MHT CET 2024 Mathematics Question Paper with Answer and Solution

(D) The word $BARRACK$ consists of $7$ letters: $A, A, R, R, B, C, K$. The distinct letters are ${A, R, B, C, K}$.
We need to form $4$-letter words. The cases are:
$(i)$ All $4$ letters are distinct: We choose $4$ letters from ${A, R, B, C, K}$ in $^5C_4 = 5$ ways. Each set can be arranged in $4! = 24$ ways. Total $= 5 \times 24 = 120$.
(ii) $2$ pairs of identical letters: The pairs are ${A, A}$ and ${R, R}$. We choose both pairs in $^2C_2 = 1$ way. The number of arrangements is $\frac{4!}{2!2!} = 6$.
(iii) $2$ identical letters and $2$ distinct letters: We choose $1$ pair from ${A, A}$ or ${R, R}$ in $^2C_1 = 2$ ways. We then choose $2$ distinct letters from the remaining $4$ letters in $^4C_2 = 6$ ways. The number of arrangements for each selection is $\frac{4!}{2!} = 12$. Total $= 2 \times 6 \times 12 = 144$.
Total number of $4$-letter words $= 120 + 6 + 144 = 270$.

(C) Total members available are $8$ males and $5$ females,so total persons = $13$. We need to select $11$ members.
For $m$ (at least $6$ males):
Possible cases are ($6$ males,$5$ females),($7$ males,$4$ females),($8$ males,$3$ females).
$m = \binom{8}{6} \times \binom{5}{5} + \binom{8}{7} \times \binom{5}{4} + \binom{8}{8} \times \binom{5}{3} = (28 \times 1) + (8 \times 5) + (1 \times 10) = 28 + 40 + 10 = 78$.
For $n$ (at least $3$ females):
Possible cases are ($8$ males,$3$ females),($7$ males,$4$ females),($6$ males,$5$ females).
$n = \binom{5}{3} \times \binom{8}{8} + \binom{5}{4} \times \binom{8}{7} + \binom{5}{5} \times \binom{8}{6} = (10 \times 1) + (5 \times 8) + (1 \times 28) = 10 + 40 + 28 = 78$.
Thus,$m = n = 78$.

(B) The implication $P \Rightarrow (q \vee r)$ is false only when the antecedent is true and the consequent is false.
That is,$P = T$ and $(q \vee r) = F$.
For the disjunction $(q \vee r)$ to be false,both $q$ and $r$ must be false.
Therefore,$p = T, q = F, r = F$.

(B) Given equation: $(81)^{\sin ^{2} x} + (81)^{\cos ^{2} x} = 30$.
Since $\cos ^{2} x = 1 - \sin ^{2} x$,we have $(81)^{\sin ^{2} x} + (81)^{1 - \sin ^{2} x} = 30$.
$(81)^{\sin ^{2} x} + \frac{81}{(81)^{\sin ^{2} x}} = 30$.
Let $t = (81)^{\sin ^{2} x}$. Then $t + \frac{81}{t} = 30$,which implies $t^{2} - 30t + 81 = 0$.
$(t - 3)(t - 27) = 0$,so $t = 3$ or $t = 27$.
Case $1$: $(81)^{\sin ^{2} x} = 3 \implies 3^{4 \sin ^{2} x} = 3^{1} \implies 4 \sin ^{2} x = 1 \implies \sin ^{2} x = \frac{1}{4}$.
In $[0, \pi]$,$\sin x = \frac{1}{2}$ or $\sin x = -\frac{1}{2}$. Since $\sin x \ge 0$ in $[0, \pi]$,$\sin x = \frac{1}{2}$ gives $x = \frac{\pi}{6}, \frac{5\pi}{6}$ ($2$ solutions).
Case $2$: $(81)^{\sin ^{2} x} = 27 \implies 3^{4 \sin ^{2} x} = 3^{3} \implies 4 \sin ^{2} x = 3 \implies \sin ^{2} x = \frac{3}{4}$.
In $[0, \pi]$,$\sin x = \frac{\sqrt{3}}{2}$ or $\sin x = -\frac{\sqrt{3}}{2}$. Since $\sin x \ge 0$ in $[0, \pi]$,$\sin x = \frac{\sqrt{3}}{2}$ gives $x = \frac{\pi}{3}, \frac{2\pi}{3}$ ($2$ solutions).
Total number of solutions = $2 + 2 = 4$.

(D) Let the number of questions selected from sections $A, B,$ and $C$ be $n_1, n_2,$ and $n_3$ respectively,such that $n_1 + n_2 + n_3 = 5$ and $n_i \ge 1$.
The possible distributions $(n_1, n_2, n_3)$ are:
$1. (1, 2, 2)$ and its permutations: $(1, 2, 2), (2, 1, 2), (2, 2, 1)$ (Total $3$ ways).
$2. (1, 1, 3)$ and its permutations: $(1, 1, 3), (1, 3, 1), (3, 1, 1)$ (Total $3$ ways).
Number of ways for case $(1, 2, 2) = \binom{5}{1} \times \binom{5}{2} \times \binom{5}{2} = 5 \times 10 \times 10 = 500$.
Since there are $3$ such permutations,total ways $= 3 \times 500 = 1500$.
Number of ways for case $(1, 1, 3) = \binom{5}{1} \times \binom{5}{1} \times \binom{5}{3} = 5 \times 5 \times 10 = 250$.
Since there are $3$ such permutations,total ways $= 3 \times 250 = 750$.
Total number of ways $= 1500 + 750 = 2250$.

(A) Given curves are $y^2=6x$ $(i)$ and $9x^2+by^2=16$ $(ii)$.
For $(i)$,differentiating with respect to $x$: $2y \frac{dy}{dx} = 6 \Rightarrow \frac{dy}{dx} = \frac{3}{y}$.
For $(ii)$,differentiating with respect to $x$: $18x + 2by \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{9x}{by}$.
Since the curves intersect at right angles,the product of their slopes at the point of intersection $(x, y)$ is $-1$:
$(\frac{3}{y}) \times (-\frac{9x}{by}) = -1$
$\frac{27x}{by^2} = 1 \Rightarrow by^2 = 27x$.
Substitute $y^2 = 6x$ into this equation:
$b(6x) = 27x \Rightarrow 6b = 27 \Rightarrow b = \frac{27}{6} = \frac{9}{2}$.

(D) The given equations of the lines are $4x + 3y - 20 = 0$ and $4x + 3y + 15 = 0$.
Since the coefficients of $x$ and $y$ are the same,the lines are parallel.
The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \left| \frac{c_1 - c_2}{\sqrt{a^2 + b^2}} \right|$.
Here,$a = 4$,$b = 3$,$c_1 = -20$,and $c_2 = 15$.
$d = \left| \frac{-20 - 15}{\sqrt{4^2 + 3^2}} \right| = \left| \frac{-35}{\sqrt{16 + 9}} \right| = \left| \frac{-35}{5} \right| = 7$ units.
Since the distance between two opposite sides of a square is equal to its side length $s$,we have $s = 7$ units.
The area of the square is $s^2 = 7^2 = 49$ sq. units.

(B) To determine the correct system of inequalities for the shaded region,we analyze the boundary lines shown in the figure:
$1$. The line passing through $(0, 3)$ and $(6, 0)$ has the equation $\frac{x}{6} + \frac{y}{3} = 1$,which simplifies to $x + 2y = 6$. Since the shaded region is above this line,the inequality is $x + 2y \geq 6$.
$2$. The line passing through $(0, 5)$ and $(3, 0)$ has the equation $\frac{x}{3} + \frac{y}{5} = 1$,which simplifies to $5x + 3y = 15$. Since the shaded region is above this line,the inequality is $5x + 3y \geq 15$.
$3$. The vertical line is $x = 7$. Since the shaded region is to the left of this line,the inequality is $x \leq 7$.
$4$. The horizontal line is $y = 6$. Since the shaded region is below this line,the inequality is $y \leq 6$.
$5$. The region is in the first quadrant,so $x \geq 0$ and $y \geq 0$.
Combining these,the system is $x + 2y \geq 6, 5x + 3y \geq 15, x \leq 7, y \leq 6, x, y \geq 0$. This corresponds to option $(B)$.

(D) Given equation of circle $C_1$: $x^2+y^2-6x-4y-12=0$.
Rewriting in standard form: $(x^2-6x+9) + (y^2-4y+4) = 12+9+4$.
$(x-3)^2 + (y-2)^2 = 25$.
Thus,the center is $(3,2)$ and the radius $r = 5$.
Area of $C_1 = \pi r^2 = 25\pi$.
Let the radius of the required concentric circle be $R$.
Given that the area of the required circle is double the area of $C_1$:
$\pi R^2 = 2 \times (25\pi) = 50\pi$.
$R^2 = 50$.
The equation of the concentric circle with center $(3,2)$ is $(x-3)^2 + (y-2)^2 = R^2$.
$(x-3)^2 + (y-2)^2 = 50$.
$x^2-6x+9 + y^2-4y+4 = 50$.
$x^2+y^2-6x-4y+13 = 50$.
$x^2+y^2-6x-4y = 37$.

(D) The required circle is concentric with $2x^2+2y^2-8x-12y-9=0$.
Dividing by $2$,we get $x^2+y^2-4x-6y-\frac{9}{2}=0$.
Thus,the centre of the required circle is $(2, 3)$.
The circle passes through the centre of $x^2+y^2+8x+10y-7=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,the centre is $(-4, -5)$.
The radius $r$ is the distance between $(2, 3)$ and $(-4, -5)$:
$r = \sqrt{(-4-2)^2 + (-5-3)^2} = \sqrt{(-6)^2 + (-8)^2} = \sqrt{36+64} = 10$.
The equation of the circle is $(x-2)^2 + (y-3)^2 = 10^2$.
$x^2-4x+4 + y^2-6y+9 = 100$.
$x^2+y^2-4x-6y-87=0$.

(B) The center of the circle $x^2+y^2+6x-14y+5=0$ is $(-3, 7)$.
The center of the circle $x^2+y^2-4x+10y-4=0$ is $(2, -5)$.
The equation of a circle with diameter end points $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.
Substituting the centers $(-3, 7)$ and $(2, -5)$:
$(x - (-3))(x - 2) + (y - 7)(y - (-5)) = 0$
$(x+3)(x-2) + (y-7)(y+5) = 0$
$x^2 - 2x + 3x - 6 + y^2 + 5y - 7y - 35 = 0$
$x^2 + y^2 + x - 2y - 41 = 0$.

(C) The given circle is $x^2+y^2-6x-4y-12=0$.
Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3$ and $f=-2$.
The center of the circle is $(-g, -f) = (3, 2)$.
Since the required circle is concentric,its center is also $(3, 2)$.
Because the circle touches the $X$-axis,its radius $r$ is equal to the absolute value of the $y$-coordinate of its center,so $r = |2| = 2$.
The equation of the circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.
Substituting the values,we get $(x-3)^2 + (y-2)^2 = 2^2$.
Expanding this,we get $(x^2-6x+9) + (y^2-4y+4) = 4$.
Simplifying,we get $x^2+y^2-6x-4y+9 = 0$.

(D) Let $A \equiv (x_1, y_1)$ and $B \equiv (x_2, y_2)$.
From the given equations,by the properties of roots of a quadratic equation:
$x_1+x_2 = -2a$ and $x_1x_2 = -b^2$.
$y_1+y_2 = -2p$ and $y_1y_2 = -q^2$.
The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.
Expanding this,we get $x^2 - (x_1+x_2)x + x_1x_2 + y^2 - (y_1+y_2)y + y_1y_2 = 0$.
Substituting the values of the sum and product of roots:
$x^2 - (-2a)x + (-b^2) + y^2 - (-2p)y + (-q^2) = 0$.
$x^2 + y^2 + 2ax + 2py - (b^2+q^2) = 0$.

(A) The given equation of the circle is $x^2+y^2-ax-by=0$.
Completing the square,we get:
$(x^2-ax+\frac{a^2}{4})+(y^2-by+\frac{b^2}{4}) = \frac{a^2}{4}+\frac{b^2}{4}$
$(x-\frac{a}{2})^2+(y-\frac{b}{2})^2 = (\frac{\sqrt{a^2+b^2}}{2})^2$.
This is in the form $(x-h)^2+(y-k)^2=r^2$,where the center $(h, k) = (\frac{a}{2}, \frac{b}{2})$ and the radius $r = \frac{\sqrt{a^2+b^2}}{2}$.
The parametric equations are given by $x = h + r \cos \theta$ and $y = k + r \sin \theta$.
Substituting the values,we get $x = \frac{a}{2} + \frac{\sqrt{a^2+b^2}}{2} \cos \theta$ and $y = \frac{b}{2} + \frac{\sqrt{a^2+b^2}}{2} \sin \theta$.

