AP EAMCET 2023 Mathematics Question Paper with Answer and Solution

(B) Given the equation $z^3+\bar{z}=0$.
Taking the modulus on both sides,we get $|z^3| = |-\bar{z}|$,which implies $|z|^3 = |z|$.
This gives $|z|(|z|^2 - 1) = 0$.
Case $1$: $|z| = 0$,which implies $z = 0$. This is $1$ solution.
Case $2$: $|z|^2 = 1$,which implies $z\bar{z} = 1$,so $\bar{z} = \frac{1}{z}$.
Substituting this into the original equation: $z^3 + \frac{1}{z} = 0$,which gives $z^4 + 1 = 0$.
The equation $z^4 = -1$ has $4$ distinct roots.
Thus,the total number of solutions is $1 + 4 = 5$.

(C) The given equation is $|z - 1 + i| = 1$.
This can be rewritten as $|z - (1 - i)| = 1$.
Comparing this with the standard equation of a circle in the complex plane,$|z - z_0| = r$,where $z_0$ is the centre and $r$ is the radius,we get $z_0 = 1 - i$ and $r = 1$.
In Cartesian coordinates,$z_0 = 1 - i$ corresponds to the point $(1, -1)$.
Thus,$S$ represents a circle with centre $(1, -1)$ and radius $1$ unit.

(C) Given $\left|z-\frac{2}{z}\right|=2$.
Using the triangle inequality property $||z_1|-|z_2|| \leq |z_1-z_2|$,we have:
$||z|-\left|\frac{2}{z}\right|| \leq \left|z-\frac{2}{z}\right| = 2$.
Let $|z| = r$. Then $|r - \frac{2}{r}| \leq 2$.
This implies $-2 \leq r - \frac{2}{r} \leq 2$.
Considering the right inequality: $r - \frac{2}{r} \leq 2
\Rightarrow r^2 - 2r - 2 \leq 0$.
Solving the quadratic equation $r^2 - 2r - 2 = 0$ using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$r = \frac{2 \pm \sqrt{4 - 4(1)(-2)}}{2} = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}$.
Since $r = |z| > 0$,we have $r \leq 1 + \sqrt{3}$.
Thus,the greatest value of $|z|$ is $\sqrt{3} + 1$.

(D) Let $z = -3-4i$. We are given $\sqrt{z} = re^{i\theta}$.
Squaring both sides,we get $z = r^2 e^{i2\theta} = r^2(\cos 2\theta + i \sin 2\theta)$.
Comparing the real and imaginary parts with $z = -3-4i$:
$r^2 \cos 2\theta = -3$ and $r^2 \sin 2\theta = -4$.
We know that $r^2 = |z| = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9+16} = 5$.
Using the double angle formula for tangent: $\tan 2\theta = \frac{r^2 \sin 2\theta}{r^2 \cos 2\theta} = \frac{-4}{-3} = \frac{4}{3}$.
Using $\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} = \frac{4}{3}$,we get $6 \tan \theta = 4 - 4 \tan^2 \theta$,or $4 \tan^2 \theta + 6 \tan \theta - 4 = 0$.
Dividing by $2$,$2 \tan^2 \theta + 3 \tan \theta - 2 = 0$.
Factoring gives $(2 \tan \theta - 1)(\tan \theta + 2) = 0$,so $\tan \theta = \frac{1}{2}$ or $\tan \theta = -2$.
Since $z$ is in the third quadrant,its square root $\sqrt{z}$ lies in the second or fourth quadrant. For $\sqrt{z} = x+iy$,$xy = -2$,so $x$ and $y$ have opposite signs.
Testing $\tan \theta = -2$,$r^2 \tan \theta = 5 \times (-2) = -10$.

(D) Given $z = \sqrt{\frac{1-i}{1+i}}$.
Multiply numerator and denominator by $(1-i)$:
$z = \sqrt{\frac{(1-i)^2}{1^2+1^2}} = \sqrt{\frac{(1-i)^2}{2}} = \pm \frac{1-i}{\sqrt{2}}$.
Thus,$z_1 = \frac{1-i}{\sqrt{2}}$ and $z_2 = \frac{-1+i}{\sqrt{2}}$.
For $z_1 = \frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}$,the argument is $\arg(z_1) = \tan^{-1}\left(\frac{-1/\sqrt{2}}{1/\sqrt{2}}\right) = -\frac{\pi}{4}$.
For $z_2 = -\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}$,the argument is $\arg(z_2) = \pi + \tan^{-1}\left(\frac{1/\sqrt{2}}{-1/\sqrt{2}}\right) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
Since $-\frac{\pi}{2} < -\frac{\pi}{4} < \frac{3\pi}{4} < \pi$,the condition is satisfied.
Therefore,$\arg(z_1) + \arg(z_2) = -\frac{\pi}{4} + \frac{3\pi}{4} = \frac{2\pi}{4} = \frac{\pi}{2}$.

(B) The points are $A(-2, 1)$ and $B(3, -4)$.
Using the section formula,point $C$ divides $AB$ in the ratio $m:n = 1:2$.
$C = \left( \frac{1(3) + 2(-2)}{1+2}, \frac{1(-4) + 2(1)}{1+2} \right) = \left( \frac{3-4}{3}, \frac{-4+2}{3} \right) = \left( -\frac{1}{3}, -\frac{2}{3} \right)$.
Since $C$ lies in the third quadrant,its argument is $\theta = \tan^{-1}\left( \frac{y}{x} \right) - \pi$.
$\text{arg}(C) = \tan^{-1}\left( \frac{-2/3}{-1/3} \right) - \pi = \tan^{-1}(2) - \pi$.

(B) Given $z_1 = 4+3 i$.
The modulus $\alpha = |z_1| = \sqrt{4^2+3^2} = \sqrt{16+9} = 5$.
The region is defined by $|z - \overline{z_1}| \leq 5$.
Since $\overline{z_1} = 4-3 i$,the inequality is $|z - (4-3 i)| \leq 5$.
We test the options by checking if they satisfy $|z - (4-3 i)| \leq 5$:
For option $B$,$z = z_1 = 4+3 i$:
$|(4+3 i) - (4-3 i)| = |6 i| = 6$.
Since $6 > 5$,the point $z_1$ does not satisfy the inequality.
Thus,$z_1$ does not lie in the region.

(A) Given that $z_1, z_2, z_3$ are the vertices of an equilateral triangle and $z$ is its circumcentre.
Since $z$ is the circumcentre,the distance from $z$ to each vertex is equal to the circumradius $R$.
Therefore,$|z-z_1| = |z-z_2| = |z-z_3| = R$.
Now,consider the ratio $\frac{|z-z_1|}{|z-z_2|} = \frac{R}{R} = 1$.
Also,$\frac{|z-z_3|}{|z-z_1|} = \frac{R}{R} = 1$.
Thus,$\frac{|z-z_1|}{|z-z_2|} = \frac{|z-z_3|}{|z-z_1|} = 1$.

(C) Given the equation: $4a + i(3a - b) = b - 6i$
Equating the real and imaginary parts on both sides:
Real part: $4a = b$
Imaginary part: $3a - b = -6$
Substitute $b = 4a$ into the second equation:
$3a - 4a = -6$ $\Rightarrow -a = -6$ $\Rightarrow a = 6$
Then,$b = 4(6) = 24$
Now,substitute $a$ and $b$ into $z$:
$z = 6 + \frac{24}{4}i = 6 + 6i$
The modulus $|z|$ is:
$|z| = \sqrt{6^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}$
Finally,calculate $\frac{|z|}{a}$:
$\frac{|z|}{a} = \frac{6\sqrt{2}}{6} = \sqrt{2}$

(B) Given $z = \frac{-2+i}{(1-2i)^2}$.
First,simplify the denominator: $(1-2i)^2 = 1^2 + (2i)^2 - 2(1)(2i) = 1 - 4 - 4i = -3 - 4i$.
So,$z = \frac{-2+i}{-3-4i} = \frac{2-i}{3+4i}$.
Multiply the numerator and denominator by the conjugate of the denominator $(3-4i)$:
$z = \frac{(2-i)(3-4i)}{(3+4i)(3-4i)} = \frac{6 - 8i - 3i + 4i^2}{3^2 + 4^2} = \frac{6 - 11i - 4}{9 + 16} = \frac{2 - 11i}{25}$.
The conjugate is $\bar{z} = \frac{2 + 11i}{25}$.
The modulus is $|\bar{z}| = |z| = \sqrt{(\frac{2}{25})^2 + (-\frac{11}{25})^2} = \sqrt{\frac{4 + 121}{625}} = \sqrt{\frac{125}{625}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}$.

(B) Given that the amplitude (argument) of the complex number $z = x + iy$ is $\theta$.
By definition,$\arg(z) = \tan^{-1}\left(\frac{y}{x}\right) = \theta$.
Taking the tangent of both sides,we get $\frac{y}{x} = \tan \theta$.
Therefore,the locus of the point is $y = x \tan \theta$,which represents a straight line passing through the origin.

(D) The given inequality is $|z - (1 - i)| \leq 2$. This represents a disk in the Argand plane with center $(1, -1)$ and radius $r = 2$.
We check each point $z$ to see if $|z - (1 - i)| \leq 2$ holds:
$(A)$ For $z = \frac{1 - i}{2}$,$|z - (1 - i)| = |\frac{1 - i}{2} - 1 + i| = |\frac{1 - i - 2 + 2i}{2}| = |\frac{-1 + i}{2}| = \frac{\sqrt{(-1)^2 + 1^2}}{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \approx 0.707 \leq 2$. (Inside)
$(B)$ For $z = 1$,$|z - (1 - i)| = |1 - 1 + i| = |i| = 1 \leq 2$. (Inside)
$(C)$ For $z = \frac{1 - i}{4}$,$|z - (1 - i)| = |\frac{1 - i}{4} - 1 + i| = |\frac{1 - i - 4 + 4i}{4}| = |\frac{-3 + 3i}{4}| = \frac{3}{4} \sqrt{1^2 + (-1)^2} = \frac{3\sqrt{2}}{4} \approx 1.06 \leq 2$. (Inside)
$(D)$ For $z = i$,$|z - (1 - i)| = |i - 1 + i| = |-1 + 2i| = \sqrt{(-1)^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \approx 2.236 > 2$. (Outside)
Thus,the point $i$ is not in the region.

(B) Given that $-3+ix^2y$ and $x^2+y+4i$ are complex conjugates.
Therefore,$-3-ix^2y = x^2+y+4i$.
Comparing the real and imaginary parts on both sides:
$x^2+y = -3$ $(i)$
$-x^2y = 4$ $(ii)$
From $(ii)$,$y = -\frac{4}{x^2}$.
Substituting $y$ in $(i)$:
$x^2 - \frac{4}{x^2} = -3$
Let $x^2 = t$,then $t - \frac{4}{t} = -3 \implies t^2 + 3t - 4 = 0$.
$(t+4)(t-1) = 0$.
Since $t = x^2$ must be positive,$t = 1$.
$x^2 = 1 \implies x = \pm 1$.

(D) Given $z = 1 + \cos \theta - i \sin \theta$.
$z - 1 = \cos \theta - i \sin \theta$.
$|z - 1| = \sqrt{\cos^2 \theta + \sin^2 \theta} = 1$.
$|z|^2 = (1 + \cos \theta)^2 + (-\sin \theta)^2 = 1 + 2 \cos \theta + \cos^2 \theta + \sin^2 \theta = 2 + 2 \cos \theta$.
Now,substitute these into the expression:
$\left[|z - 1|^2 - \frac{|z|^2}{4}\right]^{1/2} = \left[1^2 - \frac{2 + 2 \cos \theta}{4}\right]^{1/2} = \left[1 - \frac{1 + \cos \theta}{2}\right]^{1/2} = \left[\frac{2 - 1 - \cos \theta}{2}\right]^{1/2} = \left[\frac{1 - \cos \theta}{2}\right]^{1/2}$.
Using the identity $1 - \cos \theta = 2 \sin^2 \left(\frac{\theta}{2}\right)$,we get:
$\left[\frac{2 \sin^2 \left(\frac{\theta}{2}\right)}{2}\right]^{1/2} = \sqrt{\sin^2 \left(\frac{\theta}{2}\right)} = \sin \left(\frac{\theta}{2}\right)$ (since $0 < \frac{\theta}{2} < \frac{\pi}{2}$,$\sin \left(\frac{\theta}{2}\right) > 0$).

(A) Given the equation: $|z-1|=|i(z+1)|$
Since $|i|=1$,we can write: $|z-1|=|z+1|$
Let $z = x+iy$. Then: $|x+iy-1|=|x+iy+1|$
Squaring both sides: $|(x-1)+iy|^2 = |(x+1)+iy|^2$
$(x-1)^2 + y^2 = (x+1)^2 + y^2$
$x^2 - 2x + 1 + y^2 = x^2 + 2x + 1 + y^2$
$-2x = 2x$
$4x = 0 \Rightarrow x = 0$
The equation $x=0$ represents the $Y$-axis in the Argand plane.

