Quadratic Equation and Inequalities

The given equation is $\lambda x^2-(\lambda+3) x+3=0$. We can find the roots by factoring:

$ \lambda x^2-\lambda x-3 x+3=0 $

$\Rightarrow $ $ \lambda x(x-1)-3(x-1)=0 $

$\Rightarrow $ $ (x-1)(\lambda x-3)=0 $

The two roots are : $ x=1, \quad x=\frac{3}{1}$

We are given that $\alpha$ and $\beta$ are the roots such that $\alpha<\beta$ and the following condition holds:

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

Because we don't know if 1 is smaller or larger than $\frac{3}{\lambda}$, we must consider two distinct cases.

Case A : $\alpha=1$ and $\beta=\frac{3}{\lambda}$

Substitute these into the condition $\frac{1}{\alpha}-\frac{1}{\beta}=\frac{1}{3}$ :

$ \begin{array}{r} \frac{1}{1}-\frac{1}{3 / \lambda}=\frac{1}{3} \\ 1-\frac{\lambda}{3}=\frac{1}{3} \\ 1-\frac{1}{3}=\frac{\lambda}{3} \\ 1-\frac{1}{3}=\frac{\lambda}{3} \\ \frac{2}{3}=\frac{\lambda}{3} \Longrightarrow \lambda=\mathbf{2} \end{array} $

Check the constraint $\alpha<\beta$ : If $\lambda=2$, then $\beta=3 / 2=1.5$. Since $1<1.5$, this value is valid.

Case B : $\alpha=\frac{3}{\lambda}$ and $\beta=1$

Substitute these into the condition $\frac{1}{\alpha}-\frac{1}{\beta}=\frac{1}{3}$ :

$ \begin{aligned} & \frac{1}{3 / \lambda}-\frac{1}{1}=\frac{1}{3} \\ & \frac{\lambda}{3}-1=\frac{1}{3} \\ & \frac{\lambda}{3}=1+\frac{1}{3} \\ & \frac{\lambda}{3}=\frac{4}{3} \Longrightarrow \lambda=4 \end{aligned} $

Check the constraint $\alpha<\beta$ : If $\lambda=4$, then $\alpha=3 / 4=0.75$. Since $0.75<1$, this value is valid.

The possible values for $\lambda$ are 2 and 4 .

$ \text { Sum }=2+4=6 $

The sum of all possible values of $\lambda$ is 6.

$ S=\left\{x^3+a x^2+b x+c: a, b, c \in \mathbb{N} \text { and } a, b, c \leq 20\right\} \text { so } a, b, c \in\{1,2,3, \ldots 20\} $

$x^3+a x^2+b x+c$ is divisible by $x^2+2$

$\therefore $ $ x^3+a x^2+b x+c=\left(x^2+2\right) L(x)$

Let $L(x)=x+n$ because L.H.S is cubic have leading coefficient 1 .

$ \begin{aligned} x^3+a x^2+b x+c & =\left(x^2+2\right)(x+n) \\ & =x^3+x^2 n+2 x+2 n \end{aligned} $

Compare $n=a, b=2,2 n=c \Longrightarrow c=2 a$

so, $c=2 a$ and $b=2$

Finding the Number of Possible Polynomials

• For $b$ : There is only $\mathbf{1}$ choice $(b=2)$, which is within the range $[1,20]$.

• For $a$ and $c$ : Since $c=2 a$, we must ensure $c \leq 20$.

• If $a=1, c=2$

• If $a=2, c=4$

• …

• If $a=10, c=20$
  
  There are 10 possible values for the pair ( $a, c$ ).

Total polynomials $=($ Choices for $b) \times($ Choices for $a, c$ pairs $)=1 \times 10=\mathbf{1 0}$

Given equation

$ x^4-a x^2+9=0--- \text { (1) } $

let $x^2=t$, equation (1) become

$ t^2-a t+9=0--- \text { (2) } $

If $t_1$ & $t_2$ are roots of equation (2) then four distinct roots of equation (1) is $\left\{-\sqrt{t_1},-\sqrt{t_2}, \sqrt{t_1}, \sqrt{t_2}\right\}$

roots of this equation are real and distinct if $\mathrm{D}>0$

$ \begin{aligned} & (-a)^2-4 \times 9>0 \\\\ & \Rightarrow a^2-36>0 \end{aligned} $

so $a \in(-\infty,-6) \cup(6, \infty)$

so smallest positive integral value of $a$ is 7 as $a>6$.

given $x|x+3|+|x-1|-2=0$.

critical points $x=-3,1$.

two critical points means 3 cases.

Case 1 : $ x<-3,|x+3|=-(x+3)$ and $|x-1|=-(x-1)$. equation become

$ \begin{aligned} & -x(x+3)-(x-1)-2=0 \\ & -x^2-3 x-x+1-2=0 \\ & -x^2-4 x-1=0 \Longrightarrow x^2+4 x+1=0 \\ & x=\frac{-4 \pm \sqrt{16-4}}{2}=\frac{-4 \pm 2 \sqrt{3}}{2}=-2 \pm \sqrt{3} \end{aligned} $

since $x<-3$ so $x=-2-\sqrt{3}$ is only solution.

Case 2 : $-3 \leq x \leq 1,|x+3|=x+3$ and $|x-1|=-(x-1)$. equation become

$ \begin{aligned} & x(x+3)-(x-1)-2=0 . \\ & x^2+3 x-x+1-2=0 . \\ & x^2+2 x-1=0 . \\ & x=\frac{-2 \pm \sqrt{4+4}}{2}=\frac{-2 \pm 2 \sqrt{2}}{2}=-1 \pm \sqrt{2} . \end{aligned} $

since $-3 \leq x \leq 1$ so $x=-1+\sqrt{2},-1-\sqrt{2}$ both accepted.

Case 3 : $x>1,|x+3|=x+3 \text { and }|x-1|=x-1 \text {. } $

equation become

$ x(x+3)+x-1-2=0 . $

$ x^2+4 x-3=0 . $

$x=\frac{-4 \pm \sqrt{16+12}}{2}=\frac{-4 \pm 2 \sqrt{7}}{2}=-2 \pm \sqrt{7}$. Checking positive solution. since $x>1$, so $x=\sqrt{7}-2 \approx 0.645$, so no solution.

so final solution is

$ x=-2-\sqrt{3},-1+\sqrt{2},-1-\sqrt{2} . $

Number of solution = 3

$ \begin{aligned} & \text { } f(x)=\log _{\left(10 x^2-17 x+7\right)}\left(18 x^2-11 x+1\right) \\ & \log _b a \text { is defined if } a>0 \text { and } b>0 \text { and } b \neq 1 \text {. } \\ & 18 x^2-11 x+1>0 \\ & 18 x^2-9 x-2 x+1>0 \\ & 9 x(2 x-1)-1(2 x-1)>0 \\ & (2 x-1)(9 x-1)>0 \\ & S_1: x \in(-\infty, 1 / 9) \cup(1 / 2, \infty) \text {. } \\ & 10 x^2-17 x+7>0 \\ & 10 x^2-10 x-7 x+7>0 \\ & 10 x(x-1)-7(x-1)>0 \\ & (10 x-7)(x-1)>0 \\ & S_2: x \in(-\infty, 7 / 10) \cup(1, \infty) \text {. } \\ & \text { And } 10 x^2-17 x+7 \neq 1 \end{aligned} $

$ \begin{aligned} & 10 x^2-17 x+6 \neq 0 \\ & 10 x^2-12 x-5 x+6 \neq 0 \\ & 2 x(5 x-6)-1(5 x-6) \neq 0 \\ & (2 x-1)(5 x-6) \neq 0 \\ & S_3: x \neq 1 / 2,6 / 5 \Rightarrow x \neq 0.5,1.2 . \\ & \text { Final solution } x \in\left[S_1 \cap S_2\right]-\{0.5,1.2\} . \\ & x \in[(-\infty, 1 / 9) \cup(1 / 2, \infty)] \cap[(-\infty, 7 / 10) \cup(1, \infty)]-\{0.5,1.2\} . \\ & x \in(-\infty, 1 / 9) \cup(1 / 2,7 / 10) \cup(1, \infty)-\{1.2\} . \end{aligned} $

Since 0.5 is not included, it need not to be subtracted.

