Quadratic Equations

Given quadratic equation does not have real roots.

\(2 x^2+(a-5) x+15=3\)

so \(D<0\)

\((a-5)^2-8(15-3 a)<0\)

\(a^2+14 a+25-120<0\)

\(a^2+14 a-95<0\)

\((a+19)(a-5)<0\)

\(a \in(-19,5)\)

\(\therefore-19<x<5\)

\( \therefore \sum_{x \in X} x^2= \left(1^2+2^2+\ldots+4^2\right) +\left(1^2+2^2+\ldots+18^2\right)\)

\(=\frac{4 \times 5 \times 9}{6}+\frac{18 \times 19 \times 37}{6}\)

\(=2139\)

\(S=\alpha+\beta\)

\(p=\alpha\beta\)

\(P_1=\alpha+\beta=1\)

\(S=1\)

\(P_n=sP_{n-1}-pP_{n-2}\)

\(P_{10}=123,\quad P_9=76,\quad P_8=47\)

\(123=P_9-pP_8\)

\(123=76-p(47)\)

\(47p=-47\)

\(p=-1\)

\(\alpha+\beta=1\)

\(\alpha\beta=-1\).

Therefore \(\alpha\) and \(\beta\) are roots of

\(x^2-x-1=0\)

Now the roots required are \(\dfrac1\alpha\) and \(\dfrac1\beta\)

Their sum is \(\dfrac{\alpha+\beta}{\alpha\beta}
=\dfrac{1}{-1}=-1\)

Their product is

\(\dfrac1{\alpha\beta}=-1\)

Hence the required quadratic equation is

\(x^2-(\text{sum})x+(\text{product})=0\)

\(x^2+x-1=0\)

Given

\((a+b-2c)x^2+(b+c-2a)x+(c+a-2b)=0\)

Observe that

\((a+b-2c)+(b+c-2a)+(c+a-2b)=0\)

Hence \(x=1\) is always a root.

Since \(\alpha\ne1\) is the other root,

\(\alpha=\dfrac{c+a-2b}{a+b-2c}\).

Let

\(p=a-b>0,\quad q=b-c>0\).

Then

\(a=b+p,\quad c=b-q\).

Substituting,

\(\alpha=\dfrac{(b-q)+(b+p)-2b}{(b+p)+b-2(b-q)}
=\dfrac{p-q}{p+2q}\).

Since \(p,q>0\),

\(-1<\dfrac{p-q}{p+2q}<1\).

Thus \(\alpha\in(-1,1)\).

Now let \(b\) be the geometric mean of \(a\) and \(c\):

\(b^2=ac=(b+p)(b-q)\).

This gives

\(b(p-q)=pq\).

Since \(b>0\),

\(p-q=\dfrac{pq}{b}>0\).

Hence

\(\alpha=\dfrac{p-q}{p+2q}>0\).

Therefore, if \(b\) is the geometric mean of \(a\) and \(c\), then necessarily

\(\alpha\in(0,1)\).

So:

(I) If \(\alpha\in(-1,0)\), then \(b\) cannot be the geometric mean of \(a\) and \(c\). True.

(II) If \(\alpha\in(0,1)\), then \(b\) may be the geometric mean of \(a\) and \(c\). True.

For example, take \(a=4,\ b=2,\ c=1\) \((b=\sqrt{ac})\).

Then

\(\alpha=\dfrac{4+1-4}{4+2-2}
=\dfrac14\in(0,1)\).

Hence both statements are true

\( f(x)=(\lambda+2) x^2-3 \lambda x+4 \lambda\)

Case - 1: \(\frac{c}{a}>0\) or \(a c>0\)

\( (\lambda+2) 4 \lambda>0\)

\( \Rightarrow \lambda<-2 \text { or } \lambda>0\)

Case - 2: \(-\frac{\mathrm{b}}{2 \mathrm{a}}>0 \)
\( \frac{3 \lambda}{2(\lambda+2)}>0 \)
\( \Rightarrow \lambda<-2 \text { or } \lambda>0\)
Case - 3: \(\mathrm{D} \geq 0\)
\( (-3 \lambda)^2-4(\lambda+2) \times 4 \lambda \geq 0 \)
\( \lambda(7 \lambda+32) \leq 0 \)
\( \Rightarrow \lambda \in\left[\frac{-32}{7}, 0\right]\)
Intersection of Case-1, Case-2 and Case-3

\( \Rightarrow \lambda \in\left[\frac{-32}{7},-2\right)\)
\( \Rightarrow \lambda \in[-4.57,-2)\)
\( \Rightarrow \lambda=-4,-3\)
Number of values of \(\lambda=2\)

x² + 4x + 4 = n + 4
(x + 2)² = n + 4
x = -2 ± √n + 4
∵ 20 ≤ n ≤ 100
√24 ≤ √n + 4 ≤ √104
⇒ √n + 4 ∈ {5,6,7,8,9,10}
∴ '6' integral values of 'n' are possible

Since \(\alpha, \beta\) be the roots

of the quadratic equation \(\lambda x^2-(\lambda+3) x+3=0\)

$\alpha +\beta =\frac{\lambda +3}{\lambda },\alpha \beta =\frac{3}{\lambda }$

Now, $\frac{\beta -\alpha }{\alpha \beta }=\frac{1}{3}$

$\Rightarrow \beta -\alpha =\frac{\alpha \beta }{3}=\frac{1}{\lambda }$

on squaring

${\alpha }^2+{\beta }^2-2\alpha \beta =\frac{1}{{\lambda }^2} ...\left(1\right)$

${\alpha }^2+{\beta }^2+2\alpha \beta =\frac{{(\lambda +3)}^2}{{\lambda }^2} ...\left(2\right)$

eq(2) –eq(1)