(A) First,simplify the expression inside the argument:
$\frac{4+2 i}{1-2 i} = \frac{(4+2 i)(1+2 i)}{(1-2 i)(1+2 i)} = \frac{4+8 i+2 i-4}{1+4} = \frac{10 i}{5} = 2 i$
$\frac{3+4 i}{2+3 i} = \frac{(3+4 i)(2-3 i)}{(2+3 i)(2-3 i)} = \frac{6-9 i+8 i+12}{4+9} = \frac{18-i}{13}$
Now,add the two results:
$2 i + \frac{18-i}{13} = \frac{26 i + 18 - i}{13} = \frac{18+25 i}{13} = \frac{18}{13} + \frac{25}{13} i$
The argument is $\tan^{-1}\left(\frac{25/13}{18/13}\right) = \tan^{-1}\left(\frac{25}{18}\right)$.
Since $\frac{25}{18} > 1$,we have $\tan^{-1}\left(\frac{25}{18}\right) > \tan^{-1}(1) = \frac{\pi}{4}$.
Also,since $\frac{25}{18} > 0$,the argument lies in the first quadrant,i.e.,$(0, \frac{\pi}{2})$.
Thus,the value lies in the interval $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$.

(B) We know that $(1+i)^2 = 1+i^2+2i = 1-1+2i = 2i$ and $(1-i)^2 = 1+i^2-2i = 1-1-2i = -2i$.
Substituting these into the expression:
$(1+i)^{2024} + (1-i)^{2024} = [(1+i)^2]^{1012} + [(1-i)^2]^{1012}$
$= (2i)^{1012} + (-2i)^{1012}$
$= 2^{1012} \cdot i^{1012} + (-2)^{1012} \cdot i^{1012}$
Since $1012$ is an even number,$(-2)^{1012} = 2^{1012}$ and $i^{1012} = (i^4)^{253} = 1^{253} = 1$.
$= 2^{1012} \cdot 1 + 2^{1012} \cdot 1$
$= 2 \cdot 2^{1012} = 2^{1013}$.

(C) The $n^{\text{th}}$ roots of a complex number $z = r(\cos \theta + i \sin \theta)$ are given by $z_k = r^{1/n} \operatorname{cis} \left( \frac{\theta + 2k\pi}{n} \right)$ for $k = 0, 1, \dots, n-1$.
For $z = -1$,we have $r = 1$ and $\theta = \pi$.
Thus,the $15^{\text{th}}$ roots are $\operatorname{cis} \left( \frac{\pi + 2k\pi}{15} \right)$ for $k = 0, 1, \dots, 14$.
For $k = 6$,the root is $\operatorname{cis} \left( \frac{\pi + 12\pi}{15} \right) = \operatorname{cis} \left( \frac{13\pi}{15} \right)$.

(A) Given that $\alpha$ and $\beta$ are the roots of $x^2+x+1=0$.
These roots are the cube roots of unity,specifically $\omega$ and $\omega^2$.
Let $\alpha = \omega = e^{i \frac{2\pi}{3}}$ and $\beta = \omega^2 = e^{i \frac{4\pi}{3}}$.
We need to find the equation with roots $\alpha^{2023}$ and $\beta^{1012}$.
First,calculate $\alpha^{2023} = (\omega)^{2023} = \omega^{2023 \pmod 3} = \omega^{2022+1} = \omega^1 = \alpha$.
Next,calculate $\beta^{1012} = (\omega^2)^{1012} = \omega^{2024} = \omega^{2022+2} = \omega^2 = \beta$.
Since the roots of the new equation are $\alpha$ and $\beta$,the required quadratic equation is the same as the original equation: $x^2+x+1=0$.

(C) Given $|z|=1$ and $\operatorname{Arg}(z)=\frac{\pi}{6}$.
$z = |z|e^{i \operatorname{Arg}(z)} = 1 \cdot e^{i \frac{\pi}{6}} = e^{i \frac{\pi}{6}}$.
Then $z^6 = (e^{i \frac{\pi}{6}})^6 = e^{i \pi} = -1$.
And $z^{12} = (e^{i \frac{\pi}{6}})^{12} = e^{2 \pi i} = 1$.
Substituting these values into the expression:
$\frac{z^{12}+1-z^6}{z^{12}+i z^6-1} = \frac{1+1-(-1)}{1+i(-1)-1} = \frac{3}{-i} = \frac{3}{-i} \cdot \frac{i}{i} = \frac{3i}{-i^2} = \frac{3i}{1} = 3i$.

(D) Using De Moivre's Theorem,$[r(\cos \theta + i \sin \theta)]^n = r^n(\cos n\theta + i \sin n\theta)$.
Given expression: $[\sqrt{2}(\cos 56^{\circ} 15^{\prime} + i \sin 56^{\circ} 15^{\prime})]^8$.
Here,$r = \sqrt{2}$,$\theta = 56^{\circ} 15^{\prime} = 56.25^{\circ}$,and $n = 8$.
Applying the theorem:
$= (\sqrt{2})^8 [\cos(8 \times 56.25^{\circ}) + i \sin(8 \times 56.25^{\circ})]$.
$= 2^4 [\cos(450^{\circ}) + i \sin(450^{\circ})]$.
Since $450^{\circ} = 360^{\circ} + 90^{\circ}$,we have $\cos(450^{\circ}) = \cos(90^{\circ}) = 0$ and $\sin(450^{\circ}) = \sin(90^{\circ}) = 1$.
$= 16(0 + i(1)) = 16i$.

(A) We have to calculate $\cos \left(\sum_{k=1}^7(k-\omega)(k-\omega^2) \frac{\pi}{175}\right)$.
Since $\omega^2+\omega+1=0$ and $\omega^3=1$,we have $(k-\omega)(k-\omega^2) = k^2 - k(\omega+\omega^2) + \omega^3 = k^2 - k(-1) + 1 = k^2+k+1$.
Thus,the expression becomes $\cos \left(\frac{\pi}{175} \sum_{k=1}^7 (k^2+k+1)\right)$.
Calculating the sum: $\sum_{k=1}^7 k^2 = \frac{7(8)(15)}{6} = 140$,$\sum_{k=1}^7 k = \frac{7(8)}{2} = 28$,and $\sum_{k=1}^7 1 = 7$.
Sum $= 140 + 28 + 7 = 175$.
Therefore,$\cos \left(\frac{\pi}{175} \times 175\right) = \cos(\pi) = -1$.

(A) Given that $1+\omega+\omega^2 = 0$,we have $1+\omega^2 = -\omega$ and $1+\omega = -\omega^2$.
Substituting these into the expression:
$(1-\omega+\omega^2)^{3k} + (1-\omega^2+\omega)^{3k} = (1-\omega+\omega^2)^{3k+1} + (1+\omega-\omega^2)^{3k+1}$
$(-2\omega)^{3k} + (-2\omega^2)^{3k} = (-2\omega)^{3k+1} + (-2\omega^2)^{3k+1}$
$(-2)^{3k} \omega^{3k} + (-2)^{3k} \omega^{6k} = (-2)^{3k+1} \omega^{3k+1} + (-2)^{3k+1} \omega^{2(3k+1)}$
Since $\omega^3 = 1$,$\omega^{3k} = 1$ and $\omega^{6k} = 1$:
$(-2)^{3k}(1+1) = (-2)^{3k} \cdot (-2) \cdot (\omega + \omega^2)$
$2 = -2(\omega + \omega^2)$
Since $\omega + \omega^2 = -1$,we get $2 = -2(-1) = 2$.
This identity holds for all $k \in N$. Thus,$k = r$ where $r \in N$.

(A) Given that $\omega$ is a complex cube root of unity,we have $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
First,simplify the powers of $\omega$:
$\omega^{10} = (\omega^3)^3 \cdot \omega = 1^3 \cdot \omega = \omega$
$\omega^{23} = (\omega^3)^7 \cdot \omega^2 = 1^7 \cdot \omega^2 = \omega^2$
Now substitute these into the expression:
$\sin \left[ (\omega + \omega^2) \pi - \frac{\pi}{4} \right]$
Since $1 + \omega + \omega^2 = 0$,it follows that $\omega + \omega^2 = -1$.
Substituting this value:
$\sin \left[ (-1) \pi - \frac{\pi}{4} \right] = \sin \left( -\pi - \frac{\pi}{4} \right)$
$= \sin \left( -\frac{5\pi}{4} \right) = -\sin \left( \frac{5\pi}{4} \right)$
$= -\sin \left( \pi + \frac{\pi}{4} \right) = -(-\sin \frac{\pi}{4}) = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$

(B) Given equation: $8z^3 - 12z^2 + 6z - 28 = 0$ ...$(i)$
By inspection,$z = 2$ is a root because $8(8) - 12(4) + 6(2) - 28 = 64 - 48 + 12 - 28 = 0$.
Dividing the equation by $(z - 2)$,we get:
$8z^3 - 16z^2 + 4z^2 - 8z + 14z - 28 = 0$
$8z^2(z - 2) + 4z(z - 2) + 14(z - 2) = 0$
$(z - 2)(8z^2 + 4z + 14) = 0$
$(z - 2)(4z^2 + 2z + 7) = 0$
For $4z^2 + 2z + 7 = 0$,the roots are $z = \frac{-2 \pm \sqrt{4 - 112}}{8} = \frac{-2 \pm \sqrt{-108}}{8} = \frac{-2 \pm 6\sqrt{3}i}{8} = \frac{-1 \pm 3\sqrt{3}i}{4}$.
We know $\omega = \frac{-1 + i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$.
Consider $\frac{3\omega + 1}{2} = \frac{3(\frac{-1 + i\sqrt{3}}{2}) + 1}{2} = \frac{-3 + 3i\sqrt{3} + 2}{4} = \frac{-1 + 3i\sqrt{3}}{4}$.
Similarly,$\frac{3\omega^2 + 1}{2} = \frac{3(\frac{-1 - i\sqrt{3}}{2}) + 1}{2} = \frac{-3 - 3i\sqrt{3} + 2}{4} = \frac{-1 - 3i\sqrt{3}}{4}$.
Thus,the roots are $2, \frac{3\omega + 1}{2}, \frac{3\omega^2 + 1}{2}$.

(B) Given $f(x)=a_0+a_1 x+a_2 x^2+\ldots+a_n x^n$ with $a_i \in \mathbb{Z}$.
Since the coefficients are integers,complex roots must occur in conjugate pairs and irrational roots of the form $a+\sqrt{b}$ must occur with their conjugates $a-\sqrt{b}$.
Given roots are $x_1 = 3+i$ and $x_2 = 2-\sqrt{3}$.
Therefore,their conjugates $x_3 = 3-i$ and $x_4 = 2+\sqrt{3}$ must also be roots.
Since there are at least $4$ roots,the minimum degree $n$ is $4$.
For a polynomial of degree $4$,the constant term $a_0$ is given by the product of the roots multiplied by $(-1)^n$ (if the leading coefficient is $1$): $a_0 = x_1 x_2 x_3 x_4$.
$a_0 = (3+i)(3-i) \times (2-\sqrt{3})(2+\sqrt{3})$
$a_0 = (3^2 - i^2) \times (2^2 - (\sqrt{3})^2)$
$a_0 = (9+1) \times (4-3)$
$a_0 = 10 \times 1 = 10$.
Thus,$n=4$ and $a_0=10$.

(A) Given $z_1 = 2 - i$ and $z_2 = 6 + 3i$.
We need to find $\operatorname{amp}\left(\frac{z_1-z_2}{z_1+z_2}\right)$.
First,calculate $z_1 - z_2 = (2 - 6) + (-1 - 3)i = -4 - 4i$.
Next,calculate $z_1 + z_2 = (2 + 6) + (-1 + 3)i = 8 + 2i$.
Now,$\frac{z_1-z_2}{z_1+z_2} = \frac{-4-4i}{8+2i} = \frac{-2-2i}{4+i}$.
Multiply numerator and denominator by the conjugate $(4-i)$:
$\frac{(-2-2i)(4-i)}{(4+i)(4-i)} = \frac{-8 + 2i - 8i + 2i^2}{16 - i^2} = \frac{-8 - 6i - 2}{16 + 1} = \frac{-10 - 6i}{17} = -\frac{10}{17} - \frac{6}{17}i$.
Alternatively,using the property $\operatorname{amp}\left(\frac{z_1-z_2}{z_1+z_2}\right) = \operatorname{amp}(z_1-z_2) - \operatorname{amp}(z_1+z_2)$:
$\operatorname{amp}(-4-4i) = -\pi + \tan^{-1}\left(\frac{-4}{-4}\right) = -\pi + \frac{\pi}{4} = -\frac{3\pi}{4}$.
$\operatorname{amp}(8+2i) = \tan^{-1}\left(\frac{2}{8}\right) = \tan^{-1}\left(\frac{1}{4}\right)$.
Thus,the result is $-\frac{3\pi}{4} - \tan^{-1}\left(\frac{1}{4}\right)$.

