MHT CET 2023 Mathematics Question Paper with Answer and Solution

(A) The number of diagonals in a polygon with $n$ sides is given by the formula $\frac{n(n - 3)}{2}$.
Given that the number of diagonals is $54$,we have:
$\frac{n(n - 3)}{2} = 54$
$n(n - 3) = 108$
$n^2 - 3n - 108 = 0$
Factoring the quadratic equation:
$n^2 - 12n + 9n - 108 = 0$
$n(n - 12) + 9(n - 12) = 0$
$(n - 12)(n + 9) = 0$
Since the number of sides $n$ must be positive,$n = 12$.

(C) Given the curve equation $3x^2 - y^2 = 8$.
Differentiating with respect to $x$,we get $6x - 2y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = \frac{3x}{y}$.
The slope of the tangent at any point $(x, y)$ is $\frac{3x}{y}$.
The slope of the normal is the negative reciprocal of the tangent slope,which is $-\frac{y}{3x}$.
The normal is parallel to the line $x + 3y = 10$,which has a slope of $-\frac{1}{3}$.
Setting the slopes equal: $-\frac{y}{3x} = -\frac{1}{3} \Rightarrow y = x$.
Substituting $y = x$ into the curve equation: $3x^2 - x^2 = 8 \Rightarrow 2x^2 = 8 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$.
Thus,the points are $(2, 2)$ and $(-2, -2)$.
For point $(2, 2)$,the equation of the normal is $y - 2 = -\frac{1}{3}(x - 2) \Rightarrow 3y - 6 = -x + 2 \Rightarrow x + 3y - 8 = 0$.
For point $(-2, -2)$,the equation of the normal is $y + 2 = -\frac{1}{3}(x + 2) \Rightarrow 3y + 6 = -x - 2 \Rightarrow x + 3y + 8 = 0$.
Comparing with the options,$x + 3y + 8 = 0$ is the correct choice.

(B) Let $a$ and $b$ be the lengths of two sides of a triangle.
According to the given condition,$a+b=x$ and $ab=y$.
Given $x^2-c^2=y$,we substitute $x=a+b$:
$(a+b)^2-c^2=ab$
$a^2+b^2+2ab-c^2=ab$
$a^2+b^2-c^2=-ab$
Dividing by $2ab$,we get:
$\frac{a^2+b^2-c^2}{2ab} = -\frac{1}{2}$
By the Law of Cosines,$\cos C = \frac{a^2+b^2-c^2}{2ab}$.
Therefore,$\cos C = -\frac{1}{2}$.
Since $C$ is an angle in a triangle,$C = 120^\circ$ or $\frac{2\pi}{3}$ radians.
The circumradius $R$ is given by $R = \frac{c}{2\sin C}$.
$R = \frac{c}{2\sin(120^\circ)} = \frac{c}{2(\frac{\sqrt{3}}{2})} = \frac{c}{\sqrt{3}}$.

(A) Given $f(x) = 5 - |x - 2|$. Since $|x - 2| \geq 0$,the maximum value of $f(x)$ is $5$,which occurs when $|x - 2| = 0$,i.e.,$x = 2$. Thus,$\alpha = 2$.
Given $g(x) = |x + 1|$. Since $|x + 1| \geq 0$,the minimum value of $g(x)$ is $0$,which occurs when $|x + 1| = 0$,i.e.,$x = -1$. Thus,$\beta = -1$.
We need to evaluate $\lim _{x \rightarrow -\alpha \beta} \frac{(x - 1)(x^2 - 5x + 6)}{x^2 - 6x + 8}$.
Since $-\alpha \beta = -(2)(-1) = 2$,the limit is $\lim _{x \rightarrow 2} \frac{(x - 1)(x - 2)(x - 3)}{(x - 2)(x - 4)}$.
Canceling the common factor $(x - 2)$,we get $\lim _{x \rightarrow 2} \frac{(x - 1)(x - 3)}{x - 4}$.
Substituting $x = 2$,we get $\frac{(2 - 1)(2 - 3)}{2 - 4} = \frac{(1)(-1)}{-2} = \frac{-1}{-2} = \frac{1}{2}$.

(B) We are given the limit: $\lim _{x \rightarrow a} \frac{g(x) f(a)-g(a) f(x)}{x-a}$.
Since $f(a)=2, f^{\prime}(a)=1, g(a)=-1, g^{\prime}(a)=2$,substituting $x=a$ gives $\frac{g(a)f(a)-g(a)f(a)}{a-a} = \frac{0}{0}$,which is an indeterminate form.
Applying $L$-Hospital's rule by differentiating the numerator and denominator with respect to $x$:
$\lim _{x \rightarrow a} \frac{\frac{d}{dx}[g(x) f(a)-g(a) f(x)]}{\frac{d}{dx}[x-a]}$
$= \lim _{x \rightarrow a} \frac{g^{\prime}(x) f(a)-g(a) f^{\prime}(x)}{1}$
$= g^{\prime}(a) f(a)-g(a) f^{\prime}(a)$
$= (2)(2)-(-1)(1)$
$= 4+1 = 5$.

(D) The formula for compound interest is $A = P(1 + r)^n$, where $r$ is the rate of interest per period.
Given $P = 200$, $A = 400$, and $n = 6$ years:
$400 = 200(1 + r)^6$
$(1 + r)^6 = 2$
$(1 + r) = 2^{\frac{1}{6}}$
Now, we need to find the amount $A$ after $n = 33$ years:
$A = 200(1 + r)^{33}$
$A = 200(2^{\frac{1}{6}})^{33}$
$A = 200(2^{\frac{33}{6}})$
$A = 200(2^{\frac{11}{2}})$
$A = 200(2^5 \cdot 2^{\frac{1}{2}})$
$A = 200(32 \sqrt{2})$
$A = 6400 \sqrt{2}$

(C) Given the equation: $\log _2 x + \log _4 x + \log _8 x + \log _{16} x = \frac{25}{36}$
Using the change of base formula $\log _a b = \frac{\log b}{\log a}$:
$\frac{\log x}{\log 2} + \frac{\log x}{\log 4} + \frac{\log x}{\log 8} + \frac{\log x}{\log 16} = \frac{25}{36}$
Since $\log 4 = 2 \log 2$,$\log 8 = 3 \log 2$,and $\log 16 = 4 \log 2$:
$\frac{\log x}{\log 2} + \frac{\log x}{2 \log 2} + \frac{\log x}{3 \log 2} + \frac{\log x}{4 \log 2} = \frac{25}{36}$
Factor out $\frac{\log x}{\log 2}$ (which is $\log _2 x$):
$\log _2 x \left( 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) = \frac{25}{36}$
Calculate the sum in the bracket: $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{12+6+4+3}{12} = \frac{25}{12}$
So,$\log _2 x \left( \frac{25}{12} \right) = \frac{25}{36}$
$\log _2 x = \frac{25}{36} \times \frac{12}{25} = \frac{1}{3}$
Since $x = 2^k$,then $\log _2 x = k$.
Therefore,$k = \frac{1}{3}$.

(A) The greatest value of ${}^{n}C_{r}$ occurs at $r = \frac{n}{2}$ if $n$ is even,and at $r = \frac{n-1}{2}$ or $r = \frac{n+1}{2}$ if $n$ is odd.
For ${}^{10}C_{r}$,$n=10$ (even),so the greatest value is at $r = \frac{10}{2} = 5$.
For ${}^{11}C_{r}$,$n=11$ (odd),so the greatest values are at $r = \frac{11-1}{2} = 5$ and $r = \frac{11+1}{2} = 6$.
Taking $r=5$ for both,we have:
$\frac{{}^{10}C_{5}}{{}^{11}C_{5}} = \frac{\frac{10!}{5!5!}}{\frac{11!}{5!6!}} = \frac{10!}{5!5!} \times \frac{5!6!}{11!} = \frac{10!}{11!} \times \frac{6!}{5!} = \frac{1}{11} \times 6 = \frac{6}{11}$.

(A) The given equation of the circle is $x^2+y^2+2x-4y-4=0$.
Completing the square for $x$ and $y$ terms:
$(x^2+2x+1) + (y^2-4y+4) - 4 - 1 - 4 = 0$
$(x+1)^2 + (y-2)^2 = 9$
$(x+1)^2 + (y-2)^2 = 3^2$
Comparing this with the standard form $(x-h)^2 + (y-k)^2 = r^2$,we get the center $(h, k) = (-1, 2)$ and radius $r = 3$.
The parametric equations are given by $x = h + r \cos \theta$ and $y = k + r \sin \theta$.
Substituting the values,we get $x = -1 + 3 \cos \theta$ and $y = 2 + 3 \sin \theta$.

(B) Let the centre of the circle be $(0, y)$.
Since the circle passes through $(4, 0)$ and $(0, 2)$,the distance from the centre to these points must be equal to the radius $r$.
$\sqrt{(4-0)^2 + (0-y)^2} = \sqrt{(0-0)^2 + (2-y)^2}$
$16 + y^2 = (2-y)^2$
$16 + y^2 = 4 - 4y + y^2$
$16 = 4 - 4y$
$4y = -12$
$y = -3$
Thus,the centre is $(0, -3)$.
The radius $r$ is the distance from $(0, -3)$ to $(0, 2)$:
$r = \sqrt{(0-0)^2 + (2 - (-3))^2} = \sqrt{0^2 + 5^2} = 5$.
Now,calculate $r^2 - r + 1$:
$r^2 - r + 1 = 5^2 - 5 + 1 = 25 - 5 + 1 = 21$.

(B) The given equation of the circle is $x^2+y^2+2x+2y-3=0$.
Completing the square,we get $(x+1)^2+(y+1)^2=5$.
This is a circle with center $C(-1, -1)$ and radius $r = \sqrt{5}$.
The distance $d$ from the center $C(-1, -1)$ to the line $2x+y+13=0$ is given by $d = \frac{|2(-1) + (-1) + 13|}{\sqrt{2^2 + 1^2}} = \frac{|-2-1+13|}{\sqrt{5}} = \frac{10}{\sqrt{5}} = 2\sqrt{5}$.
The maximum distance of a point on the circle from the line is $d + r$.
Thus,$\lambda_{max} = 2\sqrt{5} + \sqrt{5} = 3\sqrt{5}$.

(D) The rectangle is bounded by $x=-2, x=4, y=-2, y=5$. The vertices are $A(-2, -2), D(4, -2), B(4, 5), C(-2, 5)$.
The centre of the rectangle is the intersection of the diagonals,which is the midpoint of $AC$ or $BD$.
Centre $P = \left(\frac{-2+4}{2}, \frac{-2+5}{2}\right) = \left(1, \frac{3}{2}\right)$.
The width of the rectangle is $4 - (-2) = 6$ units,so the radius for touching vertical sides is $r_1 = 3$ units.
The height of the rectangle is $5 - (-2) = 7$ units,so the radius for touching horizontal sides is $r_2 = 3.5 = \frac{7}{2}$ units.
Case $1$: Radius $r = 3$. The equation is $(x-1)^2 + (y-1.5)^2 = 3^2 \implies x^2 - 2x + 1 + y^2 - 3y + 2.25 = 9 \implies x^2 + y^2 - 2x - 3y - 5.75 = 0$.
Case $2$: Radius $r = 3.5$. The equation is $(x-1)^2 + (y-1.5)^2 = (3.5)^2 \implies x^2 - 2x + 1 + y^2 - 3y + 2.25 = 12.25 \implies x^2 + y^2 - 2x - 3y - 9 = 0$.
Comparing with the options,the equation $x^2+y^2-2x-3y-9=0$ is present.
Thus,Option $(D)$ is correct.

(D) Let $A \equiv (x_1, y_1)$ and $B \equiv (x_2, y_2)$.
According to the given condition,the roots of $x^2+2ax-b^2=0$ are $x_1, x_2$,so $x_1+x_2 = -2a$ and $x_1x_2 = -b^2$.
Similarly,the roots of $y^2+2py-q^2=0$ are $y_1, y_2$,so $y_1+y_2 = -2p$ and $y_1y_2 = -q^2$.
The equation of the circle with $A(x_1, y_1)$ and $B(x_2, y_2)$ as the endpoints of the diameter is given by $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.
Expanding this,we get $x^2 - x(x_1+x_2) + x_1x_2 + y^2 - y(y_1+y_2) + y_1y_2 = 0$.
Substituting the values,we get $x^2 - x(-2a) - b^2 + y^2 - y(-2p) - q^2 = 0$.
Therefore,the equation is $x^2+y^2+2ax+2py-(b^2+q^2) = 0$.

(B) The equation of the circle is $x^2 - 2x + y^2 - 4y = 0$.
Completing the square,we get $(x-1)^2 + (y-2)^2 = 5$.
The centre of the circle is $(1, 2)$ and the radius $r = \sqrt{5}$.
For the line $x - 2y - m = 0$ to intersect the circle at two distinct points,the perpendicular distance from the centre $(1, 2)$ to the line must be less than the radius $r$.
Distance $d = \frac{|1 - 2(2) - m|}{\sqrt{1^2 + (-2)^2}} = \frac{|1 - 4 - m|}{\sqrt{5}} = \frac{|-3 - m|}{\sqrt{5}} = \frac{|m + 3|}{\sqrt{5}}$.
Setting $d < r$,we have $\frac{|m + 3|}{\sqrt{5}} < \sqrt{5}$.
$|m + 3| < 5$.
$-5 < m + 3 < 5$.
$-8 < m < 2$.
Since $m \in \mathbb{Z}$,the possible values for $m$ are $\{-7, -6, -5, -4, -3, -2, -1, 0, 1\}$.
The total number of such values is $9$.