$4\alpha \beta =\frac{{(\lambda +3)}^2-1}{{\lambda }^2}$

$\Rightarrow \frac{12}{\lambda }=\frac{{\lambda }^2+6\lambda +8}{{\lambda }^2}$

$\Rightarrow {\lambda }^2-6\lambda +8=0$

$\Rightarrow \lambda =2,4$

Sum of possible values of $\lambda$ is $=6$

Given. equation has no real roots. i.e. $D<0$

Now, $$\begin{gathered} {(a-5)}^2-4.2.(15-3a)<0 \\ {a}^2+14a-95<0 \\ (a+19)(a-5)<0 \\ a\in (-19,5) \\ \alpha =-19, \beta =5 \\ and \left|4\right(\alpha +\beta \left)\right|=\left|4\right(-19+5\left)\right|=56 \end{gathered}$$

\(\begin{aligned}
f(x) & =5 x^3-15 x-a \\
f^{\prime}(x) & =15 x^2-15 \\
f^{\prime}(x) & =15\left(x^2-1\right) \\
& =15(x-1)(x+1)
\end{aligned}\)

\((-1) \rightarrow\) point of maxima \(1 \rightarrow\) point of minima

\(
\begin{aligned}
& f(1)=-10-9 \\
& f(-1)=10-9
\end{aligned}
\)

For 3 distinet meakerw)

\(
\begin{array}{lll}
-10-a<0 \Rightarrow & a>-10 \\
a \text { ard } \quad 10 a>0 & a<10
\end{array}
\)

\(\begin{aligned}
& =1 \quad a \in(-10,10) \\
& \alpha=-10, \quad \beta=10 \\
& \Rightarrow \beta-2 \alpha=10-2 \times(-10) \\
& =10+20 \\
& =30
\end{aligned}\)

Given \(\frac{a x^2+2(a+1) x+9 a+4}{x^2-8 x+32}<0 \quad \forall x \in \mathbb{R}\)
In \(x^2-8 x+32\), we have \(D=64-128<0\)

\( \therefore x^2-8 x+32>0\) \(\forall x \in \mathbb{R} \)
\( \Rightarrow a x^2+2(a+1) x+9 a+4<0\) \(\forall x \in R \)
\( \Rightarrow a<0\) and \(D<0\)

Since Question has asked for positive integral values of \(a\)

\( \therefore|S|=0\)

but we need positive integral value of a.
So, No solution

$$\begin{gathered} P_{10}=P_8+P_9 \\ \Rightarrow x^2=x+1\mathit{ }\mathrm{has} \mathrm{roots} \mathrm{\alpha } \mathrm{and} \mathrm{\beta } \\ \Rightarrow \mathrm{Required} \mathrm{equation} \mathrm{is} \\ {\mathrm{x}}^2+\mathrm{x}-1=0 \end{gathered}$$

For integral roots, the discriminant must be a perfect square.

\(D=16+4n\)

\(=4(n+4)\)

Let

\(n+4=m^2\)

where \(m\) is a positive integer.

Then

\(n=m^2-4\)

Given

\(20\le n\le100\)

\(20\le m^2-4\le100\)

\(24\le m^2\le104\)

Hence

\(5\le m\le10\)

Thus,

\(m=5,6,7,8,9,10\)

giving

\(n=21,32,45,60,77,96\)

All these values produce integral roots.

Therefore, the number of distinct values of \(n\) is

\(6\)

Given: $(-2+\sqrt{3})\left(\right|\sqrt{x}-3\left|\right)+(x-6\sqrt{x})+(9-2\sqrt{3})=0$

\(\Rightarrow (x-6 \sqrt{x}+9)-(2-\sqrt{3})|\sqrt{x}-3|-2 \sqrt{3}=0 \)

\( \Rightarrow|\sqrt{x}-3|^2-(2-\sqrt{3})|\sqrt{x}-3|-2 \sqrt{3}=0 \)

\(\Rightarrow (|\sqrt{x}-3|-2)(|\sqrt{x}-3|+\sqrt{3})=0\)
\( \Rightarrow|\sqrt{x}-3|=2 \text { or }|\sqrt{x}-3|=-\sqrt{3}\) (not possible)
\(\Rightarrow \sqrt{x}=1\) or \(5 \)
\( \Rightarrow x=1 \) or \(25\)
\(\Rightarrow \alpha=1\) and \(\beta=25\)

\(\sqrt{\frac{\beta}{\alpha}}+\sqrt{\alpha \beta}=\sqrt{\frac{25}{1}}+\sqrt{25 \cdot 1}=10\)

$$\begin{gathered} 2{\mathrm{x}}^2+\left(\cos \theta \right)\mathrm{x}-1=0 \\ {\alpha }_{\theta }+{\beta }_{\theta }=-\frac{\cos \theta }{2}\text{ and} {\alpha }_{\theta }.{\beta }_{\theta }=-\frac{1}{2} \end{gathered}$$

$$\begin{gathered} \text{Now, } \\ {{\alpha }_{\theta }}^4+{{\beta }_{\theta }}^4={\left({{\alpha }_{\theta }}^2+{{\beta }_{\theta }}^2\right)}^2-2{{\alpha }_{\theta }}^2{{\beta }_{\theta }}^2 \\ ={\left[({\alpha }_{\theta }+{\beta }_{\theta }{)}^2-2{\alpha }_{\theta }{\beta }_{\theta }\right]}^2-2({\alpha }_{\theta }{\beta }_{\theta }{)}^2 \\ ={\left[\frac{{\cos }^2\theta }{4}+1\right]}^2-2.\frac{1}{4} \\ ={\left(\frac{{\cos }^2\theta }{4}+1\right)}^2-\frac{1}{2} \\ \mathrm{M}=\frac{25}{16}-\frac{1}{2}=\frac{17}{16} \left(\text{when} {\cos }^2\theta =1\right) \\ \mathrm{m}=1-\frac{1}{2}=\frac{1}{2} \left(\text{when} {\cos }^2\theta =0\right) \\ \Rightarrow 16(\mathrm{M}+\mathrm{m})=25 \end{gathered}$$

Let $\alpha$ and $\beta$ be the roots of $x^2 + \sqrt{3}x - 16 = 0$.