(A) The $4^{th}$ roots of unity are $1, -1, i, -i$. The real roots are $\alpha=1, \beta=-1$ and the other roots are $\gamma=i, \delta=-i$.
For the first conic $|z-1|+|z+1|=4$,this represents an ellipse with foci at $(\pm 1, 0)$ and major axis $2a=4$,so $a=2$. The distance between foci is $2ae_1=2$,so $e_1=\frac{1}{2}$.
For the second conic $|z-i|+|z+i|=6$,this represents an ellipse with foci at $(0, \pm 1)$ and major axis $2a=6$,so $a=3$. The distance between foci is $2ae_2=2$,so $e_2=\frac{1}{3}$.
The sum of the eccentricities is $e_1+e_2 = \frac{1}{2}+\frac{1}{3} = \frac{5}{6}$.

(D) Given $x^3-a^3=0 \Rightarrow (x-a)(x^2+ax+a^2)=0$.
Since $a>0$,the real root is $\alpha = a$.
The other roots are $\beta = a \omega = a(-\frac{1}{2} + i\frac{\sqrt{3}}{2})$ and $\gamma = a \omega^2 = a(-\frac{1}{2} - i\frac{\sqrt{3}}{2})$.
The first curve is $|z - \beta| = \frac{\sqrt{3}a}{2}$,which represents a circle with center $\beta = (-\frac{a}{2}, \frac{a\sqrt{3}}{2})$ and radius $R = \frac{\sqrt{3}a}{2}$.
The second curve is $|z - \gamma| = \frac{\sqrt{3}a}{2}$,which represents a circle with center $\gamma = (-\frac{a}{2}, -\frac{a\sqrt{3}}{2})$ and radius $R = \frac{\sqrt{3}a}{2}$.
The distance between the centers is $d = \sqrt{(-\frac{a}{2} - (-\frac{a}{2}))^2 + (\frac{a\sqrt{3}}{2} - (-\frac{a\sqrt{3}}{2}))^2} = \sqrt{0 + (a\sqrt{3})^2} = a\sqrt{3}$.
The sum of the radii is $R_1 + R_2 = \frac{\sqrt{3}a}{2} + \frac{\sqrt{3}a}{2} = a\sqrt{3}$.
Since the distance between the centers $d$ is equal to the sum of the radii $R_1 + R_2$,the two circles touch each other externally at exactly one point.
Thus,the number of common points is $1$.

(A) $z_3 = \frac{z_1+z_2}{2} = \frac{(2+3i)+(4-5i)}{2} = \frac{6-2i}{2} = 3-i$.
Given $5z_1+xz_2+yz_3=0$.
Substituting the values: $5(2+3i) + x(4-5i) + y(3-i) = 0$.
$(10+15i) + (4x-5xi) + (3y-yi) = 0$.
Grouping real and imaginary parts: $(10+4x+3y) + i(15-5x-y) = 0$.
Equating real and imaginary parts to zero:
$4x+3y = -10$ $(i)$
$5x+y = 15$ (ii)
From (ii),$y = 15-5x$.
Substituting into $(i)$: $4x + 3(15-5x) = -10$.
$4x + 45 - 15x = -10$.
$-11x = -55 \Rightarrow x = 5$.
$y = 15 - 5(5) = 15-25 = -10$.
Therefore,$x+y = 5-10 = -5$.

(B) number is divisible by $3$ if and only if the sum of its digits is divisible by $3$.
Given digits are $4, 6, 9, 5, 3, x, y$.
Sum of digits $= 4 + 6 + 9 + 5 + 3 + x + y = 27 + x + y$.
For the number to be divisible by $3$,$(27 + x + y)$ must be a multiple of $3$.
Since $27$ is a multiple of $3$,$(x + y)$ must be a multiple of $3$.
The digits must be distinct,so $x, y \in \{0, 1, 2, 7, 8\}$.
Possible pairs $(x, y)$ such that $x + y$ is a multiple of $3$ (where $x \neq y$ and $x, y \notin \{4, 6, 9, 5, 3\}$):
If $x = 0$,$y$ can be $3$ (not allowed) or $6$ (not allowed) or $9$ (not allowed). No value.
If $x = 1$,$y$ can be $2, 5$ (not allowed),$8$. Pairs: $(1, 2), (1, 8)$.
If $x = 2$,$y$ can be $1, 4$ (not allowed),$7$. Pairs: $(2, 1), (2, 7)$.
If $x = 7$,$y$ can be $2, 5$ (not allowed),$8$. Pairs: $(7, 2), (7, 8)$.
If $x = 8$,$y$ can be $1, 4$ (not allowed),$7$. Pairs: $(8, 1), (8, 7)$.
Total pairs are $(1, 2), (1, 8), (2, 1), (2, 7), (7, 2), (7, 8), (8, 1), (8, 7)$.
There are $8$ such ordered pairs.

(D) number is divisible by $5$ if its last digit is $0$ or $5$. We consider numbers with $1, 2, 3,$ or $4$ digits without repetition.
Case $1$: Numbers ending in $0$.
- $1$-digit: ${0}$ is not a natural number,so $0$.
- $2$-digits: $9$ choices for the first digit $(1-9)$,$1$ choice for the last $(0)$. Total $= 9 \times 1 = 9$.
- $3$-digits: $9$ choices for the first,$8$ for the second,$1$ for the last. Total $= 9 \times 8 \times 1 = 72$.
- $4$-digits: $9$ choices for the first,$8$ for the second,$7$ for the third,$1$ for the last. Total $= 9 \times 8 \times 7 \times 1 = 504$.
Sum for Case $1 = 9 + 72 + 504 = 585$.
Case $2$: Numbers ending in $5$.
- $1$-digit: ${5}$. Total $= 1$.
- $2$-digits: $8$ choices for the first (cannot be $0$ or $5$),$1$ choice for the last. Total $= 8 \times 1 = 8$.
- $3$-digits: $8$ choices for the first (cannot be $0$ or $5$),$7$ for the second (cannot be $5$ or the first digit),$1$ for the last. Total $= 8 \times 7 \times 1 = 56$.
- $4$-digits: $7$ choices for the first (cannot be $0$ or $5$),$7$ for the second,$6$ for the third,$1$ for the last. Total $= 7 \times 7 \times 6 \times 1 = 294$.
Sum for Case $2 = 1 + 8 + 56 + 294 = 359$.
Total $= 585 + 359 = 944$.

(D) We need to form a six-digit number using the digits $\{0, 2, 3, 4, 5, 6, 7, 8\}$.
There are $8$ available digits in total.
For a six-digit number,the first digit (at the hundred-thousands place) cannot be $0$. Thus,there are $7$ choices for the first digit $(\{2, 3, 4, 5, 6, 7, 8\})$.
Since repetition of digits is allowed,each of the remaining $5$ positions can be filled by any of the $8$ available digits.
Therefore,the total number of six-digit numbers is:
$7 \times 8 \times 8 \times 8 \times 8 \times 8 = 7 \times 8^5$
Since $8 = 2^3$,we have $8^5 = (2^3)^5 = 2^{15}$.
Thus,the total number of such six-digit numbers is $7 \times 2^{15}$.

(D) Using the identity ${ }^n C_r + { }^n C_{r-1} = { }^{n+1} C_r$,we have ${ }^{(n-1)} C_2 + { }^{(n-1)} C_3 = { }^n C_3$.
Given the inequality ${ }^n C_3 > { }^n C_2$.
Expanding the combinations: $\frac{n!}{3!(n-3)!} > \frac{n!}{2!(n-2)!}$.
Simplifying the expression: $\frac{1}{3(n-3)!} > \frac{1}{2(n-2)(n-3)!}$.
$\frac{1}{3} > \frac{1}{2(n-2)}$.
$2(n-2) > 3$.
$2n - 4 > 3$.
$2n > 7$.
$n > 3.5$.
Since $n$ must be an integer and $n \geq 3$ for the combinations to be defined,the smallest integer $n$ satisfying the inequality is $n = 4$. However,checking the original expression for $n=4$: ${ }^3 C_2 + { }^3 C_3 = 3 + 1 = 4$ and ${ }^4 C_2 = 6$. $4 > 6$ is false.
Checking $n=5$: ${ }^4 C_2 + { }^4 C_3 = 6 + 4 = 10$ and ${ }^5 C_2 = 10$. $10 > 10$ is false.
Checking $n=6$: ${ }^5 C_2 + { }^5 C_3 = 10 + 10 = 20$ and ${ }^6 C_2 = 15$. $20 > 15$ is true.
Thus,the least value is $n = 6$.

(B) Given the ratio: $\frac{^{2n}C_4}{^nC_3} = \frac{99}{4}$
Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$:
$\frac{\frac{(2n)!}{4!(2n-4)!}}{\frac{n!}{3!(n-3)!}} = \frac{99}{4}$
$\frac{(2n)(2n-1)(2n-2)(2n-3)}{4 \times 3 \times 2 \times 1} \times \frac{3 \times 2 \times 1}{n(n-1)(n-2)} = \frac{99}{4}$
$\frac{(2n)(2n-1) \times 2(n-1)(2n-3)}{4 \times n(n-1)(n-2)} = \frac{99}{4}$
$\frac{2(2n-1)(2n-3)}{n-2} = 99$
$4(4n^2 - 6n - 2n + 3) = 99(n-2)$
$16n^2 - 32n + 12 = 99n - 198$
$16n^2 - 131n + 210 = 0$
$(n-6)(16n-35) = 0$
Since $n$ must be an integer,$n = 6$.

(C) The word $ARRANGEMENT$ has $11$ letters: $A(2), R(2), N(2), E(2), G(1), M(1), T(1)$.
Total arrangements $= \frac{11!}{2! \cdot 2! \cdot 2! \cdot 2!} = \frac{11!}{16}$.
To find the arrangements where two $E$s do not occur together,we subtract the arrangements where they do occur together from the total.
Treating the two $E$s as a single unit $(EE)$,we have $10$ items to arrange: $A(2), R(2), N(2), G(1), M(1), T(1), (EE)(1)$.
Number of arrangements with $E$s together $= \frac{10!}{2! \cdot 2! \cdot 2!} = \frac{10!}{8}$.
Number of arrangements with $E$s not together $= \frac{11!}{16} - \frac{10!}{8} = \frac{11 \cdot 10!}{16} - \frac{2 \cdot 10!}{16} = \frac{9 \cdot 10!}{16} = \frac{9}{16}(10!)$.

(B) The word $VOWEL$ consists of $5$ distinct letters: $V, O, W, E, L$.
There are $2$ vowels: $O$ and $E$.
Since the vowels must always remain together,we treat $(OE)$ as a single unit.
Now,we have $4$ units: $V, W, L, (OE)$.
These $4$ units can be arranged in $4! = 24$ ways.
The $2$ vowels within the unit $(OE)$ can be arranged in $2! = 2$ ways.
Therefore,the total number of words is $24 \times 2 = 48$.

(A) An $8$-digit number is formed by $8$ positions.
For the number to be odd,the last digit (units place) must be one of $\{1, 3, 5, 7, 9\}$,which gives $5$ choices.
The first digit cannot be $0$,so it has $9$ choices $(\{1, 2, 3, 4, 5, 6, 7, 8, 9\})$.
The remaining $6$ positions (from the $2^{nd}$ to the $7^{th}$ digit) can each be filled by any of the $10$ digits $(\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\})$,giving $10^6$ ways.
Therefore,the total number of $8$-digit odd numbers is $9 \times 10^6 \times 5 = 45 \times 10^6$.

(A) four-digit number has four places: $ABCD$.
Given that the first digit $A$ is fixed as $4$,there is only $1$ way to fill this place.
The last digit $D$ can be either $0$ or $5$,so there are $2$ ways to fill this place.
The second digit $B$ can be any digit from $0$ to $9$,so there are $10$ ways.
The third digit $C$ can be any digit from $0$ to $9$,so there are $10$ ways.
Therefore,the total number of such four-digit numbers is $1 \times 10 \times 10 \times 2 = 200$.

(D) The word $KANGAROO$ contains $8$ letters: $K, A, N, G, A, R, O, O$. The letter $A$ appears $2$ times and $O$ appears $2$ times.
Total number of arrangements $= \frac{8!}{2!2!} = \frac{40320}{4} = 10080$.
To find the number of arrangements where $A$'s do not appear together,we subtract the arrangements where $A$'s are together from the total.
Treating the two $A$'s as a single unit $(AA)$,we have $7$ units: $(AA), K, N, G, R, O, O$.
The number of arrangements where $A$'s are together $= \frac{7!}{2!} = \frac{5040}{2} = 2520$.
Therefore,the number of required arrangements $= 10080 - 2520 = 7560$.

(C) To find the number of natural numbers less than $500$ with no repeated digits,we consider $1, 2,$ and $3$-digit numbers separately.
$1$. $1$-digit numbers: The possible numbers are $1, 2, 3, 4, 5, 6, 7, 8, 9$. Total = $9$.
$2$. $2$-digit numbers: The first digit can be any of $9$ digits $(1-9)$ and the second digit can be any of the remaining $9$ digits (including $0$). Total = $9 \times 9 = 81$.
$3$. $3$-digit numbers less than $500$: The hundreds place can be filled by $1, 2, 3,$ or $4$ ($4$ ways). The tens place can be filled by any of the remaining $9$ digits,and the units place by any of the remaining $8$ digits. Total = $4 \times 9 \times 8 = 288$.
Total count = $9 + 81 + 288 = 378$.

