MHT CET 2023 Mathematics Question Paper with Answer and Solution

(D) Given that $p = T$,$q = T$,$r = F$,and $s = F$.
For the first statement pattern $(p \wedge q) \vee r$:
$(T \wedge T) \vee F \equiv T \vee F \equiv T$.
For the second statement pattern $(p \vee s) \leftrightarrow (q \wedge r)$:
$(T \vee F) \leftrightarrow (T \wedge F) \equiv T \leftrightarrow F \equiv F$.
Thus,the truth values are $T$ and $F$ respectively.

(B) We use the logical equivalence $p \rightarrow q \equiv \sim p \vee q$.
$p \rightarrow \sim(p \wedge \sim q)$
$\equiv \sim p \vee \sim(p \wedge \sim q)$
$\equiv \sim p \vee (\sim p \vee \sim(\sim q))$
$\equiv \sim p \vee (\sim p \vee q)$
$\equiv (\sim p \vee \sim p) \vee q$
$\equiv \sim p \vee q$

(C) The implication $(p \leftrightarrow \sim q) \rightarrow (\sim p \wedge q)$ is false only when the antecedent is true and the consequent is false.
So,$(p \leftrightarrow \sim q) \equiv T$ and $(\sim p \wedge q) \equiv F$.
From $(\sim p \wedge q) \equiv F$,we know that the conjunction is false.
Now,consider the antecedent $(p \leftrightarrow \sim q) \equiv T$. This means $p$ and $\sim q$ must have the same truth value.
If we test option $C$ $(p=T, q=F)$:
$\sim q = \sim F = T$.
Then $(p \leftrightarrow \sim q) = (T \leftrightarrow T) = T$.
And $(\sim p \wedge q) = (\sim T \wedge F) = (F \wedge F) = F$.
Since the antecedent is $T$ and the consequent is $F$,the implication is $T \rightarrow F = F$.
Thus,the truth values are $p=T$ and $q=F$.

(D) Let $p$ be the statement: "$x$ and $y$ are integers such that $x y$ is odd".
Let $q$ be the statement: "both $x$ and $y$ are odd".
The given statement is $p \rightarrow q$.
The contrapositive of $p \rightarrow q$ is $\sim q \rightarrow \sim p$.
Here,$\sim q$ is "it is not the case that both $x$ and $y$ are odd",which means "$x$ and $y$ are not both odd".
$\sim p$ is "$x y$ is not odd".
Therefore,the contrapositive is "If $x$ and $y$ are not both odd,then $x y$ is not odd".
Thus,option $D$ is correct.

(A) Let $p$ : The surface area increases.
Let $q$ : The pressure decreases.
The given statement is $p \rightarrow q$.
The inverse of a conditional statement $p \rightarrow q$ is defined as $\sim p \rightarrow \sim q$.
Here,$\sim p$ is "The surface area does not increase" and $\sim q$ is "The pressure does not decrease".
Therefore,the inverse statement is "If the surface area does not increase,then the pressure does not decrease.".
Thus,Option $(A)$ is correct.

(D) Let $p$: $A$ quadrilateral is a square.
Let $q$: All sides of the quadrilateral are equal.
Statement $1$ is $p \rightarrow q$.
Statement $2$ is $q \rightarrow p$.
By definition,the converse of a conditional statement $p \rightarrow q$ is $q \rightarrow p$.
Therefore,Statement $2$ is the converse of Statement $1$.

(B) Let $S$ be the statement $(p \wedge q) \rightarrow (p \vee \sim q)$.
The inverse of $S$ is $\sim(p \wedge q) \rightarrow \sim(p \vee \sim q)$.
Using the logical equivalence $A \rightarrow B \equiv \sim A \vee B$,the inverse is:
$\sim[\sim(p \wedge q)] \vee \sim(p \vee \sim q)$
$\equiv (p \wedge q) \vee (\sim p \wedge q)$ (By De Morgan's Law)
$\equiv (q \wedge p) \vee (q \wedge \sim p)$ (By Commutative Law)
$\equiv q \wedge (p \vee \sim p)$ (By Distributive Law)
$\equiv q \wedge T$ (By Complement Law,where $T$ is a tautology)
$\equiv q$ (By Identity Law).
Now,the negation of the inverse is $\sim(q) = \sim q$.