By the property of roots, we have:

$\alpha^2 + \sqrt{3}\alpha - 16 = 0$ $\beta^2 + \sqrt{3}\beta - 16 = 0$

$\alpha^n + \sqrt{3}\alpha^{n-1} - 16\alpha^{n-2} = 0$ $\beta^n + \sqrt{3}\beta^{n-1} - 16\beta^{n-2} = 0$

For $n = 25$, we get: $P_{25} + \sqrt{3}P_{24} = 16P_{23}$

So, the first part of the expression is: $\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} = \frac{16P_{23}}{2P_{23}} = 8$

Let $\gamma$ and $\delta$ be the roots of $x^2 + 3x - 1 = 0$.

Similarly, by Newton's Sums, we have:

$\gamma^n + 3\gamma^{n-1} - \gamma^{n-2} = 0$ $\delta^n + 3\delta^{n-1} - \delta^{n-2} = 0$

For $n = 25$, we get: $Q_{25} - Q_{23} = -3Q_{24}$

So, the second part of the expression is: $\frac{Q_{25} - Q_{23}}{Q_{24}} = \frac{-3Q_{24}}{Q_{24}} = -3$

The total expression is: $\frac{P_{25}+\sqrt{3}{P}_{24}}{2P_{23}}+\frac{Q_{25}-Q_{23}}{Q_{24}}=8+(-3)=5$

$$\begin{gathered} {\left(x^2-9x+11\right)}^2-\left(x^2-9x+20\right)=3 \\ {\left(x^2-9x+11\right)}^2-\left(x^2-9x\right)-20=3 \\ {\left(x^2-9x+11\right)}^2-\left(x^2-9x\right)-23=0 \\ \Rightarrow x^2-9x=p \\ {\left(p+11\right)}^2-\left(p\right)-23=0 \\ \Rightarrow p^2+22p+121-p-23=0 \\ \Rightarrow (p+14)(p+7)=0 \\ p=-7, -14 \\ {x}^2-9x=-7, -14 \\ {x}^2-9x+7=0 \text{ or } x^2-9x+14=0 \\ x=\frac{9\pm \sqrt{53}}{2} \text{or x}=\frac{9\pm 5}{2} \\ \text{ Product of all rational roots }=7\times 2=14 \end{gathered}$$

Given. equation has no real roots. i.e. $D<0$

Now, $$\begin{gathered} {(a-5)}^2-4.2.(15-3a)<0 \\ {a}^2+14a-95<0 \\ (a+19)(a-5)<0 \\ a\in (-19,5) \\ \alpha =-19, \beta =5 \\ and \left|4\right(\alpha +\beta \left)\right|=\left|4\right(-19+5\left)\right|=56 \end{gathered}$$

Since p, q, r...., G.P.

$\therefore p=A,q=AR,r=A{R}^2$

$\Rightarrow A{x}^2+ARx-A{R}^2=0\Rightarrow x^2+Rx-R^2=0$

$\alpha +\beta =-R,\alpha \beta =-R^2$

$\frac{1}{\alpha }+\frac{1}{\beta }=\frac{3}{4}$

$\therefore \frac{\alpha +\beta }{\alpha \beta }=\frac{3}{4}\Rightarrow \frac{-R}{-R^2}=\frac{3}{4}\Rightarrow R=\frac{4}{3}$

${(\alpha -\beta )}^2={(\alpha +\beta )}^2-4\alpha \beta =R^2-4(-R^2)=\frac{80}{9}$

Since both the roots of the equation are negative real numbers, therefore

\(2 p+9>0\)

\(\Rightarrow p>\frac{-9}{2}\)

\(p+2<0\)

\(\Rightarrow p<-2\)

and \(\mathrm{D} \geq 0\)

\((p+2)^2-4(2 p+9) \geq 0\)

\((p+4)(p-8) \geq 0\)

\(\Rightarrow p \in(-\infty,-4] \cup[8, \infty)\)

So, \(p \in\left(-\frac{9}{2},-4\right]\)

\(\alpha=-\frac{9}{2} ; \beta=-4\)

$\beta -2\alpha =-4+9=5$

\((x^2-9x+11)^2-(x-4)(x-5)=3\)

\((x^2-9x+11)^2-(x^2-9x+20)=3\)

Let \(t=x^2-9x+11\)

Then, \(t^2-(t+9)=3\)

\(t^2-t-12=0\)

\((t-4)(t+3)=0\)

So, \(t=4\) or \(t=-3\)

For \(t=4\),

\(x^2-9x+11=4\)

\(x^2-9x+7=0\)

Its roots are \(x=\dfrac{9\pm\sqrt{53}}{2}\)

These are irrational.

