Past JEE PYQs (Integer Type)

Hint

(c) Given:
\(
(1-a) x^2+2(a-3) x+9=0
\)
We are told that the roots are positive, and we must find the set of \(a \in \mathbb{R}-\{1\}\) for which that’s true.
The set is given as:
\(
(-\infty,-\alpha] \cup[\beta, \gamma)
\)
and we need to find \(2 \alpha+\beta+\gamma\).
Step 1: Write down coefficients
\(
A=1-a, \quad B=2(a-3), \quad C=9
\)
Step 2: Conditions for real roots
Discriminant \(\Delta \geq 0\) :
\(
\Delta=B^2-4 A C=[2(a-3)]^2-4(1-a)(9)
\)
Simplify:
\(
\begin{gathered}
\Delta=4(a-3)^2-36(1-a) \\
\Delta=4\left(a^2-6 a+9\right)-36+36 a \\
\Delta=4 a^2-24 a+36-36+36 a \\
\Delta=4 a^2+12 a \\
\Delta=4 a(a+3)
\end{gathered}
\)
So, for real roots:
\(
a(a+3) \geq 0 \Rightarrow a \leq-3 \text { or } a \geq 0
\)
Step 3: Roots are positive
For a quadratic \(A x^2+B x+C=0\) :
Let the roots be \(x_1, x_2\).
\(
x_1+x_2=-\frac{B}{A}, \quad x_1 x_2=\frac{C}{A}
\)
We need both roots positive:
\(
x_1+x_2>0, \quad x_1 x_2>0
\)
Step 4: Compute sums and products
\(
\begin{gathered}
x_1+x_2=-\frac{B}{A}=-\frac{2(a-3)}{1-a}=\frac{2(3-a)}{1-a} \\
x_1 x_2=\frac{C}{A}=\frac{9}{1-a}
\end{gathered}
\)
Step 5: Analyze signs
Case 1: \(1-a>0 \Rightarrow a<1\)
Then:
\(
x_1 x_2=\frac{9}{1-a}>0 \text { (always positive) }
\)
Now, \(x_1+x_2=\frac{2(3-a)}{1-a}>0\)
Denominator \(>0\), so we need numerator \(>0 \Rightarrow 3-a>0 \Rightarrow a<3\)
Thus for \(a<1\), sum positive automatically holds because \(a<1<3\).
So both conditions hold for all \(a<1\), provided real roots exist.
But we must also have \(\Delta \geq 0 \Rightarrow a \leq-3\) or \(a \geq 0\).
Intersect with \(a<1\) :
\(
a \leq-3 \text { or } 0 \leq a<1
\)
Case 2: \(1-a<0 \Rightarrow a>1\)
Then denominator is negative.
\(
x_1 x_2=\frac{9}{1-a}<0
\)
\(\rightarrow\) Product negative, so one root positive and one negative.
Thus no solution for \(a>1\).
Step 6: Combine valid intervals
\(
a \in(-\infty,-3] \cup[0,1)
\)
Step 7: Match with given form
\(
(-\infty,-\alpha] \cup[\beta, \gamma)
\)
Hence:
\(
\alpha=3, \quad \beta=0, \quad \gamma=1
\)
Step 8: Compute \(2 \alpha+\beta+\gamma\)
\(
2(3)+0+1=6+1=7
\)

Hint

(a) The given quadratic equation is \(a(b-c) x^2+b(c-a) x+c(a-b)=0\). We are told that the roots are equal. We can observe that \(x=1\) is a root of the equation because substituting \(x=1\) gives:
\(
a(b-c)+b(c-a)+c(a-b)=a b-a c+b c-a b+a c-b c=0
\)
Since the roots are equal, both roots must be 1. The sum of the roots, \(\alpha+\beta\), is therefore \(1+1=2\).
Step 1: Use the sum of roots formula
The sum of the roots of a quadratic equation \(\boldsymbol{A} \boldsymbol{x}^2+\boldsymbol{B} \boldsymbol{x}+\boldsymbol{C}=\mathbf{0}\) is given by \(-\frac{\boldsymbol{B}}{\boldsymbol{A}}\). For the given equation, the sum of the roots is \(-\frac{b(c-a)}{a(b-c)}\). Since the sum of the roots is 2, we can set up the following equation:
\(
\begin{gathered}
-\frac{b(c-a)}{a(b-c)}=2 \\
-b(c-a)=2 a(b-c) \\
-b c+a b=2 a b-2 a c \\
2 a c=2 a b-a b+b c \\
2 a c=a b+b c \\
2 a c=b(a+c)
\end{gathered}
\)
Step 2: Substitute given values
We are given that \(a+c=15\) and \(b=\frac{36}{5}\). Substitute these values into the equation from Step 1:
\(
\begin{gathered}
2 a c=b(a+c) \\
2 a c=\left(\frac{36}{5}\right)(15) \\
2 a c=36 \times 3 \\
2 a c=108 \\
a c=54
\end{gathered}
\)
Step 3: Calculate \(a^2+c^2\)
We can use the algebraic identity \(a^2+c^2=(a+c)^2-2 a c\). Substitute the known values for \((a+c)\) and \(a c\) :
\(
\begin{gathered}
a^2+c^2=(15)^2-2(54) \\
a^2+c^2=225-108 \\
a^2+c^2=117
\end{gathered}
\)

Hint

(b) Step 1: Analyze the equation by splitting the domain
The equation is \(|x+1||x+3|-4|x+2|+5=0\). The absolute value functions change their signs at \(x=-1, x=-2\), and \(x=-3\). We need to consider four cases based on these values.
Step 2: Solve the equation for each case
Case 1: \(x<-3\)
In this interval, \(|x+1|=-(x+1),|x+2|=-(x+2)\), and \(|x+3|=-(x+3)\). The equation becomes:
\(
\begin{gathered}
(-x-1)(-x-3)-4(-x-2)+5=0 \\
(x+1)(x+3)+4(x+2)+5=0 \\
x^2+4 x+3+4 x+8+5=0 \\
x^2+8 x+16=0 \\
(x+4)^2=0
\end{gathered}
\)
This gives \(x=-4\). Since \(-4<-3\), this is a valid root.
Case 2: \(-3 \leq x<-2\)
In this interval, \(|x+1|=-(x+1),|x+2|=-(x+2)\), and \(|x+3|=x+3\).
The equation becomes:
\(
\begin{gathered}
(-x-1)(x+3)-4(-x-2)+5=0 \\
-\left(x^2+4 x+3\right)+4 x+8+5=0 \\
-x^2-4 x-3+4 x+13=0 \\
-x^2+10=0 \\
x^2=10
\end{gathered}
\)
This gives \(x= \pm \sqrt{10}\). We check if these roots are in the interval \([-3,-2)\).
Since \(\sqrt{10} \approx 3.16\) and \(-\sqrt{10} \approx-3.16\), neither root is in the interval \([-3,-2)\). So, there are no roots in this case.
Case 3: \(-2 \leq x<-1\)
In this interval, \(|x+1|=-(x+1),|x+2|=x+2\), and \(|x+3|=x+3\).
The equation becomes:
\(
\begin{gathered}
(-x-1)(x+3)-4(x+2)+5=0 \\
-\left(x^2+4 x+3\right)-4 x-8+5=0 \\
-x^2-4 x-3-4 x-3=0 \\
-x^2-8 x-6=0 \\
x^2+8 x+6=0
\end{gathered}
\)
Using the quadratic formula,
\(
x=\frac{-8 \pm \sqrt{8^2-4(1)(6)}}{2}=\frac{-8 \pm \sqrt{64-24}}{2}=\frac{-8 \pm \sqrt{40}}{2}=\frac{-8 \pm 2 \sqrt{10}}{2}=-4 \pm \sqrt{10}
\)
We need to check if these roots are in the interval \([-2,-1)\).
\(x_1=-4+\sqrt{10} \approx-4+3.16=-0.84\). This is not in the interval.
\(x_2=-4-\sqrt{10} \approx-4-3.16=-7.16\). This is not in the interval.
So, there are no roots in this case.
Case 4: \(x \geq-1\)
In this interval, \(|x+1|=x+1,|x+2|=x+2\), and \(|x+3|=x+3\). The equation becomes:
\(
\begin{gathered}
(x+1)(x+3)-4(x+2)+5=0 \\
x^2+4 x+3-4 x-8+5=0 \\
x^2=0
\end{gathered}
\)
This gives \(x=0\). Since \(0 \geq-1\), this is a valid root.
The distinct real roots are \(x=-4\) and \(x=0\). There are two distinct real roots.
The number of distinct real roots of the equation is 2.

Hint

(c) Step 1: Use the properties of the roots of the quadratic equation
The given quadratic equation is \(x^2+\sqrt{2} x-8=0\), and its roots are \(\alpha\) and \(\beta\). By definition of a root, if \(\alpha\) is a root, it must satisfy the equation. Therefore:
\(
\alpha^2+\sqrt{2} \alpha-8=0
\)
This can be rearranged to find an expression for \(\alpha^2\) :
\(
\alpha^2=-\sqrt{2} \alpha+8
\)
Similarly, for the root \(\boldsymbol{\beta}\).
\(
\beta^2=-\sqrt{2} \beta+8
\)
Also, we can manipulate the original equation to find a different relationship: From \(\alpha^2+\sqrt{2} \alpha-8=0\), we can express \(\alpha+\sqrt{2}\) in terms of \(\alpha\).
\(
\alpha+\sqrt{2}=\frac{8}{\alpha}
\)
And for \(\boldsymbol{\beta}\) :
\(
\beta+\sqrt{2}=\frac{8}{\beta}
\)
Step 2: Simplify the numerator of the expression
The expression to evaluate is \(\frac{\mathrm{U}_{10}+\sqrt{2} \mathrm{U}_9}{2 \mathrm{U}_8}\).
First, let’s work on the numerator, \(\mathrm{U}_{10}+\sqrt{2} \mathrm{U}_9\).
The sequence is defined as \(\mathrm{U}_{\mathrm{n}}=\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}\).
Substitute this into the numerator:
\(
\mathrm{U}_{10}+\sqrt{2} \mathrm{U}_9=\left(\alpha^{10}+\beta^{10}\right)+\sqrt{2}\left(\alpha^9+\beta^9\right)
\)
Group the terms with the same base:
\(
\mathrm{U}_{10}+\sqrt{2} \mathrm{U}_9=\alpha^{10}+\sqrt{2} \alpha^9+\beta^{10}+\sqrt{2} \beta^9
\)
Factor out \(\alpha^9\) and \(\beta^9\) :
\(
\mathrm{U}_{10}+\sqrt{2} \mathrm{U}_9=\alpha^9(\alpha+\sqrt{2})+\beta^9(\beta+\sqrt{2})
\)
Now, substitute the relationships found in Step 1, where \(\alpha+\sqrt{2}=\frac{8}{\alpha}\) and \(\beta+\sqrt{2}=\frac{8}{\beta}:\)
\(
\alpha^9\left(\frac{8}{\alpha}\right)+\beta^9\left(\frac{8}{\beta}\right)=8 \alpha^{9-1}+8 \beta^{9-1}=8 \alpha^8+8 \beta^8
\)
Factor out the 8:
\(
8\left(\alpha^8+\beta^8\right)
\)
Recall the definition of \(\mathrm{U}_{\mathrm{n}}\) : \(\mathrm{U}_8=\boldsymbol{\alpha}^8+\boldsymbol{\beta}^8\).
So, the numerator simplifies to \(8 \mathrm{U}_8\).
The numerator of the expression is \(8 \mathrm{U}_8\), and the denominator is \(2 \mathrm{U}_8\). Therefore, the value of the expression is:
\(
\frac{8 U_8}{2 U_8}=\frac{8}{2}=4
\)
The value of the expression is \(\mathbf{4}\).