(C) We simplify the expression using logical laws:
$(p \wedge \sim q) \vee q \vee (\sim p \wedge q)$
Using the distributive law on the first two terms:
$\equiv ((p \vee q) \wedge (\sim q \vee q)) \vee (\sim p \wedge q)$
Since $(\sim q \vee q) \equiv T$ (Complement law):
$\equiv ((p \vee q) \wedge T) \vee (\sim p \wedge q)$
$\equiv (p \vee q) \vee (\sim p \wedge q)$
Applying the distributive law again:
$\equiv (p \vee q \vee \sim p) \wedge (p \vee q \vee q)$
Since $(p \vee \sim p) \equiv T$ and $(q \vee q) \equiv q$ (Idempotent law):
$\equiv (T \vee q) \wedge (p \vee q)$
Since $(T \vee q) \equiv T$:
$\equiv T \wedge (p \vee q)$
$\equiv p \vee q$
Therefore,the expression is equivalent to $p \vee q$.

(B) Given that $q$ is false and $p \wedge q \leftrightarrow r$ is true.
Since $q \equiv F$,then $p \wedge q \equiv F$.
For the biconditional $p \wedge q \leftrightarrow r$ to be true,$r$ must have the same truth value as $p \wedge q$.
Therefore,$r \equiv F$.
Now,let us evaluate the options:
$(A)$ $p \vee r \equiv p \vee F \equiv p$,which is not a tautology.
$(B)$ $(p \wedge r)$ $\rightarrow (p \vee r) \equiv (p \wedge F)$ $\rightarrow (p \vee F) \equiv F$ $\rightarrow p$. Since $F \rightarrow p$ is always true for any truth value of $p$,this is a tautology.
$(C)$ $(p \vee r)$ $\rightarrow (p \wedge r) \equiv (p \vee F)$ $\rightarrow (p \wedge F) \equiv p$ $\rightarrow F$,which is not a tautology.
$(D)$ $p \wedge r \equiv p \wedge F \equiv F$,which is a contradiction.

(B) The contrapositive of the statement $(p \vee \sim q) \rightarrow (p \wedge \sim q)$ is given by $\sim (p \wedge \sim q) \rightarrow \sim (p \vee \sim q)$.
Using the equivalence $A \rightarrow B \equiv \sim A \vee B$,we get:
$\sim (p \wedge \sim q) \vee \sim (p \vee \sim q)$
Applying De Morgan's law:
$(\sim p \vee q) \vee (\sim p \wedge q)$
Now,we find the negation of this contrapositive:
$\sim [(\sim p \vee q) \vee (\sim p \wedge q)]$
Using De Morgan's law again:
$\sim (\sim p \vee q) \wedge \sim (\sim p \wedge q)$
$(p \wedge \sim q) \wedge (p \vee \sim q)$
Wait,let us re-evaluate the negation of the contrapositive directly:
The contrapositive is $C = \sim (p \wedge \sim q) \rightarrow \sim (p \vee \sim q)$.
The negation of $A \rightarrow B$ is $A \wedge \sim B$.
Thus,$\sim C = \sim (p \wedge \sim q) \wedge \sim [\sim (p \vee \sim q)]$
$\sim C = (\sim p \vee q) \wedge (p \vee \sim q)$.

(D) To determine when the statement $[(p$ $\rightarrow q) \wedge \sim q]$ $\rightarrow r$ is a tautology,we analyze the expression $[(p \rightarrow q) \wedge \sim q]$.
Note that $(p \rightarrow q) \equiv (\sim p \vee q)$.
Thus,$[(p \rightarrow q) \wedge \sim q] \equiv [(\sim p \vee q) \wedge \sim q]$.
Using the distributive law,this becomes $(\sim p \wedge \sim q) \vee (q \wedge \sim q)$.
Since $(q \wedge \sim q) \equiv F$ (a contradiction),the expression simplifies to $(\sim p \wedge \sim q) \vee F \equiv \sim p \wedge \sim q$.
Now,the original statement is $(\sim p \wedge \sim q) \rightarrow r$.
For this to be a tautology,the implication must be true for all truth values of $p$ and $q$.
If we set $r \equiv \sim p \wedge \sim q$,the statement becomes $(\sim p \wedge \sim q) \rightarrow (\sim p \wedge \sim q)$,which is always true.
However,looking at the options,we check if $r \equiv \sim q$ works.
If $r \equiv \sim q$,the statement is $(\sim p \wedge \sim q) \rightarrow \sim q$.
Since $(\sim p \wedge \sim q)$ is a subset of $\sim q$ (i.e.,if $\sim p \wedge \sim q$ is true,then $\sim q$ must be true),the implication is always true.
Therefore,the statement is a tautology when $r \equiv \sim q$.