(C) The given equations of the sides are $x=-2, x=6, y=-2$ and $y=5$.
The vertices of the rectangle are $A(-2, -2)$,$B(6, -2)$,$C(6, 5)$,and $D(-2, 5)$.
The diagonal of the rectangle acts as the diameter of the circle.
Taking diagonal $AC$ with endpoints $A(-2, -2)$ and $C(6, 5)$,the equation of the circle in diameter form is given by:
$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$
$(x - (-2))(x - 6) + (y - (-2))(y - 5) = 0$
$(x + 2)(x - 6) + (y + 2)(y - 5) = 0$
$x^2 - 6x + 2x - 12 + y^2 - 5y + 2y - 10 = 0$
$x^2 + y^2 - 4x - 3y - 22 = 0$

(A) The equation of the tangent to the circle $x^2+y^2=5$ at the point $(x_1, y_1) = (1, -2)$ is given by $x x_1 + y y_1 = r^2$.
Substituting the values,we get $x(1) + y(-2) = 5$,which simplifies to $x - 2y = 5$.
For the second circle $x^2 + y^2 - 8x + 6y + 20 = 0$,the center is $(4, -3)$ and the radius is $\sqrt{4^2 + (-3)^2 - 20} = \sqrt{16 + 9 - 20} = \sqrt{5}$.
The point of contact on the second circle must lie on the line $x - 2y = 5$.
Checking the options:
For $(3, -1)$: $3 - 2(-1) = 3 + 2 = 5$. This point satisfies the tangent equation.
For $(3, 1)$: $3 - 2(1) = 1 \neq 5$.
For $(-3, -1)$: $-3 - 2(-1) = -1 \neq 5$.
For $(-3, 1)$: $-3 - 2(1) = -5 \neq 5$.
Thus,the point of contact is $(3, -1)$.

(B) The equation of the tangent to the circle $x^2+y^2=a^2$ at the point $P(\theta)$ is given by $x \cos \theta + y \sin \theta = a$.
Here,the radius $a = 5$ and the angle $\theta = \frac{\pi}{3}$.
Substituting these values into the equation:
$x \cos \frac{\pi}{3} + y \sin \frac{\pi}{3} = 5$
$x \left( \frac{1}{2} \right) + y \left( \frac{\sqrt{3}}{2} \right) = 5$
Multiplying the entire equation by $2$,we get:
$x + \sqrt{3} y = 10$.

(C) For the circle $x^2+y^2-x=0$,the center $C_1 = (\frac{1}{2}, 0)$ and radius $r_1 = \sqrt{(\frac{1}{2})^2 + 0^2 - 0} = \frac{1}{2}$.
For the circle $x^2+y^2+x=0$,the center $C_2 = (-\frac{1}{2}, 0)$ and radius $r_2 = \sqrt{(-\frac{1}{2})^2 + 0^2 - 0} = \frac{1}{2}$.
The distance between the centers is $C_1C_2 = \sqrt{(\frac{1}{2} - (-\frac{1}{2}))^2 + (0-0)^2} = \sqrt{1^2} = 1$.
The sum of the radii is $r_1 + r_2 = \frac{1}{2} + \frac{1}{2} = 1$.
Since $C_1C_2 = r_1 + r_2$,the circles touch each other externally.
When two circles touch each other externally,the number of common tangents is $3$.

(C) The given circle equation is $x^2+y^2-6x-5y-1=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3$ and $f=-2.5$.
The center of the circle is $(-g, -f) = (3, 2.5)$.
Let the other end of the diameter be $(x_2, y_2)$. Since the center $(3, 2.5)$ is the midpoint of the diameter with endpoints $(-1, 3)$ and $(x_2, y_2)$,we have:
$\frac{-1+x_2}{2} = 3 \Rightarrow x_2 = 7$
$\frac{3+y_2}{2} = 2.5 \Rightarrow y_2 = 2$
So,the other end is $(7, 2)$.
The equation of the tangent to the circle $x^2+y^2+2gx+2fy+c=0$ at point $(x_1, y_1)$ is given by $xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0$.
Substituting $(x_1, y_1) = (7, 2)$,$g=-3$,$f=-2.5$,and $c=-1$:
$7x + 2y - 3(x+7) - 2.5(y+2) - 1 = 0$
$7x + 2y - 3x - 21 - 2.5y - 5 - 1 = 0$
$4x - 0.5y - 27 = 0$
Multiplying by $2$,we get $8x - y - 54 = 0$.

(C) The circle is $x^2 + y^2 = 25$,so its center $O$ is $(0, 0)$ and radius $r = 5$.
Length of the tangent segment from $P(1, 7)$ to the circle is given by $\sqrt{x_1^2 + y_1^2 - r^2} = \sqrt{1^2 + 7^2 - 25} = \sqrt{1 + 49 - 25} = \sqrt{25} = 5$.
In the quadrilateral $PQOR$,$PQ$ and $PR$ are tangents,so $PQ = PR = 5$.
Also,$OQ = OR = 5$ (radii of the circle).
Since the tangent is perpendicular to the radius at the point of contact,$\angle OQP = \angle ORP = 90^{\circ}$.
The quadrilateral $PQOR$ consists of two congruent right-angled triangles,$\triangle PQO$ and $\triangle PRO$.
Area of $\triangle PQO = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times PQ \times OQ = \frac{1}{2} \times 5 \times 5 = 12.5 \text{ sq. units}$.
Total area of quadrilateral $PQOR = 2 \times \text{Area of } \triangle PQO = 2 \times 12.5 = 25 \text{ sq. units}$.

(C) The given circle is $2x^2+2y^2-6x+8y+1=0$.
Dividing by $2$,we get $x^2+y^2-3x+4y+\frac{1}{2}=0$.
The center is $(\frac{3}{2}, -2)$ and the radius $r_1$ is given by $r_1^2 = (\frac{3}{2})^2 + (-2)^2 - \frac{1}{2} = \frac{9}{4} + 4 - \frac{1}{2} = \frac{9+16-2}{4} = \frac{23}{4}$.
Let the required circle be $x^2+y^2-3x+4y+k=0$.
Its radius $r_2$ satisfies $r_2^2 = (\frac{3}{2})^2 + (-2)^2 - k = \frac{25}{4} - k$.
Since the area is double,$\pi r_2^2 = 2(\pi r_1^2)$,so $r_2^2 = 2r_1^2$.
$\frac{25}{4} - k = 2(\frac{23}{4}) = \frac{23}{2}$.
$k = \frac{25}{4} - \frac{46}{4} = -\frac{21}{4}$.
Substituting $k$ back,$x^2+y^2-3x+4y-\frac{21}{4}=0$.
Multiplying by $4$,we get $4x^2+4y^2-12x+16y-21=0$.

(A) The radius $r$ of the circle is the perpendicular distance from the center $(3, 4)$ to the line $5x + 12y - 11 = 0$.
$r = \left| \frac{5(3) + 12(4) - 11}{\sqrt{5^2 + 12^2}} \right| = \left| \frac{15 + 48 - 11}{\sqrt{25 + 144}} \right| = \frac{52}{13} = 4$.
The equation of the circle with center $(h, k) = (3, 4)$ and radius $r = 4$ is given by $(x - h)^2 + (y - k)^2 = r^2$.
$(x - 3)^2 + (y - 4)^2 = 4^2$
$x^2 - 6x + 9 + y^2 - 8y + 16 = 16$
$x^2 + y^2 - 6x - 8y + 9 = 0$.

(D) Given $\left(-2-\frac{1}{3} i\right)^3=\frac{x+i y}{27}$.
We can rewrite the left side as $\left(\frac{-6-i}{3}\right)^3 = \frac{(-6-i)^3}{27}$.
Equating the numerators,we have $x+iy = (-6-i)^3$.
Using the identity $(a+b)^3 = a^3+3a^2b+3ab^2+b^3$:
$(-6-i)^3 = (-6)^3 + 3(-6)^2(-i) + 3(-6)(-i)^2 + (-i)^3$.
$= -216 + 3(36)(-i) + 3(-6)(-1) - (-i)$.
$= -216 - 108i + 18 + i$.
$= -198 - 107i$.
Comparing with $x+iy$,we get $x = -198$ and $y = -107$.
Therefore,$y-x = -107 - (-198) = -107 + 198 = 91$.

(B) Given $z = \frac{(1 + i)^2}{a - i}$.
Since $(1 + i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$,we have $z = \frac{2i}{a - i}$.
Multiply numerator and denominator by the conjugate $(a + i)$:
$z = \frac{2i(a + i)}{a^2 + 1} = \frac{2ai - 2}{a^2 + 1} = \frac{-2 + 2ai}{a^2 + 1}$.
The magnitude is $|z| = \sqrt{\frac{(-2)^2 + (2a)^2}{(a^2 + 1)^2}} = \sqrt{\frac{4(1 + a^2)}{(a^2 + 1)^2}} = \sqrt{\frac{4}{a^2 + 1}} = \frac{2}{\sqrt{a^2 + 1}}$.
Given $|z| = \sqrt{\frac{2}{5}}$,so $\frac{2}{\sqrt{a^2 + 1}} = \sqrt{\frac{2}{5}}$.
Squaring both sides: $\frac{4}{a^2 + 1} = \frac{2}{5}$ $\Rightarrow 20 = 2(a^2 + 1)$ $\Rightarrow 10 = a^2 + 1$ $\Rightarrow a^2 = 9$.
Since $a > 0$,$a = 3$.
Substituting $a = 3$ into $z$: $z = \frac{-2 + 2(3)i}{3^2 + 1} = \frac{-2 + 6i}{10} = -\frac{1}{5} + \frac{3}{5}i$.
Therefore,the conjugate $\overline{z} = -\frac{1}{5} - \frac{3}{5}i$.

(A) Given $w = \frac{z-1}{z+1}$.
Since $|z|=1$,we can write $z = x+iy$ where $x^2+y^2=1$.
Then $w = \frac{(x-1)+iy}{(x+1)+iy}$.
To find the real part,multiply the numerator and denominator by the conjugate of the denominator: $(x+1)-iy$.
$w = \frac{((x-1)+iy)((x+1)-iy)}{(x+1)^2+y^2} = \frac{(x^2-1) + y^2 + i(y(x+1) - y(x-1))}{(x+1)^2+y^2}$.
$w = \frac{(x^2+y^2-1) + 2iy}{(x+1)^2+y^2}$.
Since $|z|=1$,we have $x^2+y^2=1$,so $x^2+y^2-1=0$.
Therefore,$\operatorname{Re}(w) = \frac{0}{(x+1)^2+y^2} = 0$.

(A) Given $Z = \frac{-2}{1 + \sqrt{3}i}$.
Rationalizing the denominator:
$Z = \frac{-2(1 - \sqrt{3}i)}{(1 + \sqrt{3}i)(1 - \sqrt{3}i)}$
$Z = \frac{-2(1 - \sqrt{3}i)}{1^2 - (\sqrt{3}i)^2}$
$Z = \frac{-2(1 - \sqrt{3}i)}{1 + 3} = \frac{-2(1 - \sqrt{3}i)}{4}$
$Z = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$.
Here,the real part $a = -\frac{1}{2}$ and the imaginary part $b = \frac{\sqrt{3}}{2}$.
Since $a < 0$ and $b > 0$,the complex number lies in the second quadrant.
$\arg(Z) = \pi - \tan^{-1}\left|\frac{b}{a}\right|$
$\arg(Z) = \pi - \tan^{-1}\left|\frac{\sqrt{3}/2}{-1/2}\right|$
$\arg(Z) = \pi - \tan^{-1}(\sqrt{3})$
$\arg(Z) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

(D) Let $Z = a + ib$.
Then $|Z| = \sqrt{a^2 + b^2}$.
Given $|Z| + Z = 2 + i$.
Substituting the values,we get $\sqrt{a^2 + b^2} + a + ib = 2 + i$.
Comparing the real and imaginary parts,we get $b = 1$ and $\sqrt{a^2 + b^2} + a = 2$.
Since $b = 1$,$\sqrt{a^2 + 1} = 2 - a$.
Squaring both sides,$a^2 + 1 = (2 - a)^2$.
$a^2 + 1 = 4 - 4a + a^2$.
$4a = 3$,so $a = \frac{3}{4}$.
Now,$|Z| = \sqrt{a^2 + b^2} = \sqrt{(\frac{3}{4})^2 + 1^2} = \sqrt{\frac{9}{16} + 1} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.

(B) Given $|z|+z=2+i$.
Let $z=x+iy$,where $x, y \in \mathbb{R}$.
Then $|z|=\sqrt{x^2+y^2}$.
Substituting these into the equation: $\sqrt{x^2+y^2}+x+iy=2+i$.
Equating real and imaginary parts,we get:
$y=1$ and $\sqrt{x^2+y^2}+x=2$.
Substituting $y=1$ into the second equation: $\sqrt{x^2+1}=2-x$.
Squaring both sides: $x^2+1=(2-x)^2 = 4-4x+x^2$.
$1=4-4x$ $\Rightarrow 4x=3$ $\Rightarrow x=\frac{3}{4}$.
Thus,$z=\frac{3}{4}+i$.
Therefore,$|z|=\sqrt{(\frac{3}{4})^2+1^2} = \sqrt{\frac{9}{16}+1} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.