Compare final solution with $(-\infty, a) \cup(b, c) \cup(d, \infty)-\{e\}$.

$ \begin{aligned} & a=1 / 9, b=1 / 2, c=7 / 10, d=1 \text { and } e=1.2 . \\ & 90(a+b+c+d+e)=90(1 / 9+1 / 2+7 / 10+1+1.2) \\ & =10+45+63+90+108=316 . \end{aligned} $

$ \begin{aligned} &\begin{aligned} & (-2+\sqrt{3})(|\sqrt{x}-3|)+(x-6 \sqrt{x}+9)-2 \sqrt{3}=0, x \geq 0 \\ & \text { put }|\sqrt{x}-3|=t \\ & \Rightarrow(-2+\sqrt{3}) t+t^2-2 \sqrt{3}=0 \\ & \Rightarrow t(t-2)+\sqrt{3}(t-2)=0 \\ & \Rightarrow(t+\sqrt{3})(t-2)=0 \\ & t+\sqrt{3} \neq 0 \Rightarrow t=2 \\ & \Rightarrow|\sqrt{x}-3|=2 \\ & \sqrt{x}-3= \pm 2 \\ & \sqrt{x}=5,1 \\ & x=25,1 \\ & \Rightarrow \alpha=1, \beta=25 \\ & \Rightarrow \sqrt{\frac{\beta}{\alpha}}+\sqrt{\alpha \beta}=10 \end{aligned}\\ &\text { Option (2) is Correct. } \end{aligned} $

$ \text { Let B's time }=x \text { days } $

$ \begin{aligned} & \Rightarrow \text { As time }=(x-24) \text { day } \\ & \text { Rate of } \mathrm{A}=\frac{1}{x-24} \\ & \text { Rate of } \mathrm{B}=\frac{1}{x} \\ & \frac{1}{x-24}+\frac{1}{x}=\frac{1}{22.5} \\ & \frac{2 x-24}{x(x-24)}=\frac{1}{22.5} \\ & 45 x-540=x^2-24 x \\ & x^2-69 x+540=0 \\ & (x-60)(x-9)=0 \\ & \Rightarrow x=60 \\ & \text { At time }=60-24 \\ & =36 \text { days } \end{aligned} $

$ \begin{aligned} & \frac{1}{2} \leq|\alpha-\beta| \leq \frac{3}{2} \\ & \frac{1}{4} \leq|\alpha-\beta|^2 \leq \frac{9}{4} \\ & \frac{1}{4} \leq(\alpha+\beta)^2-4 \alpha \beta \leq \frac{9}{4} \\ & \frac{1}{4} \leq \frac{25}{9}-4 \times \frac{\lambda}{4} \leq \frac{9}{4} \\ & -\frac{91}{36} \leq-\lambda \leq \frac{-19}{36} \\ & \frac{19}{36} \leq \lambda \leq \frac{91}{36} \\ & \lambda=1,2 \\ & \text { Sum }=3 \end{aligned} $

To find the number of distinct real solutions for the given equation, we need to understand how the expressions inside the absolute values behave. The absolute value expressions are $|x+4|$ and $|x+2|$. The signs of $x+4$ and $x+2$ change at $x = -4$ and $x = -2$. These two critical points divide the real number line into three intervals:

1. $x \ge -2$

2. $-4 \le x < -2$

3. $x < -4$

Let's analyze the equation $x|x+4| + 3|x+2| + 10 = 0$ for each case separately.

Case 1: When $x \ge -2$

In this interval, both $x+4$ and $x+2$ are positive (or zero). So, $|x+4| = x+4$ and $|x+2| = x+2$.

Substitute these into the equation:

$x(x+4) + 3(x+2) + 10 = 0$

$x^2 + 4x + 3x + 6 + 10 = 0$

$x^2 + 7x + 16 = 0$

Now, we check the discriminant $D$ (where $D = b^2 - 4ac$) of this quadratic equation:

$D = 7^2 - 4(1)(16) = 49 - 64 = -15$

Since $D < 0$, there are no real roots in this interval.

Case 2: When $-4 \le x < -2$

Here, $x+4$ is positive (or zero), but $x+2$ is negative. So, $|x+4| = x+4$ and $|x+2| = -(x+2)$.

The equation becomes:

$x(x+4) - 3(x+2) + 10 = 0$

$x^2 + 4x - 3x - 6 + 10 = 0$

$x^2 + x + 4 = 0$

Let's check the discriminant for this equation:

$D = 1^2 - 4(1)(4) = 1 - 16 = -15$

Again, $D < 0$, so there are no real roots in this interval either.

Case 3: When $x < -4$

In this case, both $x+4$ and $x+2$ are negative. So, $|x+4| = -(x+4)$ and $|x+2| = -(x+2)$.

The equation changes to:

$-x(x+4) - 3(x+2) + 10 = 0$

$-x^2 - 4x - 3x - 6 + 10 = 0$

$-x^2 - 7x + 4 = 0$

Multiplying by $-1$ makes it easier to solve:

$x^2 + 7x - 4 = 0$

Let's find the discriminant:

$D = 7^2 - 4(1)(-4) = 49 + 16 = 65$

Since $D > 0$, this quadratic equation gives two real roots. Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$, the roots are:

$x = \frac{-7 \pm \sqrt{65}}{2}$

Now we must verify if these roots actually lie in our current interval, which is $x < -4$.

We know that $\sqrt{65}$ is slightly larger than $8$ (because $\sqrt{64} = 8$). Let's roughly estimate it as $8.06$.

For the first root (taking the positive sign):

$x_1 = \frac{-7 + \sqrt{65}}{2} \approx \frac{-7 + 8.06}{2} \approx 0.53$

This value is greater than $-4$, so it does not lie in the given interval $x < -4$. We reject this root.

For the second root (taking the negative sign):

$x_2 = \frac{-7 - \sqrt{65}}{2} \approx \frac{-7 - 8.06}{2} \approx -7.53$

This value is strictly less than $-4$, so it falls correctly within our interval.

This means there is exactly $1$ valid real solution for the entire equation.

Option D is correct.

$ \begin{aligned} & \because \alpha<1<\beta \\ & f(1)<0 \\ & \Rightarrow 1+2 a+(3 a+10)<0 \\ & \Rightarrow 5 a+11<0 \\ & a<\frac{-11}{5} \\ & \therefore a \in\left(-\infty, \frac{-11}{5}\right) \end{aligned} $

$ \begin{aligned} & |x-1| 2-5|x-1|+6=0 \\ & \text { Let }|x-1|=t \\ & t^2-5+6=0 \\ & (t-3)(t-2)=0 \\ & t=2,3 \\ & |x-1|=2 \text { or }|x-1|=3 \\ & x-1= \pm 2 \text { or } x-1= \pm 3 \\ & \Rightarrow x=3,-1,4,-2 \\ & \Rightarrow \text { Sum }=3-1+4-2 \\ & \quad=4 \end{aligned} $

Let the roots of the quadratic equation be $ \alpha $ and $ \beta $.