(A) Let $p$: The number is an odd number.
Let $q$: The number is divisible by $3$.
The given statement is $p \leftrightarrow q$.
The negation of a biconditional statement is $\sim(p \leftrightarrow q) \equiv (p \wedge \sim q) \vee (q \wedge \sim p)$.
Thus,the negation is: "The number is an odd number but not divisible by $3$ $OR$ the number is divisible by $3$ but not an odd number."

(B) Given expression: $[p \wedge (q \vee r)] \vee [\sim r \wedge \sim q \wedge p]$
Using the commutative law on the second part: $[p \wedge (q \vee r)] \vee [p \wedge \sim q \wedge \sim r]$
Applying De Morgan's law to the second part: $[p \wedge (q \vee r)] \vee [p \wedge \sim (q \vee r)]$
Using the distributive law: $p \wedge [(q \vee r) \vee \sim (q \vee r)]$
Using the complement law: $p \wedge T$ (where $T$ is a tautology)
Using the identity law: $p$
Therefore,the statement is equivalent to $p$.

(B) The statement $p \leftrightarrow (q \rightarrow p)$ is false.
This biconditional is false if the truth values of $p$ and $(q \rightarrow p)$ are different.
If $p$ is $T$,then $(q \rightarrow T)$ is $T$,so $T \leftrightarrow T$ is $T$.
If $p$ is $F$,then $(q \rightarrow F)$ must be $T$ for the biconditional to be false.
For $(q \rightarrow F)$ to be $T$,$q$ must be $F$.
Thus,$p \equiv F$ and $q \equiv F$.
Now,check option $(B)$: $p$ $\rightarrow (p \vee \sim q) \equiv F$ $\rightarrow (F \vee \sim F) \equiv F$ $\rightarrow (F \vee T) \equiv F$ $\rightarrow T \equiv T$.
Since the result is $T$,option $(B)$ is the correct statement pattern.

(C) The normal form of the equation of a line is $x \cos \alpha + y \sin \alpha = p$,where $p$ is the perpendicular distance from the origin and $\alpha$ is the angle the perpendicular makes with the positive $X$-axis.
For line $L_1$,$\alpha = \frac{\pi}{4}$ and $p = 1$:
$x \cos \frac{\pi}{4} + y \sin \frac{\pi}{4} = 1$
$\Rightarrow \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = 1$
$\Rightarrow x + y - \sqrt{2} = 0$
For line $L_2$,$\alpha = \frac{3 \pi}{4}$ and $p = 1$:
$x \cos \frac{3 \pi}{4} + y \sin \frac{3 \pi}{4} = 1$
$\Rightarrow -\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = 1$
$\Rightarrow -x + y - \sqrt{2} = 0$
$\Rightarrow x - y + \sqrt{2} = 0$
The joint equation is the product of the two individual equations:
$(x + y - \sqrt{2})(x - y + \sqrt{2}) = 0$
$x(x - y + \sqrt{2}) + y(x - y + \sqrt{2}) - \sqrt{2}(x - y + \sqrt{2}) = 0$
$x^2 - xy + \sqrt{2}x + xy - y^2 + \sqrt{2}y - \sqrt{2}x + \sqrt{2}y - 2 = 0$
$x^2 - y^2 + 2\sqrt{2}y - 2 = 0$

(B) The joint equation of the pair of lines passing through the point $(x_1, y_1)$ and perpendicular to the lines $ax^2 + 2hxy + by^2 = 0$ is given by $b(x - x_1)^2 - 2h(x - x_1)(y - y_1) + a(y - y_1)^2 = 0$.
Here,$a = 5$,$2h = 2$ (so $h = 1$),$b = -3$,and $(x_1, y_1) = (3, -2)$.
Substituting these values into the formula:
$-3(x - 3)^2 - 2(x - 3)(y + 2) + 5(y + 2)^2 = 0$
Expanding the terms:
$-3(x^2 - 6x + 9) - 2(xy + 2x - 3y - 6) + 5(y^2 + 4y + 4) = 0$
$-3x^2 + 18x - 27 - 2xy - 4x + 6y + 12 + 5y^2 + 20y + 20 = 0$
Combining like terms:
$-3x^2 - 2xy + 5y^2 + 14x + 26y + 5 = 0$
Multiplying by $-1$ to standardize:
$3x^2 + 2xy - 5y^2 - 14x - 26y - 5 = 0$.