For \(t=-3\),

\(x^2-9x+11=-3\)

\(x^2-9x+14=0\)

\((x-2)(x-7)=0\)

Hence the rational roots are \(x=2,7\)

Therefore, the product of all rational roots is

\(2\times 7=14\)

Since \(\alpha, \beta\) be the roots

of the quadratic equation \(\lambda x^2-(\lambda+3) x+3=0\)

$\alpha +\beta =\frac{\lambda +3}{\lambda },\alpha \beta =\frac{3}{\lambda }$

Now, $\frac{\beta -\alpha }{\alpha \beta }=\frac{1}{3}$

$\Rightarrow \beta -\alpha =\frac{\alpha \beta }{3}=\frac{1}{\lambda }$

on squaring

${\alpha }^2+{\beta }^2-2\alpha \beta =\frac{1}{{\lambda }^2} ...\left(1\right)$

${\alpha }^2+{\beta }^2+2\alpha \beta =\frac{{(\lambda +3)}^2}{{\lambda }^2} ...\left(2\right)$

eq(2) –eq(1)

$4\alpha \beta =\frac{{(\lambda +3)}^2-1}{{\lambda }^2}$

$\Rightarrow \frac{12}{\lambda }=\frac{{\lambda }^2+6\lambda +8}{{\lambda }^2}$

$\Rightarrow {\lambda }^2-6\lambda +8=0$

$\Rightarrow \lambda =2,4$

Sum of possible values of $\lambda$ is $=6$

$$\begin{gathered} \text{Given quadratic equation doesnt } \\ \text{have real roots.} \\ 2x^2+(a-5)x+15=3 \\ \text{so D<0} \\ (a-5{)}^2-8(15-3a)<0 \\ {a}^2+14a+25-120<0 \\ {a}^2+14a-95<0 \\ (a+19\left)\right(a-5)<0 \\ \mathrm{a}\in (-19,5) \\ \therefore -19<x<5 \\ \therefore \sum _{\mathrm{x}\in \mathrm{X}}{\mathrm{x}}^2=\left(1^2+2^2+\dots +4^2\right) \\ +\left(1^2+2^2+\dots +{18}^2\right) \\ =\frac{4\times 5\times 9}{6}+\frac{18\times 19\times 37}{6} \\ =30+2109 \\ =2139 \end{gathered}$$

\(2 x^2+(\cos \theta) x-1=0\)

\(\alpha_\theta+\beta_\theta=-\frac{\cos \theta}{2}\) and \(\alpha_\theta \cdot \beta_\theta=-\frac{1}{2}\)

Now,

\(\alpha_\theta{ }^4+\beta_\theta{ }^4=\left(\alpha_\theta{ }^2+\beta_\theta{ }^2\right)^2-2 \alpha_\theta{ }^2 \beta_\theta{ }^2\)

\(=\left[\left(\alpha_\theta+\beta_\theta\right)^2-2 \alpha_\theta \beta_\theta\right]^2-2\left(\alpha_\theta \beta_\theta\right)^2\)

\(=\left[\frac{\cos ^2 \theta}{4}+1\right]^2-2 \cdot \frac{1}{4}\)

\(=\left(\frac{\cos ^2 \theta}{4}+1\right)^2-\frac{1}{2}\)

\(M=\frac{25}{16}-\frac{1}{2}=\frac{17}{16}\left(\right.\) when \(\left.\cos ^2 \theta=1\right)\)

\(m=1-\frac{1}{2}=\frac{1}{2}\left(\right.\) when \(\left.\cos ^2 \theta=0\right)\)

\(\Rightarrow 16(M+m)=25\)

Let

\(P_n=\alpha^n-\beta^n\).

Since \(\alpha,\beta\) are roots,

\(\alpha+\beta=\sqrt2\),

\(\alpha\beta=-\sqrt3\).

Also,

\(P_{n+2}=(\alpha+\beta)P_{n+1}-\alpha\beta\,P_n\)

\(=\sqrt2\,P_{n+1}+\sqrt3\,P_n\).

Therefore

\(P_{12}=\sqrt2\,P_{11}+\sqrt3\,P_{10}\).

Now

\((11\sqrt3-10\sqrt2)P_{10}+(11\sqrt2+10)P_{11}-11P_{12}\)

\(=(11\sqrt3-10\sqrt2)P_{10}+(11\sqrt2+10)P_{11}\)

\(\qquad-11(\sqrt2\,P_{11}+\sqrt3\,P_{10})\)

\(=-10\sqrt2\,P_{10}+10P_{11}\)

\(=10(P_{11}-\sqrt2\,P_{10})\).

Using

\(P_{11}=\sqrt2\,P_{10}+\sqrt3\,P_9\),

we get

\(10(P_{11}-\sqrt2\,P_{10})
=10\sqrt3\,P_9\).

Since \(\alpha\) and \(\beta\) satisfy

\(x^2-x-1=0\),

we have

\(\alpha^2=\alpha+1\)

\(\beta^2=\beta+1\)

Multiplying by \(\alpha^{n-2}\) and \(\beta^{n-2}\),

\(\alpha^n=\alpha^{n-1}+\alpha^{n-2}\)

\(\beta^n=\beta^{n-1}+\beta^{n-2}\)

Multiplying by \(2023\) and \(2024\) respectively and adding,

\(S_n=S_{n-1}+S_{n-2}\)

Therefore,

\(S_{12}=S_{11}+S_{10}\)

Hence option A is true.

Checking the remaining options:

B gives

\(2S_{11}=S_{12}+S_{10}\)

\(=S_{11}+2S_{10}\)

\(\Rightarrow S_{11}=2S_{10}\)

which is not generally true.

C gives

\(S_{11}=S_{10}+S_{12}\)

which contradicts \(S_{12}=S_{11}+S_{10}\).

D gives

\(2S_{12}=S_{11}+S_{10}\)

\(\Rightarrow S_{12}=0\)

which is false.