(C) The circle is $x^2+y^2=2^2$,so the radius $r = 2$.
The distance $OP = \sqrt{(-4-0)^2 + (0-0)^2} = 4$.
In the right-angled triangle $\triangle PBO$,the length of the tangent $PB = \sqrt{OP^2 - OB^2} = \sqrt{4^2 - 2^2} = \sqrt{16-4} = \sqrt{12} = 2\sqrt{3}$.
The area of the quadrilateral $PAOB$ is the sum of the areas of $\triangle PBO$ and $\triangle PAO$.
Since $\triangle PBO \cong \triangle PAO$,the area of $PAOB = 2 \times \text{Area}(\triangle PBO)$.
Area of $\triangle PBO = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times PB \times OB = \frac{1}{2} \times 2\sqrt{3} \times 2 = 2\sqrt{3}$.
Therefore,the area of the quadrilateral $PAOB = 2 \times 2\sqrt{3} = 4\sqrt{3}$ sq. units.

(D) For the circle $x^2+y^2=9$,the center $C_1 = (0,0)$ and radius $r_1 = 3$.
For the circle $x^2+y^2+2\alpha x+2y+1=0$,the center $C_2 = (-\alpha, -1)$ and radius $r_2 = \sqrt{(-\alpha)^2 + (-1)^2 - 1} = \sqrt{\alpha^2} = |\alpha|$.
Since the circles touch each other internally,the distance between their centers is equal to the absolute difference of their radii: $C_1C_2 = |r_1 - r_2|$.
$C_1C_2 = \sqrt{(-\alpha - 0)^2 + (-1 - 0)^2} = \sqrt{\alpha^2 + 1}$.
Thus,$\sqrt{\alpha^2 + 1} = |3 - |\alpha||$.
Squaring both sides,$\alpha^2 + 1 = (3 - |\alpha|)^2 = 9 + \alpha^2 - 6|\alpha|$.
$1 = 9 - 6|\alpha|$ $\Rightarrow 6|\alpha| = 8$ $\Rightarrow |\alpha| = \frac{4}{3}$.
Therefore,$\alpha^3 = (\pm \frac{4}{3})^3 = \pm \frac{64}{27}$. Given the options,we consider $\alpha = \frac{4}{3}$,so $\alpha^3 = \frac{64}{27}$.

(B) For the circle $x^2+y^2+2ax+c=0$,the center is $C_1 = (-a, 0)$ and the radius is $r_1 = \sqrt{a^2-c}$.
For the circle $x^2+y^2+2by+c=0$,the center is $C_2 = (0, -b)$ and the radius is $r_2 = \sqrt{b^2-c}$.
Since the circles touch each other externally,the distance between their centers must be equal to the sum of their radii:
$d(C_1, C_2) = r_1 + r_2$
$\sqrt{(-a-0)^2 + (0-(-b))^2} = \sqrt{a^2-c} + \sqrt{b^2-c}$
$\sqrt{a^2+b^2} = \sqrt{a^2-c} + \sqrt{b^2-c}$
Squaring both sides:
$a^2+b^2 = (a^2-c) + (b^2-c) + 2\sqrt{(a^2-c)(b^2-c)}$
$a^2+b^2 = a^2+b^2-2c + 2\sqrt{(a^2-c)(b^2-c)}$
$2c = 2\sqrt{(a^2-c)(b^2-c)}$
$c = \sqrt{(a^2-c)(b^2-c)}$
Squaring again:
$c^2 = (a^2-c)(b^2-c)$
$c^2 = a^2b^2 - a^2c - b^2c + c^2$
$0 = a^2b^2 - c(a^2+b^2)$
$c(a^2+b^2) = a^2b^2$
Dividing both sides by $a^2b^2c$:
$\frac{a^2+b^2}{a^2b^2} = \frac{1}{c}$
$\frac{1}{b^2} + \frac{1}{a^2} = \frac{1}{c}$

(C) For the circle $x^2+y^2-6x-14y+48=0$,the center $C_1$ is $(3, 7)$ and the radius $r_1 = \sqrt{3^2+7^2-48} = \sqrt{9+49-48} = \sqrt{10}$.
For the circle $x^2+y^2-6x=0$,the center $C_2$ is $(3, 0)$ and the radius $r_2 = \sqrt{3^2+0^2-0} = 3$.
The distance between the centers $C_1$ and $C_2$ is $d = \sqrt{(3-3)^2+(7-0)^2} = 7$.
Since $r_1 + r_2 = \sqrt{10} + 3 \approx 3.16 + 3 = 6.16$,we observe that $d > r_1 + r_2$ because $7 > 6.16$.
Since the distance between the centers is greater than the sum of the radii,the circles are disjoint and lie outside each other.
Therefore,the number of common tangents is $4$.

(A) The given circle is $x^2+y^2-4x-6y-12=0$. Its centre $C_1$ is $(2, 3)$ and radius $R = \sqrt{2^2+3^2-(-12)} = \sqrt{4+9+12} = \sqrt{25} = 5$.
Let the centre of the required circle be $C_2(h, k)$ and its radius be $r = 3$.
Since the circles touch internally at $P(-1, -1)$,the point $P$ divides the line segment $C_1C_2$ externally in the ratio $R:r = 5:3$.
Using the section formula for external division:
$P = \left(\frac{R h - r x_1}{R - r}, \frac{R k - r y_1}{R - r}\right)$
$-1 = \frac{5h - 3(2)}{5-3} \implies -2 = 5h - 6 \implies 5h = 4 \implies h = \frac{4}{5}$
$-1 = \frac{5k - 3(3)}{5-3} \implies -2 = 5k - 9 \implies 5k = 7 \implies k = \frac{7}{5}$
Thus,the centre is $\left(\frac{4}{5}, \frac{7}{5}\right)$.

(B) Given $Z_1 = 4i^{40} - 5i^{35} + 6i^{17} + 2$.
Since $i^4 = 1$,$i^{40} = (i^4)^{10} = 1$.
$i^{35} = i^{32} \times i^3 = 1 \times (-i) = -i$.
$i^{17} = i^{16} \times i = 1 \times i = i$.
Substituting these values: $Z_1 = 4(1) - 5(-i) + 6(i) + 2 = 4 + 5i + 6i + 2 = 6 + 11i$.
Given $Z_2 = -1 + i$.
$Z_1 + Z_2 = (6 + 11i) + (-1 + i) = (6 - 1) + (11i + i) = 5 + 12i$.
The modulus $|Z_1 + Z_2| = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.

(D) Given expression: $\frac{i^{248}+i^{246}+i^{244}+i^{242}+i^{240}}{i^{249}+i^{247}+i^{245}+i^{243}+i^{241}}$
Factor out $i^{240}$ from the numerator and $i^{241}$ from the denominator:
$= \frac{i^{240}(i^8+i^6+i^4+i^2+1)}{i^{241}(i^8+i^6+i^4+i^2+1)}$
Cancel the common term $(i^8+i^6+i^4+i^2+1)$:
$= \frac{i^{240}}{i^{241}}$
$= \frac{1}{i}$
Multiply numerator and denominator by $i$:
$= \frac{i}{i^2} = \frac{i}{-1} = -i$

(A) Given $x = \frac{5}{1-2i}$. Multiplying numerator and denominator by the conjugate $(1+2i)$:
$x = \frac{5(1+2i)}{(1-2i)(1+2i)} = \frac{5(1+2i)}{1+4} = \frac{5(1+2i)}{5} = 1+2i$.
Now,calculate $x^2$:
$x^2 = (1+2i)^2 = 1^2 + (2i)^2 + 2(1)(2i) = 1 - 4 + 4i = -3 + 4i$.
Now,calculate $x^3$:
$x^3 = x^2 \cdot x = (-3+4i)(1+2i) = -3 - 6i + 4i + 8i^2 = -3 - 2i - 8 = -11 - 2i$.
Substitute these values into the expression $x^3 + x^2 - x + 22$:
$(-11 - 2i) + (-3 + 4i) - (1 + 2i) + 22$
$= (-11 - 3 - 1 + 22) + (-2i + 4i - 2i)$
$= 7 + 0i = 7$.

(B) Given $z = \frac{(1+i)^2}{a+i} = \frac{1+i^2+2i}{a+i} = \frac{2i}{a+i}$.
Magnitude $|z| = \left| \frac{2i}{a+i} \right| = \frac{|2i|}{|a+i|} = \frac{2}{\sqrt{a^2+1}}$.
Given $|z| = \frac{2}{\sqrt{5}}$,so $\frac{2}{\sqrt{a^2+1}} = \frac{2}{\sqrt{5}}$ $\Rightarrow a^2+1 = 5$ $\Rightarrow a^2 = 4$.
Since $a > 0$,we have $a = 2$.
Substituting $a = 2$ into $z$,we get $z = \frac{2i}{2+i} = \frac{2i(2-i)}{(2+i)(2-i)} = \frac{4i - 2i^2}{4+1} = \frac{2+4i}{5} = \frac{2}{5} + \frac{4}{5}i$.
Therefore,$\bar{z} = \frac{2}{5} - \frac{4}{5}i$.

(B) Given $\operatorname{Im}(z)=10$, let $z=x+10i$.
The given equation is $\frac{2z-n}{2z+n}=2i-1$.
Substituting $z=x+10i$:
$\frac{2(x+10i)-n}{2(x+10i)+n}=2i-1$
$(2x-n)+20i=(2i-1)(2x+n+20i)$
$(2x-n)+20i = 4xi + 2ni - 40 - 2x - n - 20i$
$(2x-n)+20i = (-2x-n-40) + (4x+2n-20)i$
Equating real and imaginary parts:
Real part: $2x-n = -2x-n-40$ $\Rightarrow 4x = -40$ $\Rightarrow x = -10$.
Imaginary part: $20 = 4x+2n-20$.
Substituting $x=-10$: $20 = 4(-10)+2n-20$ $\Rightarrow 20 = -40+2n-20$ $\Rightarrow 2n = 80$ $\Rightarrow n = 40$.
Thus, $n=40$ and $\operatorname{Re}(z)=x=-10$.

(B) Given $Z_1=2+i$ and $Z_2=3-4i$.
The conjugates are $\overline{Z_1}=2-i$ and $\overline{Z_2}=3+4i$.
We need to evaluate $\frac{\overline{Z_1}}{\overline{Z_2}} = \frac{2-i}{3+4i}$.
Multiply the numerator and denominator by the conjugate of the denominator $(3-4i)$:
$\frac{2-i}{3+4i} \times \frac{3-4i}{3-4i} = \frac{6-8i-3i+4i^2}{3^2-(4i)^2}$.
Since $i^2=-1$,we have $\frac{6-11i-4}{9+16} = \frac{2-11i}{25} = \frac{2}{25} - \frac{11}{25}i$.
Comparing this with $a+bi$,we get $a=\frac{2}{25}$ and $b=\frac{-11}{25}$.
Now,calculate $-7a+b = -7(\frac{2}{25}) - \frac{11}{25} = \frac{-14-11}{25} = \frac{-25}{25} = -1$.

(C) Let $z = \frac{1+i \sqrt{3}}{\sqrt{3}+i}$.
Multiply the numerator and denominator by the conjugate of the denominator $(\sqrt{3}-i)$:
$z = \frac{(1+i \sqrt{3})(\sqrt{3}-i)}{(\sqrt{3}+i)(\sqrt{3}-i)}$
$z = \frac{\sqrt{3} - i + 3i - i^2 \sqrt{3}}{3 - i^2}$
Since $i^2 = -1$,we have:
$z = \frac{\sqrt{3} + 2i + \sqrt{3}}{3 + 1} = \frac{2\sqrt{3} + 2i}{4} = \frac{\sqrt{3}}{2} + \frac{1}{2}i$
The argument of $z = a + bi$ is given by $\theta = \tan^{-1}\left(\frac{b}{a}\right)$.
$\theta = \tan^{-1}\left(\frac{1/2}{\sqrt{3}/2}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$.

(B) Two complex numbers $z_1 = a+bi$ and $z_2 = c+di$ are conjugates if $a=c$ and $b=-d$.
Given $z_1 = (3x+2) - (5y-3)i$ and $z_2 = (6x+3) + (2y-4)i$.
Equating the real parts: $3x+2 = 6x+3$ $\Rightarrow 3x = -1$ $\Rightarrow x = -\frac{1}{3}$.
Equating the imaginary parts: $-(5y-3) = 2y-4$ $\Rightarrow -5y+3 = 2y-4$ $\Rightarrow 7y = 7$ $\Rightarrow y = 1$.
Wait,let's re-evaluate the conjugate condition: $z_1 = \overline{z_2}$.
$(3x+2) - (5y-3)i = (6x+3) - (2y-4)i$.
Real parts: $3x+2 = 6x+3$ $\Rightarrow 3x = -1$ $\Rightarrow x = -\frac{1}{3}$.
Imaginary parts: $-(5y-3) = -(2y-4)$ $\Rightarrow 5y-3 = 2y-4$ $\Rightarrow 3y = -1$ $\Rightarrow y = -\frac{1}{3}$.
Now,calculate $\frac{x-y}{x+y} = \frac{-\frac{1}{3} - (-\frac{1}{3})}{-\frac{1}{3} + (-\frac{1}{3})} = \frac{0}{-\frac{2}{3}} = 0$.