(D) The student needs to select $10$ questions in total,with at least $4$ from section-$A$ ($8$ available) and at least $3$ from section-$B$ ($6$ available).
Possible combinations $(A, B)$ are:
$(i)$ $4$ from $A$ and $6$ from $B$: $\binom{8}{4} \times \binom{6}{6} = 70 \times 1 = 70$
(ii) $5$ from $A$ and $5$ from $B$: $\binom{8}{5} \times \binom{6}{5} = 56 \times 6 = 336$
(iii) $6$ from $A$ and $4$ from $B$: $\binom{8}{6} \times \binom{6}{4} = 28 \times 15 = 420$
(iv) $7$ from $A$ and $3$ from $B$: $\binom{8}{7} \times \binom{6}{3} = 8 \times 20 = 160$
Total ways $= 70 + 336 + 420 + 160 = 986$.

(B) To get a sum of $7$ with $5$ dice,each die must show at least $1$. Let the outcomes be $x_1, x_2, x_3, x_4, x_5$ where $x_i \ge 1$. The sum is $x_1 + x_2 + x_3 + x_4 + x_5 = 7$.
Since each $x_i \ge 1$,we can write $x_i = 1 + y_i$ where $y_i \ge 0$.
Substituting this,we get $(1+y_1) + (1+y_2) + (1+y_3) + (1+y_4) + (1+y_5) = 7$,which simplifies to $y_1 + y_2 + y_3 + y_4 + y_5 = 2$.
The number of non-negative integer solutions is given by the formula $\binom{n+r-1}{r-1}$ where $n=2$ and $r=5$.
This is $\binom{2+5-1}{5-1} = \binom{6}{4} = \binom{6}{2} = \frac{6 \times 5}{2} = 15$.
Alternatively,the cases are:
$(i)$ Four $1$s and one $3$: $\frac{5!}{4!1!} = 5$ ways.
(ii) Three $1$s and two $2$s: $\frac{5!}{3!2!} = 10$ ways.
Total ways = $5 + 10 = 15$.

(C) First,find the prime factorization of $67500$:
$67500 = 675 \times 100 = (25 \times 27) \times (4 \times 25) = 5^2 \times 3^3 \times 2^2 \times 5^2 = 2^2 \times 3^3 \times 5^4$.
To find the number of odd positive divisors,we consider only the odd prime factors,which are $3$ and $5$.
The odd divisors are of the form $3^a \times 5^b$,where $0 \le a \le 3$ and $0 \le b \le 4$.
The number of choices for $a$ is $(3+1) = 4$.
The number of choices for $b$ is $(4+1) = 5$.
Therefore,the total number of odd positive divisors is $4 \times 5 = 20$.

(D) First,find the prime factorization of $6300$:
$6300 = 63 \times 100 = (9 \times 7) \times (10^2) = (3^2 \times 7^1) \times (2^2 \times 5^2) = 2^2 \times 3^2 \times 5^2 \times 7^1$.
Total number of divisors is given by the product of (exponent $+ 1$) for each prime factor:
Total divisors $= (2+1)(2+1)(2+1)(1+1) = 3 \times 3 \times 3 \times 2 = 54$.
Odd divisors are formed by the prime factors excluding $2$:
Odd divisors $= (2+1)(2+1)(1+1) = 3 \times 3 \times 2 = 18$.
Therefore,the number of even divisors is the total number of divisors minus the number of odd divisors:
Even divisors $= 54 - 18 = 36$.

(B) Given that $a, b, c$ are in arithmetic progression,so $2b = a + c$.
Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$.
Thus,$a = k \sin A$,$b = k \sin B$,and $c = k \sin C$.
Since $C = 2A$,we have $\sin C = \sin 2A = 2 \sin A \cos A$.
From $2b = a + c$,we get $2 \sin B = \sin A + \sin C$.
Using $B = 180^{\circ} - (A + C) = 180^{\circ} - 3A$,we have $\sin B = \sin 3A$.
So,$2 \sin 3A = \sin A + \sin 2A$.
$2(3 \sin A - 4 \sin^3 A) = \sin A + 2 \sin A \cos A$.
Dividing by $\sin A$ (since $\sin A \neq 0$): $6 - 8 \sin^2 A = 1 + 2 \cos A$.
$6 - 8(1 - \cos^2 A) = 1 + 2 \cos A \Rightarrow 8 \cos^2 A - 2 \cos A - 3 = 0$.
$(4 \cos A - 3)(2 \cos A + 1) = 0$.
Since $A$ is an angle of a triangle,$\cos A = \frac{3}{4}$.
Then $\frac{a}{c} = \frac{\sin A}{\sin C} = \frac{\sin A}{2 \sin A \cos A} = \frac{1}{2 \cos A} = \frac{1}{2(3/4)} = \frac{1}{3/2} = \frac{2}{3}$.

(B) The given equation is $16x^3 - 44x^2 + 36x - 9 = 0$.
Let the roots be $\alpha, \beta, \gamma$ in harmonic progression ($H$.$P$.).
Then $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$ are in arithmetic progression ($A$.$P$.).
This implies $\frac{2}{\beta} = \frac{1}{\alpha} + \frac{1}{\gamma}$.
From the properties of roots,$\sum \alpha\beta = \frac{36}{16} = \frac{9}{4}$ and $\alpha\beta\gamma = \frac{9}{16}$.
Also,$\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma} = \frac{9/4}{9/16} = 4$.
Substituting $\frac{1}{\alpha} + \frac{1}{\gamma} = \frac{2}{\beta}$,we get $\frac{2}{\beta} + \frac{1}{\beta} = 4$ $\Rightarrow \frac{3}{\beta} = 4$ $\Rightarrow \beta = \frac{3}{4}$.
Now,$\alpha + \gamma = \frac{44}{16} - \frac{3}{4} = \frac{11}{4} - \frac{3}{4} = 2$ and $\alpha\gamma = \frac{9/16}{3/4} = \frac{3}{4}$.
Solving $t^2 - 2t + \frac{3}{4} = 0$,we get $4t^2 - 8t + 3 = 0 \Rightarrow (2t - 1)(2t - 3) = 0$.
Thus,the roots are $\frac{1}{2}, \frac{3}{4}, \frac{3}{2}$.
The greatest root is $\frac{3}{2}$.

(B) The exponent of a prime $p$ in the prime factorization of $m!$ is given by Legendre's Formula: $E_p(m!) = \sum_{k=1}^{\infty} \left[ \frac{m}{p^k} \right]$.
Here,$m = 16$ and $p = 2$.
$n = \left[ \frac{16}{2} \right] + \left[ \frac{16}{4} \right] + \left[ \frac{16}{8} \right] + \left[ \frac{16}{16} \right]$.
$n = 8 + 4 + 2 + 1 = 15$.
Since $2^{15}$ divides $16!$ and $2^{16}$ does not divide $16!$,the value of $n$ is $15$.

(D) We know that $\cot 18^{\circ} \cot 36^{\circ}+1 = \frac{\cos 18^{\circ}}{\sin 18^{\circ}} \cdot \frac{\cos 36^{\circ}}{\sin 36^{\circ}}+1$.
Using $\cos 36^{\circ} = 1-2\sin^2 18^{\circ}$,we get:
$\frac{\cos 18^{\circ}(1-2\sin^2 18^{\circ})}{\sin 18^{\circ} \cdot 2 \sin 18^{\circ} \cos 18^{\circ}}+1 = \frac{1-2\sin^2 18^{\circ}}{2\sin^2 18^{\circ}}+1 = \frac{1}{2\sin^2 18^{\circ}}$.
Since $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$,then $\sin^2 18^{\circ} = \frac{5+1-2\sqrt{5}}{16} = \frac{6-2\sqrt{5}}{16} = \frac{3-\sqrt{5}}{8}$.
Substituting this value:
$\frac{1}{2(\frac{3-\sqrt{5}}{8})} = \frac{4}{3-\sqrt{5}} = \frac{4(3+\sqrt{5})}{9-5} = \frac{4(3+\sqrt{5})}{4} = 3+\sqrt{5}$.

(D) Given: $\sin \theta = \frac{3}{5}$.
Since $\theta$ is not in the first quadrant and $\sin \theta$ is positive,$\theta$ must be in the second quadrant.
Therefore,$\cos \theta = -\sqrt{1 - \sin^2 \theta} = -\sqrt{1 - \frac{9}{25}} = -\frac{4}{5}$.
Now,$\sin 2 \theta = 2 \sin \theta \cos \theta = 2 \left(\frac{3}{5}\right) \left(-\frac{4}{5}\right) = -\frac{24}{25}$.
$\cos 2 \theta = 1 - 2 \sin^2 \theta = 1 - 2 \left(\frac{9}{25}\right) = 1 - \frac{18}{25} = \frac{7}{25}$.
$\tan 2 \theta = \frac{\sin 2 \theta}{\cos 2 \theta} = \frac{-24/25}{7/25} = -\frac{24}{7}$.
Substituting these values into the expression:
$15 \sin 2 \theta - 20 \cos 2 \theta - 7 \tan 2 \theta = 15 \left(-\frac{24}{25}\right) - 20 \left(\frac{7}{25}\right) - 7 \left(-\frac{24}{7}\right)$.
$= -\frac{72}{5} - \frac{28}{5} + 24 = -\frac{100}{5} + 24 = -20 + 24 = 4$.

(D) Given $f(g(x+y)) = f(g(x)) + f(g(y))$ ... $(i)$
Let $h(x) = f(g(x))$. Then the equation becomes $h(x+y) = h(x) + h(y)$,which is Cauchy's functional equation.
The solution to this is $h(x) = cx$ for some constant $c$.
Given $g(1) = 2$ and $f(2) = 1$,we have $h(1) = f(g(1)) = f(2) = 1$.
Since $h(1) = c(1) = 1$,we get $c = 1$.
Thus,$h(x) = f(g(x)) = x$.
Since $f(g(x)) = x$,$f$ and $g$ are inverse functions of each other.
Therefore,$g(f(x)) = x$ for all $x \in R$.
The function $g(f(x)) = x$ is a polynomial function,which is continuous everywhere on the set of real numbers $R$.
Hence,the set of points where $g(f(x))$ is discontinuous is the empty set,denoted by $\phi$.

(B) For the function $f(x) = \sqrt{\log_{10}\left(\frac{3-x}{x}\right)}$ to be defined,the expression inside the square root must be non-negative and the argument of the logarithm must be positive.
$1$. Condition for the logarithm argument: $\frac{3-x}{x} > 0$.
Solving this inequality,we get $x \in (0, 3)$.
$2$. Condition for the square root: $\log_{10}\left(\frac{3-x}{x}\right) \geq 0$.
This implies $\frac{3-x}{x} \geq 10^0$,so $\frac{3-x}{x} \geq 1$.
$\frac{3-x}{x} - 1 \geq 0 \implies \frac{3-x-x}{x} \geq 0 \implies \frac{3-2x}{x} \geq 0$.
Multiplying by $-1$ (and reversing the inequality),we get $\frac{2x-3}{x} \leq 0$.
The critical points are $x = 0$ and $x = \frac{3}{2}$. Testing intervals,we find $x \in (0, \frac{3}{2}]$.
Combining both conditions: $(0, 3) \cap (0, \frac{3}{2}] = (0, \frac{3}{2}]$.
Thus,the domain of $f$ is $\left(0, \frac{3}{2}\right]$.

(C) The function $f: A \rightarrow B$ is defined by $f(x)=x^2-3x+2$.
To be a bijection,the function must be both one-one and onto.
The derivative is $f'(x) = 2x - 3$.
Setting $f'(x) = 0$ gives the vertex at $x = \frac{3}{2}$.
The minimum value of the function is $f\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) + 2 = \frac{9}{4} - \frac{9}{2} + 2 = \frac{9-18+8}{4} = -\frac{1}{4}$.
For the function to be one-one,we must restrict the domain to either $A = \left(-\infty, \frac{3}{2}\right]$ or $A = \left[\frac{3}{2}, \infty\right)$.
For the function to be onto,the codomain $B$ must be equal to the range,which is $\left[-\frac{1}{4}, \infty\right)$.
Comparing this with the given options,option $B$ and $C$ are candidates. However,standard convention for such problems usually selects the interval starting from the vertex. Thus,$A = \left[\frac{3}{2}, \infty\right)$ and $B = \left[-\frac{1}{4}, \infty\right)$ is a valid bijection.

(C) The total number of functions from a set $A$ with $n$ elements to itself is $n^n$.
$A$ function is one-one if every element in the domain has a unique image in the codomain. For a set with $n$ elements,the number of one-one functions is given by the number of permutations of $n$ elements,which is $n!$.
Therefore,the number of functions that are not one-one is the total number of functions minus the number of one-one functions.
Number of functions that are not one-one $= n^n - n!$.