Hint

(b) First, observe that for a degree 4 polynomial
\(
P(x)=4 x^4+8 x^3-17 x^2-12 x+9
\)
with roots \(x_1, \ldots, x_4\), we have
\(
\prod_{i=1}^4\left(x_i^2+4\right)=\prod_{i=1}^4\left[\left(x_i-2 i\right)\left(x_i+2 i\right)\right]=\frac{P(2 i) P(-2 i)}{4^2}
\)
Compute \(P(2 i)\) :
\(
P(2 i)=4(2 i)^4+8(2 i)^3-17(2 i)^2-12(2 i)+9=64-64 i+68-24 i+9=141-88 i .
\)
By conjugation,
\(
P(-2 i)=141+88 i .
\)
Hence
\(
\prod_{i=1}^4\left(x_i^2+4\right)=\frac{(141-88 i)(141+88 i)}{16}=\frac{141^2+88^2}{16}=\frac{19881+7744}{16}=\frac{27625}{16} .
\)
We are told
\(
\prod_{i=1}^4\left(x_i^2+4\right)=\frac{125}{16} m=\frac{27625}{16}
\)
so
\(
125 m=27625 \quad \Longrightarrow \quad m=221 .
\)

Hint

(d) The equation to solve is \(x|x+5|+2|x+7|-2=0\), or \(x|x+5|+2|x+7|=2\). To find the number of real solutions, we analyze the equation in different intervals based on the critical points where the absolute value expressions change signs, which are \(x=-7\) and \(x=-5\).
Case 1: \(\boldsymbol{x}<-\mathbf{7}\)
In this interval, \(x+5<0\) and \(x+7<0\), so \(|x+5|=-(x+5)\) and \(|x+7|=-(x+7)\).
The equation becomes:
\(
\begin{gathered}
x(-(x+5))+2(-(x+7))-2=0 \\
-x^2-5 x-2 x-14-2=0 \\
-x^2-7 x-16=0 \\
x^2+7 x+16=0
\end{gathered}
\)
To check for real solutions, we use the discriminant \(\Delta=b^2-4 a c\).
\(
\Delta=7^2-4(1)(16)=49-64=-15
\)
Since the discriminant is negative, there are no real solutions in this interval.
Case 2: \(-7 \leq x<-5\)
In this interval, \(x+5<0\) and \(x+7 \geq 0\), so \(|x+5|=-(x+5)\) and \(|x+7|=x+7\).
The equation becomes:
\(
\begin{gathered}
x(-(x+5))+2(x+7)-2=0 \\
-x^2-5 x+2 x+14-2=0 \\
-x^2-3 x+12=0 \\
x^2+3 x-12=0
\end{gathered}
\)
Using the quadratic formula, \(x=\frac{-b \pm \sqrt{\Delta}}{2 a}\) :
\(
\Delta=3^2-4(1)(-12)=9+48=57
\)
The solutions are \(x=\frac{-3 \pm \sqrt{57}}{2}\).
We check if these solutions are in the interval \([-7,-5) . \sqrt{57}\) is approximately 7.5. \(x_1=\frac{-3+7.5}{2} \approx 2.25\), which is not in the interval \([-7,-5)\). \(x_2=\frac{-3-7.5}{2} \approx-5.25\), which is in the interval \([-7,-5)\).
So, \(x=\frac{-3-\sqrt{57}}{2}\) is a valid solution. This gives one solution in this interval.
Case 3: \(x \geq-5\)
In this interval, \(x+5 \geq 0\) and \(x+7 \geq 0\), so \(|x+5|=x+5\) and \(|x+7|=x+7\). The equation becomes:
\(
\begin{gathered}
x(x+5)+2(x+7)-2=0 \\
x^2+5 x+2 x+14-2=0 \\
x^2+7 x+12=0
\end{gathered}
\)
This quadratic equation can be factored as \((x+3)(x+4)=0\).
The solutions are \(x=-3\) and \(x=-4\).
Both solutions, \(x=-3\) and \(x=-4\), are in the interval \(x \geq-5\).
So, there are two solutions in this interval.
Total Number of Solutions
Combining the solutions from all cases, we have a total of \(1+2=3\) real solutions. The solutions are \(x=-4, x=-3\), and \(x=\frac{-3-\sqrt{57}}{2}\).

Hint

(c) The equation is \(|x||x+2|-5|x+1|-1=0\), which can be rewritten as \(|x(x+2)|-5|x+1|-1=0\).
Note that \(x(x+2)=x^2+2 x=(x+1)^2-1\).
Let \(y=|x+1|\). Then \(x+1= \pm y\), so \(x=-1 \pm y\). \((x+1)^2=y^2\), so \(x(x+2)=y^2-1\).
The equation becomes \(\left|y^2-1\right|-5 y-1=0\).
We consider two cases for the absolute value \(\left|y^2-1\right|\) based on the value of \(y \geq 0\)
Case 1: \(y^2-1 \geq 0\), which means \(y \geq 1\) (since \(y \geq 0\) ).
The equation becomes \(\left(y^2-1\right)-5 y-1=0\), so \(y^2-5 y-2=0\).
Using the quadratic formula,
\(
y=\frac{-(-5) \pm \sqrt{(-5)^2-4(1)(-2)}}{2(1)}=\frac{5 \pm \sqrt{25+8}}{2}=\frac{5 \pm \sqrt{33}}{2} .
\)
Since we require \(y \geq 1\), we check the values:
\(y=\frac{5+\sqrt{33}}{2} \approx \frac{5+5.74}{2} \approx 5.37 \geq 1\), which is a valid solution for \(y\).
\(y=\frac{5-\sqrt{33}}{2} \approx \frac{5-5.74}{2} \approx-0.37<1\), which is not a valid solution for this case.
So \(y=\frac{5+\sqrt{33}}{2}\) gives real roots for \(x\). Since \(y=|x+1|\), we have \(|x+1|=\frac{5+\sqrt{33}}{2}\), which leads to two distinct real roots for \(x\).
Case 2: \(y^2-1<0\), which means \(0 \leq y<1\).
The equation becomes \(-\left(y^2-1\right)-5 y-1=0\), so \(-y^2+1-5 y-1=0\), which simplifies to \(-y^2-5 y=0\), or \(y(y+5)=0\).
The solutions are \(y=0\) and \(y=-5\).
Since we require \(0 \leq y<1, y=0\) is a valid solution for \(y . y=-5\) is not.
So \(y=0\) gives a real root for \(x\). Since \(y=|x+1|\), we have \(|x+1|=0\), which leads to one real root for \(x, x=-1\).
We check if \(y=0\) satisfies the condition \(y^2-1<0: 0^2-1=-1<0\), which is true.
In total, we have two distinct real roots from Case 1 and one distinct real root from Case 2.
The distinct real roots are given by \(x+1= \pm \frac{5+\sqrt{33}}{2}\) and \(x+1=0\), so \(x=-1 \pm \frac{5+\sqrt{33}}{2}\) and \(x=-1\). There are a total of 3 distinct real roots. 

Hint

(d) Step 1: Analyze the quadratic equation
The first part of the problem involves the quadratic equation
\(
\left(a^2+b^2\right) x^2-2 b(a+c) x+b^2+c^2=0 .
\)
This equation can be rearranged by expanding and grouping terms to form the sum of two perfect squares:
\(
\left(a^2 x^2-2 a b x+b^2\right)+\left(b^2 x^2-2 b c x+c^2\right)=0
\)
This simplifies to:
\(
(a x-b)^2+(b x-c)^2=0
\)
Since the square of any real number is non-negative, the sum of two squares is zero if and only if each term is zero. Thus, we have \(a x-b=0\) and \(b x-c=0\).
From these, we can find the value of \(x\) :
\(
x=\frac{b}{a} \quad \text { and } \quad x=\frac{c}{b}
\)
For a unique solution for \(x\) to exist, we must have \(\frac{b}{a}=\frac{c}{b}\), which implies that \(b^2=a c\).
Step 2: Solve the inequalities and find the values for \(\boldsymbol{\alpha}\) and \(\boldsymbol{\beta}\)
The second part of the problem presents several inequalities. The inequalities \(x^2-x+1>0\) is always true for all real \(x\). The other two inequalities are:
\(
x^2-x-1<0 \text { and } x^2+x-1>0
\)
Solving the first inequality, \(x^2-x-1<0\), by finding the roots of \(x^2-x-1=0\) using the quadratic formula \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\) gives:
\(
x=\frac{1 \pm \sqrt{(-1)^2-4(1)(-1)}}{2}=\frac{1 \pm \sqrt{5}}{2}
\)
The inequality is satisfied for \(x\) values between the roots, so \(\frac{1-\sqrt{5}}{2}<x<\frac{1+\sqrt{5}}{2}\)
Solving the second inequality, \(x^2+x-1>0\), by finding the roots of \(x^2+x-1=0\) gives:
\(
x=\frac{-1 \pm \sqrt{1^2-4(1)(-1)}}{2}=\frac{-1 \pm \sqrt{5}}{2}
\)
The inequality is satisfied for \(x\) values outside the roots, so \(x<\frac{-1-\sqrt{5}}{2}\) or \(x>\frac{-1+\sqrt{5}}{2}\).
To satisfy both inequalities, we need to find the intersection of the two solution sets.
The value \(\frac{1-\sqrt{5}}{2}\) is approximately -0.618 , and \(\frac{-1-\sqrt{5}}{2}\) is approximately -1.618.
The value \(\frac{1+\sqrt{5}}{2}\) is approximately 1.618 , and \(\frac{-1+\sqrt{5}}{2}\) is approximately 0.618.
The intersection of the two intervals is \(\frac{-1+\sqrt{5}}{2}<x<\frac{1+\sqrt{5}}{2}\).
This is equivalent to \(\frac{\sqrt{5}-1}{2}<x<\frac{\sqrt{5}+1}{2}\).
From this, we define the bounds of the interval as \(\alpha=\frac{\sqrt{5}-1}{2}\) and \(\beta=\frac{\sqrt{5}+1}{2}\)
Step 3: Calculate the final value
The last step is to calculate \(12\left(\alpha^2+\beta^2\right)\).
We can first calculate \(\alpha^2\) and \(\beta^2\) :
\(
\begin{aligned}
& \alpha^2=\left(\frac{\sqrt{5}-1}{2}\right)^2=\frac{5-2 \sqrt{5}+1}{4}=\frac{6-2 \sqrt{5}}{4}=\frac{3-\sqrt{5}}{2} \\
& \beta^2=\left(\frac{\sqrt{5}+1}{2}\right)^2=\frac{5+2 \sqrt{5}+1}{4}=\frac{6+2 \sqrt{5}}{4}=\frac{3+\sqrt{5}}{2}
\end{aligned}
\)
Then, sum them up:
\(
\alpha^2+\beta^2=\frac{3-\sqrt{5}}{2}+\frac{3+\sqrt{5}}{2}=\frac{6}{2}=3
\)
Finally, multiply by 12 :
\(
12\left(\alpha^2+\beta^2\right)=12(3)=36
\)

Hint

(a) The equation \(x\left(x^2+3|x|+5|x-1|+6|x-2|\right)=0\) has a solution at \(x=0\). The other solutions come from solving \(x^2+3|x|+5|x-1|+6|x-2|=0\). We analyze this equation by considering the intervals where the absolute value functions change their definition.
Step 1: Case (I) \(\boldsymbol{x}<0\)
When \(x<0\), we have \(|x|=-x,|x-1|=-(x-1)=-x+1\), and \(|x-2|=-(x-2)=-x+2\).
Substituting these into the equation:
\(
\begin{gathered}
x^2+3(-x)+5(-x+1)+6(-x+2)=0 \\
x^2-3 x-5 x+5-6 x+12=0 \\
x^2-14 x+17=0
\end{gathered}
\)
\(
x=7 \pm 4 \sqrt{2}
\)
Both roots, \(7+4 \sqrt{2}\) and \(7-4 \sqrt{2}\), are positive. Since we assumed \(x<0\), there are no solutions in this case.
Step 2: Case (II) \(0 \leq x<1\)
When \(0 \leq x<1\), we have \(|x|=x,|x-1|=-(x-1)=-x+1\), and \(|x-2|=-(x-2)=-x+2\).
Substituting these into the equation:
\(
\begin{gathered}
x^2+3(x)+5(-x+1)+6(-x+2)=0 \\
x^2+3 x-5 x+5-6 x+12=0 \\
x^2-8 x+17=0
\end{gathered}
\)
The discriminant is \(\boldsymbol{D}=\boldsymbol{b}^2-4 a c=(-8)^2-4(1)(17)=64-68=-4\). Since the discriminant is negative, there are no real solutions in this case.
Step 3: Case (III) \(1 \leq x<2\)
When \(1 \leq x<2\), we have \(|x|=x,|x-1|=x-1\), and \(|x-2|=-(x-2)=-x+2\). Substituting these into the equation:
\(
\begin{gathered}
x^2+3(x)+5(x-1)+6(-x+2)=0 \\
x^2+3 x+5 x-5-6 x+12=0 \\
x^2+2 x+7=0
\end{gathered}
\)
The discriminant is \(\boldsymbol{D}=\boldsymbol{b}^2-4 a c=(2)^2-4(1)(7)=4-28=-24\). Since the discriminant is negative, there are no real solutions in this case.
Step 4: Case (IV) \(\boldsymbol{x} \boldsymbol{\geq} \mathbf{2}\)
When \(x \geq 2\), we have \(|x|=x,|x-1|=x-1\), and \(|x-2|=x-2\). Substituting these into the equation:
\(
\begin{gathered}
x^2+3(x)+5(x-1)+6(x-2)=0 \\
x^2+3 x+5 x-5+6 x-12=0 \\
x^2+14 x-17=0
\end{gathered}
\)
\(
x=-7 \pm \sqrt{66}
\)
We assumed \(x \geq 2\). One root is \(-7-\sqrt{66}\), which is negative. The other root is \(-7+\sqrt{66}\). Since \(\sqrt{66} \approx 8.12\), the value is approximately \(-7+8.12=1.12\). This is less than 2, so it is also not a solution in this case. There are no solutions in this case.
The only solution to the given equation is \(\boldsymbol{x} \boldsymbol{=} \mathbf{0}\)