(A) Given $z = \frac{(1+i)^2}{a-i} = \frac{1 + 2i + i^2}{a-i} = \frac{2i}{a-i}$.
Taking the modulus on both sides: $|z| = \left| \frac{2i}{a-i} \right| = \frac{|2i|}{|a-i|} = \frac{2}{\sqrt{a^2 + 1}}$.
Given $|z| = \frac{2}{\sqrt{5}}$,we have $\frac{2}{\sqrt{a^2 + 1}} = \frac{2}{\sqrt{5}}$.
Thus,$a^2 + 1 = 5$,which implies $a^2 = 4$. Since $a > 0$,we get $a = 2$.
Substituting $a = 2$ into $z$: $z = \frac{2i}{2-i} = \frac{2i(2+i)}{(2-i)(2+i)} = \frac{4i + 2i^2}{4+1} = \frac{-2 + 4i}{5} = -\frac{2}{5} + \frac{4}{5}i$.
Therefore,the conjugate $\bar{z} = -\frac{2}{5} - \frac{4}{5}i$.

(B) $z^{1/3} = p+iq$
$\Rightarrow z = (p+iq)^3$
$\Rightarrow x+iy = p^3 + 3p^2(iq) + 3p(iq)^2 + (iq)^3$
$\Rightarrow x+iy = (p^3 - 3pq^2) + i(3p^2q - q^3)$
$\Rightarrow x = p^3 - 3pq^2$ and $y = 3p^2q - q^3$
$\Rightarrow \frac{x}{p} = p^2 - 3q^2$ and $\frac{y}{q} = 3p^2 - q^2$
$\therefore \left(\frac{x}{p} + \frac{y}{q}\right) = (p^2 - 3q^2) + (3p^2 - q^2) = 4p^2 - 4q^2 = 4(p^2 - q^2)$

(B) Given $w = \frac{z}{z - \frac{1}{3}i}$.
Multiplying numerator and denominator by $3$,we get $w = \frac{3z}{3z - i}$.
Since $|w| = 1$,we have $|\frac{3z}{3z - i}| = 1$,which implies $3|z| = |3z - i|$.
Let $z = x + iy$. Then $3|x + iy| = |3x + i(3y - 1)|$.
Squaring both sides,we get $9(x^2 + y^2) = (3x)^2 + (3y - 1)^2$.
$9x^2 + 9y^2 = 9x^2 + 9y^2 - 6y + 1$.
$6y - 1 = 0$,which is the equation of a straight line.

(C) Given the inequality $|z-(2-i)| \leq 2$,this represents a disk in the complex plane centered at $2-i$ with radius $r=2$.
Let $z_0 = 2-i$. The distance of the center from the origin is $|z_0| = |2-i| = \sqrt{2^2+(-1)^2} = \sqrt{5}$.
The maximum value of $|z|$ is $|z_0| + r = \sqrt{5} + 2$.
The minimum value of $|z|$ is $|z_0| - r = \sqrt{5} - 2$.
The difference between the greatest and least value is $(\sqrt{5} + 2) - (\sqrt{5} - 2) = 4$.

(C) Given the equation of the circle: $x^2+y^2+ax+by=0$.
Completing the square for $x$ and $y$ terms:
$(x^2+ax+\frac{a^2}{4}) + (y^2+by+\frac{b^2}{4}) = \frac{a^2}{4} + \frac{b^2}{4}$.
This simplifies to: $(x+\frac{a}{2})^2 + (y+\frac{b}{2})^2 = \frac{a^2+b^2}{4}$.
Comparing this with the standard form of a circle $(x-h)^2 + (y-k)^2 = r^2$,we identify the center $(h, k) = (-\frac{a}{2}, -\frac{b}{2})$ and the radius $r = \sqrt{\frac{a^2+b^2}{4}}$.
The parametric equations for a circle with center $(h, k)$ and radius $r$ are given by $x = h + r \cos \theta$ and $y = k + r \sin \theta$.
Substituting the values,we get $x = -\frac{a}{2} + \sqrt{\frac{a^2+b^2}{4}} \cos \theta$ and $y = -\frac{b}{2} + \sqrt{\frac{a^2+b^2}{4}} \sin \theta$.

(C) We are given the sum $p = \sum_{r=1}^{50} \tan ^{-1} \frac{1}{2 r^2}$.
Multiply the numerator and denominator inside the $\tan^{-1}$ by $2$ to get $\tan ^{-1} \frac{2}{4 r^2}$.
We can rewrite the expression as $\tan ^{-1} \left[ \frac{(2r+1) - (2r-1)}{1 + (2r+1)(2r-1)} \right]$.
Using the identity $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \frac{x-y}{1+xy}$,this becomes $\tan^{-1}(2r+1) - \tan^{-1}(2r-1)$.
Now,the sum is a telescoping series:
$p = \sum_{r=1}^{50} [\tan^{-1}(2r+1) - \tan^{-1}(2r-1)]$
$p = (\tan^{-1} 3 - \tan^{-1} 1) + (\tan^{-1} 5 - \tan^{-1} 3) + \dots + (\tan^{-1} 101 - \tan^{-1} 99)$
All intermediate terms cancel out,leaving $p = \tan^{-1} 101 - \tan^{-1} 1$.
Using the formula $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \frac{x-y}{1+xy}$:
$p = \tan^{-1} \left( \frac{101 - 1}{1 + 101 \times 1} \right) = \tan^{-1} \left( \frac{100}{102} \right)$.
Therefore,$\tan p = \frac{100}{102} = \frac{50}{51}$.

(A) We need to evaluate the limit: $\lim _{x \rightarrow 0} \frac{(1-\cos 2 x) \sin 5 x}{x^2 \sin 3 x}$.
Using the identity $1-\cos 2x = 2\sin^2 x$,we rewrite the expression:
$= \lim _{x \rightarrow 0} \frac{2 \sin^2 x \cdot \sin 5 x}{x^2 \sin 3 x}$.
$= 2 \cdot \lim _{x \rightarrow 0} \left( \frac{\sin x}{x} \right)^2 \cdot \frac{\sin 5 x}{x} \cdot \frac{x}{\sin 3 x}$.
Multiply and divide by $5$ and $3$ to use the standard limit $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$:
$= 2 \cdot \left( \lim _{x \rightarrow 0} \frac{\sin x}{x} \right)^2 \cdot 5 \left( \lim _{x \rightarrow 0} \frac{\sin 5 x}{5 x} \right) \cdot \frac{1}{3} \left( \lim _{x \rightarrow 0} \frac{3 x}{\sin 3 x} \right)$.
$= 2 \cdot (1)^2 \cdot 5 \cdot (1) \cdot \frac{1}{3} \cdot (1) = \frac{10}{3}$.

(A) Given $f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^2 - 7$.
First,find the derivative $f'(x) = 30x^9 - 56x^7 + 30x^5 - 63x^2 + 6x$.
Evaluating at $x = 1$,$f'(1) = 30 - 56 + 30 - 63 + 6 = -53$.
Now,consider the limit $L = \lim_{\alpha \rightarrow 0} \frac{f(1-\alpha) - f(1)}{\alpha^3 + 3\alpha}$.
Rewrite the expression as $L = \lim_{\alpha \rightarrow 0} \left( \frac{f(1-\alpha) - f(1)}{-\alpha} \right) \times \left( \frac{-\alpha}{\alpha^3 + 3\alpha} \right)$.
Since $\lim_{\alpha \rightarrow 0} \frac{f(1-\alpha) - f(1)}{-\alpha} = f'(1) = -53$ and $\lim_{\alpha \rightarrow 0} \frac{-\alpha}{\alpha(\alpha^2 + 3)} = \lim_{\alpha \rightarrow 0} \frac{-1}{\alpha^2 + 3} = -\frac{1}{3}$.
Therefore,$L = (-53) \times (-1/3) = \frac{53}{3}$.

(C) We evaluate the limit $\lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}}$ by rationalizing both the numerator and the denominator.
Multiplying by the conjugates:
$\lim _{x}$ ${\rightarrow a} \frac{(\sqrt{a+2 x}-\sqrt{3 x})(\sqrt{a+2 x}+\sqrt{3 x})(\sqrt{3 a+x}+2 \sqrt{x})}{(\sqrt{3 a+x}-2 \sqrt{x})(\sqrt{3 a+x}+2 \sqrt{x})(\sqrt{a+2 x}+\sqrt{3 x})}$
Simplifying the terms using $(A-B)(A+B) = A^2 - B^2$:
$\lim _{x \rightarrow a} \frac{(a+2 x-3 x)(\sqrt{3 a+x}+2 \sqrt{x})}{(3 a+x-4 x)(\sqrt{a+2 x}+\sqrt{3 x})}$
$\lim _{x \rightarrow a} \frac{(a-x)(\sqrt{3 a+x}+2 \sqrt{x})}{3(a-x)(\sqrt{a+2 x}+\sqrt{3 x})}$
Canceling the common factor $(a-x)$:
$\lim _{x \rightarrow a} \frac{\sqrt{3 a+x}+2 \sqrt{x}}{3(\sqrt{a+2 x}+\sqrt{3 x})}$
Substituting $x = a$:
$\frac{\sqrt{4a} + 2\sqrt{a}}{3(\sqrt{3a} + \sqrt{3a})} = \frac{2\sqrt{a} + 2\sqrt{a}}{3(2\sqrt{3a})} = \frac{4\sqrt{a}}{6\sqrt{3a}} = \frac{2}{3\sqrt{3}}$

(C) We evaluate the limit: $\lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}}$
Rationalizing the numerator and denominator:
$= \lim _{x}$ ${\rightarrow a} \frac{(\sqrt{a+2 x}-\sqrt{3 x})(\sqrt{a+2 x}+\sqrt{3 x})(\sqrt{3 a+x}+2 \sqrt{x})}{(\sqrt{3 a+x}-2 \sqrt{x})(\sqrt{3 a+x}+2 \sqrt{x})(\sqrt{a+2 x}+\sqrt{3 x})}$
$= \lim _{x \rightarrow a} \frac{(a+2 x-3 x)(\sqrt{3 a+x}+2 \sqrt{x})}{(3 a+x-4 x)(\sqrt{a+2 x}+\sqrt{3 x})}$
$= \lim _{x \rightarrow a} \frac{(a-x)(\sqrt{3 a+x}+2 \sqrt{x})}{3(a-x)(\sqrt{a+2 x}+\sqrt{3 x})}$
Canceling $(a-x)$ and substituting $x = a$:
$= \frac{\sqrt{3 a+a}+2 \sqrt{a}}{3(\sqrt{a+2 a}+\sqrt{3 a})}$
$= \frac{2 \sqrt{a}+2 \sqrt{a}}{3(\sqrt{3 a}+\sqrt{3 a})}$
$= \frac{4 \sqrt{a}}{3(2 \sqrt{3 a})} = \frac{4 \sqrt{a}}{6 \sqrt{3} \sqrt{a}} = \frac{2}{3 \sqrt{3}}$

(C) We are given the limit: $\lim _{x \rightarrow 2}\left[\frac{1}{x-2}-\frac{2}{x^3-3 x^2+2 x}\right]$
First,factor the denominator of the second term: $x^3-3x^2+2x = x(x^2-3x+2) = x(x-2)(x-1)$
Now,substitute this back into the limit: $\lim _{x \rightarrow 2}\left[\frac{1}{x-2}-\frac{2}{x(x-2)(x-1)}\right]$
Find a common denominator: $\lim _{x \rightarrow 2}\left[\frac{x(x-1)-2}{x(x-2)(x-1)}\right]$
Simplify the numerator: $x^2-x-2 = (x-2)(x+1)$
Substitute the simplified numerator: $\lim _{x \rightarrow 2}\left[\frac{(x-2)(x+1)}{x(x-2)(x-1)}\right]$
Cancel the common factor $(x-2)$: $\lim _{x \rightarrow 2}\frac{x+1}{x(x-1)}$
Evaluate the limit by substituting $x=2$: $\frac{2+1}{2(2-1)} = \frac{3}{2(1)} = \frac{3}{2}$

(B) Let $I = \lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3}$
$= \lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x(1-\sin x)}{\sin x(\pi-2 x)^3}$
Substitute $x = \frac{\pi}{2}-h$,then $\pi-2x = 2h$. As $x \rightarrow \frac{\pi}{2}$,$h \rightarrow 0$.
$I = \lim _{h \rightarrow 0} \frac{\cos(\frac{\pi}{2}-h)(1-\sin(\frac{\pi}{2}-h))}{\sin(\frac{\pi}{2}-h)(2h)^3}$
$= \lim _{h \rightarrow 0} \frac{\sin h(1-\cos h)}{\cos h \cdot 8h^3}$
Using $1-\cos h = 2\sin^2(\frac{h}{2})$,
$I = \lim _{h \rightarrow 0} \frac{\sin h \cdot 2\sin^2(\frac{h}{2})}{\cos h \cdot 8h^3} = \frac{2}{8} \lim _{h \rightarrow 0} \frac{\sin h}{h} \cdot \lim _{h \rightarrow 0} \frac{\sin^2(\frac{h}{2})}{(\frac{h}{2})^2 \cdot 4} \cdot \lim _{h \rightarrow 0} \frac{1}{\cos h}$
$= \frac{1}{4} \cdot 1 \cdot (1)^2 \cdot \frac{1}{4} \cdot 1 = \frac{1}{16}$

(C) We know that $x^{\circ} = \frac{\pi}{180} x \text{ radians}$.
So,the limit becomes $\lim _{x \rightarrow 0} \frac{\cos \left(\frac{7 \pi}{180} x\right)-\cos \left(\frac{2 \pi}{180} x\right)}{x^2}$.
Using the standard limit formula $\lim _{x \rightarrow 0} \frac{\cos(ax) - \cos(bx)}{x^2} = \frac{b^2 - a^2}{2}$,where $a = \frac{7\pi}{180}$ and $b = \frac{2\pi}{180}$:
$= \frac{(\frac{2\pi}{180})^2 - (\frac{7\pi}{180})^2}{2}$
$= \frac{1}{2} \cdot \frac{\pi^2}{180^2} (2^2 - 7^2)$
$= \frac{1}{2} \cdot \frac{\pi^2}{32400} (4 - 49)$
$= \frac{1}{2} \cdot \frac{\pi^2}{32400} (-45)$
$= \frac{-45 \pi^2}{64800} = \frac{-\pi^2}{1440}$.