(C) The equation $6x^2+xy-y^2=0$ can be factored as $-(y-3x)(y+2x)=0$,which gives the lines $y=3x$ and $y=-2x$.
To find the vertices of the triangle,we find the intersection points of these lines with $x+3y=10$:
$1$. Intersection of $y=3x$ and $x+3y=10$: $x+3(3x)=10 \implies 10x=10 \implies x=1, y=3$. Vertex is $(1,3)$.
$2$. Intersection of $y=-2x$ and $x+3y=10$: $x+3(-2x)=10 \implies -5x=10 \implies x=-2, y=4$. Vertex is $(-2,4)$.
$3$. Intersection of $y=3x$ and $y=-2x$ is the origin $(0,0)$.
The centroid is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) = \left(\frac{0+1-2}{3}, \frac{0+3+4}{3}\right) = \left(-\frac{1}{3}, \frac{7}{3}\right)$.

(A) The given equation of the pair of lines is $x^2+\lambda xy-y^2 \tan^2 \theta=0$.
Comparing this with the general form $ax^2+2hxy+by^2=0$,we get $a=1$,$h=\frac{\lambda}{2}$,and $b=-\tan^2 \theta$.
The angle $\alpha$ between the pair of lines is given by $\tan \alpha = \left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$.
Here,$\alpha = 2\theta$,so $\tan 2\theta = \left|\frac{2\sqrt{(\frac{\lambda}{2})^2 - (1)(-\tan^2 \theta)}}{1-\tan^2 \theta}\right|$.
Using the identity $\tan 2\theta = \frac{2\tan \theta}{1-\tan^2 \theta}$,we have $\frac{2\tan \theta}{1-\tan^2 \theta} = \left|\frac{2\sqrt{\frac{\lambda^2}{4}+\tan^2 \theta}}{1-\tan^2 \theta}\right|$.
Squaring both sides,we get $\frac{4\tan^2 \theta}{(1-\tan^2 \theta)^2} = \frac{4(\frac{\lambda^2}{4}+\tan^2 \theta)}{(1-\tan^2 \theta)^2}$.
This simplifies to $\tan^2 \theta = \frac{\lambda^2}{4} + \tan^2 \theta$,which implies $\frac{\lambda^2}{4} = 0$.
Therefore,$\lambda = 0$.

(C) Given equation: $(x \cos \alpha + y \sin \alpha)^2 = (x^2 + y^2) \sin^2 \alpha$
Expanding the left side: $x^2 \cos^2 \alpha + y^2 \sin^2 \alpha + 2xy \sin \alpha \cos \alpha = x^2 \sin^2 \alpha + y^2 \sin^2 \alpha$
Subtracting $y^2 \sin^2 \alpha$ from both sides: $x^2 \cos^2 \alpha + 2xy \sin \alpha \cos \alpha = x^2 \sin^2 \alpha$
Rearranging: $x^2(\cos^2 \alpha - \sin^2 \alpha) + 2xy \sin \alpha \cos \alpha = 0$
This is a homogeneous equation of the form $ax^2 + 2hxy + by^2 = 0$,where $a = \cos^2 \alpha - \sin^2 \alpha$,$h = \sin \alpha \cos \alpha$,and $b = 0$.
For the lines to be perpendicular,the condition is $a + b = 0$.
Substituting the values: $(\cos^2 \alpha - \sin^2 \alpha) + 0 = 0$
$\cos^2 \alpha = \sin^2 \alpha$
$\tan^2 \alpha = 1$
Since $\alpha$ is typically in the first quadrant for such problems,$\alpha = \frac{\pi}{4}$.