From the relation between coefficients and roots, we know that:

$ \alpha + \beta = \frac{-b}{a} = k_1 \quad \Rightarrow \quad b = -a k_1 \, \text{where} \, k_1 \in \mathbb{Z} $

Also,

$ \alpha \beta = \frac{c}{a} = k_2 \quad \Rightarrow \quad c = a k_2 \, \text{where} \, k_2 \in \mathbb{Z} $

Since $ a, b, c $ are in arithmetic progression, we have:

$ c - b = b - a $

Substituting the values of $ b $ and $ c $ from above:

$ a k_2 + a k_1 = 2a k_1 + a \Rightarrow k_2 = -2k_1 - 1 $

Now, the quadratic equation becomes:

$ x^2 - k_1 x + k_2 = 0 $

Substituting the value of $ k_2 $:

$ x^2 - k_1 x - 2k_1 - 1 = 0 $

Since $ \alpha \beta = k_2 $ is odd, both roots $ \alpha $ and $ \beta $ must be odd integers.

Let us test some possible integer values. For example, if $ c = 15 $, the pairs of integer roots could be $ (-1, -15) $ or $ (-3, -5) $.

If the roots are $ -1 $ and $ -15 $, then:
Sum of roots $ = -16 \Rightarrow b = 16, \; a = 17 $
The quadratic becomes $ 17x^2 + 16x + 15 = 0 $.

This does not satisfy the condition because its roots are not the expected $ -1 $ and $ -15 $.

Now, if the roots are $ -3 $ and $ -5 $, then:
Sum of roots $ = -8, \; \text{Product} = 15 $
Hence, the quadratic is $ x^2 + 8x + 15 = 0 $.

This equation fits our condition with integer roots.

For the equation $ x^2 - k_1 x - 2k_1 - 1 = 0 $, substitute $ x = 3 $:

$ 9 - 3k_1 - 2k_1 - 1 = 0 \Rightarrow 8 - 5k_1 = 0 \Rightarrow k_1 = \frac{5}{8} $

Since $ k_1 $ must be an integer, this is not possible. Hence, 3 cannot be a root of the quadratic equation.

We are given

$ \lim_{x \to 2} \frac{\tan(x-2)\left(rx^2+(p-2)x-2p\right)}{(x-2)^2}=5. $

We have to first find the relation between $r$ and $p$, and then determine all values of $q$ for which the roots of

$ rx^2-px+q=0 $

lie in $(0,2)$.

Step 1: Simplify the limit

Let

$ f(x)=rx^2+(p-2)x-2p. $

Then the given limit becomes

$ \lim_{x\to 2}\frac{\tan(x-2)\,f(x)}{(x-2)^2}=5. $

Now, as $x\to 2$,

$ \tan(x-2)\sim x-2. $

So the limit becomes

$ \lim_{x\to 2}\frac{(x-2)f(x)}{(x-2)^2} = \lim_{x\to 2}\frac{f(x)}{x-2}. $

For this limit to be finite, we must have

$ f(2)=0. $

Now,

$ f(2)=r(2)^2+(p-2)(2)-2p=4r+2p-4-2p=4r-4. $

So,

$ 4r-4=0 \implies r=1. $

Then

$ f(x)=x^2+(p-2)x-2p. $

Since $f(2)=0$, factor $(x-2)$:

$ x^2+(p-2)x-2p=(x-2)(x+p). $

Hence the limit becomes

$ \lim_{x\to 2}\frac{\tan(x-2)(x-2)(x+p)}{(x-2)^2} = \lim_{x\to 2}\frac{\tan(x-2)}{x-2}(x+p). $

Using

$ \lim_{t\to 0}\frac{\tan t}{t}=1, $

we get

$ 1\cdot (2+p)=5. $

So,

$ p+2=5 \implies p=3. $

Thus,

$ r=1,\qquad p=3. $

Step 2: Form the quadratic equation

Now the equation becomes

$ x^2-3x+q=0. $

We want both roots to lie in $(0,2)$.

Let the roots be $\alpha_1,\alpha_2$. Then

$ \alpha_1+\alpha_2=3,\qquad \alpha_1\alpha_2=q. $

Since both roots lie in $(0,2)$, let us find the condition on $q$.

Step 3: Use condition for both roots in $(0,2)$

Let

$ f(x)=x^2-3x+q. $

Since coefficient of $x^2$ is positive, parabola opens upward.

For both roots to lie in $(0,2)$, we need:

1. Real roots

So discriminant must be non-negative:

$ \Delta=9-4q\ge 0 \implies q\le \frac94. $

2. Both roots positive

Since product of roots is $q$,

$ q>0. $

3. Both roots less than $2$

Now check value at $x=2$:

$ f(2)=4-6+q=q-2. $

If both roots are less than $2$, then for an upward parabola, point $x=2$ lies to the right of both roots, so we need

$ f(2)>0. $

Thus,

$ q-2>0 \implies q>2. $

Now combine:

$ 2

So the set of all possible values of $q$ is

$ (\alpha,\beta]=(2,\tfrac94]. $

Hence,

$ \alpha=2,\qquad \beta=\frac94. $

Therefore,

$ 4(\alpha+\beta)=4\left(2+\frac94\right) =4\left(\frac{17}{4}\right)=17. $

Final Answer

$ \boxed{17} $

So, the correct option is Option C.

For the quadratic equation

$ (\lambda+2)x^2-3\lambda x+4\lambda=0,\qquad \lambda\neq -2 $

to have two positive roots, let us use the NCERT conditions for a quadratic equation.

If the roots are $\alpha$ and $\beta$, then for both roots to be positive, we need:

1. Discriminant $\ge 0$ so that roots are real,

2. Sum of roots $>0$,

3. Product of roots $>0$.

1. Identify coefficients

Comparing with $ax^2+bx+c=0$, we get

$ a=\lambda+2,\qquad b=-3\lambda,\qquad c=4\lambda $

2. Condition for real roots

Discriminant:

$ D=b^2-4ac $

So,

$ D=(-3\lambda)^2-4(\lambda+2)(4\lambda) $

$ D=9\lambda^2-16\lambda(\lambda+2) $

$ D=9\lambda^2-16\lambda^2-32\lambda $

$ D=-7\lambda^2-32\lambda $

For real roots,

$ -7\lambda^2-32\lambda \ge 0 $

$ -\lambda(7\lambda+32)\ge 0 $

Multiplying by $-1$ reverses the inequality:

$ \lambda(7\lambda+32)\le 0 $

This gives

$ -\frac{32}{7}\le \lambda \le 0 $

So possible integral values from this condition are

$ \lambda=-4,-3,-2,-1,0 $

But given $\lambda\ne -2$, so remaining are

$ \lambda=-4,-3,-1,0 $

3. Condition for sum of roots positive

Sum of roots is

$ \alpha+\beta = -\frac{b}{a}=\frac{3\lambda}{\lambda+2} $

For two positive roots,

$ \frac{3\lambda}{\lambda+2}>0 $

Since $3>0$, this means

$ \frac{\lambda}{\lambda+2}>0 $

This is positive when numerator and denominator have same sign:

• $\lambda>0$ and $\lambda+2>0$, or

• $\lambda<0$ and $\lambda+2<0$

So,

$ \lambda>0 \quad \text{or} \quad \lambda<-2 $

From our shortlisted values $-4,-3,-1,0$, this keeps only

$ \lambda=-4,-3 $

4. Condition for product of roots positive

Product of roots is

$ \alpha\beta=\frac{c}{a}=\frac{4\lambda}{\lambda+2} $

For positive roots,

$ \frac{4\lambda}{\lambda+2}>0 $

Again,

$ \frac{\lambda}{\lambda+2}>0 $

This gives the same condition:

$ \lambda>0 \quad \text{or} \quad \lambda<-2 $

Among the shortlisted values, again we get

$ \lambda=-4,-3 $

5. Check both values

• For $\lambda=-4$:

  $ D=-7(16)-32(-4)=-112+128=16>0 $

  So roots are real and distinct.