(A) Given $z_1 = 5 - 2i$ and $z_2 = 3 + i$.
First,calculate $z_1 + z_2 = (5 + 3) + (-2 + 1)i = 8 - i$.
Next,calculate $z_1 - z_2 = (5 - 3) + (-2 - 1)i = 2 - 3i$.
Now,consider the ratio $\frac{z_1 + z_2}{z_1 - z_2} = \frac{8 - i}{2 - 3i}$.
To simplify,multiply the numerator and denominator by the conjugate of the denominator $(2 + 3i)$:
$\frac{8 - i}{2 - 3i} \times \frac{2 + 3i}{2 + 3i} = \frac{16 + 24i - 2i - 3i^2}{2^2 + 3^2} = \frac{16 + 22i + 3}{4 + 9} = \frac{19 + 22i}{13} = \frac{19}{13} + \frac{22}{13}i$.
The argument is given by $\tan^{-1}\left(\frac{\text{imaginary part}}{\text{real part}}\right) = \tan^{-1}\left(\frac{22/13}{19/13}\right) = \tan^{-1}\left(\frac{22}{19}\right)$.

(C) Given that $z$ is a complex cube root of unity,we have $z^3 = 1$ and $1+z+z^2 = 0$.
From $1+z+z^2 = 0$,we get $z^2+1 = -z$.
Now,consider the expression $\left(z^3+\frac{1}{z^3}\right)^2+\left(z^4+\frac{1}{z^4}\right)^2$.
Substituting $z^3 = 1$,we get $\left(1+\frac{1}{1}\right)^2+\left(z^3 \cdot z+\frac{1}{z^3 \cdot z}\right)^2$.
$= (1+1)^2 + \left(z+\frac{1}{z}\right)^2$.
$= 4 + \left(\frac{z^2+1}{z}\right)^2$.
Substituting $z^2+1 = -z$,we get $4 + \left(\frac{-z}{z}\right)^2$.
$= 4 + (-1)^2 = 4 + 1 = 5$.

(C) Given $w = \frac{-1 + i \sqrt{3}}{2}$,which is the complex cube root of unity,denoted as $\omega$.
We know that $1 + \omega + \omega^2 = 0$,which implies $\omega^2 = -1 - \omega$.
We need to evaluate $(3 + \omega + 3 \omega^2)^4$.
Substitute $\omega^2 = -1 - \omega$ into the expression:
$(3 + \omega + 3(-1 - \omega))^4 = (3 + \omega - 3 - 3 \omega)^4$.
$= (-2 \omega)^4$.
$= 16 \omega^4$.
Since $\omega^3 = 1$,we have $\omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega$.
Therefore,$16 \omega^4 = 16 \omega$.

(D) Given $\left|\frac{z-1}{z+2i}\right| = 1$.
This implies $|z-1| = |z+2i|$.
Substituting $z = x + iy$:
$|x + iy - 1| = |x + iy + 2i|$
$|(x-1) + iy| = |x + i(y+2)|$
Squaring both sides:
$(x-1)^2 + y^2 = x^2 + (y+2)^2$
$x^2 - 2x + 1 + y^2 = x^2 + y^2 + 4y + 4$
$-2x + 1 = 4y + 4$
$2x + 4y + 3 = 0$.
This is the equation of a straight line.

(D) Given $\left|\frac{z}{1+i}\right|=2$.
Since $|1+i| = \sqrt{1^2+1^2} = \sqrt{2}$,we have $|z| = 2|1+i| = 2\sqrt{2}$.
Substituting $z=x+iy$,we get $|x+iy| = 2\sqrt{2}$.
Squaring both sides,we get $x^2+y^2 = (2\sqrt{2})^2 = 8$.
This represents a circle $x^2+y^2 = 8$ with centre $C \equiv (0,0)$ and radius $r = \sqrt{8} = 2\sqrt{2}$.

(C) Given the condition $|z+1|=1$.
Substitute $z=x+iy$ into the equation:
$|x+iy+1|=1$
$|(x+1)+iy|=1$
Using the definition of the modulus of a complex number $|a+ib|=\sqrt{a^2+b^2}$,we get:
$\sqrt{(x+1)^2+y^2}=1$
Squaring both sides:
$(x+1)^2+y^2=1^2$
$(x+1)^2+y^2=1$
This is the standard equation of a circle $(x-h)^2+(y-k)^2=r^2$,where the centre is $(h,k)$ and the radius is $r$.
Comparing the equations,the centre is $(-1,0)$ and the radius is $1$ unit.

(A) Given,$z\bar{z}^3+\bar{z}z^3=350$
$\Rightarrow z\bar{z}(\bar{z}^2+z^2)=350$
$\Rightarrow |z|^2(x-iy)^2+(x+iy)^2=350$
$\Rightarrow (x^2+y^2)(x^2-y^2-2ixy+x^2-y^2+2ixy)=350$
$\Rightarrow (x^2+y^2)(2x^2-2y^2)=350$
$\Rightarrow 2(x^2+y^2)(x^2-y^2)=350$
$\Rightarrow x^4-y^4=175$
Since $x$ and $y$ are integers,we test values: $x^4-y^4=(x^2-y^2)(x^2+y^2)=175$.
For $x=4, y=3$: $4^4-3^4=256-81=175$.
Thus,the vertices are $(4,3), (-4,3), (-4,-3), (4,-3)$.
The length of the rectangle is $|4-(-4)|=8$ and the breadth is $|3-(-3)|=6$.
Area $= 8 \times 6 = 48 \text{ sq. units}$.

(C) The given equation of the pair of lines is $4x^2 + kxy + y^2 = 0$.
Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we get $a = 4$,$2h = k$ (so $h = k/2$),and $b = 1$.
Let the slopes of the two lines be $m_1$ and $m_2$.
We know that $m_1 + m_2 = -\frac{2h}{b} = -k$ and $m_1 m_2 = \frac{a}{b} = 4$.
According to the given condition,$m_1 = 4m_2$.
Substituting this into the sum of slopes: $4m_2 + m_2 = -k \implies 5m_2 = -k \implies m_2 = -k/5$.
Substituting into the product of slopes: $(4m_2)(m_2) = 4 \implies 4m_2^2 = 4 \implies m_2^2 = 1 \implies m_2 = \pm 1$.
Since $m_2 = -k/5$,we have $-k/5 = \pm 1$,which gives $k = \mp 5$.
Considering the magnitude or the standard form of such problems,the value is $k = \pm 5$. Given the options,$5$ is the correct magnitude.

(C) Given the limit: $\lim _{x \rightarrow 0^{-}} \frac{x([x]+|x|) \sin [x]}{|x|}$.
For $x \rightarrow 0^{-}$,we have $[x] = -1$ and $|x| = -x$.
Substituting these values into the expression:
$\lim _{x \rightarrow 0^{-}} \frac{x(-1 + (-x)) \sin(-1)}{-x}$
$= \lim _{x \rightarrow 0^{-}} \frac{x(-1 - x)(-\sin 1)}{-x}$
$= \lim _{x \rightarrow 0^{-}} (1 + x) \sin 1$
As $x \rightarrow 0$,this becomes $(1 + 0) \sin 1 = \sin 1$.

(D) To find the limit,we evaluate the left-hand limit $(L.H.L.)$ and the right-hand limit $(R.H.L.)$ at $x = 0$.
$L.H.L. = \lim _{x \rightarrow 0^{-}} \frac{x}{|x|+x^2} = \lim _{x \rightarrow 0^{-}} \frac{x}{-x+x^2} = \lim _{x \rightarrow 0^{-}} \frac{1}{-1+x} = \frac{1}{-1} = -1$.
$R.H.L. = \lim _{x \rightarrow 0^{+}} \frac{x}{|x|+x^2} = \lim _{x \rightarrow 0^{+}} \frac{x}{x+x^2} = \lim _{x \rightarrow 0^{+}} \frac{1}{1+x} = \frac{1}{1} = 1$.
Since $L.H.L. \neq R.H.L.$,the limit does not exist.

(C) Given $\lim _{x \rightarrow 1} \frac{x^2-ax+b}{x-1}=7$.
Since the limit exists and the denominator approaches $0$ as $x \rightarrow 1$,the numerator must also approach $0$ at $x=1$.
Therefore,$(1)^2 - a(1) + b = 0$,which implies $1 - a + b = 0$,or $a - b = 1 \dots (i)$.
Using $L$'$H$ôpital's rule or factoring,we have $\lim _{x \rightarrow 1} \frac{d}{dx}(x^2 - ax + b) = 7$.
$\Rightarrow \lim _{x}$ ${\rightarrow 1} (2x - a) = 7$.
Substituting $x=1$,we get $2(1) - a = 7$,so $2 - a = 7$,which gives $a = -5$.
Substituting $a = -5$ into equation $(i)$,we get $-5 - b = 1$,so $b = -6$.
Thus,$a + b = -5 + (-6) = -11$.

(B) Given $\lim _{x \rightarrow \infty}\left(\frac{x^2+x+1}{x+1}-a x-b\right)=4$.
Simplify the expression inside the limit:
$\lim _{x \rightarrow \infty}\left(\frac{x^2+x+1-ax(x+1)-b(x+1)}{x+1}\right)=4$
$\lim _{x \rightarrow \infty}\left(\frac{x^2+x+1-ax^2-ax-bx-b}{x+1}\right)=4$
$\lim _{x \rightarrow \infty}\left(\frac{(1-a)x^2+(1-a-b)x+(1-b)}{x+1}\right)=4$
For the limit to be finite,the coefficient of $x^2$ must be zero:
$1-a=0 \Rightarrow a=1$.
Now,substitute $a=1$ into the expression:
$\lim _{x \rightarrow \infty}\left(\frac{(1-1-b)x+(1-b)}{x+1}\right)=4$
$\lim _{x \rightarrow \infty}\left(\frac{-bx+(1-b)}{x+1}\right)=4$
Divide numerator and denominator by $x$:
$\lim _{x \rightarrow \infty}\left(\frac{-b+\frac{1-b}{x}}{1+\frac{1}{x}}\right)=4$
$-b=4 \Rightarrow b=-4$.
Thus,$a=1$ and $b=-4$.

(A) We evaluate the limit: $\lim _{x \rightarrow 0} \frac{9^x-4^x}{x(9^x+4^x)}$
Using the standard limit formula $\lim _{x \rightarrow 0} \frac{a^x-1}{x} = \log(a)$:
$= \lim _{x \rightarrow 0} \left( \frac{9^x-1 - (4^x-1)}{x} \right) \cdot \frac{1}{9^x+4^x}$
$= \left( \lim _{x \rightarrow 0} \frac{9^x-1}{x} - \lim _{x \rightarrow 0} \frac{4^x-1}{x} \right) \cdot \lim _{x \rightarrow 0} \frac{1}{9^x+4^x}$
$= (\log(9) - \log(4)) \cdot \frac{1}{1+1}$
$= \log \left( \frac{9}{4} \right) \cdot \frac{1}{2}$
$= \log \left( \left( \frac{3}{2} \right)^2 \right) \cdot \frac{1}{2}$
$= 2 \log \left( \frac{3}{2} \right) \cdot \frac{1}{2} = \log \left( \frac{3}{2} \right)$

(C) Let $L = \lim _{x \rightarrow 2}\left(\frac{5^x+5^{3-x}-30}{5^{3-x}-5^{\frac{x}{2}}}\right)$.
Let $t = 5^{\frac{x}{2}}$. As $x \rightarrow 2$,$t \rightarrow 5^1 = 5$.
Then $5^x = t^2$ and $5^{3-x} = \frac{125}{5^x} = \frac{125}{t^2}$.
Substituting these into the limit:
$L = \lim _{t \rightarrow 5} \frac{t^2 + \frac{125}{t^2} - 30}{\frac{125}{t^2} - t} = \lim _{t \rightarrow 5} \frac{t^4 - 30t^2 + 125}{125 - t^3}$.
Factorizing the numerator: $t^4 - 30t^2 + 125 = (t^2 - 25)(t^2 - 5) = (t-5)(t+5)(t^2-5)$.
Factorizing the denominator: $125 - t^3 = (5-t)(25 + 5t + t^2) = -(t-5)(25 + 5t + t^2)$.
Thus,$L = \lim _{t \rightarrow 5} \frac{(t-5)(t+5)(t^2-5)}{-(t-5)(25 + 5t + t^2)} = \lim _{t \rightarrow 5} \frac{-(t+5)(t^2-5)}{25 + 5t + t^2}$.
Substituting $t = 5$:
$L = \frac{-(5+5)(25-5)}{25 + 5(5) + 25} = \frac{-10 \times 20}{75} = \frac{-200}{75} = \frac{-8}{3}$.