Hint

(d) The roots \(\alpha, \beta\) of the equation \(x^2-70 x+\lambda=0\) satisfy Vieta’s formulas:
\(
\begin{gathered}
\alpha+\beta=70 \\
\alpha \beta=\lambda
\end{gathered}
\)
Since \(\alpha, \beta \in \mathbf{N}\) (natural numbers), \(\lambda\) must be a natural number and can be expressed as \(\lambda=\alpha(70-\alpha)\).
The condition is that \(\frac{\lambda}{2} \notin \mathrm{~N}\) and \(\frac{\lambda}{3} \notin \mathrm{~N}\). This means \(\lambda\) is not divisible by 2 and not divisible by 3.
Also, for \(\boldsymbol{\alpha}, \boldsymbol{\beta}\) to be natural numbers, the discriminant of the quadratic equation must be a perfect square, say \(D=b^2-4 a c=(-70)^2-4(1)(\lambda)=4900-4 \lambda=k^2\), where \(k\) is an integer.
So, \(4 \lambda=4900-k^2\), which means \(4900-k^2\) must be divisible by 4, so \(k\) must be even. Let \(k=2 m\).
\(
4 \lambda=4900-4 m^2 \Longrightarrow \lambda=1225-m^2
\)
The roots are \(\alpha, \beta=\frac{70 \pm \sqrt{D}}{2}=\frac{70 \pm k}{2}=35 \pm m\).
Since \(\alpha, \beta \in \mathbf{N}, m\) must be an integer, and \(35 \pm m \geq 1\). Thus \(m \leq 34\) and \(m \geq-34\). Also, \(\lambda=\alpha \beta=(35-m)(35+m)=35^2-m^2=1225-m^2\).
We need to find the minimum \(\lambda\) that is not divisible by 2 and not by \(3 . \lambda\) must also be positive.
We can check values of \(\lambda\) starting from small possible values or iterate through possible values of \(\alpha\) (from 1 to 69).
Let’s iterate through possible values of \(\alpha\). Then \(\beta=70-\alpha\) and \(\lambda=\alpha(70-\alpha)\).
We require \(\lambda\) to not be a multiple of 2 or 3.
If \(\alpha=1, \beta=69, \lambda=69=3 \times 23\) (divisible by 3 ).
If \(\alpha=2, \beta=68, \lambda=136=2 \times 68\) (divisible by 2 ).
If \(\alpha=3, \beta=67, \lambda=201=3 \times 67\) (divisible by 3).
If \(\alpha=4, \beta=66, \lambda=264=2 \times 132\) (divisible by 2 ).
If \(\alpha=5, \beta=65, \lambda=325 . \frac{\lambda}{2}=162.5 \notin \mathbf{N}, \frac{\lambda}{3}=108.33 \ldots \boldsymbol{\mathrm { N }}\). This is the minimum possible value of \(\lambda\) that satisfies the condition.
So, \(\lambda=325\). The roots are \(\alpha=5, \beta=65\) (or vice versa).
Now we evaluate the expression:
\(
\begin{gathered}
\frac{(\sqrt{\alpha-1}+\sqrt{\beta-1})(\lambda+35)}{|\alpha-\beta|}=\frac{(\sqrt{5-1}+\sqrt{65-1})(325+35)}{|5-65|} \\
=\frac{(\sqrt{4}+\sqrt{64})(360)}{|-60|}=\frac{(2+8)(360)}{60} \\
=\frac{10 \times 360}{60}=\frac{3600}{60}=60
\end{gathered}
\)

Hint

(d) First, let’s consider the equation \(x^2-2^y=2023\) where \(x\) and \(y\) are natural numbers. Our goal is to find all the pairs \((x, y)\) that satisfy this equation and then sum the values of \(x+y\) for each pair in set \(C\).
Since 2023 is an odd number, and \(x^2\), the square of any natural number, is even when \(x\) is even and odd when \(x\) is odd, we can determine that for the left-hand side of the equation to be odd (thus equal to 2023 ), \(x\) must be odd since the right-hand side of the equation \(\left(2^y\right)\) is always even as it represents a power of two.
Also, 2023 can be factored into prime factors to further analyze the possible solutions:
\(
2023=7 \times 17 \times 17
\)
Thus, allowing us to rewrite the equation as:
\(
x^2-2^y=7 \times 17^2
\)
The next step is to check for potential values of \(x\) that would fit the equation, keeping in mind that \(x\) must be odd. We can try to express \(x^2\) as \(7 \times 17^2\) plus a power of 2, recognizing that we are looking for the decomposition of the form:
\(
x^2=7 \times 17^2+2^y
\)
By examining the powers of 2 and keeping in mind that they grow very quickly, we can reason that \(y\) cannot be very large because \(x^2\) must not exceed 2023 by a large margin.
Let’s start by trying the lowest values for \(y\) since that would make \(2^y\) small and \(x\) has a better chance of being a natural number:
For \(y=1\) :
\(
x^2=2023+2^1=2023+2=2025
\)
Surprisingly, we find a perfect square since \(45^2=2025\). Therefore, \((x, y)=(45,1)\) is one solution.
For \(y=2\) or higher:
\(2^y\) becomes at least 4 and increases exponentially, so \(x^2\) must be at least 2027 or higher in such cases. There’s no natural number between 45 and 46, and \(46^2\) far exceeds the target (2116), making it impossible for \(x^2\) to be less than 2116 for any larger \(y\).
Hence, it appears there is only one possible solution: \((x, y)=(45,1)\).
Therefore, the sum \(\sum_{(x, y) \in C}(x+y)\) for this set will consist of only this one pair:
\(
\sum_{(x, y) \in C}(x+y)=45+1=46
\)
So the answer is 46.

Hint

(b) The sum is calculated by grouping the terms for which the value of \([\sqrt{n}]\) is constant.
The value of \([\sqrt{n}]\) is \(k\) for \(n\) such that \(k^2 \leq n<(k+1)^2\).
For \(n \in[1,3],[\sqrt{n}]=1\). There are \(3-1+1=3\) terms.
For \(n \in[4,8],[\sqrt{n}]=2\). There are \(8-4+1=5\) terms.
For \(n \in[9,15],[\sqrt{n}]=3\). There are \(15-9+1=7\) terms.
For \(n \in[16,24],[\sqrt{n}]=4\). There are \(24-16+1=9\) terms.
For \(n \in[25,35],[\sqrt{n}]=5\). There are \(35-25+1=11\) terms.
For \(n \in[36,48],[\sqrt{n}]=6\). There are \(48-36+1=13\) terms.
For \(n \in[49,63],[\sqrt{n}]=7\). There are \(63-49+1=15\) terms.
For \(n \in[64,80],[\sqrt{n}]=8\). There are \(80-64+1=17\) terms.
For \(n \in[81,99],[\sqrt{n}]=9\). There are \(99-81+1=19\) terms.
For \(n \in[100,120],[\sqrt{n}]=10\). There are \(120-100+1=21\) terms.
The sum is:
\(
S=(1 \times 3)+(2 \times 5)+(3 \times 7)+(4 \times 9)+(5 \times 11)+(6 \times 13)+(7 \times 15)+(8 \times 17)+(9 \times 19)+(10 \times 21)
\)
\(
\begin{gathered}
S=3+10+21+36+55+78+105+136+171+210 \\
S=825
\end{gathered}
\)

Hint

(b) Step 1: Substitute a variable to simplify the equation
Let \(y=\mathrm{e}^{2 x}\). Since \(x \in \mathbb{R}\), we know that \(y=\mathrm{e}^{2 x}\) will always be positive, so \(y>0\). Substituting \(y\) into the equation \(f(x)=0\) gives:
\(
y^4-y^3-3 y^2-y+1=0
\)
Step 2: Factor the polynomial
Since \(y \neq 0\), we can divide the equation by \(y^2\) :
\(
y^2-y-3-\frac{1}{y}+\frac{1}{y^2}=0
\)
Group the terms with the same powers:
\(
\left(y^2+\frac{1}{y^2}\right)-\left(y+\frac{1}{y}\right)-3=0
\)
Let \(z=y+\frac{1}{y}\). Then \(z^2=y^2+2+\frac{1}{y^2}\), which means \(y^2+\frac{1}{y^2}=z^2-2\).
Substitute this into the equation:
\(
\begin{gathered}
\left(z^2-2\right)-z-3=0 \\
z^2-z-5=0
\end{gathered}
\)
Step 3: Solve for \(\boldsymbol{z}\)
Using the quadratic formula, \(\mathrm{z}=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\) :
\(
\begin{gathered}
z=\frac{1 \pm \sqrt{(-1)^2-4(1)(-5)}}{2(1)} \\
z=\frac{1 \pm \sqrt{1+20}}{2} \\
z=\frac{1 \pm \sqrt{21}}{2}
\end{gathered}
\)
Step 4: Solve for \(\boldsymbol{y}\)
We have \(z=y+\frac{1}{y}\). Since \(y>0\), we know that \(y+\frac{1}{y} \geq 2 \sqrt{y \cdot \frac{1}{y}}=2\).
Therefore, we must have \(z \geq 2\).
Let’s check the two values for \(\mathbf{z}\) :
\(z_1=\frac{1+\sqrt{21}}{2} \approx \frac{1+4.58}{2}=2.79>2\). This value is valid.
\(z_2=\frac{1-\sqrt{21}}{2} \approx \frac{1-4.58}{2}=-1.79<2\). This value is not valid.
So we only need to solve for \(y\) using \(z=\frac{1+\sqrt{21}}{2}\) :
\(
\begin{gathered}
y+\frac{1}{y}=\frac{1+\sqrt{21}}{2} \\
2 y^2+2=(1+\sqrt{21}) y \\
2 y^2-(1+\sqrt{21}) y+2=0
\end{gathered}
\)
Using the quadratic formula to solve for \(y\) :
\(
\begin{gathered}
y=\frac{(1+\sqrt{21}) \pm \sqrt{-(1+\sqrt{21})^2-4(2)(2)}}{2(2)} \\
y=\frac{(1+\sqrt{21}) \pm \sqrt{(1+21+2 \sqrt{21})-16}}{4} \\
y=\frac{1+\sqrt{21} \pm \sqrt{22+2 \sqrt{21}-16}}{4} \\
y=\frac{1+\sqrt{21} \pm \sqrt{6+2 \sqrt{21}}}{4}
\end{gathered}
\)
Since the discriminant, \(6+2 \sqrt{21}\), is positive, there are two distinct real positive values for \(y\). Let’s call them \(y_1\) and \(y_2\).
Step 5: Solve for \(\boldsymbol{x}\)
We have \(y=\mathrm{e}^{2 x}\). Taking the natural logarithm of both sides gives \(\ln (y)=2 x\), so \(x=\frac{\ln (y)}{2}\).
Since we found two distinct positive values for \(y_1 y_1\) and \(y_2\), there will be two distinct real values for \(\boldsymbol{x}\) :
\(
x_1=\frac{\ln \left(y_1\right)}{2} \quad \text { and } \quad x_2=\frac{\ln \left(y_2\right)}{2}
\)
Therefore, there are two points where the curve cuts the \(x\)-axis.
The number of points where the curve cuts the \(x\)-axis is equal to 2.

Hint

(a) We have, \(a\) and \(b\) are the roots of the equation
\(
\begin{aligned}
& x^2-7 x-1=0 \\
& \Rightarrow a^2-7 a-1=0 \Rightarrow a^2-1=7 a \dots(i)
\end{aligned}
\)
On squaring both sides, we get \(a^4+1=51 a^2\)
Similarly, \(b^4+1=51 b^2 \dots(ii)\)
Now, \(\frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}}=\frac{a^{17}\left(a^4+1\right)+b^{17}\left(b^4+1\right)}{a^{19}+b^{19}}\)
\(
\begin{aligned}
& =\frac{a^{17}\left(51 a^2\right)+b^{17}\left(51 b^2\right)}{a^{19}+b^{19}} \quad[\because \text { From Eq. (i) and (ii) }] \\
& =\frac{51\left[a^{19}+b^{19}\right]}{a^{19}+b^{19}}=51
\end{aligned}
\)

Explanation:

Identify the roots’ properties: The problem correctly uses the fact that \(\boldsymbol{a}\) and \(\boldsymbol{b}\) are roots to establish \(a^2-7 a-1=0\) and \(b^2-7 b-1=0\).
Derive key relationships: It correctly derives \(a^2-1=7 a\) and \(b^2-1=7 b\), which is used later to get \(a^4+1=51 a^2\) and \(b^4+1=51 b^2\) (after squaring both sides: \(\left(a^2-1\right)^2=(7 a)^2 \Rightarrow a^4-2 a^2+1=49 a^2 \Rightarrow a^4+1=51 a^2\) ).