• For $\lambda=-3$:

  $ D=-7(9)-32(-3)=-63+96=33>0 $

  So roots are real and distinct.

Hence both values give two positive roots.

Therefore, the number of possible integral values of $\lambda$ is

$ \boxed{2} $

So, the correct option is B.

We need to solve

$ |x^2+x-9|=|x|+|x^2-9|. $

We will do it by considering intervals where the expressions inside modulus change sign.

The critical points are obtained from:

• $x=0$ from $|x|$

• $x=\pm 3$ from $|x^2-9|$

• roots of $x^2+x-9=0$, but for NCERT-style interval splitting here, it is enough to use the main sign-change points $-3,0,3$ and then check sign of $x^2+x-9$ in each interval.

Now solve interval-wise.

1. For $x\le -3$

Here,

• $|x|=-x$

• $|x^2-9|=x^2-9$

• Also $x^2+x-9>0$, so $|x^2+x-9|=x^2+x-9$

Thus,

$ x^2+x-9=-x+(x^2-9) $

$ x^2+x-9=x^2-x-9 $

$ x=-x $

$ 2x=0 \Rightarrow x=0 $

But $0\not\le -3$, so no solution here.

2. For $-3\le x\le 0$

Here,

• $|x|=-x$

• $|x^2-9|=9-x^2$

Now we check $x^2+x-9$ on this interval.

At $x=-3$,

$ x^2+x-9=9-3-9=-3<0 $

At $x=0$,

$ x^2+x-9=-9<0 $

So throughout this interval,

$ |x^2+x-9|=-(x^2+x-9)=-x^2-x+9. $

Hence equation becomes

$ -x^2-x+9=-x+(9-x^2). $

Simplify RHS:

$ -x^2-x+9=-x+9-x^2. $

Both sides are identical, so this is true for every $x\in[-3,0]$.

So all $x\in[-3,0]$ are solutions.

3. For $0\le x\le 3$

Here,

• $|x|=x$

• $|x^2-9|=9-x^2$

Also on this interval,

$ x^2+x-9 $

At $x=0$, it is $-9<0$, and at $x=3$, it is $9+3-9=3>0$.

So the sign changes inside this interval.

Let us find the root:

$ x^2+x-9=0 $

$ x=\frac{-1\pm\sqrt{37}}{2}. $

The positive root is

$ \frac{-1+\sqrt{37}}{2}. $

So split further.

(i) For $0\le x\le \frac{-1+\sqrt{37}}{2}$

Here $x^2+x-9\le 0$, so

$ |x^2+x-9|=-x^2-x+9. $

Equation:

$ -x^2-x+9=x+(9-x^2) $

$ -x^2-x+9=-x^2+x+9 $

$ -x=x $

$ 2x=0 \Rightarrow x=0. $

So only $x=0$, already included in previous interval.

(ii) For $\frac{-1+\sqrt{37}}{2}\le x\le 3$

Here $x^2+x-9\ge 0$, so

$ |x^2+x-9|=x^2+x-9. $

Equation:

$ x^2+x-9=x+(9-x^2) $

$ x^2-9=9-x^2 $

$ 2x^2=18 $

$ x^2=9 $

$ x=\pm 3. $

In this interval, only $x=3$ is valid.

So from $[0,3]$, only $x=0,3$ are solutions.

4. For $x\ge 3$

Here,

• $|x|=x$

• $|x^2-9|=x^2-9$

• $x^2+x-9>0$, so $|x^2+x-9|=x^2+x-9$

Thus,

$ x^2+x-9=x+(x^2-9) $

$ x^2+x-9=x^2+x-9 $

This is true for every $x\ge 3$.

So all $x\in[3,\infty)$ are solutions.

Complete solution set

Combining all intervals:

$ [-3,0]\cup[3,\infty). $

Thus,

$ \alpha=-3,\quad \beta=0,\quad \gamma=3. $

Now,

$ \alpha^2+\beta^2+\gamma^2=(-3)^2+0^2+3^2=9+0+9=18. $

So the correct answer is

$ \boxed{18} $

Hence, Option B is correct.

For the equation $x^2 - 3x + r = 0$ with roots $\alpha, \beta$, by using the standard results:

$ \begin{aligned} & \alpha+\beta=3 \\ & \frac{\alpha}{2}+2 \beta=-3 \end{aligned} $

The first line comes from “sum of roots” $= -\dfrac{\text{coefficient of }x}{\text{coefficient of }x^2} = -\dfrac{-3}{1}=3$.

The second line comes from the equation $x^2 + 3x + r = 0$, whose roots are $\dfrac{\alpha}{2}$ and $2\beta$. So, their sum is:

$\frac{\alpha}{2} + 2\beta = -\frac{3}{1} = -3$

Now solve the two equations:

$\alpha+\beta=3 \Rightarrow \alpha=3-\beta$

Substitute into $\dfrac{\alpha}{2}+2\beta=-3$:

$\frac{3-\beta}{2}+2\beta=-3$

Multiply by $2$:

$3-\beta+4\beta=-6 \Rightarrow 3+3\beta=-6 \Rightarrow 3\beta=-9 \Rightarrow \beta=-3$

Then $\alpha=3-(-3)=6$.

Now use “product of roots” for $x^2 - 3x + r = 0$:

$\alpha\beta=r \Rightarrow r=6\times(-3)=-18$

Now consider the equation $x^2 + 6x = m$. Rewrite it in standard form:

$x^2 + 6x - m = 0$

So, product of its roots $= \dfrac{-m}{1}=-m$.

The roots are given as $2\alpha + \beta + 2r$ and $\alpha - 2\beta - \dfrac{r}{2}$. Substitute $\alpha=6$, $\beta=-3$, $r=-18$:

First root:

$2\alpha + \beta + 2r = 2(6)+(-3)+2(-18)=12-3-36=-27$

Second root:

$\alpha - 2\beta - \frac{r}{2} = 6-2(-3)-\frac{-18}{2}=6+6+9=21$

Now product of roots:

$ \begin{aligned} & =(2 \alpha+\beta+2 \mathrm{r})\left(\alpha-2 \beta-\frac{\mathrm{r}}{2}\right) \\ & \Rightarrow-\mathrm{m}=(-27)(21) \\ & \Rightarrow \mathrm{m}=567 \end{aligned} $

We are given that

the quadratic equation is

$ (x)(x+2) + (x+1)(x+3) + (x+2)(x+4) + \ldots + (x + n - 1)(x + n + 1) = 4n $

and the roots are $ \alpha, \alpha + 2 $ (both integers).

We need to find the value of $ n + \alpha $.

Step 1: Expand the general term

The $k^{th}$ term of the left-hand side is:

$ (x + k)(x + k + 2) $

Expanding this:

$ (x + k)(x + k + 2) = x^2 + 2(k+1)x + k(k+2) $

Step 2: Sum over all $k$ from $0$ to $n-1$

$ \text{LHS} = \sum_{k=0}^{n-1} [x^2 + 2(k+1)x + k(k+2)] $

Let's compute each part separately.