(B) The given equation of the pair of lines is $x^2-3xy+\lambda y^2+3x-5y+2=0$.
Comparing this with the general form $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=1$,$2h=-3 \Rightarrow h=-\frac{3}{2}$,and $b=\lambda$.
The angle $\theta$ between the lines is given by $\tan \theta = \left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$.
Given $\tan \theta = \frac{1}{3}$,we have $\frac{1}{3} = \left|\frac{2\sqrt{(-\frac{3}{2})^2 - (1)(\lambda)}}{1+\lambda}\right|$.
Squaring both sides,$\frac{1}{9} = \frac{4(\frac{9}{4}-\lambda)}{(1+\lambda)^2}$.
$(1+\lambda)^2 = 36(\frac{9}{4}-\lambda) = 81 - 36\lambda$.
$\lambda^2 + 2\lambda + 1 = 81 - 36\lambda$.
$\lambda^2 + 38\lambda - 80 = 0$.
$(\lambda+40)(\lambda-2) = 0$.
Since $\lambda \geq 0$,we have $\lambda = 2$.

(A) Given that $PQR$ is a right-angled isosceles triangle at $Q(2, 1)$. The line $PR$ has the equation $2x + y = 3$,so its slope is $m = -2$.
Since $\triangle PQR$ is isosceles and right-angled at $Q$,the lines $PQ$ and $QR$ make an angle of $45^\circ$ with the line $PR$.
Let the slope of $PQ$ be $m_1$. Then,$\tan 45^\circ = |\frac{m_1 - (-2)}{1 + m_1(-2)}| = |\frac{m_1 + 2}{1 - 2m_1}|$.
$1 = |\frac{m_1 + 2}{1 - 2m_1}| \Rightarrow 1 - 2m_1 = m_1 + 2$ or $1 - 2m_1 = -(m_1 + 2)$.
Case $1$: $3m_1 = -1 \Rightarrow m_1 = -1/3$.
Case $2$: $m_1 = 3$.
Thus,the slopes of $PQ$ and $QR$ are $-1/3$ and $3$.
The equations of the lines passing through $Q(2, 1)$ are:
$y - 1 = -1/3(x - 2)$ $\Rightarrow 3y - 3 = -x + 2$ $\Rightarrow x + 3y - 5 = 0$.
$y - 1 = 3(x - 2)$ $\Rightarrow y - 1 = 3x - 6$ $\Rightarrow 3x - y - 5 = 0$.
The combined equation is $(x + 3y - 5)(3x - y - 5) = 0$.
Expanding this: $3x^2 - xy - 5x + 9xy - 3y^2 - 15y - 15x + 5y + 25 = 0$.
$3x^2 + 8xy - 3y^2 - 20x - 10y + 25 = 0$.

(A) The given equation of the pair of lines is $ax^2 + (2a + 1)xy + 2y^2 = 0$.
Comparing this with $Ax^2 + 2Hxy + By^2 = 0$,we have $A = a$,$2H = 2a + 1$,and $B = 2$.
Let the slopes of the lines be $m_1$ and $m_2$.
Given that $m_1 = \frac{1}{m_2}$,which implies $m_1 m_2 = 1$.
For a pair of lines $ax^2 + 2hxy + by^2 = 0$,the product of slopes is $m_1 m_2 = \frac{A}{B} = \frac{a}{2}$.
Equating the two,we get $\frac{a}{2} = 1$,so $a = 2$.
The sum of the slopes is $m_1 + m_2 = -\frac{2H}{B} = -\frac{2a + 1}{2}$.
Substituting $a = 2$,we get $m_1 + m_2 = -\frac{2(2) + 1}{2} = -\frac{5}{2}$.
We need to find $m_1^2 + m_2^2$.
Using the identity $m_1^2 + m_2^2 = (m_1 + m_2)^2 - 2m_1 m_2$,we get:
$m_1^2 + m_2^2 = (-\frac{5}{2})^2 - 2(1) = \frac{25}{4} - 2 = \frac{25 - 8}{4} = \frac{17}{4}$.