Simplify the expression: The core of the solution is factoring the numerator of the expression \(\frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}}\) into \(\frac{a^{17}\left(a^4+1\right)+b^{17}\left(b^4+1\right)}{a^{19}+b^{19}}\).

Substitute and cancel: Substituting \(a^4+1=51 a^2\) and \(b^4+1=51 b^2\) allows the problem to simplify the expression to \(\frac{51\left(a^{19}+b^{19}\right)}{a^{19}+b^{19}}=51\).

Hint

(d) Step 1: Find the number of real roots for the first equation
The first equation is \(x^2-12 x+[x]+31=0\).
Let \(k=[x]\). Since \(k\) must be an integer, we can write the equation as \(x^2-12 x+31=-k\).
The roots of the quadratic equation \(x^2-12 x+(31+k)=0\) are given by the quadratic formula:
\(
x=\frac{-(-12) \pm \sqrt{(-12)^2-4(1)(31+k)}}{2(1)}=6 \pm \sqrt{5-k}
\)
For the roots to be real, the discriminant must be non-negative, so \(5-k \geq 0\), which implies \(k \leq 5\).
Also, by the definition of the greatest integer function, we have \(k \leq x<k+1\). Substituting the roots into this inequality, we get \(k \leq 6 \pm \sqrt{5-k}<k+1\).
For the root \(x_1=6+\sqrt{5-k}\) :
\(
k \leq 6+\sqrt{5-k} \Rightarrow k-6 \leq \sqrt{5-k}
\)
If \(k-6 \leq 0\) (i.e., \(k \leq 6\) ), the inequality holds for all valid \(k\). This is consistent with \(k \leq 5\)
\(
6+\sqrt{5-k}<k+1 \Rightarrow \sqrt{5-k}<k-5 .
\)
Since \(\sqrt{5-k} \geq 0\), this inequality can only hold if \(k-5>0\), i.e., \(k>5\). This contradicts \(k \leq 5\), so there are no roots for \(x_1\) that satisfy the condition.
For the root \(x_2=6-\sqrt{5-k}\) :
\(
k \leq 6-\sqrt{5-k} \Rightarrow \sqrt{5-k} \leq 6-k
\)
This implies \(k \leq 6\). We need to consider both cases:
If \(k>5\), this is impossible since \(\sqrt{5-k}\) would be imaginary.
If \(k \leq 5\), squaring both sides gives
\(
5-k \leq(6-k)^2 \Rightarrow 5-k \leq 36-12 k+k^2 \Rightarrow k^2-11 k+31 \geq 0.
\)
The discriminant of \(k^2-11 k+31\) is \((-11)^2-4(1)(31)=121-124=-3\), which is negative. Since the leading coefficient is positive, the quadratic is always positive, so this inequality holds for all real \(\boldsymbol{k}\).
\(6-\sqrt{5-k}<k+1 \Rightarrow 5-k<\sqrt{5-k}\).
Let \(\boldsymbol{u}=\sqrt{5-\boldsymbol{k}}\). The inequality becomes \(\boldsymbol{u}^2<\boldsymbol{u} \Rightarrow \boldsymbol{u}(\boldsymbol{u}-1)<0\). This implies \(0<u<1\).
Substituting back, \(0<\sqrt{5-k}<1\). Squaring gives \(0<5-k<1 \Rightarrow 4<k<5\).
Since \(k\) is an integer, there are no possible values for \(k\).
The only root comes from \(k=5\) where \(x=6\). For this, \([x]=6\), which does not equal \(k=5\). Thus, there are no real roots for the first equation, and we have \(m=0\).
Step 2: Find the number of real roots for the second equation
The second equation is \(x^2-5|x+2|-4=0\).
We need to consider two cases for the absolute value:
Case 1: \(x+2 \geq 0 \Rightarrow x \geq-2\).
The equation becomes \(x^2-5(x+2)-4=0 \Rightarrow x^2-5 x-14=0\).
Factoring this gives \((x-7)(x+2)=0\), so the roots are \(x=7\) and \(x=-2\). Both of these roots satisfy the condition \(x \geq-2\).
Case 2: \(x+2<0 \Rightarrow x<-2\).
The equation becomes \(x^2-5(-(x+2))-4=0 \Rightarrow x^2+5 x+6=0\).
Factoring this gives \((x+2)(x+3)=0\), so the roots are \(x=-2\) and \(x=-3\). Only \(x=-3\) satisfies the condition \(x<-2\).
Combining both cases, the real roots are \(x=7, x=-2\), and \(x=-3\).
Therefore, the number of real roots for the second equation is \(n=3\).
We found that \(m=0\) and \(n=3\). The value of \(m^2+m n+n^2\) is:
\(
m^2+m n+n^2=(0)^2+(0)(3)+(3)^2=0+0+9=9
\)

Hint

(c) Step 1: Find a relationship between the common root and \(\boldsymbol{a}\)
Let \(k\) be the common real root of the two quadratic equations. Substituting \(x=k\) into both equations gives:
\(
\begin{gathered}
k^2-5 a k+1=0 \dots(1)\\
k^2-a k-5=0 \dots(2)
\end{gathered}
\)
Subtracting equation (2) from equation (1) to eliminate \(k^2\) :
\(
\begin{gathered}
\left(k^2-5 a k+1\right)-\left(k^2-a k-5\right)=0 \\
-4 a k+6=0 \\
4 a k=6 \\
a k=\frac{3}{2} \\
k=\frac{3}{2 a}
\end{gathered}
\)
Step 2: Solve for \(\boldsymbol{a}\)
Substitute the value of \(\boldsymbol{k}\) from Step 1 into equation (2):
\(
\begin{gathered}
k^2-a k-5=0 \\
\left(\frac{3}{2 a}\right)^2-a\left(\frac{3}{2 a}\right)-5=0 \\
\frac{9}{4 a^2}-\frac{3}{2}-5=0 \\
\frac{9}{4 a^2}-\frac{13}{2}=0 \\
\frac{9}{4 a^2}=\frac{13}{2} \\
18=52 a^2 \\
a^2=\frac{18}{52}=\frac{9}{26}
\end{gathered}
\)
Since it’s given that \(a>0\), we have:
\(
a=\sqrt{\frac{9}{26}}=\frac{3}{\sqrt{26}}
\)
Step 3: Find the value of \(\boldsymbol{\beta}\)
The problem states that \(a=\frac{3}{\sqrt{2 \beta}}\). We have already found that \(a=\frac{3}{\sqrt{26}}\).
By comparing these two expressions for \(a\) :
\(
\begin{aligned}
& \frac{3}{\sqrt{2 \beta}}=\frac{3}{\sqrt{26}} \\
& \sqrt{2 \beta}=\sqrt{26}
\end{aligned}
\)
Squaring both sides:
\(
\begin{aligned}
2 \beta & =26 \\
\beta & =13
\end{aligned}
\)

Hint

(c) Given equation:
\(
x^7+3 x^5-13 x^3-15 x=0
\)
Factor out \(x\) :
\(
x\left(x^6+3 x^4-13 x^2-15\right)=0
\)
Thus, one root is:
\(
x=0=\alpha_7
\)
Substitution:
Let \(t=x^2\), so:
\(
t^3+3 t^2-13 t-15=0
\)
Factoring the cubic:
\(
(t+1)(t+5)(t-3)=0
\)
Hence:
\(
t=-1,-5,3
\)
Back-substitute \(t=x^2\) :
\(
x^2=-1, \quad x^2=-5, \quad x^2=3
\)
So:
\(
x= \pm i, \quad x= \pm i \sqrt{5}, \quad x= \pm \sqrt{3}
\)
Labeling the roots:
\(
\alpha_1, \alpha_2= \pm i \sqrt{5}, \quad \alpha_3, \alpha_4= \pm \sqrt{3}, \quad \alpha_5, \alpha_6= \pm i, \quad \alpha_7=0
\)
Compute the given expression:
\(
\alpha_1 \alpha_2-\alpha_3 \alpha_4+\alpha_5 \alpha_6
\)
Compute each product:
\(
\begin{gathered}
\alpha_1 \alpha_2=(i \sqrt{5})(-i \sqrt{5})=5 \\
\alpha_3 \alpha_4=(\sqrt{3})(-\sqrt{3})=-3 \\
\alpha_5 \alpha_6=(i)(-i)=1
\end{gathered}
\)
Substitute:
\(
\alpha_1 \alpha_2-\alpha_3 \alpha_4+\alpha_5 \alpha_6=5-(-3)+1=5+3+1=9
\)

Hint

(b) Let the given quadratic equation be \(x^2+60^{1 / 4} x+a=0\), with roots \(\alpha\) and \(\beta\).
According to Vieta’s formulas:
The sum of the roots is: \(\alpha+\beta=-60^{1 / 4}\)
The product of the roots is: \(\alpha \beta=a\)
The problem provides the condition \(\alpha^4+\beta^4=-30\). We can express \(\alpha^4+\beta^4\) using the elementary symmetric polynomials, ( \(\alpha+\beta\) ) and \(\alpha \beta\).
First, find \(\alpha^2+\beta^2\) :
\(
\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta
\)
Substitute the values from Step 1:
\(
\alpha^2+\beta^2=\left(-60^{1 / 4}\right)^2-2 a=60^{1 / 2}-2 a
\)
Now, find \(\alpha^4+\beta^4\) :
\(
\alpha^4+\beta^4=\left(\alpha^2+\beta^2\right)^2-2(\alpha \beta)^2
\)
Substitute the expressions for \(\alpha^2+\beta^2\) and \(\alpha \beta\) :
\(
\alpha^4+\beta^4=\left(60^{1 / 2}-2 a\right)^2-2(a)^2
\)
Set this equal to the given value:
\(
\left(60^{1 / 2}-2 a\right)^2-2 a^2=-30
\)
Step 3: Solve for the possible values of ‘\(a\)‘
Expand and simplify the equation from Step 2:
\(
\begin{gathered}
\left(60^{1 / 2}\right)^2-2\left(60^{1 / 2}\right)(2 a)+(2 a)^2-2 a^2=-30 \\
60-4 a\left(60^{1 / 2}\right)+4 a^2-2 a^2=-30 \\
2 a^2-4 a \sqrt{60}+60=-30 \\
2 a^2-4 a \sqrt{60}+90=0
\end{gathered}
\)
Divide the entire equation by 2 :
\(
a^2-2 a \sqrt{60}+45=0
\)
This is a quadratic equation in the variable \(\boldsymbol{a}\). The values of \(\boldsymbol{a}\) are the roots of this equation.
The quadratic equation for \(a\) is \(a^2-2 \sqrt{60} a+45=0\). Let the two possible values of \(a\) be \(a_1\) and \(a_2\). According to Vieta’s formulas for this new equation, the product of the roots is the constant term.
\(
a_1 \cdot a_2=\frac{45}{1}=45
\)

Hint

(c)

\(D \geq 0 \Rightarrow 4-4|\lambda-3| \geq 0\)
\(|\lambda-3| \leq 1\)
\(-1 \leq \lambda-3 \leq 1\)
\(2 \leq \lambda \leq 4\)
\(|x|=\frac{2 \pm \sqrt{4-4|\lambda-3|}}{2}\)
\(=1 \pm \sqrt{1-|\lambda-3|}\)
\(x_{\text {largest }} \quad=1+1=2\), when \(\lambda=3\)
Largest element of \(S=2+3=5\)

Explanation: The equation E is \(|x|^2-2|x|+|\lambda-3|=0\). The set is \(S=\{x+\lambda: x\) is an integer solution of E\(\}\). This means \(\lambda\) is a parameter, and for a given \(\lambda\) that allows for integer solutions, we have a set of \(\boldsymbol{x}\) values. The set S contains values of \(x+\lambda\) for all possible such \(\lambda\) and their corresponding \(x\) solutions.
The possible values for \(x+\lambda\) are:
When \(\lambda=2\), the integer solutions are \(x=1,-1\).
\(1+2=3\)
\(-1+2=1\)
When \(\lambda=3\), the integer solutions are \(x=0,2,-2\).
\(0+3=3\)
\(2+3=5\)
\(-2+3=1\)
When \(\lambda=4\), the integer solutions are \(x=1,-1\).
\(1+4=5\)
\(-1+4=3\)
The set of all possible values for \(x+\lambda\) is \(\{1,3,5\}\). The largest value in this set is 5.