1. For $x^2$:

   $ \sum_{k=0}^{n-1} x^2 = n x^2 $

2. For $2(k+1)x$:

   $ \sum_{k=0}^{n-1} 2(k+1)x = 2x \sum_{k=0}^{n-1} (k+1) = 2x \cdot \frac{n(n+1)}{2} = n(n+1)x $

3. For $k(k+2)$:

   $ \sum_{k=0}^{n-1} k(k+2) = \sum_{k=0}^{n-1} (k^2 + 2k) = \sum k^2 + 2\sum k $

Now,

$ \sum_{k=0}^{n-1} k = \frac{n(n-1)}{2}, \quad \sum_{k=0}^{n-1} k^2 = \frac{(n-1)n(2n-1)}{6} $

Therefore,

$ \sum_{k=0}^{n-1} k(k+2) = \frac{(n-1)n(2n-1)}{6} + n(n-1) $

Simplify:

$ = n(n-1)\left( \frac{2n-1}{6} + 1 \right) = n(n-1)\left( \frac{2n-1+6}{6} \right) = \frac{n(n-1)(2n+5)}{6} $

Step 3: Combine all terms

$ \text{LHS} = n x^2 + n(n+1)x + \frac{n(n-1)(2n+5)}{6} $

The equation is

$ n x^2 + n(n+1)x + \frac{n(n-1)(2n+5)}{6} = 4n $

Step 4: Simplify by dividing both sides by $n$ (since $n > 0$)

$ x^2 + (n+1)x + \frac{(n-1)(2n+5)}{6} = 4 $

$ x^2 + (n+1)x + \left[\frac{(n-1)(2n+5)}{6} - 4\right] = 0 $

Simplify the constant term:

$ \frac{(n-1)(2n+5)}{6} - 4 = \frac{(n-1)(2n+5) - 24}{6} = \frac{2n^2 + 3n -29}{6} $

Thus, the equation is

$ x^2 + (n+1)x + \frac{2n^2 + 3n - 29}{6} = 0 $

Step 5: Use relation between roots and coefficients

Let roots be $ \alpha $ and $ \alpha + 2 $.

Then,

Sum of roots:

$ \alpha + (\alpha + 2) = 2\alpha + 2 = - (n+1) $

so,

$ \alpha = \frac{ - (n+1) - 2}{2} = -\frac{n+3}{2} $

Product of roots:

$ \alpha(\alpha + 2) = \frac{2n^2 + 3n - 29}{6} $

Substitute $ \alpha = -\frac{n+3}{2} $:

$ \alpha(\alpha + 2) = \left(-\frac{n+3}{2}\right)\left(-\frac{n-1}{2}\right) = \frac{(n+3)(n-1)}{4} $

Step 6: Equate the two expressions for product

$ \frac{(n+3)(n-1)}{4} = \frac{2n^2 + 3n - 29}{6} $

Simplify:

$ 3(n+3)(n-1) = 2(2n^2 + 3n - 29) $

Expand:

$ 3(n^2 + 2n - 3) = 4n^2 + 6n - 58 $

$ 3n^2 + 6n - 9 = 4n^2 + 6n - 58 $

Simplify:

$ 3n^2 + 6n - 9 - 4n^2 - 6n + 58 = 0 $

$ -n^2 + 49 = 0 \implies n^2 = 49 $

$ n = 7 \text{ (since } n \in \mathbf{N}) $

Step 7: Find $ \alpha $ and hence $ n + \alpha $

$ \alpha = -\frac{n+3}{2} = -\frac{7+3}{2} = -5 $

$ n + \alpha = 7 + (-5) = 2 $

$ \boxed{n + \alpha = 2} $

Correct Option: C (2)

$\begin{aligned} & |\mathrm{x}-2|^2+2|\mathrm{x}-2|-|\mathrm{x}-2|-2=0 \\ & \Rightarrow(|\mathrm{x}-2|+2)(|\mathrm{x}-2|-1)=0 \\ & \Rightarrow|\mathrm{x}-2|=1 \\ & \Rightarrow \mathrm{x}=2 \pm 1=3,1 \\ & \Rightarrow \text { sum of square of roots }=9+1=10 \\ & \mathrm{x}^2-2|\mathrm{x}-3|-5=0 \\ & \text { Case-I } \mathrm{x}-3 \geq 0 \\ & \Rightarrow \mathrm{x}^2-2 \mathrm{x}+1=0 \\ & \Rightarrow(\mathrm{x}-1)^2=0 \\ & \Rightarrow \mathrm{x}=1 \\ & \text { But } \mathrm{x} \geq 3 \\ & \Rightarrow \mathrm{x} \in \phi \\ & \text { Case-II } \mathrm{x}-3<0 \\ & \mathrm{x}^2+2 \mathrm{x}-11=0, \mathrm{D}>0 \Rightarrow \text { Real & distinct roots } \\ & \mathrm{f}(\mathrm{x})=\mathrm{x}^2+2 \mathrm{x}-11 \\ & \mathrm{f}(3)>0, \frac{-\mathrm{p}}{2 \mathrm{a}}=-1<3 \\ & \Rightarrow \text { both roots }<3, \text { both roots acceptable } \\ & \text { Sum of square of roots }=(\alpha+\beta)^2-2 \alpha \beta \\ & =4+22=26 \\ & \Rightarrow \text { Final sum }=10+26=36 \end{aligned}$

$ \begin{aligned} & \text { (I) } x<2 \\ & -x^2+2 x-3 x+9+1=0 \\ & \Rightarrow x^2+x-10=0 \\ & \Rightarrow x=\frac{-1+\sqrt{41}}{2}, \frac{-1-\sqrt{41}}{2} \\ & \quad \quad \qquad \quad \times \qquad \qquad \sqrt{ } \end{aligned}$

$\begin{aligned} & \text { (II) } 2 \leq x<3 \\ & \Rightarrow \mathrm{x}^2-2 \mathrm{x}-3 \mathrm{x}+9+1=0 \\ & \Rightarrow \mathrm{x}^2-5 \mathrm{x}+10=0 \\ & \mathrm{D}<0 \\ & \text { (III) } \mathrm{x} \geq 3 \\ & \mathrm{x}^2-2 \mathrm{x}+3 \mathrm{x}-9+2=0 \\ & \Rightarrow \mathrm{x}^2+\mathrm{x}-8=0 \end{aligned}$

$\mathrm{x}=\frac{-1+\sqrt{32}}{2}, \frac{-1-\sqrt{32}}{2}$

$\quad \quad \times \qquad \qquad \times$

1 real roots

To find the set of all values of $ p \in \mathbb{R} $ for which both roots of the equation $ x^2-(p+2)x+(2p+9)=0 $ are negative real numbers, follow these steps:

Discriminant Condition:

The equation's discriminant $ D $ must be non-negative for real roots:

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

Simplifying this:

$ p^2 + 4p + 4 - 8p - 36 \geq 0 \quad \Rightarrow \quad p^2 - 4p - 32 \geq 0 $

This can be factored as:

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

Meaning $ p \in (-\infty, -4] \cup [8, \infty) $. …… (1)

Sum of Roots Condition:

The sum of the roots (which is $ p+2 $) must be negative:

$ p + 2 < 0 \quad \Rightarrow \quad p < -2 $ …… (2)

Product of Roots Condition:

The product of the roots $ (2p + 9) $ must be positive:

$ 2p + 9 > 0 \quad \Rightarrow \quad p > -\frac{9}{2} $ …… (3)

Determine the Valid Interval:

Combine the results from conditions (1), (2), and (3). From conditions (1) and (2), we find $ p < -2 $:

Intersection of $ (-\infty, -4] $ and $ (-\frac{9}{2}, -2) $ gives:

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

Calculate $\beta - 2\alpha$:

With $\alpha = -\frac{9}{2}$ and $\beta = -4$, compute:

$ \beta - 2\alpha = -4 - 2\left(-\frac{9}{2}\right) = -4 + 9 = 5 $

Therefore, the difference $\beta - 2\alpha$ is 5.