(A) Given that $g(x)$ is the inverse of the function $f(x)$,we have $f(g(x)) = x$.
By differentiating both sides with respect to $x$ using the chain rule,we get:
$f^{\prime}(g(x)) \cdot g^{\prime}(x) = 1$.
This implies $g^{\prime}(x) = \frac{1}{f^{\prime}(g(x))}$.
We are given $f^{\prime}(x) = \frac{1}{h(x)}$,which means $f^{\prime}(g(x)) = \frac{1}{h(g(x))}$.
Substituting this into the expression for $g^{\prime}(x)$:
$g^{\prime}(x) = \frac{1}{1 / h(g(x))} = h(g(x))$.
Thus,the correct option is $A$.

(C) Given the functional equation $f(x+y)=f(x)+f(y)$ for all $x, y \in R$,this is Cauchy's functional equation,which implies $f(x)=ax$ for some constant $a \in R$.
Given $f(1)=7$,we substitute $x=1$ into $f(x)=ax$ to get $7=a(1)$,so $a=7$.
Thus,the function is $f(x)=7x$.
We need to calculate the sum $\sum_{r=1}^{n} f(r) = \sum_{r=1}^{n} 7r$.
Taking the constant $7$ out,we get $7 \sum_{r=1}^{n} r$.
Using the formula for the sum of the first $n$ natural numbers,$\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$.
Therefore,$\sum_{r=1}^{n} f(r) = 7 \times \frac{n(n+1)}{2} = \frac{7n(n+1)}{2}$.

(D) Given: $f(x) = x + \frac{1}{x}$
Squaring both sides,we get:
$(f(x))^2 = (x + \frac{1}{x})^2$
$(f(x))^2 = x^2 + \frac{1}{x^2} + 2 \cdot x \cdot \frac{1}{x}$
$(f(x))^2 = x^2 + \frac{1}{x^2} + 2$
Now,consider $f(x^2) = x^2 + \frac{1}{x^2}$ and $f(1) = 1 + \frac{1}{1} = 2$.
Substituting these into the expression:
$(f(x))^2 = f(x^2) + f(1)$

(A) Given functions are $f(x) = |x|$ and $g(x) = |x| + a$ with $a > 0$.
For $0 \leq x \leq b$,the region is defined by $g(x) \leq y \leq f(x)$,which implies $|x| + a \leq y \leq |x|$.
However,since $a > 0$,$|x| + a$ is always greater than $|x|$. Thus,the condition $g(x) \leq y \leq f(x)$ is impossible for $a > 0$ as it implies $|x| + a \leq y \leq |x|$,which is a contradiction.
Re-evaluating the problem statement based on the provided image,the region is bounded by $x=0$,$x=b$,$y=|x|$,and $y=|x|+a$.
For $x \geq 0$,$f(x) = x$ and $g(x) = x + a$.
The region is bounded by $x=0$,$x=b$,$y=x$,and $y=x+a$.
These lines are $x=0$,$x=b$,$y-x=0$,and $y-x=a$.
The lines $y-x=0$ and $y-x=a$ are parallel,and the lines $x=0$ and $x=b$ are parallel.
Therefore,the region is a parallelogram.

(D) Given $g(x)=1, \forall x \in R$ and the signum function $f(x)= \begin{cases}-1, & x < 0 \\ 0, & x=0 \\ 1, & x>0\end{cases}$.
For $x < 0$,$f(x)=-1$,so $g(x)=1 = 2+(-1) = 2+f(x)$.
For $x=0$,$f(x)=0$,so $g(x)=1 = 1+0 = 1+f(x)$.
For $x>0$,$f(x)=1$,so $g(x)=1 = 0+1 = 0+f(x) = f(x)$.
Thus,$g(x)= \begin{cases}f(x)+2, & x < 0 \\ 1+f(x), & x=0 \\ f(x), & x>0\end{cases}$.

(A) For $f(x)$ to be continuous at $x=1$,the left-hand limit,right-hand limit,and the value of the function at $x=1$ must be equal.
$\lim_{x \to 1^-} f(x) = f(1) = 1 + 6(1) - 3(1)^2 = 1 + 6 - 3 = 4$.
$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x + \log_2(b^2+7)) = 1 + \log_2(b^2+7)$.
Equating the limits: $1 + \log_2(b^2+7) = 4$.
$\log_2(b^2+7) = 3$.
$b^2+7 = 2^3 = 8$.
$b^2 = 8 - 7 = 1$.
$b = \pm 1$.

(A) Since $f(x)$ is continuous on $R$,the left-hand limit at $x=0$ must equal $f(0)$.
$f(0) = \lim_{x \rightarrow 0^-} f(x) = \lim_{h \rightarrow 0} f(-h)$
$f(0) = \lim_{h \rightarrow 0} \frac{\sin(-h) - \sin(-\frac{h}{2})}{-h}$
$f(0) = \lim_{h \rightarrow 0} \frac{-\sin h + \sin(\frac{h}{2})}{-h}$
$f(0) = \lim_{h \rightarrow 0} \left[ \frac{\sin h}{h} - \frac{\sin(\frac{h}{2})}{h} \right]$
$f(0) = \lim_{h \rightarrow 0} \left[ \frac{\sin h}{h} - \frac{1}{2} \cdot \frac{\sin(\frac{h}{2})}{\frac{h}{2}} \right]$
Using the standard limit $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$,we get:
$f(0) = 1 - \frac{1}{2}(1) = \frac{1}{2}$.

(C) The function $f(x) = \sqrt{\frac{3x^2-5x-2}{2x^2-7x+5}}$ is defined where the expression inside the square root is non-negative,i.e.,$\frac{3x^2-5x-2}{2x^2-7x+5} \ge 0$.
Discontinuities occur where the denominator is zero or where the function is undefined.
First,factor the denominator: $2x^2-7x+5 = 2x^2-2x-5x+5 = 2x(x-1)-5(x-1) = (2x-5)(x-1)$.
The denominator is zero at $x = 1$ and $x = 5/2$.
At these points,the function is undefined,leading to points of discontinuity.
Thus,the function has discontinuous points at $x = 1$ and $x = 5/2$.

(C) Given $f(x)$ is continuous on $R$. For $x=2$,$\lim _{x \rightarrow 2^{+}} f(x) = f(2)$.
$\lim _{x \rightarrow 2^{+}} \frac{x-[x]}{x-2}$. Since $x \rightarrow 2^{+}$,$[x]=2$,so $\lim _{x \rightarrow 2^{+}} \frac{x-2}{x-2} = 1$. Thus,$b=1$.
For $x=2$,$\lim _{x \rightarrow 2^{-}} f(x) = f(2)$.
$\lim _{x \rightarrow 2^{-}} \frac{|(x-2)(x+1)|}{a(2+x-x^2)} = \lim _{x \rightarrow 2^{-}} \frac{|(x-2)(x+1)|}{-a(x-2)(x+1)}$.
Since $x \rightarrow 2^{-}$,$x-2 < 0$,so $|x-2| = -(x-2)$.
$\lim _{x \rightarrow 2^{-}} \frac{-(x-2)(x+1)}{-a(x-2)(x+1)} = \frac{1}{a} = 1 \Rightarrow a=1$.
Now,we evaluate $\lim _{x \rightarrow 0} \frac{\sin ^2 ax+x \tan bx}{x^2}$ with $a=1, b=1$.
$\lim _{x \rightarrow 0} \frac{\sin ^2 x+x \tan x}{x^2} = \lim _{x \rightarrow 0} \left( \frac{\sin ^2 x}{x^2} + \frac{\tan x}{x} \right) = 1^2 + 1 = 2$.

(C) Given the function $f(x) = \frac{5+[x]}{\sqrt{11+[x]-6 \sqrt{2+[x]}}}$.
For the function to be discontinuous,the denominator must be zero or the expression inside the square root must be negative.
Let $n = [x]$. The denominator is zero when $11+n - 6\sqrt{2+n} = 0$.
Let $u = \sqrt{2+n}$,then $n = u^2 - 2$.
Substituting this into the equation: $11 + (u^2 - 2) - 6u = 0$.
$u^2 - 6u + 9 = 0$.
$(u-3)^2 = 0$,which gives $u = 3$.
Since $u = \sqrt{2+n} = 3$,we have $2+n = 9$,so $n = 7$.
Thus,$[x] = 7$,which implies $x \in [7, 8)$.
Also,for the square root to be defined,$11+[x]-6\sqrt{2+[x]} \ge 0$. Since $([x]-7)^2 \ge 0$,the expression is always non-negative for $n \ge -2$. The discontinuity occurs exactly when the denominator is zero,which is at $[x] = 7$.

(D) Given $f(x) = \frac{(3^x - 1)^2}{\sin x \log(1 + x)}$ for $x \neq 0$.
Since the function is continuous at $x = 0$,we have $f(0) = \lim_{x \to 0} f(x)$.
$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{(3^x - 1)^2}{\sin x \log(1 + x)}$.
Divide the numerator and denominator by $x^2$:
$\lim_{x \to 0} \frac{\left(\frac{3^x - 1}{x}\right)^2}{\left(\frac{\sin x}{x}\right) \left(\frac{\log(1 + x)}{x}\right)}$.
Using standard limits $\lim_{x \to 0} \frac{a^x - 1}{x} = \log a$,$\lim_{x \to 0} \frac{\sin x}{x} = 1$,and $\lim_{x \to 0} \frac{\log(1 + x)}{x} = 1$:
$f(0) = \frac{(\log 3)^2}{1 \times 1} = (\log 3)^2$.

(B) Let $g(x) = x^{1/x}$. We analyze the behavior of $g(x)$ for $x \in (0, \infty)$.
Taking the natural logarithm,$\ln(g(x)) = \frac{\ln(x)}{x}$.
Let $h(x) = \frac{\ln(x)}{x}$. Then $h'(x) = \frac{1 - \ln(x)}{x^2}$.
The function $h(x)$ has a maximum at $x = e$,where $h(e) = \frac{1}{e} \approx 0.367$.
Thus,$g(x) = e^{h(x)}$ has a maximum at $x = e$,where $g(e) = e^{1/e} \approx e^{0.367} \approx 1.44$.
As $x \to 0^+$,$g(x) \to 0$,and as $x \to \infty$,$g(x) \to 1$.
The range of $g(x)$ is $(0, e^{1/e}]$.
The greatest integer function $[g(x)]$ is discontinuous whenever $g(x)$ takes an integer value.
Since the range is $(0, 1.44]$,the only integer value $g(x)$ can take is $1$.
$x^{1/x} = 1$ implies $x = 1$.
At $x = 1$,$f(1) = [1^{1/1}] = [1] = 1$.
For $x$ slightly less than $1$,$x^{1/x} < 1$,so $[x^{1/x}] = 0$.
For $x$ slightly greater than $1$,$x^{1/x} > 1$ (but less than $1.44$),so $[x^{1/x}] = 1$.
Since the left-hand limit and right-hand limit are different at $x = 1$,the function is discontinuous at $x = 1$.
Thus,there is only $1$ point of discontinuity.

(C) Given $f(x)$ is continuous at $x=2$,we have $\lim_{x \to 2^-} f(x) = f(2)$.
$\lim_{x \to 2^-} (ae^x + [x-2]) = [e^{-2}] - C$
$ae^2 + [0^-] = 0 - (1 - 2e^2)$
$ae^2 - 1 = -1 + 2e^2 \Rightarrow a = 2$.
Now,check continuity at $x=0$:
$LHL = \lim_{x \to 0^-} [e^x] = [e^0] = [1] = 1$.
$RHL = \lim_{x \to 0^+} (2e^x + [x-2]) = 2(1) + [-2] = 2 - 2 = 0$.
Since $LHL \neq RHL$,$f(x)$ is discontinuous at $x=0$.
Check continuity at $x=1$:
$f(1) = 2e^1 + [1-2] = 2e - 1$.
$\lim_{x \to 1^-} f(x) = 2e - 1$.
$\lim_{x \to 1^+} f(x) = 2e + [1-2] = 2e - 1$.
Since $LHL = RHL = f(1)$,$f(x)$ is continuous at $x=1$.
Thus,$f(x)$ is discontinuous only at $x=0$.