(B) The slope of line $QR$ is $m = -2$. Let the slopes of lines $PQ$ and $PR$ be $m_1$ and $m_2$ respectively.
Since $\triangle PQR$ is an isosceles right-angled triangle,the angles $\angle PQR = \angle PRQ = 45^{\circ}$.
Using the formula $\tan \theta = \left| \frac{m - m_1}{1 + m \cdot m_1} \right|$,we have $\tan 45^{\circ} = \left| \frac{-2 - m_1}{1 + (-2)m_1} \right| = 1$.
This gives $\left| \frac{2 + m_1}{1 - 2m_1} \right| = 1$,so $2 + m_1 = 1 - 2m_1$ or $2 + m_1 = -(1 - 2m_1)$.
Solving $3m_1 = -1$ gives $m_1 = -1/3$. Solving $-m_1 = -3$ gives $m_2 = 3$.
The lines $PQ$ and $PR$ pass through $P(2, 1)$ with slopes $-1/3$ and $3$.
The equations are $y - 1 = -1/3(x - 2) \Rightarrow x + 3y - 5 = 0$ and $y - 1 = 3(x - 2) \Rightarrow 3x - y - 5 = 0$.
The joint equation is $(x + 3y - 5)(3x - y - 5) = 0$.
Expanding this: $3x^2 - xy - 5x + 9xy - 3y^2 - 15y - 15x + 5y + 25 = 0$.
Simplifying gives $3x^2 - 3y^2 + 8xy - 20x - 10y + 25 = 0$.

(B) The number of triangles that can be formed using the vertices of a regular polygon of $n$ sides is given by $T_n = {}^{n}C_3$.
Given the condition $T_{n+1} - T_n = 21$,we substitute the formula:
${}^{n+1}C_3 - {}^{n}C_3 = 21$.
Using the identity ${}^{n}C_r + {}^{n}C_{r-1} = {}^{n+1}C_r$,we know that ${}^{n+1}C_3 - {}^{n}C_3 = {}^{n}C_2$.
Therefore,${}^{n}C_2 = 21$.
Expanding the combination formula: $\frac{n(n-1)}{2} = 21$.
$n(n-1) = 42$.
$n^2 - n - 42 = 0$.
$(n-7)(n+6) = 0$.
Since $n$ must be a positive integer,$n = 7$.
Thus,option $(B)$ is correct.

(C) Total students to be selected $= 5$. Total students available $= n$.
Number of ways in which $2$ particular students are selected $= {}^{n-2}C_{5-2} = {}^{n-2}C_3$.
Number of ways in which $2$ particular students are not selected $= {}^{n-2}C_5$.
According to the given condition,$\frac{{}^{n-2}C_3}{{}^{n-2}C_5} = \frac{2}{3}$.
Using the formula ${}^nC_r = \frac{n!}{r!(n-r)!}$,we have:
$\frac{(n-2)!}{3!(n-5)!} \times \frac{5!(n-7)!}{(n-2)!} = \frac{2}{3}$
$\frac{5 \times 4}{ (n-5)(n-6)} = \frac{2}{3}$
$\frac{20}{(n-5)(n-6)} = \frac{2}{3}$
$(n-5)(n-6) = 30$
$n^2 - 11n + 30 = 30$
$n^2 - 11n = 0$
Since $n$ must be at least $7$ (to select $5$ students excluding $2$),$n = 11$.

(A) Let $n$ be the number of members in the meeting.
Since every person shakes hands with every other person exactly once,the total number of handshakes is given by the combination formula ${}^{n}C_{2}$.
Given that the total number of handshakes is $45$,we have:
${}^{n}C_{2} = 45$
$\frac{n(n-1)}{2} = 45$
$n(n-1) = 90$
$n^2 - n - 90 = 0$
$(n - 10)(n + 9) = 0$
Since the number of members $n$ must be positive,we have $n = 10$.
Therefore,the number of members present at the meeting is $10$.

(D) Case $I$: No boy is included. Selecting $4$ girls from $6$ girls is ${}^6C_4 = 15$ ways. Selecting $1$ captain from the $4$ selected members is ${}^4C_1 = 4$ ways. Total number of ways for Case $I = 15 \times 4 = 60$.
Case $II$: Exactly one boy is included. Selecting $3$ girls from $6$ girls and $1$ boy from $4$ boys is ${}^6C_3 \times {}^4C_1 = 20 \times 4 = 80$ ways. Selecting $1$ captain from the $4$ selected members is ${}^4C_1 = 4$ ways. Total number of ways for Case $II = 80 \times 4 = 320$.
Total number of ways $= 60 + 320 = 380$.