$\begin{aligned} & x^2+4 x-n=0 \text { has integer roots } \\ & \Rightarrow x=\frac{-4 \pm \sqrt{16+4 n}}{2}=-2 \pm \sqrt{4+n} \end{aligned}$

For $x$ to be integer $4+n$ must be perfect squares

$\begin{aligned} & n \in[20,100] \\ & n+4 \in[24,104]=S \end{aligned}$

$\left\{25,36, \ldots 10^2\right\} \in S \Rightarrow 5^2, 6^2, \ldots 10^2 \Rightarrow 6$ values of $n$

Given the equation $ x(x+2)(12-k) = 2 $, we want it to have equal roots. To achieve this, we need to manipulate it into a quadratic form in terms of $ x $:

$ x^2 + 2x - \frac{2}{12-k} = 0 $

For this quadratic equation to have equal (repeated) roots, the discriminant $ D $ must be zero. The discriminant $ D $ for the equation $ ax^2 + bx + c = 0 $ is given by:

$ D = b^2 - 4ac $

Substituting $ a = 1 $, $ b = 2 $, and $ c = -\frac{2}{12-k} $ into the discriminant formula, we get:

$ 4 - 4\left(-\frac{2}{12-k}\right) = 0 $

Simplifying the expression:

$ 1 + \frac{2}{12-k} = 0 $

Solving for $ k $:

$ \frac{2}{12-k} = -1 \quad \Rightarrow \quad 2 = -(12 - k) \quad \Rightarrow \quad 2 = -12 + k \quad \Rightarrow \quad k = 14 $

Now, consider the point $(k, \frac{k}{2})$, which becomes $(14, 7)$. We need to find its distance from the line $3x + 4y + 5 = 0$. The formula for the distance $ d $ from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is:

$ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} $

Substituting $ (x_1, y_1) = (14, 7) $ and the line coefficients $ A = 3 $, $ B = 4 $, $ C = 5 $:

$ d = \frac{|3(14) + 4(7) + 5|}{\sqrt{3^2 + 4^2}} $

Calculating the numerator:

$ 3 \times 14 + 4 \times 7 + 5 = 42 + 28 + 5 = 75 $

And the denominator:

$ \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 $

Therefore, the distance is:

$ d = \frac{75}{5} = 15 $

Thus, the distance is 15.

$\begin{aligned} & \Rightarrow \quad P^n=\gamma^n+\delta^n \\ & P_{25}-P_{23}=\left(\gamma^{25}-\gamma^{23}\right)+\left(\delta^{25}-\delta^{23}\right) \\ & =\gamma^{23}\left(\gamma^2-1\right)+\delta^{23}\left(\delta^2-1\right) \\ & =\gamma^{23}(-3 \gamma)+\delta^{23}(-3 \delta)=-3\left[\gamma^{24}+\delta^{24}\right] \end{aligned}$

$\begin{aligned} &\Rightarrow \frac{P_{25}-P_{23}}{P_{24}}=(-3)\\ &\text { Similarly, } \end{aligned}$

$\begin{aligned} & \Rightarrow Q_{25}+\sqrt{3} Q_{24}=\left(\alpha^{25}+\sqrt{3} \alpha^{24}\right)+\left(\beta^{25}+\sqrt{3} \beta^{24}\right) \\ & =\alpha^{23}\left(\alpha^2+\sqrt{3} \alpha\right)+\beta^{23}\left(\beta^2+\sqrt{3} \beta\right) \\ & =\alpha^{23}(16)+16 \beta^{23} \\ & \Rightarrow \frac{Q_{25}+\sqrt{3} Q_{24}}{2 \cdot Q_{23}}=\frac{16\left(\alpha^{23}+\beta^{23}\right)}{2\left(\alpha^{23}+\beta^{23}\right)}=8 \\ & \Rightarrow \frac{Q_{25}+\sqrt{3} Q_{24}}{2 Q_{23}}+\frac{\left(P_{25}-P_{23}\right)}{P_{24}}=8+(-3)=5 \end{aligned}$

Given

$P_n=\alpha^n+\beta^n$

and

$P_{10}=123,\quad P_9=76,\quad P_8=47,\quad P_1=1.$

We want the quadratic equation whose roots are $\dfrac{1}{\alpha}$ and $\dfrac{1}{\beta}$.

For numbers $\alpha,\beta$, the sequence $P_n=\alpha^n+\beta^n$ satisfies

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

Now use the given values:

$P_{10}=(\alpha+\beta)P_9-(\alpha\beta)P_8.$

So,

$123=(\alpha+\beta)\cdot 76-(\alpha\beta)\cdot 47.$

Also, since $P_1=\alpha+\beta=1$, we get

$123=76-47\alpha\beta.$

Thus,

$47=-47\alpha\beta,$

so

$\alpha\beta=-1.$

Now the roots needed are $\dfrac{1}{\alpha}$ and $\dfrac{1}{\beta}$.

Their sum is

$\frac{1}{\alpha}+\frac{1}{\beta} =\frac{\alpha+\beta}{\alpha\beta} =\frac{1}{-1}=-1.$

Their product is

$\frac{1}{\alpha\beta}=\frac{1}{-1}=-1.$

A quadratic equation with roots $r_1,r_2$ is

$x^2-(r_1+r_2)x+r_1r_2=0.$

Here,

$x^2-(-1)x+(-1)=0,$

which gives

$x^2+x-1=0.$

So the correct option is:

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

Option A.

$\begin{aligned} & (a-5)^2-8(15-3 a)<0 \\ & a^2+14 a+25-120<0 \\ & a^2+14 a-95<0 \\ & (a+19)(a-5)<0 \\ & a \in(-19,5) \\ & \therefore-19< x<5 \\ & \therefore \sum_{x \in X} x^2=\left(1^2+2^2+\ldots .+4^2\right)+\left(1^2+2^2+\ldots+18^2\right) \\ & =\frac{4 \times 5 \times 9}{6}+\frac{18 \times 19 \times 37}{6} \\ & =30+2109 \\ & =2139 \end{aligned}$

Consider $\frac{1}{\sqrt{\mathrm{x}}}=\alpha \quad \mathrm{x}>0$

$\begin{aligned} & \left(9 \alpha^2-9 \alpha+2\right)\left(2 \alpha^2-7 \alpha+3\right)=0 \\ & (3 \alpha-2)(3 \alpha-1)(\alpha-3)(2 \alpha-1)=0 \\ & \alpha=\frac{1}{3}, \frac{1}{2}, \frac{2}{3}, 3 \\ & x=9,4, \frac{9}{4}, \frac{1}{9} \end{aligned}$

So, no. of solutions $=4$

as $f(x)$ is a polynomial of degree two let it be

$f(x)=a x^2+b x+c \quad(a \neq 0)$

on satisfying given conditions we get

$C=1 \& a= \pm 1$

hence $f(x)=1 \pm x^2$

also range $\in(-\infty, 1]$ hence

$\begin{gathered} f(x)=1-x^2 \\ \text { now } f(k)=-2 k \end{gathered}$

$1-\mathrm{k}^2=-2 \mathrm{k} \rightarrow \mathrm{k}^2-2 \mathrm{k}-1=0$

let roots of this equation be $\alpha \& \beta$ then $\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta$

$=4-2(-1)=6$

To find the sum of the squares of all the roots of the equation $ x^2 + |2x - 3| - 4 = 0 $:

Case I: $ x \geq \frac{3}{2} $

For $ x \geq \frac{3}{2} $, the expression $ |2x - 3| $ becomes $ 2x - 3 $. Thus, the equation becomes:

$ x^2 + 2x - 3 - 4 = 0 $

Simplifying gives:

$ x^2 + 2x - 7 = 0 $

Solving this quadratic equation, we find:

$ x = 2\sqrt{2} - 1 $

Case II: $ x < \frac{3}{2} $

For $ x < \frac{3}{2} $, the expression $ |2x - 3| $ becomes $ -(2x - 3) = -2x + 3 $. The equation therefore becomes:

$ x^2 + 3 - 2x - 4 = 0 $

Simplifying gives:

$ x^2 - 2x - 1 = 0 $

Solving this quadratic equation, we obtain:

$ x = 1 - \sqrt{2} $

Sum of the Squares of the Roots

The sum of the squares of the roots is:

$ (2\sqrt{2} - 1)^2 + (1 - \sqrt{2})^2 $

Calculating each term:

$(2\sqrt{2} - 1)^2 = (2\sqrt{2})^2 - 2 \cdot 2\sqrt{2} \cdot 1 + 1^2 = 8 - 4\sqrt{2} + 1 = 9 - 4\sqrt{2}$

$(1 - \sqrt{2})^2 = 1^2 - 2 \cdot 1 \cdot \sqrt{2} + (\sqrt{2})^2 = 1 - 2\sqrt{2} + 2 = 3 - 2\sqrt{2}$

Adding these results:

$ (9 - 4\sqrt{2}) + (3 - 2\sqrt{2}) = 12 - 6\sqrt{2} $

This simplifies to $ 6(2 - \sqrt{2}) $.