Since \(2\) and \(6\) are roots of \(ax^2+bx+1=0\)

by Vieta's formulas

\(2+6=-\dfrac{b}{a}\)

\(\Rightarrow b=-8a\)

\(2\cdot6=\dfrac{1}{a}\)

\(\Rightarrow a=\dfrac{1}{12}\)

\(b=-\dfrac{2}{3}\)

\(2a+b=\dfrac{1}{6}-\dfrac{2}{3}=-\dfrac{1}{2}\)

\(\dfrac{1}{2a+b}=-2\)

\(6a+b=\dfrac{1}{2}-\dfrac{2}{3}=-\dfrac{1}{6}\)

\(\dfrac{1}{6a+b}=-6\)

roots are \(-2\) and \(-6\).

Hence the required quadratic equation is \((x+2)(x+6)=0\)

\(x^2+8x+12=0\)

Given

\(x^2-(t^2-5t+6)x+1=0\),

with roots \(\alpha,\beta\).

Hence

\(\alpha+\beta=t^2-5t+6\),

\(\alpha\beta=1\).

Let

\(S_n=\alpha^n+\beta^n=a_n\).

Since \(\alpha\beta=1\),

\(a_{n+1}+a_{n-1}
=(\alpha+\beta)(\alpha^n+\beta^n)\).

Therefore

\(a_{2025}+a_{2023}
=(\alpha+\beta)a_{2024}\).

Hence

\(\dfrac{a_{2023}+a_{2025}}{a_{2024}}
=\alpha+\beta
=t^2-5t+6\).

So we need the minimum value of

\(t^2-5t+6
=\left(t-\dfrac52\right)^2-\dfrac14\).

Its minimum value is

\(-\dfrac14\).

The roots are distinct when

\((t^2-5t+6)^2-4>0\),

which excludes only the values where

\(t^2-5t+6=\pm2\).

The minimum value \(-\dfrac14\) is admissible.

Given equation

$\left(\frac{9}{x}-\frac{9}{\sqrt{x}}+2\right)\left(\frac{2}{x}-\frac{7}{\sqrt{x}}+3\right)=0$

Let $\frac{1}{\sqrt{x}}=y$

$\left(9y^2-9y+2\right)\left(2y^2-7y+3\right)=0$

$\Rightarrow (3y-2)(3y-1)(y-3)(2y-1)=0$

$\Rightarrow y=\frac{1}{3},\frac{1}{2},\frac{2}{3},3$

$\Rightarrow \mathrm{x}=9,4,\frac{9}{4},\frac{1}{9}$

Number of solutions are 4

$x^2+2\sqrt{2}x-1=0$

$\alpha +\beta =-2\sqrt{2}$

$\alpha \beta =-1$

${\alpha }^4+{\beta }^4={\left({\alpha }^2+{\beta }^2\right)}^2-2{\alpha }^2{\beta }^2$

\(=\left((\alpha+\beta)^2-2 \alpha \beta\right)^2-2(\alpha \beta)^2\)

$=(8+2{)}^2-2(-1{)}^2$

$=100-2=98$

${\alpha }^6+{\beta }^6={\left({\alpha }^3+{\beta }^3\right)}^2-2{\alpha }^3{\beta }^3$

\(=\left((\alpha+\beta)\left((\alpha+\beta)^2-3 \alpha \beta\right)^2-2(\alpha \beta)^3\right.\)

$=(-2\sqrt{2}(8+3){)}^2+2$

$=\left(8\right)\left(121\right)+2=970$

$\frac{1}{10}\left({\alpha }^6+{\beta }^6\right)=97$

$x^2-(98+97)x+\left(98\right)\left(97\right)=0$

$\Rightarrow x^2-195x+9506=0$

$x^2+2\sqrt{2}x-1=0$

$\alpha +\beta =-2\sqrt{2}$

$\alpha \beta =-1$

${\alpha }^4+{\beta }^4={\left({\alpha }^2+{\beta }^2\right)}^2-2{\alpha }^2{\beta }^2$

\(=\left((\alpha+\beta)^2-2 \alpha \beta\right)^2-2(\alpha \beta)^2\)

$=(8+2{)}^2-2(-1{)}^2$

$=100-2=98$

${\alpha }^6+{\beta }^6={\left({\alpha }^3+{\beta }^3\right)}^2-2{\alpha }^3{\beta }^3$

\(=\left((\alpha+\beta)\left((\alpha+\beta)^2-3 \alpha \beta\right)^2-2(\alpha \beta)^3\right.\)

$=(-2\sqrt{2}(8+3){)}^2+2$

$=\left(8\right)\left(121\right)+2=970$

$\frac{1}{10}\left({\alpha }^6+{\beta }^6\right)=97$

$x^2-(98+97)x+\left(98\right)\left(97\right)=0$

$\Rightarrow x^2-195x+9506=0$

Given \(a, 4, \alpha, b\) are in A.P.

Let \(d\) de the common difference,

Hence \(a=4-d, \alpha=4+d, b=4+2 d\)

\(\Rightarrow(4+d) x^2-(4-d) x+2(4+d-8-4 d)=0\)

\(\Rightarrow(4+d) x^2-(4-d) x+2(-4-3 d)=0\)

Now arithmetic mean of \(\frac{1}{a}\) and \(\frac{1}{b}\) is \(\frac{5}{16}\)

Therefore, $\frac{\frac{1}{a}+\frac{1}{b}}{2}=\frac{5}{16}$

\(\Rightarrow \frac{\frac{1}{4-d}+\frac{1}{4+2 d}}{2}=\frac{5}{16}\)

\(\Rightarrow 5 d^2-6 d-8=0\)

\(\Rightarrow d=2, -\frac{4}{5}\)

For \(d=2\): \(a=4-2=2\) (Rejected)

For \(d=-\frac{4}{5}\): \(a=4+\frac{4}{5}=\frac{24}{5}\)

Equation becomes:

\(16 x^2-24 x-2=0\)