(A) Let $f(x) = \frac{3^x + 3^{3-x} - 12}{3^{3-x} - 3^{x/2}}$.
Multiply the numerator and denominator by $3^x$:
$\lim _{x \rightarrow 2} \frac{3^{2x} - 12 \cdot 3^x + 27}{3^3 - 3^{3x/2}}$
Let $t = 3^{x/2}$. As $x \rightarrow 2$,$t \rightarrow 3$.
The expression becomes $\lim _{t \rightarrow 3} \frac{t^4 - 12t^2 + 27}{27 - t^3}$.
Factor the numerator: $t^4 - 12t^2 + 27 = (t^2 - 9)(t^2 - 3) = (t-3)(t+3)(t^2-3)$.
Factor the denominator: $27 - t^3 = (3-t)(9 + 3t + t^2) = -(t-3)(t^2 + 3t + 9)$.
$\lim _{t \rightarrow 3} \frac{(t-3)(t+3)(t^2-3)}{-(t-3)(t^2 + 3t + 9)} = \lim _{t \rightarrow 3} \frac{-(t+3)(t^2-3)}{t^2 + 3t + 9}$.
Substitute $t=3$: $\frac{-(3+3)(9-3)}{9+9+9} = \frac{-6 \cdot 6}{27} = \frac{-36}{27} = -\frac{4}{3}$.

(C) Let $L = \lim _{x \rightarrow 0^+} ((\sin x)^{\frac{1}{x}} + (\frac{1}{x})^{\sin x})$.
First,consider $L_1 = \lim _{x \rightarrow 0^+} (\sin x)^{\frac{1}{x}}$. As $x \rightarrow 0^+$,$\sin x \rightarrow 0^+$ and $\frac{1}{x} \rightarrow \infty$. Since $0$ raised to any positive power is $0$,$L_1 = 0$.
Now,consider $L_2 = \lim _{x \rightarrow 0^+} (\frac{1}{x})^{\sin x}$.
Taking the natural logarithm on both sides: $\ln L_2 = \lim _{x \rightarrow 0^+} \sin x \ln(\frac{1}{x}) = \lim _{x \rightarrow 0^+} -\sin x \ln x$.
This is an indeterminate form of type $0 \times \infty$. We rewrite it as: $\ln L_2 = -\lim _{x \rightarrow 0^+} \frac{\ln x}{\csc x}$.
Applying $L'H\hat{o}pital's$ Rule: $\ln L_2 = -\lim _{x \rightarrow 0^+} \frac{1/x}{-\csc x \cot x} = \lim _{x \rightarrow 0^+} \frac{\sin x \tan x}{x} = \lim _{x \rightarrow 0^+} (\frac{\sin x}{x}) \cdot \tan x$.
Since $\lim _{x \rightarrow 0^+} \frac{\sin x}{x} = 1$ and $\lim _{x \rightarrow 0^+} \tan x = 0$,we have $\ln L_2 = 1 \times 0 = 0$.
Thus,$L_2 = e^0 = 1$.
Therefore,$L = L_1 + L_2 = 0 + 1 = 1$.

(C) To evaluate the limit,we rationalize the numerator:
$\lim _{y \rightarrow 0} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4} \times \frac{\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}}{\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}}$
$= \lim _{y \rightarrow 0} \frac{1+\sqrt{1+y^4}-2}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})} = \lim _{y \rightarrow 0} \frac{\sqrt{1+y^4}-1}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})}$
Now,rationalize the remaining square root term:
$= \lim _{y \rightarrow 0} \frac{(\sqrt{1+y^4}-1)(\sqrt{1+y^4}+1)}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})(\sqrt{1+y^4}+1)}$
$= \lim _{y \rightarrow 0} \frac{1+y^4-1}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})(\sqrt{1+y^4}+1)}$
$= \lim _{y \rightarrow 0} \frac{y^4}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})(\sqrt{1+y^4}+1)}$
$= \frac{1}{(\sqrt{1+1}+\sqrt{2})(1+1)} = \frac{1}{(2\sqrt{2})(2)} = \frac{1}{4\sqrt{2}}$

(C) We need to evaluate the limit: $\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}$.
Using the trigonometric identity $\sin(\pi - \theta) = \sin \theta$,we can rewrite the numerator:
$\sin(\pi \cos^2 x) = \sin(\pi - \pi \cos^2 x) = \sin(\pi(1 - \cos^2 x)) = \sin(\pi \sin^2 x)$.
Now the limit becomes: $\lim _{x \rightarrow 0} \frac{\sin(\pi \sin^2 x)}{x^2}$.
Multiply and divide by $\pi \sin^2 x$:
$= \lim _{x \rightarrow 0} \left( \frac{\sin(\pi \sin^2 x)}{\pi \sin^2 x} \times \frac{\pi \sin^2 x}{x^2} \right)$.
Since $\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$ and $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$,we have:
$= 1 \times \pi \times \lim _{x \rightarrow 0} \left( \frac{\sin x}{x} \right)^2 = 1 \times \pi \times 1^2 = \pi$.

(A) We know that $1 - \cos 2x = 2 \sin^2 x$.
Substituting this in the limit,we get:
$\lim _{x \rightarrow 0} \frac{2 \sin^2 x (3 + \cos x)}{x \tan 4x}$
$= \lim _{x}$ ${\rightarrow 0} \left[ 2 \cdot \left( \frac{\sin x}{x} \right)^2 \cdot \frac{x}{\tan 4x} \cdot (3 + \cos x) \right]$
$= \lim _{x}$ ${\rightarrow 0} \left[ 2 \cdot \left( \frac{\sin x}{x} \right)^2 \cdot \frac{1}{4 \cdot \frac{\tan 4x}{4x}} \cdot (3 + \cos x) \right]$
Using the standard limits $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$ and $\lim_{\theta \rightarrow 0} \frac{\tan \theta}{\theta} = 1$:
$= 2 \cdot (1)^2 \cdot \frac{1}{4 \cdot 1} \cdot (3 + \cos 0)$
$= 2 \cdot \frac{1}{4} \cdot (3 + 1)$
$= \frac{2}{4} \cdot 4 = 2$

(C) Let $L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{(1-\sin x)(8 x^3-\pi^3) \cos x}{(\pi-2 x)^4}$.
We can rewrite the expression as:
$L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{(1-\sin x)(2x-\pi)(4x^2+\pi^2+2\pi x) \cos x}{16(\frac{\pi}{2}-x)^4}$.
Let $x - \frac{\pi}{2} = h$,so $x = \frac{\pi}{2} + h$. As $x \rightarrow \frac{\pi}{2}$,$h \rightarrow 0$.
Then $\cos x = \cos(\frac{\pi}{2} + h) = -\sin h$ and $1 - \sin x = 1 - \sin(\frac{\pi}{2} + h) = 1 - \cos h$.
Also,$2x - \pi = 2(\frac{\pi}{2} + h) - \pi = 2h$.
Substituting these into the limit:
$L = \lim _{h}$ ${\rightarrow 0} \frac{(1-\cos h)(2h)(4(\frac{\pi}{2}+h)^2 + \pi^2 + 2\pi(\frac{\pi}{2}+h))(-\sin h)}{16(-h)^4}$.
$L = \lim _{h \rightarrow 0} \frac{(1-\cos h)(2h)(3\pi^2 + 8\pi h + 4h^2)(-\sin h)}{16h^4}$.
$L = \lim _{h \rightarrow 0} \frac{-(1-\cos h)}{h^2} \cdot \frac{\sin h}{h} \cdot \frac{2h(3\pi^2 + 8\pi h + 4h^2)}{16h}$.
$L = -(\frac{1}{2}) \cdot (1) \cdot \frac{2(3\pi^2)}{16} = -\frac{1}{2} \cdot \frac{6\pi^2}{16} = -\frac{3\pi^2}{16}$.

(C) We evaluate the limit: $\lim _{x \rightarrow 0} \frac{x \tan 2x - 2x \tan x}{(1 - \cos 2x)^2}$.
Using the identity $1 - \cos 2x = 2 \sin^2 x$,the denominator becomes $(2 \sin^2 x)^2 = 4 \sin^4 x$.
So,the expression is $\lim _{x \rightarrow 0} \frac{x(\tan 2x - 2 \tan x)}{4 \sin^4 x}$.
Using Taylor series expansions: $\tan \theta = \theta + \frac{\theta^3}{3} + \dots$ and $\sin x \approx x$.
$\tan 2x = 2x + \frac{(2x)^3}{3} + O(x^5) = 2x + \frac{8x^3}{3} + O(x^5)$.
$2 \tan x = 2(x + \frac{x^3}{3} + O(x^5)) = 2x + \frac{2x^3}{3} + O(x^5)$.
Numerator: $x[(2x + \frac{8x^3}{3}) - (2x + \frac{2x^3}{3})] = x[\frac{6x^3}{3}] = 2x^4$.
Denominator: $4 \sin^4 x \approx 4x^4$.
Limit: $\lim _{x \rightarrow 0} \frac{2x^4}{4x^4} = \frac{2}{4} = \frac{1}{2}$.

(B) Let $l = \lim _{x}$ ${\rightarrow \frac{\pi}{2}} \left[ \frac{1-\tan \left(\frac{x}{2}\right)}{1+\tan \left(\frac{x}{2}\right)} \right] \left[ \frac{1-\sin x}{(\pi-2 x)^3} \right]$
Since $\tan(\frac{\pi}{4} - \frac{x}{2}) = \frac{1-\tan(x/2)}{1+\tan(x/2)}$,we have:
$l = \lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)(1-\sin x)}{(\pi-2 x)^3}$
Substitute $\pi-2x = \theta$,then $x = \frac{\pi}{2} - \frac{\theta}{2}$. As $x \rightarrow \frac{\pi}{2}$,$\theta \rightarrow 0$.
Also,$\frac{\pi}{4} - \frac{x}{2} = \frac{\pi}{4} - (\frac{\pi}{4} - \frac{\theta}{4}) = \frac{\theta}{4}$ and $1-\sin x = 1-\sin(\frac{\pi}{2}-\frac{\theta}{2}) = 1-\cos(\frac{\theta}{2}) = 2\sin^2(\frac{\theta}{4})$.
Substituting these:
$l = \lim _{\theta \rightarrow 0} \frac{\tan(\frac{\theta}{4}) \cdot 2\sin^2(\frac{\theta}{4})}{\theta^3}$
$l = \lim _{\theta}$ ${\rightarrow 0} \frac{\tan(\frac{\theta}{4})}{\frac{\theta}{4} \cdot 4} \cdot \frac{2\sin^2(\frac{\theta}{4})}{(\frac{\theta}{4})^2 \cdot 16}$
$l = \frac{2}{4 \cdot 16} \lim _{\theta}$ ${\rightarrow 0} \left( \frac{\tan(\frac{\theta}{4})}{\frac{\theta}{4}} \right) \left( \frac{\sin(\frac{\theta}{4})}{\frac{\theta}{4}} \right)^2$
$l = \frac{2}{64} (1)(1)^2 = \frac{1}{32}$

(B) Let the diagonals be $\vec{d_1} = 8\hat{i} - 6\hat{j} + 0\hat{k}$ and $\vec{d_2} = 3\hat{i} + 4\hat{j} - 12\hat{k}$.
The area of a parallelogram with diagonals $\vec{d_1}$ and $\vec{d_2}$ is given by the formula $\text{Area} = \frac{1}{2} |\vec{d_1} \times \vec{d_2}|$.
First,calculate the cross product $\vec{d_1} \times \vec{d_2}$:
$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 8 & -6 & 0 \\ 3 & 4 & -12 \end{vmatrix}$
$= \hat{i}((-6)(-12) - (0)(4)) - \hat{j}((8)(-12) - (0)(3)) + \hat{k}((8)(4) - (-6)(3))$
$= \hat{i}(72 - 0) - \hat{j}(-96 - 0) + \hat{k}(32 + 18)$
$= 72\hat{i} + 96\hat{j} + 50\hat{k}$.
Now,find the magnitude of the cross product:
$|\vec{d_1} \times \vec{d_2}| = \sqrt{72^2 + 96^2 + 50^2}$
$= \sqrt{5184 + 9216 + 2500}$
$= \sqrt{16900} = 130$.
Therefore,the area of the parallelogram is $\frac{1}{2} \times 130 = 65$ sq. units.