Thus, the sum of the squares of the roots is $ 6(2 - \sqrt{2}) $.

Only 2 solutions.

To solve the given equation, start by rewriting the expression:

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

First, simplify the second part of the expression:

$ (x - 4)(x - 5) = x^2 - 9x + 20 $

Now the equation becomes:

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

Introduce a substitution for simplification:

$ \text{Let } t = x^2 - 9x $

Thus, the equation transforms to:

$ t^2 + 22t + 121 - t - 20 - 3 = 0 $

Simplify further:

$ t^2 + 21t + 98 = 0 $

Factor the quadratic:

$ (t + 14)(t + 7) = 0 $

This gives:

$ t = -7 \quad \text{or} \quad t = -14 $

Address each case where $ t = x^2 - 9x $:

$ x^2 - 9x = -7 $

$ x^2 - 9x + 7 = 0 $

Solving this quadratic equation, we find the roots:

$ x = \frac{9 \pm \sqrt{81 - 4 \times 7}}{2} = \frac{9 \pm \sqrt{53}}{2} $

$ x^2 - 9x = -14 $

$ x^2 - 9x + 14 = 0 $

Solving this quadratic equation:

$ x = \frac{9 \pm \sqrt{81 - 4 \times 14}}{2} = \frac{9 \pm \sqrt{25}}{2} $

$ x = \frac{9 \pm 5}{2} = 7 \quad \text{or} \quad x = 2 $

The rational roots from the second equation are 7 and 2. Thus, the product of all the rational roots is:

$ 7 \times 2 = 14 $

To find the sum of the fourth powers of the roots $\alpha_\theta$ and $\beta_\theta$ of the quadratic equation $2x^2 + (\cos \theta)x - 1 = 0$, we start analyzing the expression $\alpha_\theta^4 + \beta_\theta^4$.

The equation can be rewritten with its roots using:

$ \alpha + \beta = -\frac{\cos \theta}{2}, \quad \alpha \beta = -\frac{1}{2} $

We need to calculate $\alpha^2 + \beta^2$ and $\alpha^2 \beta^2$:

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

$ \alpha^2 \beta^2 = (\alpha \beta)^2 = \left(-\frac{1}{2}\right)^2 = \frac{1}{4} $

Substitute these into:

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

Maximize and minimize $\left(\frac{\cos^2 \theta}{4} + 1\right)^2$:

Zero of $\cos\theta$ leads to:

$ \left(\frac{0^2}{4} + 1\right)^2 = 1 $

Max value $\cos^2 \theta = 1$:

$ \left(\frac{1}{4} + 1\right)^2 = \left(\frac{5}{4}\right)^2 = \frac{25}{16} $

Substitute back:

$ \text{Max: } \frac{25}{16} - \frac{1}{2} = \frac{25}{16} - \frac{8}{16} = \frac{17}{16} $

$ \text{Min: } 1 - \frac{1}{2} = \frac{1}{2} $

Finally, compute $16(M + m)$:

$ 16\left(\frac{17}{16} + \frac{1}{2}\right) = 16\left(\frac{17}{16} + \frac{8}{16}\right) = 16 \times \frac{25}{16} = 25 $

To solve for the coefficient of $ x^3 $ in the function $ h(x) = f(x + 1) - g(x + 2) $, we start by expanding both $ f(x + 1) $ and $ g(x + 2) $.

First, expand $ f(x + 1) $:

$ f(x+1) = a_1 + 10(x+1) + a_2 (x+1)^2 + a_3 (x+1)^3 + (x+1)^4 $

From this expansion, the coefficient of $ x^3 $ in $ f(x+1) $ is $ a_3 + 4 $.

Next, expand $ g(x + 2) $:

$ g(x+2) = b_1 + 3(x+2) + b_2 (x+2)^2 + b_3 (x+2)^3 + (x+2)^4 $

Here, the coefficient of $ x^3 $ in $ g(x+2) $ is $ b_3 + 8 $.

Now, calculate the coefficient of $ x^3 $ in $ h(x) = f(x+1) - g(x+2) $:

$ \text{Coefficient of } x^3 \text{ in } h(x) = (a_3 + 4) - (b_3 + 8) = a_3 - b_3 - 4 $

Given that $ f(x) \neq g(x) $ for every $ x \in \mathbb{R} $, it follows that $ f(x) - g(x) \neq 0 $ for any $ x $. This means the equation:

$ (a_1 - b_1) + 7x + (a_2 - b_2)x^2 + (a_3 - b_3)x^3 = 0 $

cannot have any real roots. For this polynomial to have no real roots, the degree of the polynomial must be less than or equal to zero, implying that the coefficients of the highest degree terms must equate to zero. Thus, $ a_3 - b_3 = 0 $.

Substituting $ a_3 - b_3 = 0 $ into our earlier expression for the coefficient of $ x^3 $ in $ h(x) $ gives:

$ a_3 - b_3 - 4 = -4 $

Thus, the coefficient of $ x^3 $ in $ h(x) $ is $-4$.

$\begin{aligned} & x^2-\sqrt{2} x-\sqrt{3}=0 \\ & P_n=\alpha^n-\beta^n \end{aligned}$

$\alpha$ and $\beta$ are the roots of the equation

Using Newton's theorem

$\begin{aligned} & P_{n+2}-\sqrt{2} P_{n+1}-\sqrt{3} P_n=0 \\ & \text { Put } n=10 \\ & P_{12}-\sqrt{2} P_{11}-\sqrt{3} P_{10}=0 \end{aligned}$

$P_{12}=\sqrt{2} P_{11}+\sqrt{3} P_{10}$

Put $n=9$

$\begin{aligned} & P_{11}-\sqrt{2} P_{10}-\sqrt{3} P_9=0 \\ & P_{11}=\sqrt{2} P_{10}+\sqrt{3} P_9 \\ & (11 \sqrt{3}-10 \sqrt{2}) P_{10}+(11 \sqrt{2}+10) P_{11}-11 P_{12} \end{aligned}$

Put the value of $P_{12}$ & $P_{12}$ in above equation.