\(\Rightarrow 2 x^2-3x-2=0\)

$\Rightarrow x=2,\frac{-1}{2}$

$$\begin{gathered} \text{ Since }b,c>0 \\ \text{ Therefore} \\ \alpha +\beta =-b<0 \\ \text{ and αβ}=-c<0 \\ \text{ Since sum and product of the roots is -ve} \\ \text{therefore roots must be of opposite sign. } \end{gathered}$$

Given: \(\alpha, \beta\) be the roots of the quadratic equation \(12 x^2-20 x+3 \lambda=0\),

$\Rightarrow \alpha +\beta =\frac{5}{3}$ and $\alpha \beta =\frac{\lambda }{4}$

Now, $\frac{1}{2}\le \left|\alpha -\beta \right|\le \frac{3}{2}$

$\Rightarrow \frac{1}{4}\le |\alpha -\beta {|}^2\le \frac{9}{4}$

$\Rightarrow \frac{1}{4}\le {(\alpha +\beta )}^2-4\alpha \beta \le \frac{9}{4}$

$\Rightarrow \frac{1}{4}\le \frac{25}{9}-4\times \frac{\lambda }{4}\le \frac{9}{4}$

$\Rightarrow -\frac{91}{36}\le -\lambda \le \frac{-19}{36}$

$\Rightarrow \frac{19}{36}\le \lambda \le \frac{91}{36}$

$\Rightarrow \lambda =1,2$

Sum \(=3\)

\(p(x)=f(x)-g(x)\)

\(p(x)=(a_1-b_1)+7x+(a_2-b_2)x^2+(a_3-b_3)x^3\)

Since \(f(x)\ne g(x)\) for every real \(x\)

\(p(x)\ne0\) for every real \(x\)

But \(p(x)\) is a polynomial of degree at most \(3\)

A polynomial of odd degree must have at least one real root

Hence \(p(x)\) cannot be of degree \(3\)

\(a_3-b_3=0\)

\(p(x)=(a_1-b_1)+7x+(a_2-b_2)x^2\)

Since \(p(x)\ne0\) for every real \(x\), its discriminant must be negative

\(49-4(a_2-b_2)(a_1-b_1)<0\)

which is possible, so no further restriction is obtained

\(h(x)=f(x+1)-g(x+2)\)

The coefficient of \(x^3\) in \(f(x+1)\) is

\(a_3+4\)

\((x+1)^4=x^4+4x^3+\cdots\)

The coefficient of \(x^3\) in \(g(x+2)\) is

\(b_3+8\)

\((x+2)^4=x^4+8x^3+\cdots\)

Hence the coefficient of \(x^3\) in \(h(x)\) is

\((a_3+4)-(b_3+8)\)

\(=(a_3-b_3)-4\)

\(=0-4\)

Given \(\frac{a x^2+2(a+1) x+9 a+4}{x^2-8 x+32}<0 \quad \forall x \in \mathbb{R}\)
In \(x^2-8 x+32\), we have \(D=64-128<0\)

\( \therefore x^2-8 x+32>0 \forall x \in \mathbb{R} \)
\( \Rightarrow a x^2+2(a+1) x+9 a+4<0 \forall x \in R \)
\( \Rightarrow a<0 \text { and } D<0\)

Since Question has asked for positive integral values of a

\( \therefore|S|=0\)

but we need positive integral value of a.
So, No solution

Given: $(-2+\sqrt{3})\left(\right|\sqrt{x}-3\left|\right)+(x-6\sqrt{x})+(9-2\sqrt{3})=0$

\(\Rightarrow (x-6 \sqrt{x}+9)-(2-\sqrt{3})|\sqrt{x}-3|-2 \sqrt{3}=0 \)

\( \Rightarrow|\sqrt{x}-3|^2-(2-\sqrt{3})|\sqrt{x}-3|-2 \sqrt{3}=0 \)

\(\Rightarrow (|\sqrt{x}-3|-2)(|\sqrt{x}-3|+\sqrt{3})=0\)
\( \Rightarrow|\sqrt{x}-3|=2 \text { or }|\sqrt{x}-3|=-\sqrt{3}\) (not possible)
\(\Rightarrow \sqrt{x}=1\) or \(5 \)
\( \Rightarrow x=1 \) or \(25\)
\(\Rightarrow \alpha=1\) and \(\beta=25\)

\(\sqrt{\frac{\beta}{\alpha}}+\sqrt{\alpha \beta}=\sqrt{\frac{25}{1}}+\sqrt{25 \cdot 1}=10\)

$$\begin{gathered} \text{Given, }{x}^2+|2x-3|-4=0 \\ \text{for} x\ge \frac{3}{2}, 2x-3\ge 0 \\ \text{then} x^2+2x-7=0 \\ \text{On simplifying, we get} \\ x=2\sqrt{2}-1 \\ \left(\text{other value rejected because }x\ge \frac{3}{2} \right) \\ \text{for} x\le \frac{3}{2}, \left(2x-3\right)<0, \\ \text{then,} x^2-2x-1=0 \\ \text{ On simplifying, we get} \\ x=1-\sqrt{2} \\ \left(\text{other value rejected because} x<\frac{3}{2} \right) \\ \text{ Sum of squares }=(2\sqrt{2}-1{)}^2+(1-\sqrt{2}{)}^2 \\ =6\left(2-\sqrt{2}\right) \end{gathered}$$

$$\begin{gathered} \text{ Since }b,c>0 \\ \text{ Therefore} \\ \alpha +\beta =-b<0 \\ \text{ and αβ}=-c<0 \\ \text{ Since sum and product of the roots is -ve} \\ \text{therefore roots must be of opposite sign. } \end{gathered}$$