Hint

(c) \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(x^2-x-4=0\).
\(\therefore \alpha\) and \(\beta\) are satisfy the given equation.
\(
\begin{aligned}
& \alpha^2-\alpha-4=0 \\
& \Rightarrow \alpha^{n+1}-\alpha^n-4 \alpha^{n-1}=0 \dots(1)
\end{aligned}
\)
and \(\beta^2-\beta-4=0\)
\(
\Rightarrow \beta^{n+1}-\beta^n-4 \beta^{n-1}=0 \dots(2)
\)
Subtracting (2) from (1), we get,
\(
\begin{aligned}
& \left(\alpha^{n+1}-\beta^{n+1}\right)-\left(\alpha^n-\beta^n\right)-4\left(\alpha^{n-1}-\beta^{n-1}\right)=0 \\
& \Rightarrow P_{n+1}-P_n-4 P_{n-1}=0 \\
& \Rightarrow P_{n+1}=P_n+4 P_{n-1} \\
& \Rightarrow P_{n+1}-P_n=4 P_{n-1}
\end{aligned}
\)
For \(n=14, P_{15}-P_{14}=4 P_{13}\)
For \(n=15, P_{16}-P_{15}=4 P_{14}\)
Now, \(\frac{P_{15} P_{16}-P_{14} P_{16}-P_{15}^2+P_{14} P_{15}}{P_{13} P_{14}}\)
\(
\begin{aligned}
& =\frac{P_{16}\left(P_{15}-P_{14}\right)-P_{15}\left(P_{15}-P_{14}\right)}{P_{13} P_{14}} \\
& =\frac{\left(P_{15}-P_{14}\right)\left(P_{16}-P_{15}\right)}{P_{13} P_{14}} \\
& =\frac{\left(4 P_{13}\right)\left(4 P_{14}\right)}{P_{13} P_{14}} \\
& =16
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
&\begin{aligned}
& \frac{3 x^2-9 x+17}{x^2+3 x+10}=\frac{5 x^2-7 x+19}{3 x^2+5 x+12} \\
& \Rightarrow \frac{3 x^2-9 x+17}{5 x^2-7 x+19}=\frac{x^2+3 x+10}{3 x^2+5 x+12} \\
& \frac{-2 x^2-2 x-2}{5 x^2-7 x+19}=\frac{-2 x^2-2 x-2}{3 x^2+5 x+12}
\end{aligned}\\
&\text { Either } x^2+x+1=0 \text { or No real roots }\\
&\begin{aligned}
& \Rightarrow 5 x^2-7 x+19=3 x^2+5 x+12 \\
& 2 x^2-12 x+7=0 \\
& \text { sum of roots }=6
\end{aligned}
\end{aligned}
\)

Explanation: Step 1: Analyze the given equation
The first step in the provided solution rearranges the terms of the equation:
\(
\frac{3 x^2-9 x+17}{5 x^2-7 x+19}=\frac{x^2+3 x+10}{3 x^2+5 x+12}
\)
This is achieved by applying a property of proportions.
The next step subtracts the numerator from the denominator on each side of the equation:
\(
\frac{\left(3 x^2-9 x+17\right)-\left(5 x^2-7 x+19\right)}{5 x^2-7 x+19}=\frac{\left(x^2+3 x+10\right)-\left(3 x^2+5 x+12\right)}{3 x^2+5 x+12}
\)
This simplifies to:
\(
\frac{-2 x^2-2 x-2}{5 x^2-7 x+19}=\frac{-2 x^2-2 x-2}{3 x^2+5 x+12}
\)
Step 2: Solve for x
There are two possibilities from the equation derived in the previous step:
1. The numerators are zero.
2. The denominators are equal.
Case 1: Numerators are zero
\(
-2 x^2-2 x-2=0
\)
Dividing by -2, we get:
\(
x^2+x+1=0
\)
To find the roots, we use the quadratic formula \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\) :
\(
x=\frac{-1 \pm \sqrt{1^2-4(1)(1)}}{2(1)}=\frac{-1 \pm \sqrt{1-4}}{2}=\frac{-1 \pm \sqrt{-3}}{2}
\)
Since the discriminant ( \(b^2-4 a c\) ) is negative, there are no real roots in this case.
Case 2: Denominators are equal
\(
5 x^2-7 x+19=3 x^2+5 x+12
\)
Rearranging the terms to form a quadratic equation:
\(
2 x^2-12 x+7=0
\)
This is a quadratic equation of the form \(a x^2+b x+c=0\). According to Vieta’s formulas, the sum of the roots is given by the formula \(-\frac{b}{a}\).
The quadratic equation from the second case, \(2 x^2-12 x+7=0\), has roots whose sum is:
\(
\text { Sum of roots }=-\frac{-12}{2}=6
\)

Hint

(b) The roots of the equation \(x^2+2 x-8=0\) are found by factoring as \((x+4)(x-2)=0\), which gives \(x=2\) and \(x=-4\).
Let the given equation be
\(
E:\left(p^2+q^2\right) x^2-2 q(p+r) x+q^2+r^2=0
\)
One of the roots of \(x^2+2 x-8=0\) is also a root of \(E\).
If \(x=2\) is a root of \(E\), then:
\(
\begin{gathered}
\left(p^2+q^2\right)(2)^2-2 q(p+r)(2)+q^2+r^2=0 \\
4 p^2+4 q^2-4 p q-4 q r+q^2+r^2=0 \\
4 p^2+5 q^2-4 p q-4 q r+r^2=0
\end{gathered}
\)
This can be rewritten by rearranging terms as a sum of squares:
\(
\begin{gathered}
(2 p-q)^2+4 q^2-4 q r+r^2=0 \\
(2 p-q)^2+(2 q-r)^2=0
\end{gathered}
\)
Since \(p, q, r \in \mathrm{R}\), this requires \(2 p-q=0\) and \(2 q-r=0\). So \(q=2 p\) and \(r=2 q=4 p\). In this case, \(p, q, r\) would all have the same sign (or all be zero). The problem states that not all of \(p, q, r\) have the same sign. Thus, \(x=2\) is not the common root.
If \(x=-4\) is a root of \(E\), then:
\(
\begin{gathered}
\left(p^2+q^2\right)(-4)^2-2 q(p+r)(-4)+q^2+r^2=0 \\
16\left(p^2+q^2\right)+8 q(p+r)+q^2+r^2=0 \\
16 p^2+16 q^2+8 p q+8 q r+q^2+r^2=0 \\
16 p^2+17 q^2+8 p q+8 q r+r^2=0
\end{gathered}
\)
Rearranging terms:
\(
\begin{gathered}
(4 p+q)^2+16 q^2+8 q r+r^2=0 \\
(4 p+q)^2+(4 q+r)^2=0
\end{gathered}
\)
This requires \(4 p+q=0\) and \(4 q+r=0\). So \(q=-4 p\) and \(r=-4 q=16 p\).
In this case, \(q\) and \(p\) have opposite signs, and \(r\) and \(q\) have opposite signs. This satisfies the condition that not all of \(\boldsymbol{p}, \boldsymbol{q}, \boldsymbol{r}\) have the same sign.
Given \(q=-4 p\) and \(r=16 p\), we can find the value of \(\frac{q^2+r^2}{p^2}\) :
\(
\frac{q^2+r^2}{p^2}=\frac{(-4 p)^2+(16 p)^2}{p^2}=\frac{16 p^2+256 p^2}{p^2}=\frac{272 p^2}{p^2}=272
\)

Hint

(c) Given polynomial
\(
x^8-x^7-x^6+x^5+3 x^4-4 x^3-2 x^2+4 x-1=0
\)
Step 1: Group and factor by ( \(x-1\) )
You rewrote it as
\(
x^7(x-1)-x^5(x-1)+3 x^3(x-1)-x\left(x^2-1\right)+2 x(1-x)+(x-1)=0
\)
and hence
\(
(x-1)\left(x^7-x^5+3 x^3-x(x+1)-2 x+1\right)=0
\)
Step 2: Simplify the second factor
\(
x^7-x^5+3 x^3-x^2-3 x+1
\)
You noticed:
\(
x^7-x^5+3 x^3-x^2-3 x+1=\left(x^2-1\right)\left(x^5+3 x-1\right)
\)
Let’s check:
\(
\begin{gathered}
\left(x^2-1\right)\left(x^5+3 x-1\right)=x^7+3 x^3-x^2-x^5-3 x+1 \\
=x^7-x^5+3 x^3-x^2-3 x+1
\end{gathered}
\)
Step 3: Complete factorization
\(
\begin{gathered}
(x-1)\left(x^2-1\right)\left(x^5+3 x-1\right)=0 \\
\Rightarrow(x-1)^2(x+1)\left(x^5+3 x-1\right)=0
\end{gathered}
\)
Step 4: Roots
1. \(x=1\) (with multiplicity 2 )
2. \(x=-1\)
3. Roots of \(x^5+3 x-1=0\)
Step 5: Check monotonicity of \(x^5+3 x-1\)
\(
f^{\prime}(x)=5 x^4+3>0 \forall x
\)
\(\Rightarrow\) strictly increasing \(\Rightarrow\) exactly one real root.
\(x=1\) (double root)
\(x=-1\)
One distinct real root of \(x^5+3 x-1=0\)
Thus, 3 distinct real roots (counting \(x=1\) only once).
\(
\text { Total real roots }=3
\)

Explanation:

A function is strictly increasing if its derivative is always positive. This property, combined with the Intermediate Value Theorem, guarantees that the function has exactly one real root.

Explanation of Concepts
1. Positive Derivative and Monotonicity: The sign of the derivative, \(f^{\prime}(x)\), determines whether a function is increasing or decreasing. If the derivative of a function is positive for all values of \(x\), i.e., \(f^{\prime}(x)>0\), it means the function’s slope is always positive. This implies that as \(x\) increases, the value of the function \(f(x)\) also consistently increases, making it a strictly increasing function. For the given function, \(f^{\prime}(x)=5 x^4+3\), the term \(5 x^4\) is always non-negative, so the entire expression is always greater than 3, and thus always positive.
2. Exactly One Real Root:
1. Existence: A polynomial function like \(f(x)=x^5+3 x-1\) is continuous. We can find values of \(x\) for which the function is negative and positive. For example, \(f(0)=-1\) and \(f(1)=3\). According to the Intermediate Value Theorem, because the function is continuous and changes sign, it must cross the \(x\)-axis at least once between \(x=0\) and \(x=1\). This guarantees the existence of at least one real root.
2. Uniqueness: A strictly increasing function can only cross the \(x\)-axis once. If it were to cross the \(x\)-axis a second time, it would have to change direction and start decreasing to get back to a value of zero, which would mean its derivative would have to be negative at some point. Since we’ve established that the derivative is always positive, the function cannot change direction, so there can be only one point where it equals zero.

Hint

(b) Given equation
\(
e^{4 x}+4 e^{3 x}-58 e^{2 x}+4 e^x+1=0
\)
Dividing both sides by \(e^{2 x}\) (which is always positive):
\(
e^{2 x}+4 e^x-58+4 e^{-x}+e^{-2 x}=0
\)
Substitution
Let
\(
t=e^x+e^{-x}, \quad \text { where } t \geq 2
\)
Then
\(
e^{2 x}+e^{-2 x}=\left(e^x+e^{-x}\right)^2-2=t^2-2
\)
Substitute into the equation:
\(
\begin{gathered}
\left(t^2-2\right)+4 t-58=0 \\
\Rightarrow t^2+4 t-60=0
\end{gathered}
\)
Solve for \(t\)
\(
t=\frac{-4 \pm \sqrt{16+240}}{2}=\frac{-4 \pm 16}{2}=6 \text { or }-10
\)
Since \(t=e^x+e^{-x} \geq 2\), only \(t=6\) is valid.
Now solve for \(\boldsymbol{x}\)
\(
e^x+e^{-x}=6
\)
Multiply both sides by \(e^x\) :
\(
e^{2 x}-6 e^x+1=0
\)
Let \(p=e^x>0\) :
\(
\begin{gathered}
p^2-6 p+1=0 \\
p=3 \pm 2 \sqrt{2}
\end{gathered}
\)
(simplifying \(p=\frac{6 \pm \sqrt{36-4}}{2}=3 \pm 2 \sqrt{2}\) ).
Therefore
\(
x=\ln (3+2 \sqrt{2}) \quad \text { or } \quad \ln (3-2 \sqrt{2})
\)

