roots, and each of them gives at most two positive values of [tex]x[/tex]. Hence the original equation has four distinct positive roots if and only if [tex]g[/tex] has two distinct roots, both in the open interval [tex]\left(1 , 2\right)[/tex]. The conditions are
[tex]D=m^2-12\gt 0,\quad g(1)=4-m\gt 0,\quad g(2)=7-2m\gt 0,\quad 1\lt \frac{m}{2}\lt 2[/tex]
From the last condition [tex]m\gt 0[/tex], so [tex]D\gt 0[/tex] gives [tex]m\gt 2\sqrt{3}[/tex]; the other conditions give [tex]m\lt 4[/tex] and [tex]m\lt \frac{7}{2}[/tex]. Since [tex]2\sqrt{3}=\sqrt{12}\lt \sqrt{\frac{49}{4}}=\frac{7}{2}[/tex], we get
[tex]2\sqrt{3}\lt m\lt \frac{7}{2}[/tex]

Solution:
Domain: [tex]x\ge 3[/tex]. Put [tex]s=\sqrt{x-3}\ge 0[/tex]; then [tex]x=s^2+3[/tex], and different [tex]s\ge 0[/tex] give different [tex]x\ge 3[/tex]. The equation becomes
[tex]s+a\left(s^2+3\right)=2a+3\iff as^2+s+a-3=0\quad (\ast)[/tex]
We need [tex](\ast)[/tex] to have exactly one non-negative root [tex]s[/tex]. If [tex]a=0[/tex], then [tex]s=3[/tex], i.e. [tex]x=12[/tex]: one solution.
Let [tex]a\ne 0[/tex]. Then
[tex]s_1s_2=\frac{a-3}{a},\quad s_1+s_2=-\frac{1}{a},\quad D=1-4a(a-3)=-4a^2+12a+1[/tex]
If [tex]0\lt a\lt 3[/tex], then [tex]s_1s_2\lt 0[/tex], so the roots are real with opposite signs and exactly one of them is positive: one solution.
If [tex]a=3[/tex], then [tex]3s^2+s=0[/tex], so [tex]s=0[/tex] or [tex]s=-\frac{1}{3}[/tex], and only [tex]s=0[/tex] ([tex]x=3[/tex]) is admissible: one solution.
If [tex]a\gt 3[/tex], then [tex]s_1s_2\gt 0[/tex] and [tex]s_1+s_2\lt 0[/tex], so any real roots are negative: no solutions.
If [tex]a\lt 0[/tex], then [tex]s_1s_2\gt 0[/tex] and [tex]s_1+s_2\gt 0[/tex], so any real roots are positive: there are two solutions for [tex]D\gt 0[/tex], none for [tex]D\lt 0[/tex] and exactly one for [tex]D=0[/tex]. We have
[tex]D=0\iff 4a^2-12a-1=0\iff a=\frac{3\pm\sqrt{10}}{2}[/tex]
The condition [tex]a\lt 0[/tex] leaves only [tex]a=\frac{3-\sqrt{10}}{2}[/tex] (double root [tex]s=-\frac{1}{2a}=3+\sqrt{10}[/tex], i.e. [tex]x=22+6\sqrt{10}[/tex]). The answer is [tex]a\in\left\{\frac{3-\sqrt{10}}{2}\right\}\cup\left[0 , 3\right][/tex].

Solution:
Domain: the base [tex]\sqrt{2-x}[/tex] must be positive and different from [tex]1[/tex], so [tex]x\lt 2[/tex] and [tex]x\ne 1[/tex]; the argument must be positive, i.e. [tex]x^2+4a\ne 0[/tex]. On the domain the equation is equivalent to
[tex]\left|x^2+4a\right|=\left(\sqrt{2-x}\right)^4=(2-x)^2[/tex]
Since [tex](2-x)^2\gt 0[/tex] for [tex]x\lt 2[/tex], every root of this equation automatically satisfies [tex]x^2+4a\ne 0[/tex]. So we count the roots with [tex]x\lt 2[/tex], [tex]x\ne 1[/tex] of the two equations [tex]x^2+4a=(2-x)^2[/tex] and [tex]x^2+4a=-(2-x)^2[/tex].
The first one gives [tex]x^2+4a=x^2-4x+4\iff x=1-a[/tex]; this root is admissible if and only if [tex]1-a\lt 2[/tex] and [tex]1-a\ne 1[/tex], i.e. [tex]a\gt -1[/tex], [tex]a\ne 0[/tex].
The second one gives
[tex]x^2+4a=-x^2+4x-4\iff (x-1)^2=-1-2a[/tex]
For [tex]a\gt -\frac{1}{2}[/tex] it has no roots, and for [tex]a=-\frac{1}{2}[/tex] its only root is [tex]x=1[/tex], which is not admissible. For [tex]a\lt -\frac{1}{2}[/tex] put [tex]s=\sqrt{-1-2a}\gt 0[/tex]; the roots are [tex]1-s[/tex] (always admissible) and [tex]1+s[/tex], which is admissible if and only if [tex]s\lt 1\iff -1-2a\lt 1\iff a\gt -1[/tex].
Counting: if [tex]a\le -1[/tex], only [tex]x=1-s[/tex] remains, i.e. one root. If [tex]a\ge -\frac{1}{2}[/tex], the second equation gives no admissible root and the first gives at most one.
If [tex]-1\lt a\lt -\frac{1}{2}[/tex], we get [tex]1-s[/tex] and [tex]1+s[/tex] with [tex]0\lt s\lt 1[/tex], and also [tex]x=1-a\in\left(\frac{3}{2} , 2\right)[/tex]. Here [tex]1-a\gt 1\gt 1-s[/tex], and [tex]1-a=1+s[/tex] would mean [tex]a^2=-1-2a[/tex], i.e. [tex](a+1)^2=0[/tex], [tex]a=-1[/tex], which is excluded. Hence there are exactly three roots.
The answer is [tex]a\in\left(-1 , -\frac{1}{2}\right)[/tex].