(C) Given $f(x) = \begin{cases} \frac{5 e^{1/x} + 2}{3 - e^{1/x}}, & x \neq 0 \\ 0, & x = 0 \end{cases}$.
Then $x f(x) = \begin{cases} \frac{x(5 e^{1/x} + 2)}{3 - e^{1/x}}, & x \neq 0 \\ 0, & x = 0 \end{cases}$.
Check continuity of $x f(x)$ at $x = 0$:
$\text{L.H.L} = \lim_{h \to 0^+} \frac{(-h)(5 e^{-1/h} + 2)}{3 - e^{-1/h}} = \frac{0(0 + 2)}{3 - 0} = 0$.
$\text{R.H.L} = \lim_{h \to 0^+} \frac{h(5 e^{1/h} + 2)}{3 - e^{1/h}} = \lim_{h \to 0^+} \frac{h e^{1/h}(5 + 2e^{-1/h})}{e^{1/h}(3e^{-1/h} - 1)} = \lim_{h \to 0^+} \frac{h(5 + 0)}{0 - 1} = 0$.
Since $\text{L.H.L} = \text{R.H.L} = f(0) = 0$,$x f(x)$ is continuous at $x = 0$.
Check differentiability of $f(x)$ at $x = 0$:
$\text{L.H.D} = \lim_{h \to 0^+} \frac{f(-h) - f(0)}{-h} = \lim_{h \to 0^+} \frac{\frac{5 e^{-1/h} + 2}{3 - e^{-1/h}} - 0}{-h} = \lim_{h \to 0^+} \frac{2}{3(-h)} = -\infty$.
Since the limit is not finite,$f(x)$ is not differentiable at $x = 0$.
Thus,$x f(x)$ is continuous and $f(x)$ is not differentiable.

(B) Given $\lim_{x \rightarrow \frac{3}{2}} f(x) = 14$. Since $\frac{3}{2} > 1$,we use the definition $f(x) = cx^2 + bx + a$.
$c(\frac{3}{2})^2 + b(\frac{3}{2}) + a = 14 \Rightarrow \frac{9c}{4} + \frac{3b}{2} + a = 14 \Rightarrow 9c + 6b + 4a = 56$ ... $(i)$
Since $f(x)$ is continuous at $x = -1$,$\lim_{x \rightarrow -1^-} f(x) = \lim_{x \rightarrow -1^+} f(x) = f(-1)$.
$a(-1)^2 + b(-1) + c = 2(-1)^2 + 4(-1) + 1 \Rightarrow a - b + c = -1$ ... (ii)
Since $f(x)$ is continuous at $x = 1$,$\lim_{x \rightarrow 1^-} f(x) = \lim_{x \rightarrow 1^+} f(x) = f(1)$.
$2(1)^2 + 4(1) + 1 = c(1)^2 + b(1) + a \Rightarrow a + b + c = 7$ ... (iii)
Adding (ii) and (iii): $2a + 2c = 6 \Rightarrow a + c = 3 \Rightarrow c = 3 - a$.
Subtracting (ii) from (iii): $2b = 8 \Rightarrow b = 4$.
Substitute $b = 4$ and $c = 3 - a$ into $(i)$: $9(3 - a) + 6(4) + 4a = 56 \Rightarrow 27 - 9a + 24 + 4a = 56 \Rightarrow -5a = 5 \Rightarrow a = -1$.
Then $c = 3 - (-1) = 4$.
For $x = -2$,$f(x) = ax^2 + bx + c$.
$\lim_{x \rightarrow -2} f(x) = a(-2)^2 + b(-2) + c = 4a - 2b + c = 4(-1) - 2(4) + 4 = -4 - 8 + 4 = -8$.

(B) The given series is a geometric progression with first term $a = 1$ and common ratio $r = 3x$.
Since $-\frac{1}{3} < x < \frac{1}{3}$,we have $-1 < 3x < 1$,which implies $|r| < 1$.
The sum of an infinite geometric series is given by $S_{\infty} = \frac{a}{1-r}$.
Substituting the values,we get $f(x) = \lim_{n \rightarrow \infty} S_n = \frac{1}{1-3x}$.
The function $f(x) = \frac{1}{1-3x}$ is a rational function which is continuous everywhere except where the denominator is zero.
The denominator $1 - 3x = 0$ when $x = \frac{1}{3}$.
Therefore,$f(x)$ is discontinuous at $x = \frac{1}{3}$.

(C) The function $f(x) = [10^x]$ is discontinuous whenever $10^x$ is an integer.
Given the interval $0 < x < 10$,the range of $10^x$ is $10^0 < 10^x < 10^{10}$,which simplifies to $1 < 10^x < 10^{10}$.
The function $[10^x]$ is discontinuous at all integer values of $10^x$ in the interval $(1, 10^{10})$.
The integers in this interval are ${2, 3, 4, \dots, 10^{10}-1}$.
The number of such integers is $(10^{10}-1) - 2 + 1 = 10^{10}-2$.
Therefore,there are $10^{10}-2$ points of discontinuity in the given interval.

(D) Given that $f^{\prime}(x)$ is a constant for all $x \in R$.
Let $f^{\prime}(x) = a$,where $a$ is a constant.
Integrating both sides with respect to $x$,we get $f(x) = ax + b$,where $b$ is an arbitrary constant.
Given $f(0) = 2$,substituting $x = 0$ in $f(x) = ax + b$,we get $a(0) + b = 2$,which implies $b = 2$.
Given $f^{\prime}(0) = 1$,since $f^{\prime}(x) = a$ for all $x$,we have $a = 1$.
Thus,the function is $f(x) = x + 2$.
Since $f(x) = x + 2$ is a polynomial function,it is continuous for all real numbers $x \in R$.

(C) Given the limit expression: $\lim _{x \rightarrow m} \frac{x f(m)-m f(x)}{x-m}+f^{\prime}(m)=f(m)$.
We can rewrite the numerator by adding and subtracting $m f(m)$:
$\lim _{x \rightarrow m} \frac{x f(m)-m f(m)+m f(m)-m f(x)}{x-m}+f^{\prime}(m)=f(m)$
$\lim _{x \rightarrow m} \left[ f(m) \frac{x-m}{x-m} - m \frac{f(x)-f(m)}{x-m} \right] + f^{\prime}(m) = f(m)$
$f(m) - m f^{\prime}(m) + f^{\prime}(m) = f(m)$
$-m f^{\prime}(m) + f^{\prime}(m) = 0$
$f^{\prime}(m)(1-m) = 0$
Since it is given that $f^{\prime}(m) \neq 0$,we must have $1-m = 0$,which implies $m = 1$.

(D) Given $f(1)=2$,$f(2)=6$ and $f(x+y)=f(x)+kxy+\frac{4}{3}y^2$.
Substitute $x=1$ and $y=1$ into the functional equation:
$f(1+1) = f(1) + k(1)(1) + \frac{4}{3}(1)^2$
$f(2) = f(1) + k + \frac{4}{3}$
$6 = 2 + k + \frac{4}{3}$
$4 = k + \frac{4}{3}$
$k = 4 - \frac{4}{3} = \frac{8}{3}$.
Now,substitute $k = \frac{8}{3}$ into the original equation:
$f(x+y) = f(x) + \frac{8}{3}xy + \frac{4}{3}y^2$.
To find $f(x)$,we use the definition of the derivative:
$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \frac{\frac{8}{3}xh + \frac{4}{3}h^2}{h} = \lim_{h \to 0} (\frac{8}{3}x + \frac{4}{3}h) = \frac{8}{3}x$.
Integrating $f'(x) = \frac{8}{3}x$ with respect to $x$:
$f(x) = \int \frac{8}{3}x \, dx = \frac{8}{3} \cdot \frac{x^2}{2} + C = \frac{4}{3}x^2 + C$.
Using $f(1)=2$:
$2 = \frac{4}{3}(1)^2 + C \Rightarrow C = 2 - \frac{4}{3} = \frac{2}{3}$.
Thus,$f(x) = \frac{4}{3}x^2 + \frac{2}{3}$.

(D) Given $f(x) = |(|x| + \alpha)(|x| - 1)|$.
Let $g(x) = (|x| + \alpha)(|x| - 1)$.
The function $f(x) = |g(x)|$ is not differentiable at the points where $g(x) = 0$ (provided the roots are simple) and at $x = 0$ due to the $|x|$ term.
Case $1$: If $\alpha > 0$,then $|x| + \alpha = 0$ has no real solution. The roots of $g(x) = 0$ are $|x| = 1$,i.e.,$x = 1, -1$.
At $x = 1, -1$,the function $f(x)$ has sharp corners. Also,at $x = 0$,$f(x) = |\alpha(-1)| = |-\alpha| = \alpha$. Since $f'(0^+)$ and $f'(0^-)$ will differ due to the $|x|$ term,$f(x)$ is not differentiable at $x = 0$. Thus,for $\alpha > 0$,there are $3$ points of non-differentiability $(x = -1, 0, 1)$.
Case $2$: If $\alpha < 0$ and $\alpha \neq -1$,let $\alpha = -k$ where $k > 0$ and $k \neq 1$. Then $g(x) = (|x| - k)(|x| - 1)$.
The roots are $|x| = k$ and $|x| = 1$,which gives $x = \pm k$ and $x = \pm 1$.
These are $4$ distinct points where $f(x) = 0$. Additionally,$f(x)$ is not differentiable at $x = 0$ because of the $|x|$ term.
Thus,for $\alpha < 0$ (and $\alpha \neq -1$),there are $5$ points of non-differentiability $(x = -1, -k, 0, k, 1)$.
Comparing with the options,$D$ is correct.

(D) To check the differentiability of $f(x)$ at $x = a$,we calculate the left-hand derivative ($L$.$H$.$D$.) and right-hand derivative ($R$.$H$.$D$.).
$L$.$H$.$D$. at $x = a$ is $\lim_{h \to 0} \frac{f(a-h) - f(a)}{-h} = \lim_{h \to 0} \frac{\frac{1}{2}(b^2 - a^2) - \frac{1}{2}(b^2 - a^2)}{-h} = 0$.
Now,$R$.$H$.$D$. at $x = a$ is $\lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = \lim_{h \to 0} \frac{\frac{1}{2}b^2 - \frac{(a+h)^2}{6} - \frac{a^3}{3(a+h)} - \frac{1}{2}(b^2 - a^2)}{h}$.
Simplifying the expression: $\lim_{h \to 0} \frac{1}{h} \left[ \frac{1}{2}a^2 - \frac{(a+h)^3 + 2a^3}{6(a+h)} \right]$.
As $h \to 0$,the expression does not converge to a finite value,implying the derivative does not exist at $x = a$.
Since $f(x)$ is not differentiable at $x = a$,$f'(x)$ is also not differentiable at $x = a$.

(B) Since $f(x)$ is differentiable at every $x \in R$,it must be differentiable at $x = 3$.
For differentiability at $x = 3$,the left-hand derivative $(LHD)$ must equal the right-hand derivative $(RHD)$.
$LHD$ = $\frac{d}{dx}(ax^2 - bx + 2) = 2ax - b$ at $x = 3$,which is $6a - b$.
$RHD$ = $\frac{d}{dx}(bx^2 - 3) = 2bx$ at $x = 3$,which is $6b$.
Equating them: $6a - b = 6b \Rightarrow 6a = 7b \Rightarrow a = \frac{7b}{6}$.
Also,$f(x)$ must be continuous at $x = 3$,so $LHL$ = $RHL$.
$LHL$ = $a(3)^2 - b(3) + 2 = 9a - 3b + 2$.
$RHL$ = $b(3)^2 - 3 = 9b - 3$.
Equating them: $9a - 3b + 2 = 9b - 3 \Rightarrow 9a - 12b = -5$.
Substitute $a = \frac{7b}{6}$ into the equation: $9(\frac{7b}{6}) - 12b = -5 \Rightarrow \frac{21b}{2} - 12b = -5 \Rightarrow \frac{21b - 24b}{2} = -5 \Rightarrow -3b = -10 \Rightarrow b = \frac{10}{3}$.
Then $a = \frac{7}{6} \times \frac{10}{3} = \frac{70}{18} = \frac{35}{9}$.
The line is $\frac{x}{a} + \frac{y}{b} = 1$. The intercepts are $x = a$ and $y = b$.
The area of the triangle formed with the axes is $\frac{1}{2} \times |a| \times |b| = \frac{1}{2} \times \frac{35}{9} \times \frac{10}{3} = \frac{350}{54} = \frac{175}{27}$ sq units.

(B) function $f(x)$ is continuous at $x=\alpha$ if and only if $\lim _{x \rightarrow \alpha^{-}} f(x) = \lim _{x \rightarrow \alpha^{+}} f(x) = f(\alpha)$.
Since the function $f(x)$ is given as discontinuous at $x=\alpha \in(a, b)$,the condition for continuity is violated.
This means that either the limit does not exist,or the limit exists but is not equal to $f(\alpha)$.
Therefore,the condition $\lim _{x \rightarrow \alpha} f(x) \neq f(\alpha)$ must hold for the function to be discontinuous at that point.
Option $(A)$ represents the condition for continuity,which is false here.
Option $(B)$ correctly identifies the condition for discontinuity at $x=\alpha$.
Options $(C)$ and $(D)$ involve limits outside the domain $[a, b]$,where the function is not defined,making them irrelevant or incorrect.

(A) fundamental theorem in calculus states that if a function $f(x)$ is differentiable at a point $x=a$,then it must be continuous at that point $x=a$.
The contrapositive of this statement is: if $f(x)$ is not continuous at $x=a$,then $f(x)$ is not differentiable at $x=a$.
Since the Reason $(R)$ provides the direct theorem that justifies the Assertion $(A)$,both are true and $(R)$ is the correct explanation of $(A)$.