(B) We are given the limit $L = \lim _{x \rightarrow 0} \frac{x \cot 4x}{\sin ^2 x \cot ^2(2x)}$.
Using the identity $\cot \theta = \frac{1}{\tan \theta}$,we rewrite the expression:
$L = \lim _{x \rightarrow 0} \frac{x \tan ^2(2x)}{\sin ^2 x \tan 4x}$.
Now,we use the standard limits $\lim _{\theta \rightarrow 0} \frac{\tan \theta}{\theta} = 1$ and $\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$:
$L = \lim _{x \rightarrow 0} \left[ \frac{x \cdot (\tan 2x)^2}{(\sin x)^2 \cdot \tan 4x} \right] = \lim _{x \rightarrow 0} \left[ \frac{x \cdot \frac{(\tan 2x)^2}{(2x)^2} \cdot (2x)^2}{\frac{(\sin x)^2}{x^2} \cdot x^2 \cdot \frac{\tan 4x}{4x} \cdot 4x} \right]$.
Simplifying the terms:
$L = \lim _{x \rightarrow 0} \left[ \frac{x \cdot 1^2 \cdot 4x^2}{1^2 \cdot x^2 \cdot 1 \cdot 4x} \right] = \lim _{x \rightarrow 0} \frac{4x^3}{4x^3} = 1$.

(C) Let $L = \lim _{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{\tan ^2 \frac{x}{2}}$.
Rationalizing the numerator,we get:
$L = \lim _{x \rightarrow 0} \frac{(1+x \sin x) - \cos x}{\tan ^2 \frac{x}{2} (\sqrt{1+x \sin x} + \sqrt{\cos x})}$.
Using the identity $1 - \cos x = 2 \sin ^2 \frac{x}{2}$,we have:
$L = \lim _{x}$ ${\rightarrow 0} \frac{x \sin x + 2 \sin ^2 \frac{x}{2}}{\tan ^2 \frac{x}{2} (\sqrt{1+x \sin x} + \sqrt{\cos x})}$.
Dividing numerator and denominator by $x^2$:
$L = \lim _{x}$ ${\rightarrow 0} \frac{\frac{\sin x}{x} + 2 \left( \frac{\sin (x/2)}{x} \right)^2}{\left( \frac{\tan (x/2)}{x} \right)^2 (\sqrt{1+x \sin x} + \sqrt{\cos x})}$.
Since $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$,$\lim _{x \rightarrow 0} \frac{\sin (x/2)}{x} = \frac{1}{2}$,and $\lim _{x \rightarrow 0} \frac{\tan (x/2)}{x} = \frac{1}{2}$:
$L = \frac{1 + 2(1/2)^2}{(1/2)^2 (\sqrt{1+0} + \sqrt{1})} = \frac{1 + 1/2}{(1/4)(2)} = \frac{3/2}{1/2} = 3$.

(B) We evaluate the limit: $\lim _{x \rightarrow \infty} x^3 \left(\sqrt{x^2+\sqrt{1+x^4}}-x \sqrt{2}\right)$
Rationalizing the expression:
$= \lim _{x \rightarrow \infty} \frac{x^3 \left(x^2+\sqrt{1+x^4}-2 x^2\right)}{\sqrt{x^2+\sqrt{1+x^4}}+x \sqrt{2}} = \lim _{x \rightarrow \infty} \frac{x^3 \left(\sqrt{1+x^4}-x^2\right)}{\sqrt{x^2+\sqrt{1+x^4}}+x \sqrt{2}}$
Rationalizing the numerator again:
$= \lim _{x \rightarrow \infty} \frac{x^3 \left(1+x^4-x^4\right)}{\left(\sqrt{x^2+\sqrt{1+x^4}}+x \sqrt{2}\right) \left(\sqrt{1+x^4}+x^2\right)}$
Dividing by $x^3$ in the denominator:
$= \lim _{x \rightarrow \infty} \frac{x^3}{x \left(\sqrt{1+\sqrt{\frac{1}{x^4}+1}}+\sqrt{2}\right) \cdot x^2 \left(\sqrt{\frac{1}{x^4}+1}+1\right)}$
$= \lim _{x \rightarrow \infty} \frac{1}{\left(\sqrt{1+\sqrt{\frac{1}{x^4}+1}}+\sqrt{2}\right) \left(\sqrt{\frac{1}{x^4}+1}+1\right)}$
Substituting $x \rightarrow \infty$:
$= \frac{1}{(\sqrt{1+1}+\sqrt{2})(1+1)} = \frac{1}{(\sqrt{2}+\sqrt{2})(2)} = \frac{1}{2 \sqrt{2} \cdot 2} = \frac{1}{4 \sqrt{2}}$

(D) Let $f(x) = \frac{x}{|x| + x^2}$.
For the left-hand limit as $x \rightarrow 0^-$,we have $|x| = -x$:
$\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0} \frac{x}{-x + x^2} = \lim_{x \rightarrow 0} \frac{1}{-1 + x} = -1$.
For the right-hand limit as $x \rightarrow 0^+$,we have $|x| = x$:
$\lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0} \frac{x}{x + x^2} = \lim_{x \rightarrow 0} \frac{1}{1 + x} = 1$.
Since $\lim_{x \rightarrow 0^-} f(x) \neq \lim_{x \rightarrow 0^+} f(x)$,the limit does not exist.

(A) Let $p$ be the switch $S_1$ and $q$ be the switch $S_2$. The symbolic form of the given circuit is $(p \wedge \sim q) \vee (\sim p \wedge q) \vee (\sim p \wedge \sim q)$.
Simplifying the expression:
$= (p \wedge \sim q) \vee [\sim p \wedge (q \vee \sim q)]$
$= (p \wedge \sim q) \vee [\sim p \wedge t]$
$= (p \wedge \sim q) \vee \sim p$
$= \sim p \vee (p \wedge \sim q)$
$= (\sim p \vee p) \wedge (\sim p \vee \sim q)$
$= t \wedge (\sim p \vee \sim q)$
$= \sim p \vee \sim q$
The expression $\sim p \vee \sim q$ represents two switches $S_1'$ and $S_2'$ connected in parallel.

(D) Let $p$ : The payment will be made.
Let $q$ : The work is finished in time.
The given statement is a biconditional statement: $p \leftrightarrow q$,which is equivalent to $(p \rightarrow q) \wedge (q \rightarrow p)$.
The negation of $(p \leftrightarrow q)$ is $\sim(p \leftrightarrow q)$,which is equivalent to $(p \wedge \sim q) \vee (q \wedge \sim p)$.
This translates to: "The payment is made and the work is not finished in time,or the work is finished in time and the payment is not made."
Therefore,Option $(D)$ is correct.

(C) To find the negation of the statement $(p \wedge q) \rightarrow (\sim p \vee r)$,we use the logical equivalence $\sim(A \rightarrow B) \equiv A \wedge \sim B$.
Let $A = (p \wedge q)$ and $B = (\sim p \vee r)$.
Then,$\sim(A \rightarrow B) \equiv (p \wedge q) \wedge \sim(\sim p \vee r)$.
Applying De Morgan's Law,$\sim(\sim p \vee r) \equiv \sim(\sim p) \wedge \sim r \equiv p \wedge \sim r$.
So,the expression becomes $(p \wedge q) \wedge (p \wedge \sim r)$.
By the Associative and Idempotent Laws,$(p \wedge q) \wedge p \wedge \sim r \equiv p \wedge q \wedge \sim r$.

(B) Given: $p = T, q = T, r = F, s = F$.
$a: \sim(p \wedge \sim r) \vee(\sim q \vee s)$
$\equiv \sim(T \wedge \sim F) \vee(\sim T \vee F)$
$\equiv \sim(T \wedge T) \vee(F \vee F)$
$\equiv \sim(T) \vee(F)$
$\equiv F \vee F \equiv F$
$b: (\sim q \wedge \sim r) \leftrightarrow(p \vee s)$
$\equiv (\sim T \wedge \sim F) \leftrightarrow(T \vee F)$
$\equiv (F \wedge T) \leftrightarrow(T)$
$\equiv F \leftrightarrow T \equiv F$
$c: (\sim p \vee q) \rightarrow(r \wedge \sim s)$
$\equiv (\sim T \vee T) \rightarrow(F \wedge \sim F)$
$\equiv (F \vee T) \rightarrow(F \wedge T)$
$\equiv T \rightarrow F \equiv F$
Thus,the truth values are $F, F, F$.

(C) We analyze the logical expression $[p$ $\rightarrow (q$ $\rightarrow r)] \leftrightarrow [(p \wedge q)$ $\rightarrow r]$.
First,consider the left side: $p$ $\rightarrow (q$ $\rightarrow r)$.
Using the implication law $A \rightarrow B \equiv \neg A \vee B$,we get:
$p$ $\rightarrow (q$ $\rightarrow r) \equiv \neg p \vee (\neg q \vee r)$.
By the associative law,this is equivalent to $(\neg p \vee \neg q) \vee r$.
By De Morgan's law,$\neg p \vee \neg q \equiv \neg (p \wedge q)$.
Thus,$p$ $\rightarrow (q$ $\rightarrow r) \equiv \neg (p \wedge q) \vee r$.
Again,using the implication law,$\neg (p \wedge q) \vee r \equiv (p \wedge q) \rightarrow r$.
Since both sides of the biconditional are logically equivalent,the expression $[p$ $\rightarrow (q$ $\rightarrow r)] \leftrightarrow [(p \wedge q)$ $\rightarrow r]$ is always true.
Therefore,it is a tautology.

(A) We need to find the negation of the statement $\sim s \vee (\sim r \wedge s)$.
Applying the negation operator:
$\sim (\sim s \vee (\sim r \wedge s))$
Using De Morgan's law,$\sim (p \vee q) \equiv \sim p \wedge \sim q$:
$\equiv s \wedge \sim (\sim r \wedge s)$
Using De Morgan's law again,$\sim (p \wedge q) \equiv \sim p \vee \sim q$:
$\equiv s \wedge (r \vee \sim s)$
Using the Distributive law,$p \wedge (q \vee r) \equiv (p \wedge q) \vee (p \wedge r)$:
$\equiv (s \wedge r) \vee (s \wedge \sim s)$
Using the Complement law,$s \wedge \sim s \equiv F$:
$\equiv (s \wedge r) \vee F$
Using the Identity law,$p \vee F \equiv p$:
$\equiv s \wedge r$

(A) Let the points on the two lines be $P_1 = (\lambda + 3, 3\lambda - 1, -\lambda + 6)$ and $P_2 = (7\alpha - 5, -6\alpha + 2, 4\alpha + 3)$.
For the lines to intersect at point $R$,we equate the coordinates:
$\lambda + 3 = 7\alpha - 5 \Rightarrow \lambda - 7\alpha = -8$ (Equation $1$)
$3\lambda - 1 = -6\alpha + 2 \Rightarrow 3\lambda + 6\alpha = 3 \Rightarrow \lambda + 2\alpha = 1$ (Equation $2$)
Subtracting Equation $1$ from Equation $2$ gives $9\alpha = 9$,so $\alpha = 1$.
Substituting $\alpha = 1$ into Equation $2$,we get $\lambda + 2(1) = 1$,so $\lambda = -1$.
Substituting $\lambda = -1$ into the first line equation,we get $R = (-1 + 3, 3(-1) - 1, -(-1) + 6) = (2, -4, 7)$.
The reflection of a point $(x, y, z)$ in the $xy$-plane is $(x, y, -z)$.
Therefore,the reflection of $R(2, -4, 7)$ in the $xy$-plane is $(2, -4, -7)$.