(A) We analyze the function $f(x)$ in different intervals:
For $x \in [-1, 0)$,$|x| = -x$ and $[x] = -1$,so $f(x) = -x - 1$.
For $x \in [0, 1)$,$|x| = x$ and $[x] = 0$,so $f(x) = x + 0 = x$.
For $x \in [1, 2)$,$|x| = x$,so $f(x) = x + x = 2x$.
For $x \in [2, 3]$,$|x| = x$,so $f(x) = x + x = 2x$.
Thus,$f(x) = \begin{cases} -x-1, & -1 \leq x < 0 \\ x, & 0 \leq x < 1 \\ 2x, & 1 \leq x \leq 3 \end{cases}$.
Checking continuity at $x=0$: $\lim_{x \to 0^-} f(x) = -1$ and $f(0) = 0$. Since $-1 \neq 0$,$f$ is discontinuous at $x=0$.
Checking continuity at $x=1$: $\lim_{x \to 1^-} f(x) = 1$ and $f(1) = 2(1) = 2$. Since $1 \neq 2$,$f$ is discontinuous at $x=1$.
Checking continuity at $x=2$: $\lim_{x \to 2^-} f(x) = 2(2) = 4$ and $f(2) = 2(2) = 4$. Since the limit equals the function value,$f$ is continuous at $x=2$.
Therefore,$f$ is discontinuous at only two points,$x=0$ and $x=1$.

(B) The sum of all probabilities in a probability distribution must be equal to $1$.
$\sum P(X) = 1 \Rightarrow K^2 + 2K + K + 2K + 5K^2 = 1$
$6K^2 + 5K - 1 = 0$
Factoring the quadratic equation: $(6K - 1)(K + 1) = 0$
This gives $K = \frac{1}{6}$ or $K = -1$.
Since the probability $P(X)$ cannot be negative,we reject $K = -1$. Thus,$K = \frac{1}{6}$.
We need to find $P(X > 2) = P(X = 3) + P(X = 4) + P(X = 5)$.
$P(X > 2) = K + 2K + 5K^2 = 3K + 5K^2$.
Substituting $K = \frac{1}{6}$:
$P(X > 2) = 3(\frac{1}{6}) + 5(\frac{1}{6})^2 = \frac{1}{2} + \frac{5}{36} = \frac{18}{36} + \frac{5}{36} = \frac{23}{36}$.

(A) Let $B(t)$ be the number of bacteria at time $t$. The rate of growth is given by $\frac{dB}{dt} = \lambda B$.
Integrating this,we get $B(t) = B_0 e^{\lambda t}$,where $B_0 = 1000$.
Given that at $t = 2$,$B(2) = 1000 + 20\% \text{ of } 1000 = 1200$.
So,$1200 = 1000 e^{2\lambda} \Rightarrow e^{2\lambda} = \frac{6}{5} \Rightarrow 2\lambda = \log_{e}\left(\frac{6}{5}\right) \Rightarrow \lambda = \frac{1}{2} \log_{e}\left(\frac{6}{5}\right)$.
We are given $B(T) = 2000$ where $T = \frac{k}{\log_{e}\left(\frac{6}{5}\right)}$.
Using $B(T) = B_0 e^{\lambda T}$,we have $2000 = 1000 e^{\lambda T} \Rightarrow 2 = e^{\lambda T} \Rightarrow \log_{e} 2 = \lambda T$.
Substituting $\lambda$ and $T$: $\log_{e} 2 = \left(\frac{1}{2} \log_{e}\left(\frac{6}{5}\right)\right) \times \left(\frac{k}{\log_{e}\left(\frac{6}{5}\right)}\right) = \frac{k}{2}$.
Thus,$k = 2 \log_{e} 2$.
Finally,$\left(\frac{k}{\log_{e} 2}\right)^{2} = \left(\frac{2 \log_{e} 2}{\log_{e} 2}\right)^{2} = 2^{2} = 4$.

(C) Three vectors are coplanar if their scalar triple product is zero.
The scalar triple product is given by the determinant:
$\left|\begin{array}{ccc}-\lambda^2 & 1 & 1 \\ 1 & -\lambda^2 & 1 \\ 1 & 1 & -\lambda^2\end{array}\right| = 0$
Expanding the determinant along the first row:
$-\lambda^2(\lambda^4 - 1) - 1(-\lambda^2 - 1) + 1(1 + \lambda^2) = 0$
$-\lambda^6 + \lambda^2 + \lambda^2 + 1 + 1 + \lambda^2 = 0$
$-\lambda^6 + 3\lambda^2 + 2 = 0$
$\lambda^6 - 3\lambda^2 - 2 = 0$
Let $x = \lambda^2$. Then $x^3 - 3x - 2 = 0$.
By testing values,$x = -1$ is a root: $(-1)^3 - 3(-1) - 2 = -1 + 3 - 2 = 0$.
Dividing by $(x+1)$,we get $(x+1)(x^2 - x - 2) = 0$,which factors to $(x+1)(x-2)(x+1) = 0$.
So,$(x+1)^2(x-2) = 0$.
Since $x = \lambda^2$,we have $(\lambda^2+1)^2(\lambda^2-2) = 0$.
For real $\lambda$,$\lambda^2+1$ cannot be zero.
Thus,$\lambda^2 - 2 = 0$,which gives $\lambda = \pm \sqrt{2}$.
There are $2$ distinct real values of $\lambda$.

(C) Given the function $g(u) = 2 \tan^{-1}(e^u) - \frac{\pi}{2}$.
To check for odd/even,we evaluate $g(-u)$:
$g(-u) = 2 \tan^{-1}(e^{-u}) - \frac{\pi}{2}$.
Using the identity $\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}$,we know $\tan^{-1}(e^{-u}) = \frac{\pi}{2} - \tan^{-1}(e^u)$.
Substituting this into $g(-u)$:
$g(-u) = 2 \left( \frac{\pi}{2} - \tan^{-1}(e^u) \right) - \frac{\pi}{2} = \pi - 2 \tan^{-1}(e^u) - \frac{\pi}{2} = \frac{\pi}{2} - 2 \tan^{-1}(e^u) = -g(u)$.
Since $g(-u) = -g(u)$,the function is odd.
To check for monotonicity,we find the derivative $g'(u)$:
$g'(u) = \frac{d}{du} (2 \tan^{-1}(e^u) - \frac{\pi}{2}) = 2 \cdot \frac{1}{1 + (e^u)^2} \cdot e^u = \frac{2e^u}{1 + e^{2u}}$.
Since $e^u > 0$ for all $u \in (-\infty, \infty)$,$g'(u) > 0$ for all $u$.
Therefore,$g$ is strictly increasing in $(-\infty, \infty)$.

(A) The position vectors of the vertices are $\vec{p} = -2\hat{i} - \hat{j}$,$\vec{q} = 4\hat{i}$,$\vec{r} = 3\hat{i} + 3\hat{j}$,and $\vec{s} = -3\hat{i} + 2\hat{j}$.
First,we check the midpoints of the diagonals $PR$ and $QS$:
Midpoint of $PR = \frac{\vec{p} + \vec{r}}{2} = \frac{(-2\hat{i} - \hat{j}) + (3\hat{i} + 3\hat{j})}{2} = \frac{\hat{i} + 2\hat{j}}{2} = \frac{1}{2}\hat{i} + \hat{j}$.
Midpoint of $QS = \frac{\vec{q} + \vec{s}}{2} = \frac{(4\hat{i}) + (-3\hat{i} + 2\hat{j})}{2} = \frac{\hat{i} + 2\hat{j}}{2} = \frac{1}{2}\hat{i} + \hat{j}$.
Since the midpoints of the diagonals coincide,$PQRS$ is a parallelogram.
Next,we calculate the side vectors:
$\vec{PQ} = \vec{q} - \vec{p} = 4\hat{i} - (-2\hat{i} - \hat{j}) = 6\hat{i} + \hat{j}$.
$\vec{PS} = \vec{s} - \vec{p} = (-3\hat{i} + 2\hat{j}) - (-2\hat{i} - \hat{j}) = -\hat{i} + 3\hat{j}$.
Check for rectangle (dot product of adjacent sides):
$\vec{PQ} \cdot \vec{PS} = (6\hat{i} + \hat{j}) \cdot (-\hat{i} + 3\hat{j}) = (6)(-1) + (1)(3) = -6 + 3 = -3 \neq 0$.
Since the dot product is not zero,the sides are not perpendicular,so it is not a rectangle.
Check for rhombus (lengths of adjacent sides):
$|\vec{PQ}| = \sqrt{6^2 + 1^2} = \sqrt{36 + 1} = \sqrt{37}$.
$|\vec{PS}| = \sqrt{(-1)^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}$.
Since $|\vec{PQ}| \neq |\vec{PS}|$,the sides are not equal,so it is not a rhombus.
Thus,$PQRS$ is a parallelogram,which is neither a rhombus nor a rectangle.

(B) The given equation of the curve is $y(x-2)(x-3)=x+6$.
At the $Y$-axis,$x=0$. Substituting $x=0$ into the equation: $y(0-2)(0-3)=0+6 \Rightarrow y(-2)(-3)=6 \Rightarrow 6y=6 \Rightarrow y=1$.
So,the point of intersection is $(0, 1)$.
Now,rewrite the equation as $y = \frac{x+6}{x^2-5x+6}$.
Differentiating with respect to $x$ using the quotient rule:
$\frac{dy}{dx} = \frac{(x^2-5x+6)(1) - (x+6)(2x-5)}{(x^2-5x+6)^2}$.
At $x=0$,the slope of the tangent $m_t = \frac{(0-0+6)(1) - (0+6)(0-5)}{(0-0+6)^2} = \frac{6 - (-30)}{36} = \frac{36}{36} = 1$.
The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{1}{1} = -1$.
The equation of the normal at $(0, 1)$ is $y - 1 = -1(x - 0)$,which simplifies to $y = -x + 1$ or $x + y = 1$.
Checking the options,for $(\frac{1}{2}, \frac{1}{2})$,we have $\frac{1}{2} + \frac{1}{2} = 1$,which satisfies the equation.
Thus,the normal passes through $(\frac{1}{2}, \frac{1}{2})$.

(C) Given the parametric equations of the curve: $x = \theta + \sin \theta$ and $y = 1 + \cos \theta$.
First,find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = 1 + \cos \theta$ and $\frac{dy}{d\theta} = -\sin \theta$.
The slope of the tangent is $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-\sin \theta}{1 + \cos \theta}$.
At $\theta = \frac{\pi}{2}$,the point $(x, y)$ is:
$x = \frac{\pi}{2} + \sin \frac{\pi}{2} = \frac{\pi}{2} + 1$ and $y = 1 + \cos \frac{\pi}{2} = 1$.
The slope of the tangent at $\theta = \frac{\pi}{2}$ is $m_t = \frac{-\sin(\pi/2)}{1 + \cos(\pi/2)} = \frac{-1}{1 + 0} = -1$.
The slope of the normal is $m_n = -\frac{1}{m_t} = -\frac{1}{-1} = 1$.
The equation of the normal at $(\frac{\pi}{2} + 1, 1)$ is:
$y - 1 = 1(x - (\frac{\pi}{2} + 1))$
$y - 1 = x - \frac{\pi}{2} - 1$
$x - y - \frac{\pi}{2} = 0$,which can be written as $2x - 2y - \pi = 0$.

(D) Given curve is $y = x \log x$.
Finding the derivative with respect to $x$:
$\frac{dy}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = 1 + \log x$.
The slope of the tangent at any point $(x, y)$ is $m_t = 1 + \log x$.
The slope of the normal is $m_n = -\frac{1}{m_t} = -\frac{1}{1 + \log x}$.
The given line is $2x - 2y = 3$,which can be written as $2y = 2x - 3$ or $y = x - \frac{3}{2}$.
The slope of this line is $1$.
Since the normal is parallel to the line,their slopes must be equal:
$-\frac{1}{1 + \log x} = 1$.
$-1 = 1 + \log x \implies \log x = -2$.
$x = e^{-2}$.
Now,find the $y$-coordinate:
$y = x \log x = e^{-2} \cdot (-2) = -2e^{-2}$.
Thus,the coordinates of the point are $(e^{-2}, -2e^{-2})$.

(B) Given the parametric equations of the curve: $x=\theta+\sin \theta$ and $y=1+\cos \theta$.
First,find the derivative $\frac{dy}{dx}$ with respect to $\theta$:
$\frac{dx}{d\theta} = 1+\cos \theta$ and $\frac{dy}{d\theta} = -\sin \theta$.
Thus,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-\sin \theta}{1+\cos \theta}$.
At $\theta = \frac{\pi}{2}$,the slope of the tangent is $m_t = \frac{-\sin(\pi/2)}{1+\cos(\pi/2)} = \frac{-1}{1+0} = -1$.
The slope of the normal is $m_n = -\frac{1}{m_t} = -\frac{1}{-1} = 1$.
Now,find the point $(x, y)$ at $\theta = \frac{\pi}{2}$:
$x = \frac{\pi}{2} + \sin(\frac{\pi}{2}) = \frac{\pi}{2} + 1$ and $y = 1 + \cos(\frac{\pi}{2}) = 1 + 0 = 1$.
The equation of the normal is $(y - y_1) = m_n(x - x_1)$:
$(y - 1) = 1(x - (1 + \frac{\pi}{2}))$.
$y - 1 = x - 1 - \frac{\pi}{2}$.
$x - y - \frac{\pi}{2} = 0$,which simplifies to $2x - 2y - \pi = 0$.