(A) Let $y = \tan^{-1}(\cos \sqrt{x}) + \sec^{-1}(e^x)$.
Applying the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\tan^{-1}(\cos \sqrt{x})) + \frac{d}{dx}(\sec^{-1}(e^x))$
$= \frac{1}{1 + (\cos \sqrt{x})^2} \cdot (-\sin \sqrt{x}) \cdot \frac{1}{2\sqrt{x}} + \frac{1}{|e^x| \sqrt{(e^x)^2 - 1}} \cdot e^x$
$= \frac{-\sin \sqrt{x}}{2\sqrt{x}(1 + \cos^2 \sqrt{x})} + \frac{1}{\sqrt{e^{2x} - 1}}$.
Now,substitute $x = \frac{\pi^2}{4}$,so $\sqrt{x} = \frac{\pi}{2}$:
$\frac{dy}{dx} = \frac{-\sin(\frac{\pi}{2})}{2(\frac{\pi}{2})(1 + \cos^2(\frac{\pi}{2}))} + \frac{1}{\sqrt{e^{2(\frac{\pi^2}{4})} - 1}}$
$= \frac{-1}{\pi(1 + 0)} + \frac{1}{\sqrt{e^{\frac{\pi^2}{2}} - 1}}$
$= \frac{1}{\sqrt{e^{\frac{\pi^2}{2}} - 1}} - \frac{1}{\pi}$.

(B) Given $f^{\prime}(x) = \sqrt{f^2(x)-1}$.
We can write this as $\frac{f^{\prime}(x)}{\sqrt{f^2(x)-1}} = 1$.
Integrating both sides with respect to $x$,we get $\int \frac{f^{\prime}(x)}{\sqrt{f^2(x)-1}} dx = \int 1 dx$.
Using the standard integral $\int \frac{1}{\sqrt{t^2-1}} dt = \log |t + \sqrt{t^2-1}| + C$,we have $\log |f(x) + \sqrt{f^2(x)-1}| = x + C$.
Given $f(0) = 1$,substituting $x=0$ gives $\log |f(0) + \sqrt{f^2(0)-1}| = 0 + C$.
$\log |1 + \sqrt{1-1}| = C \Rightarrow \log(1) = C \Rightarrow C = 0$.
So,$\log |f(x) + \sqrt{f^2(x)-1}| = x$.
At $x=1$,$\log |f(1) + \sqrt{f^2(1)-1}| = 1$.
Taking the exponential of both sides,$f(1) + \sqrt{f^2(1)-1} = e^1 = e$.
$\sqrt{f^2(1)-1} = e - f(1)$.
Squaring both sides,$f^2(1) - 1 = e^2 + f^2(1) - 2ef(1)$.
$-1 = e^2 - 2ef(1) \Rightarrow 2ef(1) = e^2 + 1$.
Thus,$f(1) = \frac{e^2+1}{2e}$.

(C) Given $u=\sin \left(\frac{x}{y}\right)$,$x=e^t$,and $y=t^2$. Substituting $x$ and $y$ in $u$,we get $u=\sin \left(\frac{e^t}{t^2}\right)$.
Now,differentiating $u$ with respect to $t$ using the chain rule:
$\frac{d u}{d t}=\cos \left(\frac{e^t}{t^2}\right) \cdot \frac{d}{d t}\left(\frac{e^t}{t^2}\right)$
Using the quotient rule for $\frac{d}{d t}\left(\frac{e^t}{t^2}\right) = \frac{t^2 e^t - e^t(2t)}{(t^2)^2} = \frac{t e^t(t-2)}{t^4} = \frac{e^t(t-2)}{t^3}$.
Thus,$\frac{d u}{d t} = \cos \left(\frac{e^t}{t^2}\right) \cdot \frac{e^t(t-2)}{t^3}$.
Squaring both sides:
$\left(\frac{d u}{d t}\right)^2 = \cos^2 \left(\frac{e^t}{t^2}\right) \cdot \frac{e^{2t}(t-2)^2}{t^6}$.
Rearranging the expression $t^6\left(\frac{d u}{d t}\right)^2 \div \left(e^{2 t}(t-2)^2\right)$:
$= \frac{t^6 \cdot \cos^2 \left(\frac{e^t}{t^2}\right) \cdot e^{2t}(t-2)^2}{t^6 \cdot e^{2t}(t-2)^2} = \cos^2 \left(\frac{e^t}{t^2}\right)$.
Since $u = \sin \left(\frac{e^t}{t^2}\right)$,we have $\sin^2 \left(\frac{e^t}{t^2}\right) = u^2$.
Using the identity $\cos^2 \theta = 1 - \sin^2 \theta$,we get $\cos^2 \left(\frac{e^t}{t^2}\right) = 1 - u^2$.

(C) Given $x = t^5 + 5t^3 + 20t + 7$ and $y = 4t^3 - 3t^2 - 18t + 3$.
First,find the derivatives with respect to $t$:
$\frac{dx}{dt} = 5t^4 + 15t^2 + 20 = 5(t^4 + 3t^2 + 4)$.
Note that $t^4 + 3t^2 + 4$ is always positive for all real $t$ (since its discriminant $D = 3^2 - 4(1)(4) = 9 - 16 = -7 < 0$ and the leading coefficient is positive).
$\frac{dy}{dt} = 12t^2 - 6t - 18 = 6(2t^2 - t - 3)$.
Now,the derivative of the curve is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{6(2t^2 - t - 3)}{5(t^4 + 3t^2 + 4)}$.
The curve is decreasing when $\frac{dy}{dx} < 0$.
Since the denominator $5(t^4 + 3t^2 + 4)$ is always positive,we only need the numerator to be negative:
$6(2t^2 - t - 3) < 0$
$2t^2 - t - 3 < 0$
Factorizing the quadratic expression: $(2t - 3)(t + 1) < 0$.
The roots are $t = -1$ and $t = 3/2$.
Testing the intervals,the expression is negative for $t \in (-1, 3/2)$.

(A) Given that $f(x)$ is a differentiable function on $[2, 6]$ and $f^{\prime}(x) \geq 5$ for all $x \in [2, 6]$.
By the Mean Value Theorem,for any $x_1, x_2 \in [2, 6]$ with $x_2 > x_1$,there exists $c \in (x_1, x_2)$ such that $f^{\prime}(c) = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$.
Since $f^{\prime}(x) \geq 5$,we have $\frac{f(6) - f(2)}{6 - 2} \geq 5$.
Substituting the value $f(2) = 4$:
$\frac{f(6) - 4}{4} \geq 5$
$f(6) - 4 \geq 20$
$f(6) \geq 24$.
Thus,a possible value for $f(6)$ is $24$.

(A) Given $y = \frac{\log x}{x}$.
First,find the first derivative $\frac{dy}{dx}$ using the quotient rule:
$\frac{dy}{dx} = \frac{x(\frac{1}{x}) - \log x(1)}{x^2} = \frac{1 - \log x}{x^2}$.
Next,find the second derivative $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{x^2(-\frac{1}{x}) - (1 - \log x)(2x)}{x^4} = \frac{-x - 2x + 2x \log x}{x^4} = \frac{-3x + 2x \log x}{x^4}$.
Now,substitute these into the expression $x^2 \frac{d^2y}{dx^2} + 3x \frac{dy}{dx} + y$:
$= x^2 \left( \frac{-3x + 2x \log x}{x^4} \right) + 3x \left( \frac{1 - \log x}{x^2} \right) + \frac{\log x}{x}$
$= \frac{-3x + 2x \log x}{x^2} + \frac{3 - 3 \log x}{x} + \frac{\log x}{x}$
$= \frac{-3 + 2 \log x}{x} + \frac{3 - 3 \log x}{x} + \frac{\log x}{x}$
$= \frac{-3 + 2 \log x + 3 - 3 \log x + \log x}{x} = \frac{0}{x} = 0$.
Thus,the value is $0$ for all $x > 0$.

(B) Given the equation: $f(x+ay)+g(x-ay)=0$.
Differentiating both sides with respect to $x$ using the chain rule:
$f^{\prime}(x+ay)(1+a \frac{dy}{dx}) + g^{\prime}(x-ay)(1-a \frac{dy}{dx}) = 0$.
Expanding the terms:
$f^{\prime}(x+ay) + a \frac{dy}{dx} f^{\prime}(x+ay) + g^{\prime}(x-ay) - a \frac{dy}{dx} g^{\prime}(x-ay) = 0$.
Rearranging to isolate the terms with $\frac{dy}{dx}$:
$a \frac{dy}{dx} [f^{\prime}(x+ay) - g^{\prime}(x-ay)] = -[f^{\prime}(x+ay) + g^{\prime}(x-ay)]$.
Therefore:
$a \frac{dy}{dx} = \frac{-(f^{\prime}(x+ay) + g^{\prime}(x-ay))}{f^{\prime}(x+ay) - g^{\prime}(x-ay)} = \frac{f^{\prime}(x+ay) + g^{\prime}(x-ay)}{g^{\prime}(x-ay) - f^{\prime}(x+ay)}$.

(C) Given that $\frac{f(x)}{g(x)} = c$,where $c$ is a non-zero constant.
Taking the derivative of both sides with respect to $x$,we get $\left(\frac{f(x)}{g(x)}\right)^{\prime} = 0$.
Since $\beta(x) = \left(\frac{f(x)}{g(x)}\right)^{\prime}$,it follows that $\beta(x) = 0$.
Also,from $\frac{f(x)}{g(x)} = c$,we have $f(x) = c \cdot g(x)$.
Differentiating both sides,$f^{\prime}(x) = c \cdot g^{\prime}(x)$,which implies $\frac{f^{\prime}(x)}{g^{\prime}(x)} = c$.
Since $\alpha(x) = \frac{f^{\prime}(x)}{g^{\prime}(x)}$,we have $\alpha(x) = c$.
Now,substituting these values into the expression $\frac{\alpha(x) - \beta(x)}{\alpha(x) + \beta(x)}$,we get $\frac{c - 0}{c + 0} = \frac{c}{c} = 1$.

(C) Given equation is $x^2 \tan ^{-1}\left(\frac{y}{x}\right)-y^2 \tan ^{-1}\left(\frac{x}{y}\right)=k$.
Differentiating both sides with respect to $x$:
$x^2 \cdot \frac{1}{1+(y/x)^2} \cdot \frac{x(dy/dx)-y}{x^2} + 2x \tan ^{-1}\left(\frac{y}{x}\right) - y^2 \cdot \frac{1}{1+(x/y)^2} \cdot \frac{y-x(dy/dx)}{y^2} - 2y \tan ^{-1}\left(\frac{x}{y}\right) \frac{dy}{dx} = 0$.
Simplifying the terms:
$\frac{x^2}{x^2+y^2} \cdot \frac{x(dy/dx)-y}{1} + 2x \tan ^{-1}\left(\frac{y}{x}\right) - \frac{y^2}{x^2+y^2} \cdot \frac{y-x(dy/dx)}{1} - 2y \tan ^{-1}\left(\frac{x}{y}\right) \frac{dy}{dx} = 0$.
At the point $(1,1)$,we have $x=1, y=1$ and $dy/dx = y_1$:
$\frac{1}{2}(y_1-1) + 2(1) \tan ^{-1}(1) - \frac{1}{2}(1-y_1) - 2(1) \tan ^{-1}(1) y_1 = 0$.
Note that $\tan ^{-1}(1) = \pi/4$.
$\frac{1}{2}y_1 - \frac{1}{2} + 2(\pi/4) - \frac{1}{2} + \frac{1}{2}y_1 - 2(\pi/4)y_1 = 0$.
$y_1 - 1 + \pi/2 - \pi/2 y_1 = 0$.
$y_1(1 - \pi/2) = 1 - \pi/2$.
Thus,$y_1 = 1$.

(A) Given $x^x y^y=e^e$. Taking natural logarithm on both sides:
$\ln(x^x y^y) = \ln(e^e)$
$x \ln x + y \ln y = e$
Differentiating with respect to $x$:
$\frac{d}{dx}(x \ln x) + \frac{d}{dx}(y \ln y) = \frac{d}{dx}(e)$
$(1 + \ln x) + (1 + \ln y) \frac{dy}{dx} = 0$
At point $(e, e)$,$\ln e = 1$:
$(1 + 1) + (1 + 1) \left(\frac{dy}{dx}\right)_{(e, e)} = 0$
$2 + 2 \left(\frac{dy}{dx}\right)_{(e, e)} = 0 \Rightarrow \left(\frac{dy}{dx}\right)_{(e, e)} = -1$
Differentiating $(1 + \ln x) + (1 + \ln y) y' = 0$ again with respect to $x$:
$\frac{1}{x} + \frac{1}{y} y' \cdot y' + (1 + \ln y) y'' = 0$
$\frac{1}{x} + \frac{(y')^2}{y} + (1 + \ln y) y'' = 0$
At $(e, e)$ where $y' = -1$:
$\frac{1}{e} + \frac{(-1)^2}{e} + (1 + \ln e) y'' = 0$
$\frac{1}{e} + \frac{1}{e} + 2 y'' = 0$
$\frac{2}{e} + 2 y'' = 0 \Rightarrow y'' = -\frac{1}{e}$
Since $\left(\frac{dy}{dx}\right)_{(e, e)} = -1$,we can write $y'' = \frac{-1}{e} = \frac{1}{e} \left(\frac{dy}{dx}\right)_{(e, e)}$.