(B) Given $\frac{f'(x)}{f(x)} = 1$ for all $x \in R$.
Integrating both sides with respect to $x$,we get $\ln|f(x)| = x + C$,which implies $f(x) = Ae^x$.
Using the condition $f(1) = 2$,we have $Ae^1 = 2$,so $A = 2e^{-1}$.
Thus,$f(x) = 2e^{-1} \cdot e^x = 2e^{x-1}$.
Consequently,$f'(x) = 2e^{x-1}$.
Given $h(x) = f(f(x))$,by the chain rule,$h'(x) = f'(f(x)) \cdot f'(x)$.
At $x = 1$,$h'(1) = f'(f(1)) \cdot f'(1)$.
Since $f(1) = 2$,we have $h'(1) = f'(2) \cdot f'(1)$.
Substituting the values,$f'(2) = 2e^{2-1} = 2e$ and $f'(1) = 2e^{1-1} = 2$.
Therefore,$h'(1) = (2e) \cdot (2) = 4e$.

(D) We need to evaluate $f(3) - f(1) = \int_{1}^{3} \frac{\sqrt{x}}{(1+x)^2} dx$.
Let $\sqrt{x} = t$,then $x = t^2$ and $dx = 2t dt$.
When $x=1$,$t=1$. When $x=3$,$t=\sqrt{3}$.
Substituting these into the integral:
$f(3) - f(1) = \int_{1}^{\sqrt{3}} \frac{t \cdot 2t}{(1+t^2)^2} dt = 2 \int_{1}^{\sqrt{3}} \frac{t^2}{(1+t^2)^2} dt$.
Using integration by parts,let $u = t$ and $dv = \frac{t}{(1+t^2)^2} dt$. Then $du = dt$ and $v = -\frac{1}{2(1+t^2)}$.
$2 \int \frac{t^2}{(1+t^2)^2} dt = 2 \left[ -\frac{t}{2(1+t^2)} + \int \frac{1}{2(1+t^2)} dt \right] = -\frac{t}{1+t^2} + \tan^{-1}(t)$.
Evaluating from $1$ to $\sqrt{3}$:
$f(3) - f(1) = \left[ -\frac{\sqrt{3}}{1+3} + \tan^{-1}(\sqrt{3}) \right] - \left[ -\frac{1}{1+1} + \tan^{-1}(1) \right]$.
$= \left( -\frac{\sqrt{3}}{4} + \frac{\pi}{3} \right) - \left( -\frac{1}{2} + \frac{\pi}{4} \right) = \frac{1}{2} - \frac{\sqrt{3}}{4} + \frac{\pi}{12}$.

(A) Let $\vec{u} = \overrightarrow{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\vec{v} = \overrightarrow{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$.
First,calculate the magnitudes: $|\vec{u}| = \sqrt{2^2 + 10^2 + 11^2} = \sqrt{4 + 100 + 121} = \sqrt{225} = 15$.
$|\vec{v}| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
Since $AD'$ is in the plane of the parallelogram and perpendicular to $AB$,it must be in the direction of the projection of $AD$ onto the plane perpendicular to $AB$ within the parallelogram plane.
The vector $\vec{AD'}$ is the component of $\vec{AD}$ perpendicular to $\vec{AB}$ in the plane. Specifically,$\vec{AD'} = \vec{AD} - \text{proj}_{\vec{AB}}(\vec{AD}) = \vec{AD} - \frac{\vec{AD} \cdot \vec{AB}}{|\vec{AB}|^2} \vec{AB}$.
Calculate $\vec{AD} \cdot \vec{AB} = (-1)(2) + (2)(10) + (2)(11) = -2 + 20 + 22 = 40$.
So,$\vec{AD'} = \vec{AD} - \frac{40}{225} \vec{AB} = \vec{AD} - \frac{8}{45} \vec{AB}$.
Alternatively,using the property of rotation,$\cos \alpha = \frac{|\vec{AD} \cdot \vec{AB}|}{|\vec{AD}| |\vec{AB}|} = \frac{40}{3 \times 15} = \frac{40}{45} = \frac{8}{9}$.

(A) Given the curve is $xy + ax + by = 0$.
Since the point $(1, 1)$ lies on the curve,we substitute $x = 1$ and $y = 1$:
$1(1) + a(1) + b(1) = 0 \implies a + b = -1$ ... $(i)$
Now,differentiate the equation $xy + ax + by = 0$ with respect to $x$:
$x \frac{dy}{dx} + y + a + b \frac{dy}{dx} = 0$
$(x + b) \frac{dy}{dx} = -(y + a)$
$\frac{dy}{dx} = -\frac{y + a}{x + b}$
Given that the slope at $(1, 1)$ is $2$:
$2 = -\frac{1 + a}{1 + b}$
$2(1 + b) = -(1 + a)$
$2 + 2b = -1 - a$
$a + 2b = -3$ ... $(ii)$
Subtracting equation $(i)$ from equation $(ii)$:
$(a + 2b) - (a + b) = -3 - (-1)$
$b = -2$
Substituting $b = -2$ into equation $(i)$:
$a - 2 = -1 \implies a = 1$
Therefore,$a - b = 1 - (-2) = 3$.

(D) Given curve is $y=\sqrt{9-2x^2}$.
If ordinate and abscissa are equal,then $y=x$.
Substituting $y=x$ in the curve equation: $x^2 = 9 - 2x^2 \Rightarrow 3x^2 = 9 \Rightarrow x^2 = 3 \Rightarrow x = \pm \sqrt{3}$.
Since $y = \sqrt{9-2x^2}$,$y$ must be positive. Thus,$x$ must be $\sqrt{3}$ because if $x = -\sqrt{3}$,then $y = \sqrt{3}$,but $x \neq y$.
So,the point of contact is $(\sqrt{3}, \sqrt{3})$.
Differentiating $y^2 = 9 - 2x^2$ with respect to $x$: $2y \frac{dy}{dx} = -4x \Rightarrow \frac{dy}{dx} = -\frac{2x}{y}$.
At $(\sqrt{3}, \sqrt{3})$,the slope $m = -\frac{2(\sqrt{3})}{\sqrt{3}} = -2$.
The equation of the tangent is $y - \sqrt{3} = -2(x - \sqrt{3})$.
$y - \sqrt{3} = -2x + 2\sqrt{3} \Rightarrow 2x + y - 3\sqrt{3} = 0$.

(C) Given that the slope of the tangent to the curve $y=f(x)$ is $\frac{dy}{dx} = 2x+1$.
Integrating both sides with respect to $x$,we get:
$y = \int (2x+1) dx = x^2 + x + c$
Since the curve passes through the point $(1, 2)$,we substitute $x=1$ and $y=2$ into the equation:
$2 = (1)^2 + 1 + c$
$2 = 1 + 1 + c$
$2 = 2 + c \Rightarrow c = 0$
Thus,the equation of the curve is $y = x^2 + x$.
The area bounded by the curve,the $X$-axis,and the line $x=1$ is the integral from $x=0$ to $x=1$ (since the curve intersects the $X$-axis at $x^2+x=0$,i.e.,$x(x+1)=0$,giving $x=0$ and $x=-1$):
Area $= \int_0^1 (x^2+x) dx$
$= \left[ \frac{x^3}{3} + \frac{x^2}{2} \right]_0^1$
$= \left( \frac{1^3}{3} + \frac{1^2}{2} \right) - (0)$
$= \frac{1}{3} + \frac{1}{2} = \frac{2+3}{6} = \frac{5}{6} \text{ sq. units}$.

(A) Given curves are $y=2x^2$ and $x=2y^2$.
For the curve $y=2x^2$,the slope of the tangent is $\frac{dy}{dx} = 4x$.
At the point $(1,1)$,the slope $m_1 = 4(1) = 4$.
For the curve $x=2y^2$,differentiating with respect to $x$,we get $1 = 4y \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{1}{4y}$.
At the point $(1,1)$,the slope $m_2 = \frac{1}{4(1)} = \frac{1}{4}$.
Let $\theta$ be the angle between the two tangents. The formula for the angle is $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values,$\tan \theta = \left| \frac{4 - 1/4}{1 + 4(1/4)} \right| = \left| \frac{15/4}{1 + 1} \right| = \frac{15/4}{2} = \frac{15}{8}$.
Therefore,$\theta = \tan^{-1}\left(\frac{15}{8}\right)$.

(C) Given the parametric equations of the curve are $x = \sqrt{t}$ and $y = t - \frac{1}{\sqrt{t}}$.
First,we find the derivatives with respect to $t$:
$\frac{dx}{dt} = \frac{d}{dt}(t^{1/2}) = \frac{1}{2\sqrt{t}}$
$\frac{dy}{dt} = \frac{d}{dt}(t - t^{-1/2}) = 1 - (-\frac{1}{2})t^{-3/2} = 1 + \frac{1}{2t^{3/2}}$
Now,the slope of the tangent $\frac{dy}{dx}$ is given by $\frac{dy/dt}{dx/dt}$:
$\frac{dy}{dx} = \frac{1 + \frac{1}{2t^{3/2}}}{\frac{1}{2\sqrt{t}}} = (1 + \frac{1}{2t^{3/2}}) \times (2\sqrt{t}) = 2\sqrt{t} + \frac{2\sqrt{t}}{2t^{3/2}} = 2\sqrt{t} + \frac{1}{t} = \frac{2t\sqrt{t} + 1}{t}$
At $t=4$,the slope of the tangent is:
$\left(\frac{dy}{dx}\right)_{t=4} = \frac{2(4)\sqrt{4} + 1}{4} = \frac{2(4)(2) + 1}{4} = \frac{16+1}{4} = \frac{17}{4}$
The slope of the normal is the negative reciprocal of the slope of the tangent:
$\text{Slope of normal} = -\frac{1}{(\frac{dy}{dx})_{t=4}} = -\frac{1}{17/4} = -\frac{4}{17}$

(B) Given the curve equation: $xy + ax + by = 0$.
Since the point $(1, 1)$ lies on the curve,we substitute $x = 1$ and $y = 1$:
$1(1) + a(1) + b(1) = 0 \implies 1 + a + b = 0 \implies a + b = -1$ ... $(i)$
Now,differentiate the equation with respect to $x$:
$y + x \frac{dy}{dx} + a + b \frac{dy}{dx} = 0$
$\frac{dy}{dx}(x + b) = -(y + a)$
$\frac{dy}{dx} = -\frac{y + a}{x + b}$
Given the slope at $(1, 1)$ is $2$:
$2 = -\frac{1 + a}{1 + b}$
$2(1 + b) = -(1 + a)$
$2 + 2b = -1 - a $
$a + 2b = -3 ... (ii)$
Subtracting $(i)$ from $(ii)$:
$(a + 2b) - (a + b) = -3 - (-1)$
$b = -2$
Substituting $b = -2$ into $(i)$:
$a - 2 = -1 \implies a = 1$
Finally, calculate $3a + b$:
$3(1) + (-2) = 3 - 2 = 1$.

(B) $y = 2 e^x \sin \left(\frac{\pi}{4} - \frac{x}{2}\right) \cos \left(\frac{\pi}{4} - \frac{x}{2}\right)$
Using the identity $2 \sin \theta \cos \theta = \sin 2 \theta$,we get:
$y = e^x \sin \left(2 \left(\frac{\pi}{4} - \frac{x}{2}\right)\right) = e^x \sin \left(\frac{\pi}{2} - x\right) = e^x \cos x$
The slope of the tangent is given by $\frac{dy}{dx}$:
$\frac{dy}{dx} = e^x \cos x + e^x (-\sin x) = e^x (\cos x - \sin x)$
Let $T(x) = e^x (\cos x - \sin x)$. To find the minimum,we find $\frac{dT}{dx}$:
$\frac{dT}{dx} = e^x (\cos x - \sin x) + e^x (-\sin x - \cos x) = e^x (\cos x - \sin x - \sin x - \cos x) = -2 e^x \sin x$
Setting $\frac{dT}{dx} = 0$ for critical points:
$-2 e^x \sin x = 0 \implies \sin x = 0$
In the interval $0 \leq x \leq 2 \pi$,$x = 0, \pi, 2 \pi$.
Evaluating $T(x)$ at these points:
$T(0) = e^0 (\cos 0 - \sin 0) = 1$
$T(\pi) = e^\pi (\cos \pi - \sin \pi) = -e^\pi$
$T(2 \pi) = e^{2 \pi} (\cos 2 \pi - \sin 2 \pi) = e^{2 \pi}$
Comparing the values,the minimum value is $-e^\pi$ at $x = \pi$.

(B) The slope of the chord $AB$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - (-3)}{4 - 1} = \frac{6}{3} = 2$.
Since the tangent is parallel to the chord $AB$,the slope of the tangent must also be $2$.
Given the curve $y = x - \frac{4}{x}$,we find the derivative $\frac{dy}{dx} = 1 + \frac{4}{x^2}$.
Setting the derivative equal to the slope of the chord: $1 + \frac{4}{x^2} = 2$.
This simplifies to $\frac{4}{x^2} = 1$,which implies $x^2 = 4$,so $x = \pm 2$.
For $x = 2$,$y = 2 - \frac{4}{2} = 0$.
For $x = -2$,$y = -2 - \frac{4}{-2} = -2 + 2 = 0$.
Thus,the required points are $(2, 0)$ and $(-2, 0)$.