(A) Given the curve $\sqrt{x}+\sqrt{y}=\sqrt{a}$.
Differentiating with respect to $x$,we get $\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0$.
Thus,the slope of the tangent is $\frac{dy}{dx}=-\sqrt{\frac{y}{x}}$.
Let the point of tangency be $(x_1, y_1)$. The equation of the tangent is $(y-y_1)=-\sqrt{\frac{y_1}{x_1}}(x-x_1)$.
Multiplying by $\sqrt{x_1}$,we get $\sqrt{x_1}y - \sqrt{x_1}y_1 = -\sqrt{y_1}x + \sqrt{y_1}x_1$.
Rearranging gives $\sqrt{y_1}x + \sqrt{x_1}y = \sqrt{x_1}y_1 + \sqrt{y_1}x_1 = \sqrt{x_1y_1}(\sqrt{x_1}+\sqrt{y_1})$.
Since $\sqrt{x_1}+\sqrt{y_1}=\sqrt{a}$,the equation becomes $\sqrt{y_1}x + \sqrt{x_1}y = \sqrt{x_1y_1a}$.
Dividing by $\sqrt{x_1y_1a}$,we get $\frac{x}{\sqrt{x_1a}} + \frac{y}{\sqrt{y_1a}} = 1$.
The $x$-intercept is $\sqrt{x_1a}$ and the $y$-intercept is $\sqrt{y_1a}$.
The sum of intercepts is $\sqrt{a}(\sqrt{x_1}+\sqrt{y_1}) = \sqrt{a} \times \sqrt{a} = a$.

(B) Given equation of the curve is $y=1-e^{\frac{x}{3}} \dots (i)$.
Since the curve intersects the $Y$-axis,we set $x=0$.
Substituting $x=0$ in $(i)$,we get $y=1-e^{0}=1-1=0$.
Thus,the point of intersection is $(0, 0)$.
The slope of the tangent is given by $\frac{dy}{dx}$.
Differentiating $(i)$ with respect to $x$,we get $\frac{dy}{dx} = -\frac{1}{3} e^{\frac{x}{3}}$.
At the point $(0, 0)$,the slope $m = \left(\frac{dy}{dx}\right)_{(0,0)} = -\frac{1}{3} e^{0} = -\frac{1}{3}$.
The equation of the tangent line passing through $(0, 0)$ with slope $m = -\frac{1}{3}$ is $y - 0 = -\frac{1}{3}(x - 0)$.
Multiplying by $3$,we get $3y = -x$,which simplifies to $x + 3y = 0$.

(A) Given the curve $y(x) = 1 + \sqrt{4x-3}$.
First,we find the derivative to determine the slope of the tangent:
$\frac{dy}{dx} = \frac{1}{2\sqrt{4x-3}} \times 4 = \frac{2}{\sqrt{4x-3}}$.
Given the slope of the tangent at point $P$ is $\frac{2}{3}$,we set $\frac{2}{\sqrt{4x-3}} = \frac{2}{3}$.
This implies $\sqrt{4x-3} = 3$,so $4x-3 = 9$,which gives $4x = 12$,or $x = 3$.
Substituting $x = 3$ into the curve equation,we get $y = 1 + \sqrt{4(3)-3} = 1 + \sqrt{9} = 1 + 3 = 4$.
Thus,the point $P$ is $(3, 4)$.
The slope of the normal at $P$ is the negative reciprocal of the tangent slope: $m_{normal} = -\frac{1}{2/3} = -\frac{3}{2}$.
The equation of the normal at $(3, 4)$ is $y - 4 = -\frac{3}{2}(x - 3)$.
Multiplying by $2$,we get $2y - 8 = -3x + 9$,which simplifies to $3x + 2y - 17 = 0$.
Checking the options:
For $(1, 7)$: $3(1) + 2(7) - 17 = 3 + 14 - 17 = 0$. This satisfies the equation.
Therefore,the normal passes through $(1, 7)$.

(A) Given the parametric equations $x = a \cos^3 \theta$ and $y = a \sin^3 \theta$.
Differentiating with respect to $\theta$,we get $\frac{dx}{d\theta} = -3a \cos^2 \theta \sin \theta$ and $\frac{dy}{d\theta} = 3a \sin^2 \theta \cos \theta$.
The slope of the tangent is $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta} = -\tan \theta$.
At $\theta = \frac{\pi}{4}$,the slope is $m = -\tan(\frac{\pi}{4}) = -1$.
The coordinates of the point at $\theta = \frac{\pi}{4}$ are $x = a \cos^3(\frac{\pi}{4}) = a(\frac{1}{\sqrt{2}})^3 = \frac{a}{2 \sqrt{2}}$ and $y = a \sin^3(\frac{\pi}{4}) = a(\frac{1}{\sqrt{2}})^3 = \frac{a}{2 \sqrt{2}}$.
The equation of the tangent line is $y - y_1 = m(x - x_1)$.
$y - \frac{a}{2 \sqrt{2}} = -1(x - \frac{a}{2 \sqrt{2}})$.
$y - \frac{a}{2 \sqrt{2}} = -x + \frac{a}{2 \sqrt{2}}$.
$x + y = \frac{a}{2 \sqrt{2}} + \frac{a}{2 \sqrt{2}} = \frac{2a}{2 \sqrt{2}} = \frac{a}{\sqrt{2}}$.

(A) The given equation of the curve is $x^4-2xy^2+y^2+3x-3y=0 \dots (i)$.
To find the angle at which the curve cuts the $X$-axis,we need to find the slope of the tangent at the point $(0,0)$.
Differentiating equation $(i)$ with respect to $x$,we get:
$4x^3 - 2(y^2 + x \cdot 2y \frac{dy}{dx}) + 2y \frac{dy}{dx} + 3 - 3 \frac{dy}{dx} = 0$.
Simplifying the expression:
$4x^3 - 2y^2 - 4xy \frac{dy}{dx} + 2y \frac{dy}{dx} + 3 - 3 \frac{dy}{dx} = 0$.
Now,substitute $(x,y) = (0,0)$ into the derivative equation:
$4(0)^3 - 2(0)^2 - 4(0)(0) \frac{dy}{dx} + 2(0) \frac{dy}{dx} + 3 - 3 \frac{dy}{dx} = 0$.
$0 - 0 - 0 + 0 + 3 - 3 \frac{dy}{dx} = 0$.
$3 = 3 \frac{dy}{dx} \implies \frac{dy}{dx} = 1$.
Since the slope $m = \tan \theta = 1$,we have $\theta = \tan^{-1}(1) = \frac{\pi}{4}$.

(C) Given the curve equation is $y^2 = px^3 + q \dots (i)$.
Since the point $(2, 3)$ lies on the curve,we substitute $x=2$ and $y=3$ into $(i)$:
$3^2 = p(2)^3 + q \Rightarrow 9 = 8p + q \dots (ii)$.
Differentiating both sides of $(i)$ with respect to $x$:
$2y \frac{dy}{dx} = 3px^2 \Rightarrow \frac{dy}{dx} = \frac{3px^2}{2y}$.
The slope of the tangent at $(2, 3)$ is $\left(\frac{dy}{dx}\right)_{(2,3)} = \frac{3p(2)^2}{2(3)} = \frac{12p}{6} = 2p$.
The given tangent line is $y = 4x - 5$,which has a slope of $4$.
Equating the slopes: $2p = 4 \Rightarrow p = 2$.
Substituting $p = 2$ into $(ii)$:
$9 = 8(2) + q \Rightarrow 9 = 16 + q \Rightarrow q = -7$.
Thus,the values are $p = 2$ and $q = -7$.

(D) Given the parametric equations $x=\sqrt{t}$ and $y=t-\frac{1}{\sqrt{t}}$.
At $t=4$,$x=\sqrt{4}=2$ and $y=4-\frac{1}{\sqrt{4}}=4-\frac{1}{2}=\frac{7}{2}$.
We find the derivative $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.
$\frac{dx}{dt} = \frac{1}{2\sqrt{t}}$ and $\frac{dy}{dt} = 1 + \frac{1}{2t\sqrt{t}}$.
At $t=4$,$\frac{dx}{dt} = \frac{1}{4}$ and $\frac{dy}{dt} = 1 + \frac{1}{16} = \frac{17}{16}$.
Thus,the slope of the tangent $m_T = \frac{17/16}{1/4} = \frac{17}{4}$.
The slope of the normal $m_N = -\frac{1}{m_T} = -\frac{4}{17}$.
The equation of the normal is $y - \frac{7}{2} = -\frac{4}{17}(x - 2)$.
Multiplying by $34$,we get $34y - 119 = -8(x - 2) \implies 34y - 119 = -8x + 16$.
Rearranging gives $8x + 34y = 135$.

(C) Given the curve $y=ax^3+bx^2+cx+5$.
Since the curve touches the $X$-axis at $(-2,0)$,it passes through $(-2,0)$ and its derivative at $x=-2$ is $0$.
Substituting $(-2,0)$ into the equation: $0 = a(-8) + b(4) + c(-2) + 5 \Rightarrow -8a + 4b - 2c = -5 \Rightarrow 8a - 4b + 2c = 5 \dots (i)$.
Also,the derivative is $\frac{dy}{dx} = 3ax^2 + 2bx + c$.
At $x=-2$,$\frac{dy}{dx} = 0 \Rightarrow 3a(4) + 2b(-2) + c = 0 \Rightarrow 12a - 4b + c = 0 \dots (ii)$.
The curve cuts the $Y$-axis at $Q$. Putting $x=0$ in the curve equation,$y=5$,so $Q$ is $(0,5)$.
The gradient at $Q$ is $3$,so $\left(\frac{dy}{dx}\right)_{x=0} = 3 \Rightarrow c = 3$.
Substituting $c=3$ into $(i)$ and $(ii)$:
$8a - 4b + 6 = 5 \Rightarrow 8a - 4b = -1 \dots (iii)$.
$12a - 4b + 3 = 0 \Rightarrow 12a - 4b = -3 \dots (iv)$.
Subtracting $(iii)$ from $(iv)$: $4a = -2 \Rightarrow a = -\frac{1}{2}$.
Substituting $a = -\frac{1}{2}$ into $(iii)$: $8(-\frac{1}{2}) - 4b = -1 \Rightarrow -4 - 4b = -1 \Rightarrow -4b = 3 \Rightarrow b = -\frac{3}{4}$.
Thus,$a = -\frac{1}{2}, b = -\frac{3}{4}, c = 3$.

(A) Given curves are $y=10-x^2$ and $y=2+x^2$.
To find the point of intersection,equate the two equations: $10-x^2 = 2+x^2 \Rightarrow 2x^2 = 8 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$.
For $x=2$,$y=6$. For $x=-2$,$y=6$. Let us consider the point $(2, 6)$.
For the first curve $y=10-x^2$,the slope $m_1 = \frac{dy}{dx} = -2x$. At $x=2$,$m_1 = -4$.
For the second curve $y=2+x^2$,the slope $m_2 = \frac{dy}{dx} = 2x$. At $x=2$,$m_2 = 4$.
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: $\tan \theta = \left| \frac{-4 - 4}{1 + (-4)(4)} \right| = \left| \frac{-8}{1 - 16} \right| = \left| \frac{-8}{-15} \right| = \frac{8}{15}$.
Thus,$|\tan \theta| = \frac{8}{15}$.

(B) Given curve is $y=x \log x$ ... $(i)$
Differentiating with respect to $x$,we get $\frac{dy}{dx} = 1 + \log x$.
The slope of the normal is given by $m_n = -\frac{1}{\frac{dy}{dx}} = -\frac{1}{1+\log x}$.
The given line is $2x-2y+3=0$,which can be written as $y = x + \frac{3}{2}$.
The slope of this line is $m = 1$.
Since the normal is parallel to the given line,their slopes must be equal:
$-\frac{1}{1+\log x} = 1$
$\Rightarrow 1+\log x = -1$
$\Rightarrow \log x = -2$
$\Rightarrow x = e^{-2}$.
Substituting $x = e^{-2}$ in $(i)$,we get $y = e^{-2} \log(e^{-2}) = e^{-2}(-2) = -2e^{-2}$.
The point of contact is $(e^{-2}, -2e^{-2})$.
The equation of the normal is $y - y_1 = m(x - x_1)$:
$y - (-2e^{-2}) = 1(x - e^{-2})$
$y + 2e^{-2} = x - e^{-2}$
$x - y = 3e^{-2}$.