Let time taken by mason \(A\) alone to complete the work be in \(x\) days

so, mason \(B\) along take \(x+24\) days

work done by \(A\) in \(1\) day \(=\frac{1}{x}\)

work done by \(B\) in \(1\) day $=\frac{1}{x+24}$

work done by \(A+B\) in \(1\) day \(=\frac{1}{22.5}\)

So, $\frac{1}{x}+\frac{1}{x+24}=\frac{2}{45}$

\(\Rightarrow x^2-21 x-540=0\)

\(\Rightarrow x=36\) and \( x = –15\) (rejected)

Let $\alpha$ and $\beta$ be the roots of $x^2 + \sqrt{3}x - 16 = 0$

So, $\alpha^2 + \sqrt{3}\alpha - 16 = 0$ and $\beta^2 + \sqrt{3}\beta - 16 = 0$

$\alpha^n + \sqrt{3}\alpha^{n-1} - 16\alpha^{n-2} = 0$ and $\beta^n + \sqrt{3}\beta^{n-1} - 16\beta^{n-2} = 0$

For $n = 25$, we get: $P_{25} + \sqrt{3}P_{24} = 16P_{23}$

$\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} = \frac{16P_{23}}{2P_{23}} = 8$

Let $\gamma$ and $\delta$ be the roots of $x^2 + 3x - 1 = 0$.

So, $\gamma^n + 3\gamma^{n-1} - \gamma^{n-2} = 0$ and $\delta^n + 3\delta^{n-1} - \delta^{n-2} = 0$

For $n = 25$, we get: $Q_{25} - Q_{23} = -3Q_{24}$

$\frac{Q_{25} - Q_{23}}{Q_{24}} = \frac{-3Q_{24}}{Q_{24}} = -3$

$\Rightarrow \frac{P_{25}+\sqrt{3}{P}_{24}}{2P_{23}}+\frac{Q_{25}-Q_{23}}{Q_{24}}=8+\left(-3\right)=5$

$S_n=2023{\alpha }^n+2024{\beta }^n$

Since \(\alpha\) and $\beta$ are the roots of the equation \(x^2 - x - 1 = 0\), they must satisfy it. Therefore, we can write

$$\begin{gathered} {\alpha }^2-\alpha -1=0\Rightarrow {\alpha }^2=\alpha +1 \\ {\beta }^2-\beta -1=0\Rightarrow {\beta }^2=\beta +1 \\ {S}_n=2023({\alpha }^{n-1}+{\alpha }^{n-2})+2024({\beta }^{n-1}+{\beta }^{n-2}) \\ {S}_n=(2023{\alpha }^{n-1}+2024{\beta }^{n-1})+(2023{\alpha }^{n-2}+2024{\beta }^{n-2}) \\ {S}_n=S_{n-1}+S_{n-2} \\ {S}_{12}=S_{11}+S_{10} \end{gathered}$$

$$\begin{gathered} f\left(x\right)=x^2-(p+2)x+(2p+9) \\ 1.D\ge 0 \\ (p+2{)}^2-4(1\left)\right(2p+9)\ge 0 \\ {p}^2-4p-32\ge 0 \\ (p-8\left)\right(p+4)\ge 0 \\ p\in (-\infty ,-4]\cup [8,\infty ) \\ 2.\frac{-b}{a}<0\Rightarrow p+2<0\Rightarrow p<-2 \\ 3.\frac{c}{a}>0\Rightarrow 2p+9>0 \\ \Rightarrow p > -4.5 \\ by(1),(2) and (3) \\ p\in (-4.5,-4] \\ \Rightarrow \beta -2\alpha =5 \end{gathered}$$

Let time taken by mason \(A\) alone to complete the work be in \(x\) days

so, mason \(B\) along take \(x+24\) days

work done by \(A\) in \(1\) day \(=\frac{1}{x}\)

work done by \(B\) in \(1\) day $=\frac{1}{x+24}$

work done by \(A+B\) in \(1\) day \(=\frac{1}{22.5}\)

So, $\frac{1}{x}+\frac{1}{x+24}=\frac{2}{45}$

\(\Rightarrow x^2-21 x-540=0\)

\(\Rightarrow x=36\) and \( x = –15\) (rejected)

Let $\alpha$ and $\beta$ be the roots of $x^2 + \sqrt{3}x - 16 = 0$.

By the property of roots, we have:

$\alpha^2 + \sqrt{3}\alpha - 16 = 0$ $\beta^2 + \sqrt{3}\beta - 16 = 0$

$\alpha^n + \sqrt{3}\alpha^{n-1} - 16\alpha^{n-2} = 0$ $\beta^n + \sqrt{3}\beta^{n-1} - 16\beta^{n-2} = 0$

For $n = 25$, we get: $P_{25} + \sqrt{3}P_{24} = 16P_{23}$

So, the first part of the expression is: $\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} = \frac{16P_{23}}{2P_{23}} = 8$

Let $\gamma$ and $\delta$ be the roots of $x^2 + 3x - 1 = 0$.