(C) Given the parametric equations of the curve are $x=2(\cos t+t \sin t)$ and $y=2(\sin t-t \cos t)$.
First,we find the derivatives with respect to $t$:
$\frac{dx}{dt} = 2(-\sin t + \sin t + t \cos t) = 2t \cos t$
$\frac{dy}{dt} = 2(\cos t - \cos t + t \sin t) = 2t \sin t$
Now,the slope of the tangent is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t \sin t}{2t \cos t} = \tan t$.
The slope of the normal is $-\frac{1}{dy/dx} = -\frac{1}{\tan t} = -\frac{\cos t}{\sin t}$.
The equation of the normal at point '$t$' is given by $y - y_1 = m_n(x - x_1)$:
$y - 2(\sin t - t \cos t) = -\frac{\cos t}{\sin t} [x - 2(\cos t + t \sin t)]$
Multiplying by $\sin t$:
$y \sin t - 2 \sin^2 t + 2t \sin t \cos t = -x \cos t + 2 \cos^2 t + 2t \sin t \cos t$
$x \cos t + y \sin t = 2(\sin^2 t + \cos^2 t)$
$x \cos t + y \sin t = 2$
The distance of the line $Ax + By + C = 0$ from the origin $(0,0)$ is given by $\frac{|C|}{\sqrt{A^2 + B^2}}$.
Here,$A = \cos t$,$B = \sin t$,and $C = -2$.
Distance $= \frac{|-2|}{\sqrt{\cos^2 t + \sin^2 t}} = \frac{2}{1} = 2$ units.

(C) $y^2=px^3+q$ ... $(i)$
Differentiating both sides with respect to $x$,we get
$2y \cdot \frac{dy}{dx} = 3px^2$
$\frac{dy}{dx} = \frac{3px^2}{2y}$
At the point $(2,3)$,the slope of the tangent is:
$\left(\frac{dy}{dx}\right)_{(2,3)} = \frac{3p(2)^2}{2(3)} = \frac{12p}{6} = 2p$
The slope of the line $y=4x-5$ is $4$.
Since the line is tangent to the curve,their slopes must be equal:
$2p = 4 \Rightarrow p = 2$
Since the point $(2,3)$ lies on the curve $y^2=px^3+q$,we substitute the values:
$3^2 = p(2)^3 + q$
$9 = 8p + q$
Substituting $p=2$:
$9 = 8(2) + q$
$9 = 16 + q \Rightarrow q = -7$
Therefore,$p-q = 2 - (-7) = 2 + 7 = 9$.

(C) The slope of the tangent is given by $\frac{dy}{dx} = -y + e^{-x}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 1$ and $Q = e^{-x}$.
The Integrating Factor ($I$.$F$.) is $e^{\int P dx} = e^{\int 1 dx} = e^x$.
The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$.
$y e^x = \int e^{-x} \cdot e^x dx + C$.
$y e^x = \int 1 dx + C$.
$y e^x = x + C$.
Since the curve passes through the origin $(0,0)$,we substitute $x=0$ and $y=0$:
$0 \cdot e^0 = 0 + C \Rightarrow C = 0$.
Thus,the equation of the curve is $y e^x = x$,which can be written as $y e^x - x = 0$.

(B) The equation of the circle is $x^2 + y^2 = 1$.
Differentiating both sides with respect to time $t$,we get:
$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$
$x \frac{dx}{dt} + y \frac{dy}{dt} = 0$
Given that the $y$-coordinate is decreasing at the rate of $3 \text{ units/sec}$,we have $\frac{dy}{dt} = -3 \text{ units/sec}$.
At the point $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$,we substitute $x = \frac{1}{2}$,$y = \frac{\sqrt{3}}{2}$,and $\frac{dy}{dt} = -3$:
$\frac{1}{2} \frac{dx}{dt} + \left(\frac{\sqrt{3}}{2}\right)(-3) = 0$
$\frac{1}{2} \frac{dx}{dt} = \frac{3\sqrt{3}}{2}$
$\frac{dx}{dt} = 3\sqrt{3} \text{ units/sec}$.
Thus,the $x$-coordinate is increasing at the rate of $3\sqrt{3} \text{ units/sec}$.

(D) Let $f(x) = \log_{10} x = \frac{\log_e x}{\log_e 10} = (\log_{10} e)(\log_e x) = 0.4343(\log_e x)$.
On differentiating with respect to $x$,we get $f'(x) = \frac{0.4343}{x}$.
Let $x = 998 = 1000 - 2 = a + h$.
Here,$a = 1000$ and $h = -2$.
$f(a) = f(1000) = \log_{10}(1000) = 3 \log_{10} 10 = 3$.
Also,$f'(a) = f'(1000) = \frac{0.4343}{1000} = 0.0004343$.
Using the approximation formula $f(a+h) \approx f(a) + hf'(a)$:
$\log_{10}(998) \approx 3 + (-2)(0.0004343) = 3 - 0.0008686 = 2.9991314$.

(B) Let $P$ be the position of the kite and $PR$ be the string. Let $PQ = 120 \ m$ be the constant height.
Let $QR = x$ and $PR = y$.
By the Pythagoras theorem in $\triangle PQR$:
$y^2 = (120)^2 + x^2 \dots (i)$
Differentiating with respect to time $t$,we get:
$2y \frac{dy}{dt} = 2x \frac{dx}{dt}$
$y \frac{dy}{dt} = x \frac{dx}{dt} \dots (ii)$
Given that the kite is moving away horizontally at the rate of $\frac{dx}{dt} = 39 \ m/sec$.
From $(i)$,when $y = 130 \ m$:
$(130)^2 = (120)^2 + x^2$
$x^2 = 16900 - 14400 = 2500$
$x = 50 \ m$
Substituting the values into $(ii)$:
$130 \frac{dy}{dt} = 50 \times 39$
$\frac{dy}{dt} = \frac{50 \times 39}{130} = \frac{1950}{130} = 15 \ m/sec$.
Thus,the rate at which the string is being paid out is $15 \ m/sec$.

(B) Let the ladder be $AC = 17 \,m$. Let the wall be $AB$ of height $x$ and the ground be $BC$ of length $y$.
In $\triangle ABC$, by Pythagoras theorem:
$x^2 + y^2 = 17^2 = 289$
Given that the lower end slips away at the rate of $\frac{dy}{dt} = 1 \,m/sec$.
When $y = 8 \,m$, we have $x^2 + 8^2 = 289 \Rightarrow x^2 = 289 - 64 = 225 \Rightarrow x = 15 \,m$.
Differentiating $x^2 + y^2 = 289$ with respect to time $t$:
$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$
$x \frac{dx}{dt} + y \frac{dy}{dt} = 0$
Substituting the values $x = 15$, $y = 8$, and $\frac{dy}{dt} = 1$:
$15 \frac{dx}{dt} + 8(1) = 0$
$15 \frac{dx}{dt} = -8$
$\frac{dx}{dt} = -\frac{8}{15} \,m/sec$.
The negative sign indicates that the height $x$ is decreasing. Therefore, the upper end is coming down at the rate of $\frac{8}{15} \,m/sec$.

(D) Let $A$, $P$, and $X$ be the area, perimeter, and side length of the square at time $t$ seconds, respectively.
Then, $A = X^2$ and $P = 4X$.
From these, we have $P = 4 \sqrt{A}$.
Differentiating both sides with respect to $t$, we get:
$\frac{dP}{dt} = 4 \cdot \frac{1}{2 \sqrt{A}} \cdot \frac{dA}{dt} = \frac{2}{\sqrt{A}} \cdot \frac{dA}{dt}$.
Since $A = X^2$, we have $\sqrt{A} = X$. Thus, $\frac{dP}{dt} = \frac{2}{X} \cdot \frac{dA}{dt}$.
Given that the area is contracting at a rate of $4 \,cm^2 / sec$, we have $\frac{dA}{dt} = -4 \,cm^2 / sec$.
At $X = 20 \,cm$:
$\frac{dP}{dt} = \frac{2}{20} \times (-4) = -\frac{8}{20} = -\frac{2}{5} \,cm / sec$.
The negative sign indicates that the perimeter is decreasing at a rate of $\frac{2}{5} \,cm / sec$.

(B) The surface area of a sphere is given by $S = 4 \pi r^2$.
Differentiating with respect to time $t$,we get $\frac{dS}{dt} = 8 \pi r \frac{dr}{dt}$.
Given $\frac{dS}{dt} = 2 \text{ cm}^2/\text{sec}$ and $r = 6 \text{ cm}$,we have $2 = 8 \pi (6) \frac{dr}{dt}$.
Thus,$\frac{dr}{dt} = \frac{2}{48 \pi} = \frac{1}{24 \pi} \text{ cm/sec}$.
The volume of a sphere is $V = \frac{4}{3} \pi r^3$.
Differentiating with respect to time $t$,we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
Substituting the values,$\frac{dV}{dt} = 4 \pi (6)^2 \times \frac{1}{24 \pi} = 4 \pi \times 36 \times \frac{1}{24 \pi} = \frac{144 \pi}{24 \pi} = 6 \text{ cm}^3/\text{sec}$.

(B) Let $x$ be the depth of water at time $t$. Given that the rate of flow is proportional to the square root of the depth,we have $\frac{dx}{dt} = -k\sqrt{x}$,where $k > 0$ is a constant (negative sign indicates decreasing depth).
Separating the variables,we get $\frac{dx}{\sqrt{x}} = -k dt$.
Integrating both sides,we get $\int x^{-1/2} dx = \int -k dt$,which gives $2\sqrt{x} = -kt + C$.
At $t = 0$,the initial depth $x = 16 \ m$. Substituting these values,$2\sqrt{16} = -k(0) + C \Rightarrow C = 8$.
So,the equation becomes $2\sqrt{x} = -kt + 8$.
At $t = 2 \ hours$,the depth $x = 4 \ m$. Substituting these values,$2\sqrt{4} = -k(2) + 8 \Rightarrow 4 = -2k + 8 \Rightarrow 2k = 4 \Rightarrow k = 2$.
Thus,the equation is $2\sqrt{x} = -2t + 8$,or $\sqrt{x} = 4 - t$.
For $t = 3.5 \ hours$,$\sqrt{x} = 4 - 3.5 = 0.5$.
Squaring both sides,$x = (0.5)^2 = 0.25 \ m$.

(B) Let the length and breadth of the tank be $x \ m$ and $y \ m$ respectively. The height of the tank is $h = 4 \ m$.
The volume of the tank is $V = x \times y \times h = 36 \ m^3$.
Substituting $h = 4$,we get $4xy = 36$,which implies $xy = 9$,so $y = \frac{9}{x}$.
The cost function $C$ is given by the cost of the base plus the cost of the four sides:
$C = 100(xy) + 50(2xh + 2yh)$
Substituting $xy = 9$,$h = 4$,and $y = \frac{9}{x}$:
$C(x) = 100(9) + 50(2x(4) + 2(\frac{9}{x})(4))$
$C(x) = 900 + 50(8x + \frac{72}{x}) = 900 + 400x + \frac{3600}{x}$
To find the minimum cost,we find the derivative $C'(x)$ and set it to zero:
$C'(x) = 400 - \frac{3600}{x^2} = 0$
$400 = \frac{3600}{x^2} \Rightarrow x^2 = 9 \Rightarrow x = 3 \ m$.
Since $x = 3$,$y = \frac{9}{3} = 3 \ m$.
The minimum cost is $C(3) = 900 + 400(3) + \frac{3600}{3} = 900 + 1200 + 1200 = ₹ 3300$.

(B) Let the height and breadth of the sheet be $y \ m$ and $x \ m$ respectively.
Given area of the sheet is $18 \ m^2$,so $x y = 18$.
Converting margins to meters: top/bottom margins are $0.75 \ m$ each,side margins are $0.5 \ m$ each.
The dimensions of the printable area are $(y - 1.5) \ m$ and $(x - 1) \ m$.
The area available for printing is $A = (y - 1.5)(x - 1)$.
Since $y = \frac{18}{x}$,we have $A = (\frac{18}{x} - 1.5)(x - 1) = 18 - \frac{18}{x} - 1.5x + 1.5 = 19.5 - \frac{18}{x} - 1.5x$.
To maximize $A$,differentiate with respect to $x$: $\frac{dA}{dx} = \frac{18}{x^2} - 1.5$.
Setting $\frac{dA}{dx} = 0$,we get $\frac{18}{x^2} = 1.5 \Rightarrow x^2 = \frac{18}{1.5} = 12$.
Thus,$x = \sqrt{12} = 2 \sqrt{3} \ m$.
Then $y = \frac{18}{2 \sqrt{3}} = \frac{9}{\sqrt{3}} = 3 \sqrt{3} \ m$.
Checking the second derivative: $\frac{d^2A}{dx^2} = -\frac{36}{x^3}$,which is negative at $x = 2 \sqrt{3}$,confirming a maximum.
Therefore,the height is $3 \sqrt{3} \ m$ and the breadth is $2 \sqrt{3} \ m$.