(D) Let $\theta$ be the temperature of the ball at any time $t$. According to Newton's law of cooling,$\frac{d\theta}{dt} = -k(\theta - 20)$,where $k > 0$.
Integrating both sides,we get $\ln|\theta - 20| = -kt + C$.
At $t = 0$,$\theta = 80^{\circ} C$,so $C = \ln(80 - 20) = \ln(60)$.
Thus,$\ln|\theta - 20| = -kt + \ln(60) \dots (i)$.
At $t = 5$,$\theta = 60^{\circ} C$,so $\ln(60 - 20) = -5k + \ln(60)$.
$5k = \ln(60) - \ln(40) = \ln(\frac{60}{40}) = \ln(\frac{3}{2})$.
So,$k = \frac{1}{5} \ln(\frac{3}{2})$.
For $t = 20$,substituting $k$ into equation $(i)$:
$\ln|\theta - 20| = -20 \times \frac{1}{5} \ln(\frac{3}{2}) + \ln(60) = -4 \ln(\frac{3}{2}) + \ln(60)$.
$\ln|\theta - 20| = \ln((\frac{2}{3})^4) + \ln(60) = \ln(\frac{16}{81} \times 60) = \ln(\frac{16 \times 20}{27}) = \ln(\frac{320}{27}) \approx \ln(11.85)$.
$\theta - 20 = 11.85 \implies \theta = 31.85^{\circ} C$.

(B) Let $f(x) = x^{\frac{1}{3}}$.
Then,$f'(x) = \frac{1}{3} x^{-\frac{2}{3}} = \frac{1}{3x^{\frac{2}{3}}}$.
We choose $a = 0.027$ because $\sqrt[3]{0.027} = 0.3$,and $h = -0.001$ such that $a + h = 0.026$.
Using the linear approximation formula $f(a + h) \approx f(a) + hf'(a)$:
$f(a) = (0.027)^{\frac{1}{3}} = 0.3$.
$f'(a) = \frac{1}{3(0.027)^{\frac{2}{3}}} = \frac{1}{3(0.09)} = \frac{1}{0.27}$.
Substituting these values:
$f(0.026) \approx 0.3 + (-0.001) \times \frac{1}{0.27}$.
$f(0.026) \approx 0.3 - \frac{0.001}{0.27} \approx 0.3 - 0.0037 = 0.2963$.

(D) Let $f(x) = \cos x$. Then $f^{\prime}(x) = -\sin x$.
We need to find the value of $\cos(30^{\circ} 30^{\prime})$.
Here,$30^{\circ} 30^{\prime} = 30^{\circ} + 30^{\prime} = 30^{\circ} + (0.5)^{\circ}$.
Converting $0.5^{\circ}$ to radians: $0.5 \times 0.0175 = 0.00875 \text{ rad}$.
Let $a = 30^{\circ} = \frac{\pi}{6} \text{ rad}$ and $h = 0.00875 \text{ rad}$.
Using the linear approximation formula $f(a+h) \approx f(a) + h f^{\prime}(a)$:
$f(a+h) \approx \cos(30^{\circ}) + (0.00875)(-\sin(30^{\circ}))$.
Given $\cos 30^{\circ} = 0.8660$ and $\sin 30^{\circ} = 0.5$.
$f(a+h) \approx 0.8660 - 0.00875 \times 0.5$.
$f(a+h) \approx 0.8660 - 0.004375 = 0.861625$.
Rounding to four decimal places,we get $0.8616$.

(B) The distance equation is given by $S = 1200t - 15t^2$.
To find the velocity $v$, we differentiate $S$ with respect to $t$:
$v = \frac{dS}{dt} = \frac{d}{dt}(1200t - 15t^2) = 1200 - 30t$.
When the bullet comes to rest, its velocity $v$ becomes $0$:
$1200 - 30t = 0$
$30t = 1200$
$t = 40 \text{ seconds}$.
Now, substitute $t = 40$ into the distance equation to find the total distance covered:
$S = 1200(40) - 15(40)^2$
$S = 48000 - 15(1600)$
$S = 48000 - 24000$
$S = 24000 \text{ cm}$.

(C) Given, the rate of change of the radius is $\frac{dr}{dt} = 8 \,cm/sec$.
The area of the circular wave is $A = \pi r^2$.
Differentiating both sides with respect to time $t$, we get $\frac{dA}{dt} = 2 \pi r \frac{dr}{dt}$.
At the instant when $r = 12 \,cm$, substituting the values:
$\frac{dA}{dt} = 2 \pi (12) (8)$.
$\frac{dA}{dt} = 192 \pi \,cm^2/sec$.

(C) Let $\theta$ be the temperature of the body at any time $t$. According to Newton's Law of Cooling,the rate of change of temperature is proportional to the difference between the body temperature and the surrounding temperature.
$\frac{d\theta}{dt} = -k(\theta - 30)$
Integrating this,we get $\ln(\theta - 30) = -kt + C$.
At $t = 0$,$\theta = 80^{\circ} C$,so $\ln(80 - 30) = C \Rightarrow C = \ln(50)$.
Thus,$\ln(\theta - 30) = -kt + \ln(50) \dots (i)$.
At $t = 30 \text{ min}$,$\theta = 60^{\circ} C$:
$\ln(60 - 30) = -k(30) + \ln(50) \Rightarrow \ln(30) - \ln(50) = -30k \Rightarrow \ln(3/5) = -30k$.
So,$k = -\frac{1}{30} \ln(3/5) = \frac{1}{30} \ln(5/3)$.
Now,for $t = 60 \text{ min}$ (one hour):
$\ln(\theta - 30) = -\left(\frac{1}{30} \ln(5/3)\right)(60) + \ln(50)$
$\ln(\theta - 30) = -2 \ln(5/3) + \ln(50) = \ln((3/5)^2 \times 50)$
$\ln(\theta - 30) = \ln(9/25 \times 50) = \ln(18)$
$\theta - 30 = 18 \Rightarrow \theta = 48^{\circ} C$.

(A) Let $y$ be the amount of moisture at time $t$.
The rate of change of moisture is proportional to the moisture content:
$\frac{dy}{dt} = -ky$ (where $k > 0$ is a constant).
Separating variables and integrating:
$\int \frac{dy}{y} = -\int k dt \Rightarrow \ln y = -kt + C$.
At $t = 0$,let the initial moisture be $y_0 = 1$ (representing $100 \%$).
Then $\ln(1) = -k(0) + C \Rightarrow C = 0$.
So,$\ln y = -kt$.
Given that at $t = 1$ hour,the sheet loses half its moisture,so $y = 0.5$ (or $1/2$):
$\ln(0.5) = -k(1) \Rightarrow k = -\ln(0.5) = \ln(2)$.
Now,we need to find $t$ when $99 \%$ of the moisture is lost,meaning $1 \%$ remains:
$y = 0.01 = \frac{1}{100} = 10^{-2}$.
Substituting into the equation $\ln y = -kt$:
$\ln(10^{-2}) = -(\ln 2)t$.
$-2 \ln(10) = -(\ln 2)t$.
$t = \frac{2 \ln 10}{\ln 2} = \frac{2 \log 10}{\log 2}$.

(D) Let $x$ be the distance of the foot of the ladder from the wall and $y$ be the height of the top of the ladder from the ground.
According to the Pythagorean theorem,$x^2 + y^2 = 5^2 = 25$.
Given that the top slides downwards at a rate of $10 \ cm/sec$,so $\frac{dy}{dt} = -10 \ cm/sec = -0.1 \ m/sec$.
When $x = 4 \ m$,we have $4^2 + y^2 = 25$,which gives $y^2 = 25 - 16 = 9$,so $y = 3 \ m$.
Differentiating $x^2 + y^2 = 25$ with respect to $t$,we get:
$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$
$x \frac{dx}{dt} = -y \frac{dy}{dt}$
Substituting the values $x = 4$,$y = 3$,and $\frac{dy}{dt} = -0.1$:
$4 \frac{dx}{dt} = -3(-0.1)$
$4 \frac{dx}{dt} = 0.3$
$\frac{dx}{dt} = \frac{0.3}{4} = 0.075 \ m/sec$.
Thus,the foot of the ladder is sliding at the rate of $0.075 \ m/sec$.

(B) Let $f(x) = \tan^{-1} x$.
Then $f'(x) = \frac{1}{1+x^2}$.
We use the linear approximation formula $f(a+h) \approx f(a) + h f'(a)$.
Here,let $a = 1$ and $h = -0.001$,so $a+h = 0.999$.
$f(a) = \tan^{-1}(1) = \frac{\pi}{4}$.
$f'(a) = \frac{1}{1+1^2} = \frac{1}{2} = 0.5$.
Substituting these values:
$\tan^{-1}(0.999) \approx \frac{\pi}{4} + (-0.001)(0.5)$.
$\tan^{-1}(0.999) \approx \frac{3.1415}{4} - 0.0005$.
$\tan^{-1}(0.999) \approx 0.785375 - 0.0005$.
$\tan^{-1}(0.999) \approx 0.784875$.
Rounding to four decimal places,we get $0.7849$.

(C) Let $p$ be the population at time $t$ years.
Given that the rate of change of population is proportional to the population:
$\frac{dp}{dt} = kp$
$\Rightarrow \frac{dp}{p} = k dt$
Integrating both sides,we get:
$\log p = kt + c$
At $t = 0$,$p = 40,000$:
$\log 40,000 = k(0) + c \Rightarrow c = \log 40,000$
So,$\log p = kt + \log 40,000 \Rightarrow \log \left(\frac{p}{40,000}\right) = kt$
At $t = 40$ years,$p = 80,000$:
$\log \left(\frac{80,000}{40,000}\right) = 40k \Rightarrow \log 2 = 40k \Rightarrow k = \frac{\log 2}{40}$
We need to find the population after another $40$ years,i.e.,at $t = 80$ years:
$\log \left(\frac{p}{40,000}\right) = \left(\frac{\log 2}{40}\right) \times 80$
$\log \left(\frac{p}{40,000}\right) = 2 \log 2 = \log 2^2 = \log 4$
$\frac{p}{40,000} = 4$
$p = 4 \times 40,000 = 160,000$

(C) Let $f(x) = x^{3/2}$.
Then,$f'(x) = \frac{3}{2} x^{1/2}$.
We can write $3.978$ as $a + h$,where $a = 4$ and $h = -0.022$.
Using the linear approximation formula $f(a + h) \approx f(a) + h \cdot f'(a)$:
$f(4) = 4^{3/2} = 8$.
$f'(4) = \frac{3}{2} \cdot (4)^{1/2} = \frac{3}{2} \cdot 2 = 3$.
Therefore,$f(3.978) \approx 8 + (-0.022) \cdot 3$.
$f(3.978) \approx 8 - 0.066$.
$f(3.978) \approx 7.934$.

(B) Let $A$, $P$, and $X$ be the area, perimeter, and length of the side of the square respectively at time $t$ seconds.
Given that the area is contracting at a rate of $\frac{dA}{dt} = -3 \,cm^2 / sec$.
The area of the square is $A = X^2$ and the perimeter is $P = 4X$.
From $A = X^2$, we have $X = \sqrt{A}$.
Substituting this into the perimeter formula: $P = 4\sqrt{A}$.
Differentiating with respect to $t$:
$\frac{dP}{dt} = 4 \cdot \frac{1}{2\sqrt{A}} \cdot \frac{dA}{dt} = \frac{2}{X} \cdot \frac{dA}{dt}$.
Given $X = 15 \,cm$ and $\frac{dA}{dt} = -3 \,cm^2 / sec$ (since it is contracting):
$\frac{dP}{dt} = \frac{2}{15} \cdot (-3) = -\frac{6}{15} = -\frac{2}{5} \,cm / sec$.
The negative sign indicates that the perimeter is decreasing at a rate of $\frac{2}{5} \,cm / sec$.

(A) Let $f(x) = x^3-2x^2+3x+2$.
We need to find the approximate value at $x = 2.01$.
Let $a = 2$ and $h = 0.01$,so $x = a+h = 2.01$.
The derivative is $f'(x) = 3x^2-4x+3$.
Calculating $f(a)$ at $a = 2$:
$f(2) = (2)^3-2(2)^2+3(2)+2 = 8-8+6+2 = 8$.
Calculating $f'(a)$ at $a = 2$:
$f'(2) = 3(2)^2-4(2)+3 = 12-8+3 = 7$.
Using the linear approximation formula $f(a+h) \approx f(a) + h \cdot f'(a)$:
$f(2.01) \approx 8 + (0.01)(7) = 8 + 0.07 = 8.07$.

(B) Given,$\frac{dx}{dt} = 2 \text{ units/sec}$.
Given equation of the parabola is $y = 2x^2$.
Differentiating with respect to $t$,we get $\frac{dy}{dt} = 4x \cdot \frac{dx}{dt}$.
Substituting $\frac{dx}{dt} = 2$,we get $\frac{dy}{dt} = 4x(2) = 8x$ ... $(i)$.
Let $r$ be the distance of the point $(x, y)$ from the origin $(0, 0)$,so $r = \sqrt{x^2 + y^2}$.
The rate of change of distance is $\frac{dr}{dt} = \frac{1}{2\sqrt{x^2 + y^2}} \cdot \frac{d}{dt}(x^2 + y^2) = \frac{2x \frac{dx}{dt} + 2y \frac{dy}{dt}}{2\sqrt{x^2 + y^2}} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{\sqrt{x^2 + y^2}}$.
Substituting $\frac{dx}{dt} = 2$ and $\frac{dy}{dt} = 8x$,we get $\frac{dr}{dt} = \frac{x(2) + y(8x)}{\sqrt{x^2 + y^2}} = \frac{2x + 8xy}{\sqrt{x^2 + y^2}}$.
At the point $(1, 2)$,$x = 1$ and $y = 2$.
$\frac{dr}{dt} = \frac{2(1) + 8(1)(2)}{\sqrt{1^2 + 2^2}} = \frac{2 + 16}{\sqrt{1 + 4}} = \frac{18}{\sqrt{5}} \text{ units/sec}$.