$\begin{aligned} = & (11 \sqrt{3}-10 \sqrt{2}) P_{10}+(11 \sqrt{2}+10)\left(\sqrt{2} P_{10}+\sqrt{3} P_9\right)-11\left(\sqrt{2} P_{11}+\sqrt{3} P_{10}\right) \\ = & 11 \sqrt{3} P_{10}-10 \sqrt{2} P_{10}+22 P_{10}+10 \sqrt{2} P_{10}+11 \sqrt{6} P_9+10 \sqrt{3} P_9-11 \sqrt{2} P_{11}-11 \sqrt{3} P_{10} \\ = & 22 P_{10}+11 \sqrt{6} P_9+10 \sqrt{3} P_9-11 \sqrt{2}\left(\sqrt{2} P_{10}+\sqrt{3} P_9\right) \\ = & 22 P_{10}+11 \sqrt{6} P_9+10 \sqrt{3} P_9-22 P_{10}-11 \sqrt{6} P_9 \\ = & 10 \sqrt{3} P_9 \end{aligned}$

$\begin{aligned} & x^2+2 \sqrt{2 x}-1=0 \\ & \alpha+\beta=-2 \sqrt{2} \text { and } \alpha \beta=-1 \\ & \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta \\ & =8+2=10 \\ & \alpha^4+\beta^4=\left(\alpha^2+\beta^2\right)^2-2(\alpha \beta)^2 \\ & =100-2=98 \\ & \alpha^6+\beta^6=\left(\alpha^2+\beta^2\right)^3-3 \alpha^2 \beta^2\left(\alpha^2+\beta^2\right) \\ & =1000-3(10) \\ & =970 \\ & \therefore \quad \frac{1}{10}\left(\alpha^6+\beta^6\right)=97 \end{aligned}$

Equation whose roots are $\alpha^4+\beta^4$ and $\frac{1}{10}\left(\alpha^6+\beta^6\right)$ is

$\begin{aligned} & x^2-(98+97) x+98 \times 97=0 \\ & x^2-195 x+9506=0 \end{aligned}$

First, let's start by substituting $y = (8)^x$ in the given equation. By substituting, the equation $8^{2x} - 16 \cdot 8^x + 48 = 0$ will be transformed into

$ y^2 - 16y + 48 = 0 $

Now, we have a quadratic equation in $y$. To find the roots of this quadratic equation, we can use the quadratic formula:

$ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $

In this case, $a = 1$, $b = -16$, and $c = 48$. Substituting these values into the formula, we get:

$ y = \frac{16 \pm \sqrt{256 - 192}}{2} $

$ y = \frac{16 \pm \sqrt{64}}{2} $

$ y = \frac{16 \pm 8}{2} $

Solving for the two possible values of $y$, we have:

$ y = \frac{16 + 8}{2} = 12 $

$ y = \frac{16 - 8}{2} = 4 $

Now, recall that we substituted $y = (8)^x$. So, we need to solve for $x$ when $y = 12$ and $y = 4$:

$ 8^x = 12 $

$ x = \log_8(12) $

$ 8^x = 4 $

$ x = \log_8(4) $

Therefore, the solutions for $x$ are $\log_8(12)$ and $\log_8(4)$. The sum of these solutions is:

$ \log_8(12) + \log_8(4) $

Using the logarithmic property that $\log_b(m) + \log_b(n) = \log_b(m \cdot n)$, we get:

$ \log_8(12) + \log_8(4) = \log_8(12 \cdot 4) $

$ = \log_8(48) $

Now, we note that:

$ 48 = 8 \cdot 6 $

Thus,

$ \log_8(48) = \log_8(8 \cdot 6) = \log_8(8) + \log_8(6) = 1 + \log_8(6) $

Therefore, the sum of all the solutions of the equation is:

Option A $1 + \log_8(6)$.

$\begin{aligned} & x^2-\left(t^2-5 t+6\right) x+1=0 \\ & \therefore a_{2025}-\left(t^2-5 t+6\right) a_{2024}+a_{2023}=0 \\ & \Rightarrow \frac{a_{2025}+a_{2023}}{a_{2024}}=t^2-5 t+6 \\ & =\left(t+\frac{5}{2}\right)^2+\left(\frac{-1}{4}\right) \\ & \text { Minimum value }=\frac{-1}{4} \end{aligned}$

Given that the roots of the quadratic equation are $2$ and $6$, we can use Vieta's formulas which relate the coefficients of the polynomial to sums and products of its roots.

The given quadratic equation is:

$a x^2 + b x + 1 = 0$

By Vieta's formulas, the sum of the roots is:

$2 + 6 = -\frac{b}{a}$

So:

$8 = -\frac{b}{a} \Rightarrow b = -8a$

And the product of the roots is:

$2 \times 6 = \frac{1}{a} \Rightarrow 12 = \frac{1}{a} \Rightarrow a = \frac{1}{12}$

Therefore, $b = -8a = -8 \left( \frac{1}{12} \right) = -\frac{2}{3}$.

Given the roots of the new quadratic equation are:

$\frac{1}{2a+b}$ and $\frac{1}{6a+b}$

We know $a = \frac{1}{12}$ and $b = -\frac{2}{3}$, so:

$2a + b = 2 \left(\frac{1}{12}\right) - \frac{2}{3} = \frac{1}{6} - \frac{2}{3} = \frac{1 - 4}{6} = -\frac{3}{6} = -\frac{1}{2}$

and:

$6a + b = 6 \left(\frac{1}{12}\right) - \frac{2}{3} = \frac{1}{2} - \frac{2}{3} = \frac{3 - 4}{6} = -\frac{1}{6}$

Thus, the roots of the new quadratic equation are:

$\frac{1}{-\frac{1}{2}} = -2$

and:

$\frac{1}{-\frac{1}{6}} = -6$

The new quadratic equation with roots $-2$ and $-6$ can be formulated as:

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

The sum of the roots is:

$-2 + (-6) = -8$

The product of the roots is:

$(-2) \times (-6) = 12$

Thus, the quadratic equation becomes:

$x^2 - (-8)x + 12 = x^2 + 8x + 12 = 0$

Hence, the correct option is:

Option A

$x^2 + 8 x + 12 = 0$

Notice that $(\sqrt{3} + \sqrt{2})$ and $(\sqrt{3} - \sqrt{2})$ are reciprocals of each other because :

• $(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1$

• This means $(\sqrt{3} - \sqrt{2})^x = \frac{1}{(\sqrt{3} + \sqrt{2})^x} $

• $a + \frac{1}{a} = 10$

Multiplying both sides by *a*, we get a quadratic equation :

• $a^2 + 1 = 10a$


• $a^2 - 10a + 1 = 0$

Using the quadratic formula, we find :

• $a = \frac{10 \pm \sqrt{96}}{2} = 5 \pm 2\sqrt{6}$

Since $a = (\sqrt{3} + \sqrt{2})^x $, we have two cases :

1. $(\sqrt{3} + \sqrt{2})^x = 5 + 2\sqrt{6}$ = $(\sqrt{3} + \sqrt{2})^2$. There is one real solution for x in this case which is x = 2.


2. $(\sqrt{3} + \sqrt{2})^x = 5 - 2\sqrt{6}$ = $(\sqrt{3} - \sqrt{2})^2= \frac{1}{(\sqrt{3} + \sqrt{2})^2} $ = $(\sqrt{3} + \sqrt{2})^{-2}$. There is one real solution for x in this case which is x = - 2.

There are two real solutions for *x*. Therefore, the number of elements in the set S is 2. This corresponds with option (C).

$\begin{aligned} & x^2-x-1=0 \\ & S_n=2023 \alpha^n+2024 \beta^n \\ & S_{n-1}+S_{n-2}=2023 \alpha^{n-1}+2024 \beta^{n-1}+2023 \alpha^{n-2}+2024 \beta^{n-2} \\ & =2023 \alpha^{n-2}[1+\alpha]+2024 \beta^{n-2}[1+\beta] \\ & =2023 \alpha^{n-2}\left[\alpha^2\right]+2024 \beta^{n-2}\left[\beta^2\right] \\ & =2023 \alpha^n+2024 \beta^n \\ & S_{n-1}+S_{n-2}=S_n \\ & P_{u t} n=12 \\ & S_{11}+S_{10}=S_{12} \end{aligned}$