(B) For the conical vessel, height $H = 9 \text{ cm}$ and base radius $R = 5 \text{ cm}$.
Full volume of the vessel is $V_{total} = \frac{1}{3} \pi R^2 H = \frac{1}{3} \pi (25)(9) = 75\pi \text{ cm}^3$.
Let $h$ be the height of water and $r$ be the radius of the water surface at time $t$.
By similar triangles, $\frac{r}{h} = \frac{R}{H} = \frac{5}{9}$, so $r = \frac{5h}{9}$.
The area of the water surface is $A = \pi r^2 = \pi \left(\frac{5h}{9}\right)^2 = \frac{25\pi h^2}{81}$.
Given the rate of change of water level is $\frac{dh}{dt} = \frac{\pi}{A} = \frac{\pi}{\frac{25\pi h^2}{81}} = \frac{81}{25h^2}$.
Separating variables, $h^2 \, dh = \frac{81}{25} \, dt$.
Integrating both sides, $\int h^2 \, dh = \int \frac{81}{25} \, dt \implies \frac{h^3}{3} = \frac{81}{25}t + C$.
Since $h=0$ at $t=0$, we have $C=0$, so $h^3 = \frac{243}{25}t$.
The volume of water at height $h$ is $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \left(\frac{25h^2}{81}\right) h = \frac{25\pi h^3}{243}$.
Substituting $h^3 = \frac{243}{25}t$, we get $V = \frac{25\pi}{243} \left(\frac{243}{25}t\right) = \pi t$.
For the cone to be completely filled, $V = V_{total} = 75\pi$.
Therefore, $\pi t = 75\pi \implies t = 75 \text{ seconds}$.

(C) Radius of the hemispherical bowl $R = 180 \text{ cm}$.
Rate of flow $\frac{dV}{dt} = 108 \text{ dm}^3/\text{min}$.
Since $1 \text{ dm} = 10 \text{ cm}$,$1 \text{ dm}^3 = 1000 \text{ cm}^3$.
$\frac{dV}{dt} = 108 \times 1000 \text{ cm}^3 / 60 \text{ sec} = 1800 \text{ cm}^3/\text{sec}$.
Let the depth of water be $x$. The volume of water in the hemispherical bowl is given by $V = \frac{\pi}{3} x^2(3R - x)$.
Substituting $R = 180$,$V = \frac{\pi}{3} x^2(540 - x) = 180 \pi x^2 - \frac{\pi}{3} x^3$.
Differentiating with respect to $t$,we get $\frac{dV}{dt} = (360 \pi x - \pi x^2) \frac{dx}{dt}$.
At $x = 120 \text{ cm}$,$1800 = (360 \pi(120) - \pi(120)^2) \frac{dx}{dt}$.
$1800 = (43200 \pi - 14400 \pi) \frac{dx}{dt} = 28800 \pi \frac{dx}{dt}$.
$\frac{dx}{dt} = \frac{1800}{28800 \pi} = \frac{18}{288 \pi} = \frac{1}{16 \pi} \text{ cm/sec}$.

(A) Let $A = (h, 2h)$.
The distance $OA = \sqrt{h^2 + (2h)^2} = \sqrt{5}h$.
Given that the point $A$ moves at a rate of $2 \text{ units/sec}$,we have $\frac{d(OA)}{dt} = 2$.
Thus,$\sqrt{5} \frac{dh}{dt} = 2$,which implies $\frac{dh}{dt} = \frac{2}{\sqrt{5}}$.
The area $\alpha$ of $\triangle ABC$ with vertices $A(h, 2h)$,$B(0, 3)$,and $C(4, 0)$ is given by the determinant formula:
$\alpha = \frac{1}{2} |h(3-0) + 0(0-2h) + 4(2h-3)| = \frac{1}{2} |3h + 8h - 12| = \frac{1}{2} |11h - 12|$.
Assuming the area is increasing,we consider $\alpha = \frac{11h - 12}{2}$.
Differentiating with respect to $t$: $\frac{d\alpha}{dt} = \frac{11}{2} \frac{dh}{dt}$.
Substituting $\frac{dh}{dt} = \frac{2}{\sqrt{5}}$,we get $\frac{d\alpha}{dt} = \frac{11}{2} \cdot \frac{2}{\sqrt{5}} = \frac{11}{\sqrt{5}} \text{ (units)}^2/\text{sec}$.

(B) Using the Taylor series expansion of $P(x)$ about $x=2$:
$P(x) = P(2) + P^{\prime}(2)(x-2) + \frac{P^{\prime \prime}(2)}{2!}(x-2)^2$
Given $P(2)=-1, P^{\prime}(2)=0, P^{\prime \prime}(2)=2$:
$P(x) = -1 + 0(x-2) + \frac{2}{2}(x-2)^2$
$P(x) = -1 + (x-2)^2$
Now,substitute $x=1.001$:
$P(1.001) = -1 + (1.001-2)^2$
$P(1.001) = -1 + (-0.999)^2$
$P(1.001) = -1 + 0.998001$
$P(1.001) = -0.001999$
Rounding to the nearest provided option,we get $-0.002$.

(B) Let $x$ be the length of the diagonal of the square and $A$ be its area.
Given that the rate of change of the diagonal is $\frac{dx}{dt} = 0.5 \text{ cm/sec}$.
The area of a square in terms of its diagonal $x$ is given by $A = \frac{x^2}{2}$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dA}{dt} = \frac{d}{dt} \left( \frac{x^2}{2} \right) = \frac{2x}{2} \cdot \frac{dx}{dt} = x \cdot \frac{dx}{dt}$.
Given that the area $A = 400 \text{ cm}^2$,we find the diagonal $x$ using $A = \frac{x^2}{2}$:
$400 = \frac{x^2}{2} \implies x^2 = 800 \implies x = \sqrt{800} = 20\sqrt{2} \text{ cm}$.
Now,substitute $x = 20\sqrt{2}$ and $\frac{dx}{dt} = 0.5$ into the expression for $\frac{dA}{dt}$:
$\frac{dA}{dt} = (20\sqrt{2}) \cdot (0.5) = 10\sqrt{2} \text{ cm}^2/\text{sec}$.

(A) Given $f(x) = x e^{x(1-x)}$.
Applying the product rule,$f'(x) = x \cdot e^{x(1-x)} \cdot \frac{d}{dx}(x-x^2) + e^{x(1-x)} \cdot 1$.
$f'(x) = x e^{x(1-x)}(1-2x) + e^{x(1-x)}$.
$f'(x) = e^{x(1-x)} [x(1-2x) + 1] = e^{x(1-x)} (x - 2x^2 + 1)$.
$f'(x) = e^{x(1-x)} (-2x^2 + x + 1) = -e^{x(1-x)} (2x^2 - x - 1)$.
$f'(x) = -e^{x(1-x)} (2x+1)(x-1) = e^{x(1-x)} (2x+1)(1-x)$.
For $f(x)$ to be increasing,$f'(x) \geq 0$.
Since $e^{x(1-x)} > 0$ for all $x \in R$,we need $(2x+1)(1-x) \geq 0$.
This inequality holds when $x \in \left[-\frac{1}{2}, 1\right]$.
Thus,$f(x)$ is increasing in $\left[-\frac{1}{2}, 1\right]$.

(B) Given $f(x) = \sin^4 x + \cos^4 x$.
We can rewrite this as $f(x) = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}(2\sin x \cos x)^2 = 1 - \frac{1}{2}\sin^2 2x$.
Differentiating with respect to $x$,we get $f'(x) = -\frac{1}{2} \cdot 2\sin 2x \cdot \cos 2x \cdot 2 = -2\sin 2x \cos 2x = -\sin 4x$.
For $f(x)$ to be increasing,$f'(x) > 0$,so $-\sin 4x > 0$,which implies $\sin 4x < 0$.
The sine function is negative in the third and fourth quadrants,so $\pi < 4x < 2\pi$.
Dividing by $4$,we get $\frac{\pi}{4} < x < \frac{\pi}{2}$.

(B) Given $f(x) = x^3 + bx^2 + cx + d$.
Taking the derivative with respect to $x$,we get $f'(x) = 3x^2 + 2bx + c$.
For $f(x)$ to be strictly increasing,we need $f'(x) > 0$ for all $x \in \mathbb{R}$.
The quadratic expression $3x^2 + 2bx + c$ is always positive if its discriminant $D < 0$ and the coefficient of $x^2$ is positive.
Here,$D = (2b)^2 - 4(3)(c) = 4b^2 - 12c = 4(b^2 - 3c)$.
Since $0 < b^2 < c$,it follows that $b^2 - 3c < c - 3c = -2c < 0$ (assuming $c > 0$).
More directly,since $b^2 < c$,then $3b^2 < 3c$,so $b^2 - 3c < -2b^2 < 0$.
Thus,$D < 0$,which implies $f'(x) > 0$ for all $x \in \mathbb{R}$.
Therefore,$f(x)$ is a strictly increasing function.

(C) Given $f(x) = \int \frac{x^2-3x+2}{x^4+1} \, dx$.
By the Fundamental Theorem of Calculus,the derivative is $f'(x) = \frac{x^2-3x+2}{x^4+1}$.
For the function $f(x)$ to be decreasing,we must have $f'(x) < 0$.
Since $x^4+1 > 0$ for all real $x$,the condition $f'(x) < 0$ is equivalent to $x^2-3x+2 < 0$.
Factoring the quadratic expression,we get $(x-1)(x-2) < 0$.
The roots of the quadratic are $x=1$ and $x=2$.
Testing the intervals,the expression $(x-1)(x-2)$ is negative when $x \in (1, 2)$.
Therefore,the function decreases in the interval $(1, 2)$.

(D) Given the constraint $x+2y=8$.
We can express $y$ in terms of $x$ as $2y = 8-x$,which gives $y = \frac{8-x}{2}$.
Let the function to be maximized be $f(x) = xy$.
Substituting the value of $y$,we get $f(x) = x \cdot \frac{8-x}{2} = 4x - \frac{x^2}{2}$.
To find the maximum value,we differentiate $f(x)$ with respect to $x$:
$f'(x) = \frac{d}{dx}(4x - \frac{x^2}{2}) = 4 - x$.
Setting the derivative to zero for critical points:
$4 - x = 0 \implies x = 4$.
To confirm it is a maximum,we check the second derivative:
$f''(x) = -1$.
Since $f''(4) = -1 < 0$,the function has a local maximum at $x = 4$.
Substituting $x = 4$ back into the expression for $y$:
$y = \frac{8-4}{2} = 2$.
The maximum value of $xy$ is $4 \times 2 = 8$.

(C) Applying Lagrange's Mean Value Theorem on the interval $[0, 2]$,we know there exists at least one $c \in (0, 2)$ such that:
$\frac{f(2) - f(0)}{2 - 0} = f^{\prime}(c)$
$\therefore f(2) - f(0) = 2 f^{\prime}(c)$
$\therefore f(2) = f(0) + 2 f^{\prime}(c)$
Given $f^{\prime}(0) = -3$,but the problem implies we are evaluating the change from $0$ to $2$. Assuming $f(0)$ is a reference point,we analyze the inequality:
Since $f^{\prime}(x) \leq 5$ for all $x$,we have $f^{\prime}(c) \leq 5$.
Thus,$f(2) - f(0) = 2 f^{\prime}(c) \leq 2(5) = 10$.
If we consider the change relative to $f(0)$,then $f(2) \leq f(0) + 10$.
Given the specific constraint $f^{\prime}(0) = -3$ is a local condition,and using the Mean Value Theorem over $[0, 2]$,the maximum increase is $10$.
Therefore,the maximum value of $f(2)$ relative to $f(0)$ is $f(0) + 10$. Given the options provided and standard interpretation of such problems where $f(0)$ is often taken as $f(0) = -3$ (or similar context),the calculated maximum is $7$.

(A) First,we determine the set $S$ by solving the inequality $x^2 + 30 \leq 11x$.
$x^2 - 11x + 30 \leq 0$
$(x - 5)(x - 6) \leq 0$
Thus,$x \in [5, 6]$.
Now,consider the function $f(x) = 3x^3 - 18x^2 + 27x - 40$.
Find the derivative $f'(x)$:
$f'(x) = 9x^2 - 36x + 27 = 9(x^2 - 4x + 3) = 9(x - 1)(x - 3)$.
For $x \in [5, 6]$,both $(x - 1)$ and $(x - 3)$ are positive,so $f'(x) > 0$.
Since $f'(x) > 0$ for all $x \in [5, 6]$,the function $f(x)$ is strictly increasing on the interval $[5, 6]$.
The maximum value occurs at the right endpoint $x = 6$.
$f(6) = 3(6)^3 - 18(6)^2 + 27(6) - 40$
$f(6) = 3(216) - 18(36) + 162 - 40$
$f(6) = 648 - 648 + 162 - 40 = 122$.

(A) Given function: $f(x) = x^3 + 6x^2 - 36x + 7$
Find the derivative: $f'(x) = 3x^2 + 12x - 36$
Factor the derivative: $f'(x) = 3(x^2 + 4x - 12) = 3(x + 6)(x - 2)$
For the function to be increasing,we require $f'(x) > 0$:
$3(x + 6)(x - 2) > 0$
$(x + 6)(x - 2) > 0$
Using the sign scheme for the quadratic inequality,the expression is positive when $x < -6$ or $x > 2$.
Thus,the range of values is $x \in (-\infty, -6) \cup (2, \infty)$.

(C) Given equation is $f(x) = x^3+x-1 = 0$.
To determine the number of real roots,we analyze the derivative of the function $f(x)$.
The derivative is $f'(x) = 3x^2 + 1$.
Since $x^2 \geq 0$ for all $x \in \mathbb{R}$,it follows that $3x^2 + 1 \geq 1 > 0$ for all $x \in \mathbb{R}$.
Because $f'(x) > 0$ for all $x$,the function $f(x)$ is strictly increasing on the entire real line.
$A$ strictly increasing function can cross the $x$-axis at most once.
We observe the values of the function at specific points:
$f(0) = 0^3 + 0 - 1 = -1$
$f(1) = 1^3 + 1 - 1 = 1$
Since $f(0) < 0$ and $f(1) > 0$,by the Intermediate Value Theorem,there must exist at least one real root $x \in (0, 1)$ such that $f(x) = 0$.
Since the function is strictly increasing,this root is unique.
Therefore,the equation has exactly one real root.