Hint

(d) Step 1: Set up the equations and relationships
(a) For the equation \(x^2-4 \lambda x+5=0\) with roots \(\alpha\) and \(\beta\) :
1. \(\alpha+\beta=4 \lambda\)
2. \(\alpha \beta=5\)
(b) For the equation \(x^2-(3 \sqrt{2}+2 \sqrt{3}) x+7+3 \sqrt{3} \lambda=0\) with roots \(\alpha\) and \(\gamma\).
1. \(\alpha+\gamma=3 \sqrt{2}+2 \sqrt{3}\)
2. \(\alpha \gamma=7+3 \sqrt{3} \lambda\)
(c) We are also given the condition: \(\beta+\gamma=3 \sqrt{2}\).
Step 2: Determine the value of \(\boldsymbol{\alpha}\)
By adding the sums of the roots from both equations, we have:
\(
\begin{gathered}
(\alpha+\beta)+(\alpha+\gamma)=4 \lambda+(3 \sqrt{2}+2 \sqrt{3}) \\
2 \alpha+(\beta+\gamma)=4 \lambda+3 \sqrt{2}+2 \sqrt{3}
\end{gathered}
\)
Now, substitute the given value \(\beta+\gamma=3 \sqrt{2}\) into this equation:
\(
\begin{aligned}
2 \alpha+3 \sqrt{2} & =4 \lambda+3 \sqrt{2}+2 \sqrt{3} \\
2 \alpha & =4 \lambda+2 \sqrt{3} \\
\alpha & =2 \lambda+\sqrt{3}
\end{aligned}
\)
Step 3: Find the value of \(\lambda\)
We can use the product of the roots from the first quadratic equation, \(\alpha \beta=5\). We know that \(\beta=4 \lambda-\alpha\).
Substitute \(\alpha=2 \lambda+\sqrt{3}\) :
\(
\beta=4 \lambda-(2 \lambda+\sqrt{3})=2 \lambda-\sqrt{3}
\)
Now we can use the product of the roots from the second quadratic equation, \(\alpha \gamma=7+3 \sqrt{3} \lambda\). We know that \(\gamma=3 \sqrt{2}+2 \sqrt{3}-\alpha\).
Substitute \(\alpha=2 \lambda+\sqrt{3}\) :
\(
\gamma=3 \sqrt{2}+2 \sqrt{3}-(2 \lambda+\sqrt{3})=3 \sqrt{2}+\sqrt{3}-2 \lambda
\)
Substitute \(\beta\) and \(\gamma\) into the given condition \(\beta+\gamma=3 \sqrt{2}\) :
\(
\begin{gathered}
(2 \lambda-\sqrt{3})+(3 \sqrt{2}+\sqrt{3}-2 \lambda)=3 \sqrt{2} \\
3 \sqrt{2}=3 \sqrt{2}
\end{gathered}
\)
This confirms that the relations are consistent but does not help us find \(\lambda\).
We can use the common root \(\boldsymbol{\alpha}\) to find \(\boldsymbol{\lambda}\). Substitute the expression for \(\boldsymbol{\alpha}\) into the first quadratic equation:
\(
\begin{gathered}
(2 \lambda+\sqrt{3})^2-4 \lambda(2 \lambda+\sqrt{3})+5=0 \\
4 \lambda^2+4 \sqrt{3} \lambda+3-8 \lambda^2-4 \sqrt{3} \lambda+5=0 \\
-4 \lambda^2+8=0 \\
\lambda^2=2
\end{gathered}
\)
Since \(\lambda>0\), we get \(\lambda=\sqrt{2}\).
Step 4: Calculate the final expression
We need to find the value of \((\alpha+2 \beta+\gamma)^2\).
We can rewrite the expression as \((\alpha+\beta+\beta+\gamma)^2\).
Using the relations from Step 1 , we know that \(\alpha+\beta=4 \lambda\) and \(\beta+\gamma=3 \sqrt{2}\).
Substitute the value of \(\lambda=\sqrt{2}\) :
\(
\alpha+\beta=4 \sqrt{2}
\)
Now, substitute these values back into the expression:
\(
(\alpha+\beta+\beta+\gamma)^2=(4 \sqrt{2}+3 \sqrt{2})^2=(7 \sqrt{2})^2=49 \times 2=98
\)

Hint

(a) Step 1: Simplify the equation
The given equation is \(e^{2 x}-11 e^x-45 e^{-x}+\frac{81}{2}=0\).
To simplify, let \(y=e^x\). Since \(e^x\) is always positive, \(y>0\). The equation becomes:
\(
y^2-11 y-45 y^{-1}+\frac{81}{2}=0
\)
To eliminate the fractions, multiply the entire equation by \(2 y\) :
\(
\begin{aligned}
& 2 y^3-22 y^2-90+81 y=0 \\
& 2 y^3-22 y^2+81 y-90=0
\end{aligned}
\)
This is a cubic equation in \(y\). Let the roots of this cubic equation be \(y_1, y_2\), and \(y_3\).
Step 2: Find the sum of the roots of the original equation
The roots of the original equation are the values of \(\boldsymbol{x}\). If \(y_i\) are the roots of the cubic equation, then \(e^{x_i}=y_i\). This means \(x_i=\log _e\left(y_i\right)\). The sum of the roots of the original equation is:
\(
\begin{gathered}
\operatorname{Sum}=x_1+x_2+x_3 \\
\operatorname{Sum}=\log _e\left(y_1\right)+\log _e\left(y_2\right)+\log _e\left(y_3\right)
\end{gathered}
\)
Using the property of logarithms, \(\log A+\log B+\log C=\log (A B C)\) :
\(
\operatorname{Sum}=\log _e\left(y_1 y_2 y_3\right)
\)
According to Vieta’s formulas, the product of the roots of a cubic equation \(a y^3+b y^2+c y+d=0\) is given by \(-\frac{d}{a}\). For our cubic equation \(2 y^3-22 y^2+81 y-90=0\), we have \(a=2\) and \(d=-90\).
The product of the roots \(y_1 y_2 y_3\) is:
\(
y_1 y_2 y_3=-\frac{-90}{2}=45
\)
Substitute this value back into the sum of the roots:
\(
\operatorname{Sum}=\log _e(45)
\)
The problem states that the sum of all the roots is \(\log _e p\). Therefore:
\(
\log _e p=\log _e 45
\)
This implies that \(\boldsymbol{p}=45\).

Hint

(d) We have to find the value of \(\left(\frac{1}{p}+\frac{1}{q}\right)^{-2}\) given that \(p+q=3\) and \(p^4+q^4=369\).
Express \(p^4+q^4\) in terms of \(p+q\) and \(p q\). The equation \(\left(p^4+q^4\right)=\left(p^2+q^2\right)^2-2 p^2 q^2=\left((p+q)^2-2 p q\right)^2-2(p q)^2\) is used.
Substitute known values: Substituting \(p+q=3\) into the equation yields \(\left(3^2-2 p q\right)^2-2(p q)^2=369\), which simplifies to \((9-2 p q)^2-2(p q)^2=369\).
Form a quadratic equation for \(p q\). Expanding and simplifying results in the quadratic equation \((p q)^2-18 p q-144=0\).
Solve for \(p q\) : The quadratic equation is solved, giving two possible values for \(p q\) : -6 or 24.
Eliminate the extraneous solution: The text states that \(p q=24\) is not possible (this would result in non-real values for \(p\) and \(q\) when combined with \(p+q=3\) ), leaving \(p q=-6\) as the valid solution.
Calculate the final expression: The expression \(\left(\frac{1}{p}+\frac{1}{q}\right)^{-2}\) is simplified to \(\left(\frac{p+q}{p q}\right)^{-2}=\left(\frac{p q}{p+q}\right)^2\).
Substitute the final values: Plugging in \(p+q=3\) and \(p q=-6\) gives
\(
\left(\frac{-6}{3}\right)^2=(-2)^2=4
\)

Hint

(c) The given equation is \(x^4-3 x^3-2 x^2+3 x+1=0\), which factors into \(\left(x^2-1\right)\left(x^2-3 x-1\right)=0\).
The roots of the equation are:
1. The roots of \(x^2-1=0\), which are 1 and -1.
2. The roots of \(x^2-3 x-1=0\), denoted as \(\alpha\) and \(\beta\).

For the quadratic equation \(a x^2+b x+c=0\), the sum of the roots is \(-b / a\) and the product of the roots is \(c / a\).
For the equation \(x^2-3 x-1=0\) :
Sum of roots: \(\alpha+\beta=-(-3) / 1=3\)
Product of roots: \(\alpha \beta=-1 / 1=-1\)
The sum of the cubes of the roots of the original equation is:
\(
1^3+(-1)^3+\alpha^3+\beta^3
\)
This simplifies to:
\(
1-1+\alpha^3+\beta^3=\alpha^3+\beta^3
\)
Using the identity \(\alpha^3+\beta^3=(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta)\) :
\(
3^3-3(-1)(3)=27-(-9)=27+9=36
\)

Hint

(c) Step 1: Construct a new polynomial
Let’s define a new polynomial, \(g(x)\), such that \(g(x)=x f(x)+2\). Since \(f(x)\) is a polynomial of degree \(3, x f(x)\) is a polynomial of degree 4. Thus, \(g(x)\) is a polynomial of degree 4.
From the given condition, we know that for \(k=2,3,4,5, g(k)=k f(k)+2=0\). This means that \(2,3,4,5\) are the roots of the polynomial \(g(x)\).
Step 2: Express \(\boldsymbol{g}(\boldsymbol{x})\) in factored form
Since we know the roots of \(g(x)\), we can write it in factored form with a constant factor, C.
\(
g(x)=C(x-2)(x-3)(x-4)(x-5)
\)
Step 3: Find the value of the constant C
We have the relationship \(g(x)=x f(x)+2\), which can be rearranged to get \(x f(x)=g(x)-2\). Dividing by \(x\) gives \(f(x)=\frac{g(x)-2}{x}\).
Since \(f(x)\) is a polynomial, the expression \(g(x)-2\) must have a root at \(x=0\).
\(
g(0)-2=0 \Longrightarrow g(0)=2
\)
Using the factored form of \(g(x)\), we can solve for \(C\) by substituting \(x=0\) :
\(
g(0)=C(0-2)(0-3)(0-4)(0-5)=C(-2)(-3)(-4)(-5)=120 C
\)
Since \(g(0)=2\), we have \(120 C=2\), which gives us \(C=\frac{2}{120}=\frac{1}{60}\).
So, the full expression for \(g(x)\) is:
\(
g(x)=\frac{1}{60}(x-2)(x-3)(x-4)(x-5)
\)
Step 4: Calculate \(\boldsymbol{f}(\mathbf{1 0})\)
We need to find the value of \(f(10)\). Using the relationship \(x f(x)=g(x)-2\), we can find \(10 f(10)\) by substituting \(x=10\) :
\(
10 f(10)=g(10)-2
\)
First, calculate \(g(10)\) :
\(
\begin{gathered}
g(10)=\frac{1}{60}(10-2)(10-3)(10-4)(10-5) \\
g(10)=\frac{1}{60}(8)(7)(6)(5) \\
g(10)=\frac{8 \cdot 7 \cdot 6 \cdot 5}{60}=\frac{1680}{60}=28
\end{gathered}
\)
Now, substitute this value back to find \(10 f(10)\) :
\(
10 f(10)=28-2=26
\)
The value of \(52-10 f(10)\) is:
\(
52-10 f(10)=52-26=26
\)