(C) Given the equation $x^3+y^3=3axy$.
Differentiating with respect to $x$:
$3x^2+3y^2 \frac{dy}{dx} = 3ay + 3ax \frac{dy}{dx}$
$(y^2-ax) \frac{dy}{dx} = ay-x^2$
$\frac{dy}{dx} = \frac{ay-x^2}{y^2-ax}$
At point $\left(\frac{3a}{2}, \frac{3a}{2}\right)$,$\frac{dy}{dx} = \frac{a(\frac{3a}{2}) - (\frac{3a}{2})^2}{(\frac{3a}{2})^2 - a(\frac{3a}{2})} = \frac{\frac{3a^2}{2} - \frac{9a^2}{4}}{\frac{9a^2}{4} - \frac{3a^2}{2}} = \frac{-\frac{3a^2}{4}}{\frac{3a^2}{4}} = -1$.
Differentiating $\frac{dy}{dx}(y^2-ax) = ay-x^2$ again with respect to $x$:
$(2y \frac{dy}{dx} - a) \frac{dy}{dx} + (y^2-ax) \frac{d^2y}{dx^2} = a \frac{dy}{dx} - 2x$
At $\left(\frac{3a}{2}, \frac{3a}{2}\right)$ with $\frac{dy}{dx} = -1$:
$(2(\frac{3a}{2})(-1) - a)(-1) + ((\frac{3a}{2})^2 - a(\frac{3a}{2})) y^{\prime \prime} = a(-1) - 2(\frac{3a}{2})$
$(-3a-a)(-1) + (\frac{9a^2}{4} - \frac{6a^2}{4}) y^{\prime \prime} = -a - 3a$
$4a + \frac{3a^2}{4} y^{\prime \prime} = -4a$
$\frac{3a^2}{4} y^{\prime \prime} = -8a$
$3ay^{\prime \prime} = -32$
Therefore,$3ay^{\prime \prime} + 40 = -32 + 40 = 8$.

(D) Given $f(x)=\sqrt{x+\sin x}$.
For $f^{\prime}(x)=0$,we differentiate $f(x)$ with respect to $x$:
$f^{\prime}(x) = \frac{1}{2\sqrt{x+\sin x}} \cdot (1+\cos x) = 0$.
This implies $1+\cos x = 0$,so $\cos x = -1$.
Thus,$x = (2n+1)\pi$ for $n \in \mathbb{Z}$.
Now,we find the value of $f(x)$ at these points:
$f((2n+1)\pi) = \sqrt{(2n+1)\pi + \sin((2n+1)\pi)}$.
Since $\sin((2n+1)\pi) = 0$ for all $n \in \mathbb{Z}$,we have $f((2n+1)\pi) = \sqrt{(2n+1)\pi}$.
Let $x = (2n+1)\pi$ and $y = f(x) = \sqrt{x}$.
Squaring both sides,we get $y^2 = x$,which is the equation of a parabola.

(A) Given the function $f(x)=|x-5|+|x+5|+|x-4|+|x+4|$.
We can define $f(x)$ in different intervals:
$f(x) = \begin{cases} -4x & x \leq -5 \\ -2x+10 & -5 < x \leq -4 \\ 18 & -4 < x \leq 4 \\ 2x+10 & 4 < x \leq 5 \\ 4x & x > 5 \end{cases}$
Now,the derivative $f^{\prime}(x)$ is:
$f^{\prime}(x) = \begin{cases} -4 & x < -5 \\ -2 & -5 < x < -4 \\ 0 & -4 < x < 4 \\ 2 & 4 < x < 5 \\ 4 & x > 5 \end{cases}$
Evaluating the required values:
$f^{\prime}(1) = 0$ (since $-4 < 1 < 4$)
$f^{\prime}(-6) = -4$ (since $-6 < -5$)
$f^{\prime}(-1) = 0$ (since $-4 < -1 < 4$)
$f^{\prime}(6) = 4$ (since $6 > 5$)
Substituting these into the expression:
$\frac{f^{\prime}(1)-f^{\prime}(-6)}{f^{\prime}(-1)+f^{\prime}(6)} = \frac{0-(-4)}{0+4} = \frac{4}{4} = 1$.

(B) Given $f(t) = \frac{t}{2} + \frac{1}{4} \log(2t - 1)$.
First,find the derivative $f^{\prime}(t)$:
$f^{\prime}(t) = \frac{d}{dt} \left( \frac{t}{2} + \frac{1}{4} \log(2t - 1) \right) = \frac{1}{2} + \frac{1}{4} \cdot \frac{1}{2t - 1} \cdot 2 = \frac{1}{2} + \frac{1}{2(2t - 1)}$.
Now,substitute $t$ with $\frac{t+1}{2t+1}$ in $f^{\prime}(t)$:
$f^{\prime}\left(\frac{t+1}{2t+1}\right) = \frac{1}{2} + \frac{1}{2\left(2\left(\frac{t+1}{2t+1}\right) - 1\right)}$.
Simplify the denominator term:
$2\left(\frac{t+1}{2t+1}\right) - 1 = \frac{2t + 2 - (2t + 1)}{2t + 1} = \frac{1}{2t + 1}$.
Substitute this back into the expression:
$f^{\prime}\left(\frac{t+1}{2t+1}\right) = \frac{1}{2} + \frac{1}{2\left(\frac{1}{2t+1}\right)} = \frac{1}{2} + \frac{2t+1}{2} = \frac{1 + 2t + 1}{2} = \frac{2t + 2}{2} = t + 1$.

(B) Given the differential equation $f(x) f^{\prime}(-x) - f(-x) f^{\prime}(x) = 0$.
This can be rewritten as $\frac{d}{dx} [f(x) f(-x)] = f(x) f^{\prime}(-x) (-1) + f^{\prime}(x) f(-x) = 0$.
Wait,let us re-evaluate: $\frac{d}{dx} [f(x) f(-x)] = f^{\prime}(x) f(-x) + f(x) f^{\prime}(-x) (-1) = f^{\prime}(x) f(-x) - f(x) f^{\prime}(-x) = 0$.
Since $f(x) f^{\prime}(-x) - f(-x) f^{\prime}(x) = 0$,we have $f^{\prime}(x) f(-x) - f(x) f^{\prime}(-x) = 0$.
Thus,$\frac{d}{dx} [f(x) f(-x)] = 0$.
This implies $f(x) f(-x) = c$,where $c$ is a constant.
At $x = 0$,$f(0) f(0) = c \Rightarrow 3 \times 3 = 9$,so $c = 9$.
Thus,$f(x) f(-x) = 9$ for all $x \in R$.
For $x = 3$,$f(3) f(-3) = 9 \Rightarrow 9 \times f(-3) = 9 \Rightarrow f(-3) = 1$.
Finally,$(1 + f(-3))^3 + 1 = (1 + 1)^3 + 1 = 2^3 + 1 = 8 + 1 = 9$.

(A) Point $A(0, \alpha)$ lies on the curve $y=f(x)$,so $f(0)=\alpha$.
Given that $f(0)=2$,we have $\alpha=2$.
The slope of the chord $AB$ connecting points $A(0, 2)$ and $B(8, \beta)$ is given by:
$m_{chord} = \frac{\beta - 2}{8 - 0} = \frac{\beta - 2}{8}$.
The slope of the tangent to the curve $y=f(x)$ at $x=4$ is $f^{\prime}(4) = \frac{-3}{4}$.
Since the chord $AB$ is parallel to the tangent at $x=4$,their slopes must be equal:
$\frac{\beta - 2}{8} = \frac{-3}{4}$
Multiplying both sides by $8$:
$\beta - 2 = -3 \times 2$
$\beta - 2 = -6$
$\beta = -6 + 2 = -4$
Thus,the value of $\beta$ is $-4$.

(D) The slope of the tangent to the curve is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t+3}{3t^2-8t-3}$.
Since the normal is parallel to the $X$-axis,the tangent at that point must be perpendicular to the $X$-axis,which means the tangent is parallel to the $Y$-axis.
Thus,the slope of the tangent $\frac{dy}{dx}$ must be undefined,or $\frac{dx}{dt} = 0$.
Setting the denominator to zero: $3t^2-8t-3 = 0$.
Solving the quadratic equation: $(3t+1)(t-3) = 0$.
This gives $t = 3$ and $t = -\frac{1}{3}$.
For these values of $t$,the slope of the tangent is undefined,implying the normal is parallel to the $X$-axis.
Therefore,there are $2$ such points.

(B) Given $A=\{x : 9x \geq x^2+20\}$.
Solving the inequality $x^2-9x+20 \leq 0$,we get $(x-4)(x-5) \leq 0$,which implies $x \in [4, 5]$.
Thus,$A=[4, 5]$.
Given $f(x)=2x^3-15x^2+36x-48$.
Finding the derivative,$f'(x)=6x^2-30x+36=6(x^2-5x+6)=6(x-3)(x-2)$.
For critical points,set $f'(x)=0$,which gives $x=2$ and $x=3$.
Since both critical points $x=2$ and $x=3$ lie outside the interval $A=[4, 5]$,the function $f(x)$ is monotonic in this interval.
Checking the sign of $f'(x)$ for $x \in [4, 5]$: $f'(4)=6(4-3)(4-2)=12 > 0$.
Since $f'(x) > 0$ for all $x \in [4, 5]$,$f(x)$ is an increasing function on $[4, 5]$.
Therefore,the maximum value occurs at the right endpoint $x=5$.
$f(5)=2(5)^3-15(5)^2+36(5)-48 = 2(125)-15(25)+180-48 = 250-375+180-48 = 7$.

(A) Given the curves $y=e^{2(1+x)-4}$ and $x^2 y=1$.
At the point $(1,1)$,we find the slopes of the tangents to both curves.
For the first curve $y=e^{2(1+x)-4}$,differentiating with respect to $x$ gives $y' = e^{2(1+x)-4} \cdot 2$.
At $(1,1)$,the slope $m_1 = y'(1) = e^{2(1+1)-4} \cdot 2 = e^0 \cdot 2 = 2$.
For the second curve $x^2 y = 1$,we have $y = x^{-2}$.
Differentiating with respect to $x$ gives $y' = -2x^{-3} = -\frac{2}{x^3}$.
At $(1,1)$,the slope $m_2 = y'(1) = -\frac{2}{1^3} = -2$.
The angle $\theta$ between the curves is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values,$\tan \theta = \left| \frac{2 - (-2)}{1 + (2)(-2)} \right| = \left| \frac{4}{1 - 4} \right| = \left| \frac{4}{-3} \right| = \frac{4}{3}$.
Since $\tan \theta = \frac{4}{3}$,we can consider a right-angled triangle with opposite side $4$ and adjacent side $3$. The hypotenuse is $\sqrt{4^2 + 3^2} = 5$.
Thus,$\sin \theta = \frac{4}{5}$ and $\cos \theta = \frac{3}{5}$.
Therefore,$|\sin \theta| + |\cos \theta| = \frac{4}{5} + \frac{3}{5} = \frac{7}{5}$.

(C) Let the intercepts of the line on the $x$-axis and $y$-axis be $a$ and $b$ respectively. The equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$.
Given that the sum of the intercepts is $a + b = 12$,so $b = 12 - a$.
The area of the triangle formed by the line with the coordinate axes is $A = \frac{1}{2}ab$.
Substituting $b = 12 - a$,we get $A = \frac{1}{2}a(12 - a) = 6a - \frac{1}{2}a^2$.
To find the maximum area,we differentiate $A$ with respect to $a$ and set it to zero:
$\frac{dA}{da} = 6 - a = 0 \Rightarrow a = 6$.
Since $a = 6$,then $b = 12 - 6 = 6$.
The equation of the line is $\frac{x}{6} + \frac{y}{6} = 1$,which simplifies to $x + y = 6$.

(B) Given the curve $y=2x^3-15x^2+36x+c$.
Stationary points occur where $\frac{dy}{dx} = 0$.
$\frac{dy}{dx} = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x-2)(x-3)$.
Setting $\frac{dy}{dx} = 0$,we get $x=2$ and $x=3$.
Since $(2, a)$ is a point on the curve,$a = 2(2)^3 - 15(2)^2 + 36(2) + c = 16 - 60 + 72 + c = 28 + c$.
Since $(b, 19)$ is a point on the curve,$b$ must be the other $x$-coordinate,so $b=3$.
Substituting $x=3$ and $y=19$ into the curve equation: $19 = 2(3)^3 - 15(3)^2 + 36(3) + c$.
$19 = 54 - 135 + 108 + c \Rightarrow 19 = 27 + c \Rightarrow c = -8$.
Now,$a = 28 + (-8) = 20$.
Therefore,$a+b+c = 20 + 3 - 8 = 15$.