(B) Let the side of the square base be $x \,m$ and the height of the tank be $y \,m$.
The volume $V = x^2 y = 500$.
Thus,$y = \frac{500}{x^2}$.
The surface area $S$ of the open tank is $S = x^2 + 4xy$.
Substituting $y$,we get $S = x^2 + 4x \left(\frac{500}{x^2}\right) = x^2 + \frac{2000}{x}$.
Differentiating with respect to $x$: $\frac{dS}{dx} = 2x - \frac{2000}{x^2}$.
Setting $\frac{dS}{dx} = 0$,we get $2x = \frac{2000}{x^2} \Rightarrow x^3 = 1000 \Rightarrow x = 10 \,m$.
The second derivative $\frac{d^2S}{dx^2} = 2 + \frac{4000}{x^3}$. At $x = 10$,$\frac{d^2S}{dx^2} = 2 + 4 = 6 > 0$,so the area is minimum.
The height $y = \frac{500}{10^2} = 5 \,m$.
Thus,the dimensions are $10 \,m, 10 \,m, 5 \,m$.

(A) Given $f(x) = \alpha \log x + \beta x^2 + x$.
Since $x=1$ and $x=2$ are extreme points,the derivative $f'(x)$ must be zero at these points.
$f'(x) = \frac{\alpha}{x} + 2\beta x + 1$.
At $x=1$,$f'(1) = \frac{\alpha}{1} + 2\beta(1) + 1 = 0 \implies \alpha + 2\beta = -1$ (Equation $1$).
At $x=2$,$f'(2) = \frac{\alpha}{2} + 2\beta(2) + 1 = 0 \implies \frac{\alpha}{2} + 4\beta = -1 \implies \alpha + 8\beta = -2$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(\alpha + 8\beta) - (\alpha + 2\beta) = -2 - (-1) \implies 6\beta = -1 \implies \beta = -\frac{1}{6}$.
Substituting $\beta = -\frac{1}{6}$ into Equation $1$: $\alpha + 2(-\frac{1}{6}) = -1 \implies \alpha - \frac{1}{3} = -1 \implies \alpha = -\frac{2}{3}$.
Now,calculate $\alpha^2 + 2\beta$: $(-\frac{2}{3})^2 + 2(-\frac{1}{6}) = \frac{4}{9} - \frac{1}{3} = \frac{4-3}{9} = \frac{1}{9}$.

(A) Given function: $f(x) = x^3 - 6x^2 + 9x + 2$
First,find the derivative $f'(x)$:
$f'(x) = 3x^2 - 12x + 9$
To find the critical points,set $f'(x) = 0$:
$3(x^2 - 4x + 3) = 0$
$3(x - 1)(x - 3) = 0$
So,the critical points are $x = 1$ and $x = 3$.
Now,find the second derivative $f''(x)$:
$f''(x) = 6x - 12$
Check the nature of the function at critical points using the second derivative test:
At $x = 1$: $f''(1) = 6(1) - 12 = -6$. Since $f''(1) < 0$,the function has a local maximum at $x = 1$.
At $x = 3$: $f''(3) = 6(3) - 12 = 6$. Since $f''(3) > 0$,the function has a local minimum at $x = 3$.
Therefore,the function has a maximum value when $x = 1$.

(A) Given $f(x) = \sqrt{25-x^2}$.
First,we find the derivative $f'(x) = \frac{1}{2\sqrt{25-x^2}} \cdot (-2x) = \frac{-x}{\sqrt{25-x^2}}$.
According to Lagrange's Mean Value Theorem,there exists $c \in (1, 5)$ such that $f'(c) = \frac{f(5) - f(1)}{5 - 1}$.
Calculate $f(5) = \sqrt{25 - 25} = 0$ and $f(1) = \sqrt{25 - 1} = \sqrt{24} = 2\sqrt{6}$.
So,$f'(c) = \frac{0 - \sqrt{24}}{5 - 1} = \frac{-\sqrt{24}}{4} = \frac{-2\sqrt{6}}{4} = \frac{-\sqrt{6}}{2}$.
Equating the derivative: $\frac{-c}{\sqrt{25-c^2}} = \frac{-\sqrt{6}}{2}$.
Squaring both sides: $\frac{c^2}{25-c^2} = \frac{6}{4} = \frac{3}{2}$.
$2c^2 = 3(25 - c^2) \implies 2c^2 = 75 - 3c^2 \implies 5c^2 = 75 \implies c^2 = 15$.
Since $c \in (1, 5)$,we take $c = \sqrt{15}$.

(A) Given function: $f(x) = x \sqrt{x+6}$ on $[-6, 0]$.
First,check the conditions of Rolle's Theorem:
$1$. $f(x)$ is continuous on $[-6, 0]$.
$2$. $f(x)$ is differentiable on $(-6, 0)$.
$3$. $f(-6) = -6 \sqrt{-6+6} = 0$ and $f(0) = 0 \sqrt{0+6} = 0$. Since $f(-6) = f(0)$,all conditions are satisfied.
Now,find $f'(x)$:
$f'(x) = x \cdot \frac{1}{2\sqrt{x+6}} + \sqrt{x+6} \cdot 1 = \frac{x + 2(x+6)}{2\sqrt{x+6}} = \frac{3x + 12}{2\sqrt{x+6}}$.
According to Rolle's Theorem,there exists $c \in (-6, 0)$ such that $f'(c) = 0$.
$\frac{3c + 12}{2\sqrt{c+6}} = 0$
$3c + 12 = 0$
$3c = -12$
$c = -4$.
Since $-4 \in (-6, 0)$,the value is $-4$.

(C) Given $f(x)=x^3-3(a-2)x^2+3ax+7$.
Since $f(x)$ is increasing in $(0,1]$ and decreasing in $[1,5)$,$f(x)$ must have a local maximum at $x=1$.
Thus,$f'(1)=0$.
$f'(x)=3x^2-6(a-2)x+3a$.
Substituting $x=1$: $3(1)^2-6(a-2)(1)+3a=0$.
$3-6a+12+3a=0 \Rightarrow -3a+15=0 \Rightarrow a=5$.
Now,substitute $a=5$ into $f(x)$: $f(x)=x^3-3(5-2)x^2+3(5)x+7 = x^3-9x^2+15x+7$.
We need to solve $\frac{f(x)-14}{(x-1)^2}=0$.
$f(x)-14 = x^3-9x^2+15x+7-14 = x^3-9x^2+15x-7$.
By synthetic division or polynomial division,$x^3-9x^2+15x-7 = (x-1)^2(x-7)$.
So,$\frac{(x-1)^2(x-7)}{(x-1)^2} = 0 \Rightarrow x-7=0 \Rightarrow x=7$.

(C) Given function is $f(x) = \sin x + \cos x + 6$ on the interval $[0, 2\pi]$.
According to Rolle's theorem,there exists at least one $c \in (0, 2\pi)$ such that $f'(c) = 0$.
First,we find the derivative: $f'(x) = \cos x - \sin x$.
Setting $f'(c) = 0$,we get $\cos c - \sin c = 0$.
This implies $\cos c = \sin c$,which simplifies to $\tan c = 1$.
In the interval $[0, 2\pi]$,the values of $c$ for which $\tan c = 1$ are $c = \frac{\pi}{4}$ and $c = \frac{\pi}{4} + \pi = \frac{5\pi}{4}$.
Thus,the values are $\frac{\pi}{4}, \frac{5\pi}{4}$.

(C) Since $f(x)$ satisfies Rolle's theorem on $[1,3]$,we have $f(1)=f(3)$.
$1+b+a+5 = 27+9b+3a+5$
$a+b+6 = 3a+9b+32$
$2a+8b = -26 \Rightarrow a+4b = -13$ ... $(i)$
Given $f(x) = x^3+bx^2+ax+5$,the derivative is $f'(x) = 3x^2+2bx+a$.
By Rolle's theorem,$f'(c) = 0$ for $c = 2+\frac{1}{\sqrt{3}}$.
$3(2+\frac{1}{\sqrt{3}})^2 + 2b(2+\frac{1}{\sqrt{3}}) + a = 0$
$3(4 + \frac{4}{\sqrt{3}} + \frac{1}{3}) + 4b + \frac{2b}{\sqrt{3}} + a = 0$
$12 + 4\sqrt{3} + 1 + 4b + \frac{2b}{\sqrt{3}} + a = 0$
$a + 4b + 13 + \frac{12+2b}{\sqrt{3}} = 0$
Since $a+4b = -13$,we substitute this into the equation:
$-13 + 13 + \frac{12+2b}{\sqrt{3}} = 0$
$\frac{12+2b}{\sqrt{3}} = 0 \Rightarrow 2b = -12 \Rightarrow b = -6$.
Substituting $b = -6$ into $(i)$: $a + 4(-6) = -13 \Rightarrow a - 24 = -13 \Rightarrow a = 11$.
Thus,the values are $a=11$ and $b=-6$.

(A) Given the curve $y = x|x|$.
Since $x|x|$ is an odd function,we analyze the integral $\int_{-1}^{1} x|x| dx$.
However,the question asks for the area bounded by the curve and the $X$-axis,which is given by $\int_{-1}^{1} |y| dx = \int_{-1}^{1} |x|x|| dx = \int_{-1}^{1} |x|^2 dx = \int_{-1}^{1} x^2 dx$.
$\text{Area} = \int_{-1}^{1} x^2 dx = 2 \int_{0}^{1} x^2 dx$.
$= 2 \left[ \frac{x^3}{3} \right]_{0}^{1}$.
$= 2 \times \left( \frac{1}{3} - 0 \right) = \frac{2}{3} \text{ sq. units}$.

(B) The given curve is $y = \sqrt{49 - x^2}$,which implies $y^2 = 49 - x^2$ or $x^2 + y^2 = 7^2$. This represents a circle centered at the origin $(0, 0)$ with a radius $r = 7$. Since $y = \sqrt{49 - x^2}$ is always non-negative,it represents the upper semi-circle.
The area of the region bounded by the curve and the $X$-axis is the area of the semi-circle.
Area $= \int_{-7}^{7} \sqrt{49 - x^2} \, dx = 2 \int_{0}^{7} \sqrt{49 - x^2} \, dx$
Using the formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \left( \frac{x}{a} \right)$:
Area $= 2 \left[ \frac{x}{2} \sqrt{49 - x^2} + \frac{49}{2} \sin^{-1} \left( \frac{x}{7} \right) \right]_{0}^{7}$
$= 2 \left[ \left( \frac{7}{2} \sqrt{49 - 49} + \frac{49}{2} \sin^{-1} (1) \right) - (0 + 0) \right]$
$= 2 \left[ 0 + \frac{49}{2} \times \frac{\pi}{2} \right] = \frac{49 \pi}{2} \text{ sq. units}$.

(A) For the $X$-axis,$y=0$.
Therefore,$x(x-2)(x+1)=0$,which gives $x=0, x=2, x=-1$.
The required area is given by the integral of $|y|$ with respect to $x$ from $-1$ to $2$.
$\text{Area} = \int_{-1}^0 y \, dx + \left| \int_0^2 y \, dx \right|$
$\text{Area} = \int_{-1}^0 (x^3-x^2-2x) \, dx + \left| \int_0^2 (x^3-x^2-2x) \, dx \right|$
Evaluating the integrals:
$\int (x^3-x^2-2x) \, dx = \frac{x^4}{4} - \frac{x^3}{3} - x^2$
For the interval $[-1, 0]$:
$\left[ \frac{x^4}{4} - \frac{x^3}{3} - x^2 \right]_{-1}^0 = 0 - \left( \frac{1}{4} - \frac{-1}{3} - 1 \right) = - \left( \frac{3+4-12}{12} \right) = - \left( \frac{-5}{12} \right) = \frac{5}{12}$
For the interval $[0, 2]$:
$\left[ \frac{x^4}{4} - \frac{x^3}{3} - x^2 \right]_0^2 = \left( \frac{16}{4} - \frac{8}{3} - 4 \right) - 0 = 4 - \frac{8}{3} - 4 = -\frac{8}{3}$
Taking the absolute value,we get $|-\frac{8}{3}| = \frac{8}{3}$.
Total Area = $\frac{5}{12} + \frac{8}{3} = \frac{5+32}{12} = \frac{37}{12}$ sq. units.

(C) The required area is given by the integral $\int_{1}^{3} |x-2| dx$.
Since the function $|x-2|$ changes its definition at $x=2$,we split the integral:
$\text{Area} = \int_{1}^{2} -(x-2) dx + \int_{2}^{3} (x-2) dx$.
Evaluating the first part: $\int_{1}^{2} (2-x) dx = [2x - \frac{x^2}{2}]_{1}^{2} = (4 - 2) - (2 - \frac{1}{2}) = 2 - 1.5 = 0.5$.
Evaluating the second part: $\int_{2}^{3} (x-2) dx = [\frac{x^2}{2} - 2x]_{2}^{3} = (4.5 - 6) - (2 - 4) = -1.5 - (-2) = 0.5$.
Adding both parts: $0.5 + 0.5 = 1 \text{ sq. unit}$.