Similarly, by Newton's Sums, we have:

$\gamma^n + 3\gamma^{n-1} - \gamma^{n-2} = 0$ $\delta^n + 3\delta^{n-1} - \delta^{n-2} = 0$

For $n = 25$, we get: $Q_{25} - Q_{23} = -3Q_{24}$

So, the second part of the expression is: $\frac{Q_{25} - Q_{23}}{Q_{24}} = \frac{-3Q_{24}}{Q_{24}} = -3$

The total expression is: $\frac{P_{25}+\sqrt{3}{P}_{24}}{2P_{23}}+\frac{Q_{25}-Q_{23}}{Q_{24}}=8+(-3)=5$

$\begin{array}{ll}\alpha \beta >0& \\ \Rightarrow 2p+9>0& \\ \Rightarrow p>\frac{-9}{2}& \end{array}$

$\begin{array}{ll}\text{and }\alpha +\beta <0& \\ p+2<0& \\ \Rightarrow p<-2& \end{array}$

$$\begin{array}{l} (p + 2)^{2} - 4(2p + 9) \geq 0 \\ (p + 4)(p - 8) \geq 0 \\ p \in (-\infty, -4) \cup (8, \infty) \\ p \in \left(- \frac{9}{2}, -4\right) = (\alpha, \beta) \\ \end{array}$$

$$\beta -2\alpha =-4+9=5$$

Given equation $\left(\frac{9}{x}-\frac{9}{\sqrt{x}}+2\right)\left(\frac{2}{x}-\frac{7}{\sqrt{x}}+3\right)=0$

let $\frac{1}{\sqrt{x}}=y \left[\mathrm{x}>0\right]$

$$\begin{gathered} \left(9y^2-9y+2\right)\left(2y^2-7y+3\right)=0 \\ \Rightarrow (3y-2)(3y-1)(y-3)(2y-1)=0 \\ \Rightarrow y=\frac{1}{3},\frac{1}{2},\frac{2}{3},3 \\ \Rightarrow \mathrm{x}=9,4,\frac{9}{4},\frac{1}{9} \\ \text{ no. of solutions} = 4 \end{gathered}$$

\(\begin{aligned}
f(x) & =5 x^3-15 x-a \\
f^{\prime}(x) & =15 x^2-15 \\
f^{\prime}(x) & =15\left(x^2-1\right) \\
& =15(x-1)(x+1)
\end{aligned}\)

\((-1) \rightarrow\) point of maxima \(1 \rightarrow\) point of minima

\(
\begin{aligned}
& f(1)=-10-9 \\
& f(-1)=10-9
\end{aligned}
\)

For 3 distinet meakerw)

\(
\begin{array}{lll}
-10-a<0 \Rightarrow & a>-10 \\
a \text { ard } \quad 10 a>0 & a<10
\end{array}
\)

\(\begin{aligned}
& =1 \quad a \in(-10,10) \\
& \alpha=-10, \quad \beta=10 \\
& \Rightarrow \beta-2 \alpha=10-2 \times(-10) \\
& =10+20 \\
& =30
\end{aligned}\)

Given, \(x^2+|2 x-3|-4=0\)

(i) \(2 x-3 \geq 0 \Rightarrow x \geq \frac{3}{2}\)

\(\Rightarrow x^2+2 x-3-4=0\)

\(x^2+2 x-7=0 \Rightarrow (x+1)^2=8\)

\(\Rightarrow x= \pm 2 \sqrt{2}-1\)

\(\Rightarrow x=(2 \sqrt{2}-1)\) as \(-2 \sqrt{2}-1<\frac{3}{2}\)

(ii) \(x \leq \frac{3}{2} \Rightarrow x^2-(2 x-3)-4=0\)

\(\Rightarrow x^2-2 x-1-0=0 \Rightarrow(x-1)^2=2\)

\(\Rightarrow x= \pm \sqrt{2}+1 \Rightarrow x=- \sqrt{2}+1\) as \(\sqrt{2}+1>\frac{3}{2}\)

⇒ two roots are \(x=-\sqrt{2}+1,2 \sqrt{2}-1\)

⇒ Sum of squares \(=12-6 \sqrt{2}=6(2-\sqrt{2})\)

Given \(a, 4, \alpha, b\) are in A.P.

Let \(d\) de the common difference,

Hence \(a=4-d, \alpha=4+d, b=4+2 d\)

\(\Rightarrow(4+d) x^2-(4-d) x+2(4+d-8-4 d)=0\)

\(\Rightarrow(4+d) x^2-(4-d) x+2(-4-3 d)=0\)

Now arithemetic mean of \(\frac{1}{a}\) and \(\frac{1}{b}\) is \(\frac{5}{16}\)

Therefore, $\frac{\frac{1}{a}+\frac{1}{b}}{2}=\frac{5}{16}$

\(\Rightarrow \frac{\frac{1}{4-d}+\frac{1}{4+2 d}}{2}=\frac{5}{16}\)

\(\Rightarrow 5 d^2-6 d-8=0\)

\(\Rightarrow d=2, -\frac{4}{5}\)

For \(d=2\): \(a=4-2=2\) (Rejected)

For \(d=-\frac{4}{5}\): \(a=4+\frac{4}{5}=\frac{24}{5}\)

Equation becomes:

\(16 x^2-24 x-2=0\)

\(\Rightarrow 2 x^2-3x-2=0\)

$\Rightarrow x=2,\frac{-1}{2}$

Given: \(\alpha, \beta\) be the roots of the quadratic equation \(12 x^2-20 x+3 \lambda=0\),

$\Rightarrow \alpha +\beta =\frac{5}{3}$ and $\alpha \beta =\frac{\lambda }{4}$

Now, $\frac{1}{2}\le \left|\alpha -\beta \right|\le \frac{3}{2}$

$\Rightarrow \frac{1}{4}\le |\alpha -\beta {|}^2\le \frac{9}{4}$

$\Rightarrow \frac{1}{4}\le {(\alpha +\beta )}^2-4\alpha \beta \le \frac{9}{4}$

$\Rightarrow \frac{1}{4}\le \frac{25}{9}-4\times \frac{\lambda }{4}\le \frac{9}{4}$

$\Rightarrow -\frac{91}{36}\le -\lambda \le \frac{-19}{36}$

$\Rightarrow \frac{19}{36}\le \lambda \le \frac{91}{36}$

$\Rightarrow \lambda =1,2$

Sum \(=3\)