Hint

(c) Step 1: Analyze the given equations and roots
We are given two quadratic equations.
For the first equation, \(x^2-x+2 \lambda=0\), the roots are \(\alpha\) and \(\beta\).
Using Vieta’s formulas, we have:
Sum of roots: \(\alpha+\beta=-(-1) / 1=1 \dots(1)\)
Product of roots: \(\alpha \beta=2 \lambda / 1=2 \lambda \dots(2)\)
For the second equation, \(3 x^2-10 x+27 \lambda=0\), the roots are \(\alpha\) and \(\gamma\).
Using Vieta’s formulas, we have:
Sum of roots: \(\alpha+\gamma=-(-10) / 3=10 / 3 \dots(3)\)
Product of roots: \(\alpha \gamma=27 \lambda / 3=9 \lambda \dots(4)\)
Step 2: Solve for the common root \(\boldsymbol{\alpha}\)
From equation (1), we can express \(\beta\) in terms of \(\alpha: \beta=1-\alpha\).
From equation (3), we can express \(\gamma\) in terms of \(\alpha: \gamma=\frac{10}{3}-\alpha\).
Since \(\alpha\) is a root of both equations, it must satisfy both. We can use the relationships from the product of roots to find the value of \(\alpha\).
From equation (2), \(\beta=\frac{2 \lambda}{\alpha}\).
From equation (4), \(\gamma=\frac{9 \lambda}{\alpha}\).
Substitute these into the expressions for \(\beta\) and \(\gamma\) from the sum of roots:
\(
\begin{aligned}
& \frac{2 \lambda}{\alpha}=1-\alpha \Longrightarrow 2 \lambda=\alpha-\alpha^2 \dots(5) \\
& \frac{9 \lambda}{\alpha}=\frac{10}{3}-\alpha \Longrightarrow 9 \lambda=\alpha\left(\frac{10}{3}-\alpha\right)=\frac{10}{3} \alpha-\alpha^2 \dots(6)
\end{aligned}
\)
Now, we can solve for \(\alpha\). From equation (5), \(\alpha^2-\alpha=-2 \lambda\). Substitute this into equation (6):
\(
\begin{gathered}
9 \lambda=\frac{10}{3} \alpha-(\alpha-2 \lambda)=\frac{10}{3} \alpha-\alpha+2 \lambda \\
7 \lambda=\frac{7}{3} \alpha \Longrightarrow \lambda=\frac{\alpha}{3}
\end{gathered}
\)
Substitute \(\lambda=\frac{\alpha}{3}\) back into equation (5):
\(
\begin{gathered}
2\left(\frac{\alpha}{3}\right)=\alpha-\alpha^2 \\
\frac{2 \alpha}{3}=\alpha-\alpha^2 \\
\alpha^2-\alpha+\frac{2 \alpha}{3}=0 \\
\alpha^2-\frac{\alpha}{3}=0 \\
\alpha\left(\alpha-\frac{1}{3}\right)=0
\end{gathered}
\)
The roots are \(\alpha=0\) and \(\alpha=1 / 3\). Since \(\lambda \neq 0\), from \(\lambda=\frac{\alpha}{3}\), we know \(\alpha \neq 0\). Therefore, \(\alpha=\frac{1}{3}\).
Step 3: Find the value of \(\frac{\beta \gamma}{\lambda}\)
Using the value of \(\alpha=\frac{1}{3}\), we can find \(\lambda\) :
\(
\lambda=\frac{\alpha}{3}=\frac{1 / 3}{3}=\frac{1}{9}
\)
Now we find \(\boldsymbol{\beta}\) and \(\gamma\).
From equation (2): \(\beta=\frac{2 \lambda}{\alpha}=\frac{2(1 / 9)}{1 / 3}=\frac{2}{9} \times 3=\frac{2}{3}\)
From equation (4): \(\gamma=\frac{9 \lambda}{\alpha}=\frac{9(1 / 9)}{1 / 3}=\frac{1}{1 / 3}=3\)
Finally, we can calculate the value of the expression \(\frac{\beta \gamma}{\lambda}\) :
\(
\frac{\beta \gamma}{\lambda}=\frac{(2 / 3)(3)}{1 / 9}=\frac{2}{1 / 9}=2 \times 9=18
\)

Hint

(d) Step 1: Simplify the given equation
The given equation is \(\frac{2}{x-1}-\frac{1}{x-2}=\frac{2}{k}\). To solve for \(x\), we first combine the fractions on the left side. The domain of \(x\) is all real numbers except \(x=1\) and \(x=2\)
\(
\begin{gathered}
\frac{2(x-2)-1(x-1)}{(x-1)(x-2)}=\frac{2}{k} \\
\frac{2 x-4-x+1}{x^2-3 x+2}=\frac{2}{k} \\
\frac{x-3}{x^2-3 x+2}=\frac{2}{k}
\end{gathered}
\)
Cross-multiplying, we get:
\(
\begin{gathered}
k(x-3)=2\left(x^2-3 x+2\right) \\
k x-3 k=2 x^2-6 x+4
\end{gathered}
\)
Rearranging the terms into a standard quadratic form \(\boldsymbol{A} \boldsymbol{x}^2+\boldsymbol{B} \boldsymbol{x}+\boldsymbol{C}=\mathbf{0}\) :
\(
2 x^2-(6+k) x+(4+3 k)=0
\)
Step 2: Determine the condition for no real roots
A quadratic equation \(\boldsymbol{A} \boldsymbol{x}^2+\boldsymbol{B} \boldsymbol{x}+\boldsymbol{C}=\mathbf{0}\) has no real roots if its discriminant, \(\Delta=B^2-4 A C\), is less than zero. In our equation, \(A=2, B=-(6+k)\), and \(C=4+3 k\).
\(
\begin{gathered}
\Delta=[-(6+k)]^2-4(2)(4+3 k)<0 \\
(6+k)^2-8(4+3 k)<0 \\
36+12 k+k^2-32-24 k<0 \\
k^2-12 k+4<0
\end{gathered}
\)
Step 3: Find the integer values of \(k\)
To find the values of \(k\) that satisfy the inequality \(k^2-12 k+4<0\), we first find the roots of the corresponding quadratic equation \(k^2-12 k+4=0\) using the quadratic formula \(k=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\).
\(
k=\frac{-(-12) \pm \sqrt{(-12)^2-4(1)(4)}}{2(1)}
\)
\(
k=6 \pm 4 \sqrt{2}
\)
The roots are approximately \(k_1=6-4 \sqrt{2} \approx 6-4(1.414)=6-5.656=0.344\) and \(k_2=6+4 \sqrt{2} \approx 6+5.656=11.656\).
Since the parabola \(y=k^2-12 k+4\) opens upwards, the inequality \(k^2-12 k+4<0\) is satisfied for values of \(k\) between the two roots:
\(
\begin{gathered}
6-4 \sqrt{2}<k<6+4 \sqrt{2} \\
0.344<k<11.656
\end{gathered}
\)
The integral values of \(\boldsymbol{k}\) that satisfy this condition are \(\mathbf{1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 1 0 , 1 1}\). The problem also states that \(k \neq 0\). All these values meet this condition.
Step 4: Calculate the sum of the integral values of \(k\)
The sum of the integral values of \(\boldsymbol{k}\) from 1 to 11 is:
\(
\text { Sum }=1+2+3+4+5+6+7+8+9+10+11=66
\)
This can also be calculated using the formula for the sum of an arithmetic series: \(\frac{n(n+1)}{2}\), where \(n=11\).
\(
\operatorname{Sum}=\frac{11(11+1)}{2}=\frac{11 \times 12}{2}=66
\)
The sum of all integral values of \(\boldsymbol{k}\) is \(\mathbf{6 6}\).

Hint

(b) Step 1: Divide the original equation by \(t^2\)
The given equation is \(t^4-t^3-4 t^2-t+1=0\). Since \(t=0\) is not a solution, we can divide the entire equation by \(t^2\) :
\(
\frac{t^4}{t^2}-\frac{t^3}{t^2}-\frac{4 t^2}{t^2}-\frac{t}{t^2}+\frac{1}{t^2}=0
\)
This simplifies to:
\(
t^2-t-4-\frac{1}{t}+\frac{1}{t^2}=0
\)
Step 2: Group terms and substitute
Rearrange the terms to group similar powers of \(t\) :
\(
\left(t^2+\frac{1}{t^2}\right)-\left(t+\frac{1}{t}\right)-4=0
\)
Introduce the substitution \(\alpha=t+\frac{1}{t}\). We know that \(\alpha^2=\left(t+\frac{1}{t}\right)^2=t^2+2+\frac{1}{t^2}\), so we can express \(t^2+\frac{1}{t^2}\) as \(\alpha^2-2\).
Substitute these into the equation:
\(
\left(\alpha^2-2\right)-\alpha-4=0
\)
This simplifies to a quadratic equation in \(\alpha\).
\(
\alpha^2-\alpha-6=0
\)
Step 3: Solve for \(\boldsymbol{\alpha}\)
We solve the quadratic equation \(\alpha^2-\alpha-6=0\) by factoring:
\(
(\alpha-3)(\alpha+2)=0
\)
This gives two possible values for \(\alpha\).
\(
\alpha=3 \text { or } \alpha=-2
\)
Given that \(t=e^x>0\), we know that \(t+\frac{1}{t} \geq 2\) for all \(t>0\). Therefore, we must reject the solution \(\alpha=-2\). We only use \(\alpha=3\).
Step 4: Solve for \(\boldsymbol{t}\)
Using the valid value \(\alpha=3\), we substitute back into the expression for \(\alpha\) :
\(
t+\frac{1}{t}=3
\)
To solve for \(t\), multiply the entire equation by \(t\) :
\(
t^2+1=3 t
\)
Rearranging gives another quadratic equation in \(t\) :
\(
t^2-3 t+1=0
\)
We can solve this using the quadratic formula \(t=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\) :
\(
t=\frac{3 \pm \sqrt{(-3)^2-4(1)(1)}}{2(1)}=\frac{3 \pm \sqrt{9-4}}{2}=\frac{3 \pm \sqrt{5}}{2}
\)
The two positive solutions for \(t\) are \(t_1=\frac{3+\sqrt{5}}{2}\) and \(t_2=\frac{3-\sqrt{5}}{2}\). Since both values are positive, we can find real roots for \(x\).
The number of real roots is 2.

Hint

(b) Step 1: Find the value of \(a^2+b^2+c^2\)
The value of \(a^2+b^2+c^2\) can be found using the identity \((a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)\). Rearranging this, we get:
\(
a^2+b^2+c^2=(a+b+c)^2-2(a b+b c+c a)
\)
Given \(a+b+c=1\) and \(a b+b c+c a=2\), we substitute these values:
\(
a^2+b^2+c^2=(1)^2-2(2)=1-4=-3
\)
Step 2: Find the value of \(a^4+b^4+c^4\)
We can find the value of \(a^4+b^4+c^4\) using the result from the previous step. The formula for the sum of the fourth powers is given by:
\(
a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2\left(a^2 b^2+b^2 c^2+c^2 a^2\right)
\)
We already have the value for \(a^2+b^2+c^2\). Now, we need to find the value of \(a^2 b^2+b^2 c^2+c^2 a^2\). We can find this value using the following identity:
\(
a^2 b^2+b^2 c^2+c^2 a^2=(a b+b c+c a)^2-2 a b c(a+b+c)
\)
Given \(a b+b c+c a=2, a b c=3\), and \(a+b+c=1\), we substitute these values:
\(
a^2 b^2+b^2 c^2+c^2 a^2=(2)^2-2(3)(1)=4-6=-2
\)
Now, substitute the values of \(a^2+b^2+c^2\) and \(a^2 b^2+b^2 c^2+c^2 a^2\) into the expression for \(a^4+b^4+c^4\) :
\(
a^4+b^4+c^4=(-3)^2-2(-2)=9-(-4)=9+4=13
\)

Hint

(a) Step 1. Given quadratic equation
\(
x^2+5 \sqrt{2} x+10=0
\)
The roots are \(\alpha, \beta\).
Step 2. Relations between roots
\(
\alpha+\beta=-5 \sqrt{2}, \quad \alpha \beta=10
\)
Step 3. Definition of \(P_n\)
\(
P_n=\alpha^n-\beta^n
\)
This satisfies a recurrence relation (standard for symmetric quadratics):
\(
P_n=-5 \sqrt{2} P_{n-1}-10 P_{n-2}
\)
with initial values:
\(
P_0=0, \quad P_1=\alpha-\beta
\)
Step 4. We need to find
\(
\frac{P_{17} P_{20}+5 \sqrt{2} P_{17} P_{19}}{P_{18} P_{19}+5 \sqrt{2} P_{18}^2}
\)
Let’s try to simplify using recurrence relations.
Step 5. Express \(P_{20}\) and \(P_{19}\)
From recurrence:
\(
\begin{aligned}
& P_{20}=-5 \sqrt{2} P_{19}-10 P_{18} \\
& P_{19}=-5 \sqrt{2} P_{18}-10 P_{17}
\end{aligned}
\)
Step 6. Substitute \(P_{20}\) into the numerator
\(
\begin{gathered}
\text { Numerator }=P_{17} P_{20}+5 \sqrt{2} P_{17} P_{19} \\
=P_{17}\left[\left(-5 \sqrt{2} P_{19}-10 P_{18}\right)+5 \sqrt{2} P_{19}\right]
\end{gathered}
\)
Simplify:
\(
=P_{17}\left(-10 P_{18}\right)=-10 P_{17} P_{18}
\)
So numerator \(=-10 P_{17} P_{18}\).
Step 7. Simplify the denominator
\(
\text { Denominator }=P_{18} P_{19}+5 \sqrt{2} P_{18}^2
\)
Substitute \(P_{19}=-5 \sqrt{2} P_{18}-10 P_{17}\) :
\(
\begin{gathered}
=P_{18}\left(-5 \sqrt{2} P_{18}-10 P_{17}\right)+5 \sqrt{2} P_{18}^2 \\
=-5 \sqrt{2} P_{18}^2-10 P_{17} P_{18}+5 \sqrt{2} P_{18}^2 \\
=-10 P_{17} P_{18}
\end{gathered}
\)
So denominator \(=-10 P_{17} P_{18}\).
Step 8. Ratio
\(
\frac{\text { Numerator }}{\text { Denominator }}=\frac{-10 P_{17} P_{18}}{-10 P_{17} P_{18}}=1
\)