(C) Radius of the hemispherical bowl $R = 180 \text{ cm}$.
Rate of flow of water $\frac{dV}{dt} = 108 \text{ dm}^3/\text{min} = 108 \times 1000 \text{ cm}^3/\text{min} = 108000 \text{ cm}^3/\text{min}$.
Converting to seconds: $\frac{dV}{dt} = \frac{108000}{60} \text{ cm}^3/\text{s} = 1800 \text{ cm}^3/\text{s}$.
Let the depth of water be $x$. The volume of water in a 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 time $t$: $\frac{dV}{dt} = (360 \pi x - \pi x^2) \frac{dx}{dt}$.
At $x = 120 \text{ cm}$,we have $1800 = (360 \pi(120) - \pi(120)^2) \frac{dx}{dt}$.
$1800 = (43200 \pi - 14400 \pi) \frac{dx}{dt} = 28800 \pi \frac{dx}{dt}$.
$\frac{dx}{dt} = \frac{1800}{28800 \pi} = \frac{18}{288 \pi} = \frac{1}{16 \pi} \text{ cm/s}$.

(B) Given,the volume of a ball $V = \frac{4}{3} \pi r^3$.
When $V = 288 \pi$,we have:
$288 \pi = \frac{4}{3} \pi r^3$
$r^3 = 288 \times \frac{3}{4} = 216$
$r = 6 \text{ cm}$.
Now,differentiating $V$ with respect to $t$:
$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$
Given $\frac{dV}{dt} = 4 \pi \text{ cc/sec}$,substituting the values:
$4 \pi = 4 \pi (6)^2 \frac{dr}{dt}$
$1 = 36 \frac{dr}{dt}$
$\frac{dr}{dt} = \frac{1}{36} \text{ cm/sec}$.

(C) The distance covered by the body is given by $s = 3t^2 - 8t + 5$.
To find the velocity $v$,we differentiate $s$ with respect to time $t$:
$v = \frac{ds}{dt} = \frac{d}{dt}(3t^2 - 8t + 5) = 6t - 8$.
The body stops when its velocity becomes zero,i.e.,$v = 0$.
Setting the velocity to zero: $6t - 8 = 0$.
Solving for $t$: $6t = 8 \Rightarrow t = \frac{8}{6} = \frac{4}{3} \text{ sec}$.
Therefore,the body will stop after $\frac{4}{3} \text{ sec}$.

(C) Let the radius of the sphere be $r$.
Volume of the sphere $V = \frac{4}{3} \pi r^3$.
Surface area of the sphere $A = 4 \pi r^2$.
We need to find the rate of change of volume with respect to surface area,which is $\frac{dV}{dA}$.
Using the chain rule,$\frac{dV}{dA} = \frac{dV/dr}{dA/dr}$.
First,differentiate $V$ with respect to $r$: $\frac{dV}{dr} = \frac{d}{dr}(\frac{4}{3} \pi r^3) = 4 \pi r^2$.
Next,differentiate $A$ with respect to $r$: $\frac{dA}{dr} = \frac{d}{dr}(4 \pi r^2) = 8 \pi r$.
Now,calculate $\frac{dV}{dA} = \frac{4 \pi r^2}{8 \pi r} = \frac{r}{2}$.
Given the radius $r = 2 \text{ cm}$,substitute this value into the expression:
$\frac{dV}{dA} = \frac{2}{2} = 1 \text{ cm}^3 / \text{ cm}^2$.

(A) Let $f(x) = x^{\frac{3}{2}}$.
Then,$f'(x) = \frac{3}{2} x^{\frac{1}{2}}$.
Using the linear approximation formula $f(a+h) \approx f(a) + h \cdot f'(a)$,where $a = 4$ and $h = -0.022$:
$f(4 - 0.022) \approx f(4) + (-0.022) \cdot f'(4)$.
$f(4) = 4^{\frac{3}{2}} = (2^2)^{\frac{3}{2}} = 2^3 = 8$.
$f'(4) = \frac{3}{2} \cdot 4^{\frac{1}{2}} = \frac{3}{2} \cdot 2 = 3$.
Therefore,$f(3.978) \approx 8 + (-0.022) \cdot 3$.
$f(3.978) \approx 8 - 0.066 = 7.934$.

(D) Let $f(x) = 3^x$.
Then,the derivative is $f'(x) = 3^x \log 3$.
We need to find the value of $f(2.001)$.
Here,$a = 2$ and $h = 0.001$.
Using the linear approximation formula $f(a+h) \approx f(a) + h f'(a)$:
$f(2.001) \approx f(2) + (0.001) f'(2)$.
Substituting the values:
$f(2.001) \approx 3^2 + (0.001)(3^2 \log 3)$.
Given $\log 3 = 1.0986$:
$f(2.001) \approx 9 + (0.001)(9 \times 1.0986)$.
$f(2.001) \approx 9 + (0.001)(9.8874)$.
$f(2.001) \approx 9 + 0.0098874$.
Rounding to the given options,we get $9.00989$.

(D) Let $x$ be the distance of the foot of the ladder from the wall and $y$ be the height of the top of the ladder on the wall.
Given that the length of the ladder is $5 \ m$,by the Pythagorean theorem,we have $x^2 + y^2 = 5^2 = 25$.
Differentiating both sides with respect to time $t$,we get:
$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$
$\frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}$
Given $\frac{dx}{dt} = 2 \ m/sec$ and at the instant when $x = 4 \ m$,we have $y = \sqrt{25 - 4^2} = \sqrt{9} = 3 \ m$.
Substituting these values into the derivative equation:
$\frac{dy}{dt} = -\frac{4}{3} \times 2 = -\frac{8}{3} \ m/sec$.
The negative sign indicates that the height is decreasing.
Therefore,the height on the wall is decreasing at the rate of $\frac{8}{3} \ m/sec$.

(C) The perimeter of the circular sector is given by $P = r + r + r\theta = 2r + r\theta$.
Given that the total length of the wire is $20 \ m$,we have $2r + r\theta = 20$.
From this,we can express $\theta$ in terms of $r$: $\theta = \frac{20 - 2r}{r}$.
The area $A$ of a circular sector is given by $A = \frac{1}{2}r^2\theta$.
Substituting the expression for $\theta$: $A = \frac{1}{2}r^2 \left( \frac{20 - 2r}{r} \right) = \frac{1}{2}r(20 - 2r) = 10r - r^2$.
To find the maximum area,we differentiate $A$ with respect to $r$: $\frac{dA}{dr} = 10 - 2r$.
Setting $\frac{dA}{dr} = 0$,we get $10 - 2r = 0$,which implies $r = 5$.
To confirm this is a maximum,we check the second derivative: $\frac{d^2A}{dr^2} = -2$.
Since $\frac{d^2A}{dr^2} < 0$,the area is maximum at $r = 5$.
The maximum area is $A = 10(5) - (5)^2 = 50 - 25 = 25 \ sq.m$.

(B) Given that the rate of change of volume is $\frac{dV}{dt} = 36 \ m^3/min$ and the radius of the base is $r = 3 \ m$.
The volume of a cylinder is given by $V = \pi r^2 h$.
Differentiating both sides with respect to time $t$,we get $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$.
Substituting the given values,$36 = \pi \times (3)^2 \times \frac{dh}{dt}$.
$36 = 9 \pi \times \frac{dh}{dt}$.
Therefore,$\frac{dh}{dt} = \frac{36}{9 \pi} = \frac{4}{\pi} \ m/min$.

(C) Volume of a sphere $(V) = \frac{4}{3} \pi r^3$.
Surface area of a sphere $(A) = 4 \pi r^2$.
Differentiating both with respect to $r$:
$\frac{dV}{dr} = 4 \pi r^2$ and $\frac{dA}{dr} = 8 \pi r$.
We need to find the rate of change of volume with respect to surface area,which is $\frac{dV}{dA}$.
Using the chain rule: $\frac{dV}{dA} = \frac{dV/dr}{dA/dr} = \frac{4 \pi r^2}{8 \pi r} = \frac{r}{2}$.
Given the radius $r = 2 \text{ cm}$,we substitute this value:
$\frac{dV}{dA} = \frac{2}{2} = 1 \text{ cm}^3 / \text{cm}^2$.

(A) First,determine the set $A$ by solving the inequality $x^2 + 20 \le 9 x$.
$x^2 - 9 x + 20 \le 0$
$(x - 4)(x - 5) \le 0$
Thus,$A = [4, 5]$.
Next,analyze the function $f(x) = 2 x^3 - 15 x^2 + 36 x - 48$.
Find the derivative: $f'(x) = 6 x^2 - 30 x + 36 = 6(x^2 - 5 x + 6) = 6(x - 2)(x - 3)$.
For $x \in [4, 5]$,both $(x - 2)$ and $(x - 3)$ are positive,so $f'(x) > 0$.
Since $f'(x) > 0$ on the interval $[4, 5]$,the function $f(x)$ is strictly increasing on $A = [4, 5]$.
The minimum value occurs at the left endpoint $x = 4$.
$f(4) = 2(4)^3 - 15(4)^2 + 36(4) - 48 = 2(64) - 15(16) + 144 - 48 = 128 - 240 + 144 - 48 = -16$.

(A) 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 discriminant of the quadratic $f'(x)$ is $D = (2b)^2 - 4(3)(c) = 4b^2 - 12c$.
Since $b^2 < c$,we have $4b^2 < 4c$.
Thus,$D = 4b^2 - 12c < 4c - 12c = -8c$.
Since $0 < b^2 < c$,it implies $c > 0$. Therefore,$D < 0$.
Since the leading coefficient of $f'(x)$ is $3 > 0$ and $D < 0$,$f'(x) > 0$ for all $x \in \mathbb{R}$.
Hence,$f(x)$ is a strictly increasing function on $(-\infty, \infty)$.

(C) Let $f(x) = \frac{\ln(\pi+x)}{\ln(e+x)}$.
Using the quotient rule,$f'(x) = \frac{\ln(e+x) \cdot \frac{1}{\pi+x} - \ln(\pi+x) \cdot \frac{1}{e+x}}{[\ln(e+x)]^2}$.
Simplifying the numerator,we get $f'(x) = \frac{(e+x)\ln(e+x) - (\pi+x)\ln(\pi+x)}{(\pi+x)(e+x)[\ln(e+x)]^2}$.
Consider the function $g(t) = t \ln(t)$. Its derivative is $g'(t) = 1 + \ln(t)$. For $t > 1$,$g'(t) > 0$,so $g(t)$ is an increasing function.
Since $\pi > e > 1$,for $x > 0$,we have $\pi+x > e+x > e > 1$.
Because $g(t)$ is increasing,$g(\pi+x) > g(e+x)$,which implies $(\pi+x)\ln(\pi+x) > (e+x)\ln(e+x)$.
Therefore,$(e+x)\ln(e+x) - (\pi+x)\ln(\pi+x) < 0$.
Since the denominator is always positive for $x > 0$,$f'(x) < 0$ for all $x \in (0, \infty)$.
Thus,$f(x)$ is decreasing on $(0, \infty)$.

(A) Given the function $f(x) = \frac{\log x}{x}$.
To determine where the function is increasing,we find its derivative $f'(x)$ using the quotient rule:
$f'(x) = \frac{x \cdot \frac{d}{dx}(\log x) - \log x \cdot \frac{d}{dx}(x)}{x^2} = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2}$.
For the function to be increasing,we require $f'(x) > 0$.
Since $x^2 > 0$ for all $x > 0$,the condition $f'(x) > 0$ implies $1 - \log x > 0$.
This simplifies to $1 > \log x$,which means $\log x < 1$.
Taking the exponential of both sides,we get $x < e^1$,or $x < e$.
Given the domain $x > 0$,the function is increasing in the interval $(0, e)$.

(C) Given the function $f(x) = 2x^3 - 6x + 5$.
To find the intervals where the function is increasing,we first find the derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(2x^3 - 6x + 5) = 6x^2 - 6$.
$A$ function is increasing when $f'(x) > 0$.
So,$6x^2 - 6 > 0$.
Dividing by $6$,we get $x^2 - 1 > 0$,which factors as $(x - 1)(x + 1) > 0$.
Using the sign scheme for the inequality $(x - 1)(x + 1) > 0$,the expression is positive when $x > 1$ or $x < -1$.
Thus,the function is increasing for $x \in (-\infty, -1) \cup (1, \infty)$.