Hint

(c) The real numbers \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(x^2-(\alpha+\beta) x+\alpha \beta=0\).
Given \(\alpha+\beta=1\) and \(\alpha \beta=-1\), the equation is \(x^2-x-1=0\).
The power sums \(p_n=\alpha^n+\beta^n\) satisfy the linear recurrence relation derived from the coefficients of the quadratic equation:
\(
p_n=(\alpha+\beta) p_{n-1}-(\alpha \beta) p_{n-2}
\)
Substituting the given values, we get:
\(
\begin{gathered}
p_n=1 \cdot p_{n-1}-(-1) \cdot p_{n-2} \\
p_n=p_{n-1}+p_{n-2}
\end{gathered}
\)
This recurrence relation holds for all \(n \geq 2\). We are given \(p_{n-1}=11\) and \(p_{n+1}=29\) for some integer \(n \geq 1\).
Using the recurrence relation for \(p_{n+1}\), we can express it in terms of the preceding terms \(p_n\) and \(p_{n-1}\) :
\(
p_{n+1}=p_n+p_{n-1}
\)
We can now substitute the given values into this equation:
\(
29=p_n+11
\)
Solving for \(\boldsymbol{p}_{\boldsymbol{n}}\) :
\(
\begin{gathered}
p_n=29-11 \\
p_n=18
\end{gathered}
\)
The question asks for the value of \(p_n^2\).
\(
p_n^2=(18)^2 =324
\)

Hint

(c) The provided solution correctly finds the roots of the equation \(x^3-2 x^2+2 x-1=0\) and calculates the sum of their \(162^{\text {nd }}\) powers.
Step 1: Find the roots of the equation
The equation is \(x^3-2 x^2+2 x-1=0\).
By inspection, \(x=1\) is a root, since \((1)^3-2(1)^2+2(1)-1=1-2+2-1=0\).
This means ( \(x-1\) ) is a factor. Dividing the polynomial by ( \(x-1\) ) gives the factorization:
\(
(x-1)\left(x^2-x+1\right)=0
\)
The roots are \(x=1\) and the roots of the quadratic equation \(x^2-x+1=0\). Using the quadratic formula, the roots are:
\(
x=\frac{-(-1) \pm \sqrt{(-1)^2-4(1)(1)}}{2(1)}=\frac{1 \pm \sqrt{1-4}}{2}=\frac{1 \pm \sqrt{-3}}{2}=\frac{1 \pm i \sqrt{3}}{2}
\)
These roots are related to the complex cube roots of unity. The primitive cube roots of unity are \(\omega=e^{i 2 \pi / 3}\) and \(\omega^2=e^{i 4 \pi / 3}\). The roots of \(x^2-x+1=0\) are \(-\omega^2=\frac{1+i \sqrt{3}}{2}\) and \(-\omega=\frac{1-i \sqrt{3}}{2}\).
Step 2: Calculate the sum of the \(162^{\text {nd }}\) power of the roots
The roots are \(1,-\omega\), and \(-\omega^2\). The sum of their \(162^{\text {nd }}\) powers is:
\(
\begin{gathered}
(1)^{162}+(-\omega)^{162}+\left(-\omega^2\right)^{162} \\
=1+(\omega)^{162}+\left(\omega^2\right)^{162}
\end{gathered}
\)
Since \(\omega^3=1\), we can simplify the exponents.
\(
\begin{aligned}
& 162=3 \times 54 \text {, so } \omega^{162}=\left(\omega^3\right)^{54}=1^{54}=1 \\
& \begin{aligned}
2 \times 162=324=3 \times 108, \text { so } \omega^{324} & =\left(\omega^3\right)^{108}=1^{108}=1 \\
& =1+1+1=3
\end{aligned}
\end{aligned}
\)
The sum of the \(162^{\text {th }}\) power of the roots is 3.

Hint

(b) The equation is solved by considering two cases based on the absolute value term, \(|x-5|:\)
Case 1: \(\boldsymbol{x} \boldsymbol{\geq} \mathbf{5}\)
In this case, \(|x-5|=x-5\). The equation becomes:
\(
\begin{gathered}
(x+1)^2+(x-5)=\frac{27}{4} \\
x^2+2 x+1+x-5=\frac{27}{4} \\
x^2+3 x-4=\frac{27}{4} \\
4 x^2+12 x-16=27 \\
4 x^2+12 x-43=0
\end{gathered}
\)
\(
\text { Using the quadratic formula, } x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \text {, where } a=4, b=12, c=-43 \text { : }
\)
\(
\begin{aligned}
&\text { The two possible roots are approximately: }\\
&\begin{gathered}
x_1 \approx \frac{-12+28.84}{8} \approx 2.1 \\
x_2 \approx \frac{-12-28.84}{8} \approx-5.1
\end{gathered}
\end{aligned}
\)
Since we assumed \(x \geq 5\), neither of these roots are valid for this case.
Case 2: \(x<5\)
In this case, \(|x-5|=-(x-5)=5-x\). The equation becomes:
\(
\begin{gathered}
(x+1)^2+(5-x)=\frac{27}{4} \\
x^2+2 x+1+5-x=\frac{27}{4} \\
x^2+x+6=\frac{27}{4} \\
4 x^2+4 x+24=27 \\
4 x^2+4 x-3=0
\end{gathered}
\)
Using the quadratic formula, where \(a=4, b=4, c=-3\) :
\(
\begin{gathered}
x=\frac{-4 \pm \sqrt{4^2-4(4)(-3)}}{2(4)} \\
x=\frac{-4 \pm \sqrt{16+48}}{8} \\
x=\frac{-4 \pm \sqrt{64}}{8} \\
x=\frac{-4 \pm 8}{8}
\end{gathered}
\)
The two possible roots are:
\(
\begin{gathered}
x_3=\frac{-4+8}{8}=\frac{4}{8}=\frac{1}{2}=0.5 \\
x_4=\frac{-4-8}{8}=\frac{-12}{8}=-\frac{3}{2}=-1.5
\end{gathered}
\)
Since we assumed \(x<5\), both of these roots ( \(x=0.5\) and \(x=-1.5\) ) are valid solutions.
Thus, the equation has exactly 2 real roots.

Hint

(d) Step 1: Rewrite the equation and determine the condition for real roots
The given quadratic equation is \(2 x^2+(a-10) x+\frac{33}{2}=2 a\). To find the values of \(a\) for which the equation has real roots, first, we must rewrite the equation in the standard form \(A x^2+B x+C=0\).
\(
2 x^2+(a-10) x+\left(\frac{33}{2}-2 a\right)=0
\)
For a quadratic equation to have real roots, the discriminant ( \(\Delta=B^2-4 A C\) ) must be greater than or equal to zero. In this case, \(A=2, B=(a-10)\), and \(C=\left(\frac{33}{2}-2 a\right)\)
Step 2: Formulate and solve the inequality
The condition for real roots is:
\(
(a-10)^2-4(2)\left(\frac{33}{2}-2 a\right) \geq 0
\)
Expand and simplify the inequality:
\(
\begin{gathered}
\left(a^2-20 a+100\right)-8\left(\frac{33}{2}-2 a\right) \geq 0 \\
a^2-20 a+100-132+16 a \geq 0 \\
a^2-4 a-32 \geq 0
\end{gathered}
\)
To solve this inequality, we find the roots of the quadratic equation \(a^2-4 a-32=0\). The roots can be found by factoring:
\(
(a-8)(a+4)=0
\)
The roots are \(a=8\) and \(a=-4\). The parabola \(y=a^2-4 a-32\) opens upwards, so the inequality \(a^2-4 a-32 \geq 0\) is satisfied for values of \(a\) outside the roots.
\(
a \leq-4 \text { or } a \geq 8
\)
The problem asks for the least positive value of \(a\). From the solution set, the positive values of \(a\) are in the interval \(a \geq 8\). The least value in this interval is 8.
The least positive value of \(\boldsymbol{a}\) for which the equation has real roots is \(\mathbf{8}\).

Hint

(b) Given polynomial is
\(
\begin{aligned}
& (x-1)(x-2)(x-3) \cdots(x-100) \\
& =x^{100}-(1+2+3+\cdots+100) x^{99}+(\cdots) x^{98} \cdots
\end{aligned}
\)
Hence, coefficient of \(x^{99}\) is
\(
\begin{aligned}
-(1+2+3+\cdots+100) & =\frac{-100 \times 101}{2} \\
& =-5050
\end{aligned}
\)

Hint

(a) As \(p\) and \(q\) are real and one root is \(2+i \sqrt{3}\), so the other root must be \(2-i \sqrt{3}\). Then,
\(
\begin{aligned}
& p=-(\text { sum of roots })=-4 \\
& q=\text { product of roots }=(2+i \sqrt{3})(2-i \sqrt{3})=4+3=7
\end{aligned}
\)

Hint

(d) Given equation is
\(
\begin{aligned}
& x^2-3 k x+2 e^{2 \ln k}-1=0 \\
\Rightarrow \quad & x^2-3 k x+\left(2 k^2-1\right)=0
\end{aligned}
\)
Here, product of roots is \(2 k^2-1\).
\(
\therefore \quad 2 k^2-1=7 \Rightarrow k^2=4 \Rightarrow k=2,-2
\)
Now for real roots, we must have
\(
\begin{aligned}
& D \geq 0 \\
\Rightarrow & 9 k^2-4\left(2 k^2-1\right) \geq 0 \\
\Rightarrow & k^2+4 \geq 0
\end{aligned}
\)
which is true for all \(k\). Thus, \(k=2,-2\). But for \(k=-2, \ln k\) is not defined. Therefore, rejecting \(k=-2\), we get \(k=2\).

Hint

(b) Let the common root of the two equations, \(x^2+a x+b=0\) and \(x^2+b x+a=0\), be \(r\)
Since \(r\) is a root of both equations, it must satisfy both equations:
\(
\begin{aligned}
& r^2+a r+b=0 \dots(1) \\
& r^2+b r+a=0 \dots(2)
\end{aligned}
\)
Subtract equation (2) from equation (1):
\(
\begin{gathered}
\left(r^2+a r+b\right)-\left(r^2+b r+a\right)=0 \\
a r+b-b r-a=0 \\
r(a-b)+(b-a)=0 \\
r(a-b)-(a-b)=0 \\
(a-b)(r-1)=0
\end{gathered}
\)
Since it is given that \(a \neq b\), we must have \(a-b \neq 0\). Therefore, we can divide both sides by \((a-b)\) :
\(
\begin{gathered}
r-1=0 \\
r=1
\end{gathered}
\)
Now substitute the value of \(\boldsymbol{r}=1\) back into either of the original equations. Using equation (1):
\(
\begin{gathered}
1^2+a(1)+b=0 \\
1+a+b=0 \\
a+b=-1
\end{gathered}
\)
Thus, the numerical value of \(a+b\) is -1.

Hint

(c) Given \(x<0, y<0\).
\(
x+y+\frac{x}{y}=\frac{1}{2} \text { and }(x+y) \frac{x}{y}=-\frac{1}{2}
\)
Let,
\(
x+y=a \text { and } \frac{x}{y}=b \dots(1)
\)
Therefore, we get
\(
a+b=\frac{1}{2}, a b=-\frac{1}{2}
\)
Solving these two, we get
\(
\begin{aligned}
& a+\left(-\frac{1}{2 a}\right)=\frac{1}{2} \\
\Rightarrow & 2 a^2-a-1=0
\end{aligned}
\)
\(
\begin{array}{ll}
\Rightarrow & a=1,-1 / 2 \\
\Rightarrow & b=-1 / 2,1
\end{array}
\)
\(\therefore \quad(1) \Rightarrow x+y=1\) and \(\frac{x}{y}=-\frac{1}{2}\)
\(
x+y=-\frac{1}{2} \text { and } \frac{x}{y}=1
\)
But \(x, y<0\)
\(
\therefore \quad x+y<0 \Rightarrow x+y=\frac{-1}{2} \text { and } \frac{x}{y}=1
\)
On solving, we get \(x=-1 / 4\) and \(y=-1 / 4\).

Hint

(d) 

\(
\begin{aligned}
&\begin{aligned}
& |x-2|^2+|x-2|-2=0 \\
\Rightarrow & (|x-2|+2)(|x-2|-1)=0 \\
\Rightarrow & |x-2|-1=0 \\
\Rightarrow & x-2= \pm 1 \\
\Rightarrow & x=1,3
\end{aligned}\\
&\text { Therefore, the sum of the roots is } 3+1=4 \text {. }
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
&\begin{aligned}
& \log _7 \log _5(\sqrt{x+5}+\sqrt{x})=0 \\
\Rightarrow & \log _5(\sqrt{x+5}+\sqrt{x})=1 \\
\Rightarrow & (\sqrt{x+5}+\sqrt{x})=5 \\
\Rightarrow & x+5=25+x-10 \sqrt{x} \\
\Rightarrow & 2=\sqrt{x} \\
\Rightarrow & x=4
\end{aligned}\\
&\text { which satisfies the given equation }
\end{aligned}
\)