(C) There are $5$ positions in total. The boy $B_1$ is fixed at the $2^{\text{nd}}$ position.
Remaining students are $4$ (including $G_1$ and $G_2$).
Since $G_1$ and $G_2$ must be adjacent,we treat $(G_1, G_2)$ as a single unit.
Now we have $3$ units to arrange in the remaining $4$ positions: the unit $(G_1, G_2)$ and the $2$ other students.
However,the $2^{\text{nd}}$ position is already occupied by $B_1$. The available positions are $1^{\text{st}}, 3^{\text{rd}}, 4^{\text{th}}, 5^{\text{th}}$.
If we place the unit $(G_1, G_2)$ in positions $(3, 4)$ or $(4, 5)$,we have $2$ ways to place the unit.
Within the unit,$G_1$ and $G_2$ can be arranged in $2! = 2$ ways.
The remaining $2$ students can be arranged in the remaining $2$ positions in $2! = 2$ ways.
Total arrangements $= 2 \times 2 \times 2 = 8$.

(C) There are $5$ positions in total. The boy $B_1$ is fixed at the $2^{\text{nd}}$ position.
Remaining positions are $1^{\text{st}}, 3^{\text{rd}}, 4^{\text{th}}, 5^{\text{th}}$.
Let the $5$ students be $B_1, G_1, G_2, S_1, S_2$.
Since $G_1$ and $G_2$ must be adjacent,we treat them as a single unit $(G_1G_2)$.
Now we have the units: $(G_1G_2), S_1, S_2$ to be placed in the remaining $4$ positions,such that $B_1$ is at the $2^{\text{nd}}$ position.
Case $1$: $(G_1G_2)$ occupies positions $(3, 4)$.
Remaining positions $1$ and $5$ are filled by $S_1, S_2$ in $2!$ ways. $G_1, G_2$ can be arranged in $2!$ ways.
Number of ways $= 2! \times 2! = 4$.
Case $2$: $(G_1G_2)$ occupies positions $(4, 5)$.
Remaining positions $1$ and $3$ are filled by $S_1, S_2$ in $2!$ ways. $G_1, G_2$ can be arranged in $2!$ ways.
Number of ways $= 2! \times 2! = 4$.
Total arrangements $= 4 + 4 = 8$.

(C) The word $CALCULATE$ has $9$ letters.
Out of which $C$ repeats $2$ times,$A$ repeats $2$ times,$L$ repeats $2$ times,and $E, U, T$ appear once.
There are $5$ consonants $(C, C, L, L, T)$ and $4$ vowels $(A, A, U, E)$.
Two consonants out of $5$ can occupy the start and end positions in $P(5, 2)$ ways.
Since there are $2$ $C$'s and $2$ $L$'s,we must account for the repetitions.
Total arrangements $= \frac{P(5, 2) \times 7!}{2! \times 2! \times 2!} = \frac{(5 \times 4) \times 7!}{8} = \frac{20 \times 7!}{8} = \frac{5 \times 7!}{2}$.

(A) Let the number of ways to colour a circle of $n$ vertices with $k$ colours such that no two adjacent vertices have the same colour be $P_n(k)$.
For a circular arrangement,the formula is given by $P_n(k) = (k-1)^n + (-1)^n(k-1)$.
Here,$n = 5$ (number of persons) and $k = 3$ (number of colours).
Substituting the values,we get $P_5(3) = (3-1)^5 + (-1)^5(3-1)$.
$P_5(3) = 2^5 - 1(2) = 32 - 2 = 30$.
Therefore,the total number of ways is $30$.

(D) Case $I$: No boy is included.
Selecting $4$ girls from $6$ girls $= {}^{6}C_{4} = 15$.
Selecting $1$ leader from the $4$ selected members $= {}^{4}C_{1} = 4$.
Total ways for Case $I = 15 \times 4 = 60$.
Case $II$: Exactly one boy is included.
Selecting $3$ girls from $6$ girls and $1$ boy from $4$ boys $= {}^{6}C_{3} \times {}^{4}C_{1} = 20 \times 4 = 80$.
Selecting $1$ leader from the $4$ selected members $= {}^{4}C_{1} = 4$.
Total ways for Case $II = 80 \times 4 = 320$.
Therefore,the total number of ways $= 60 + 320 = 380$.

(C) committee of $5$ is to be formed from $8$ boys and $5$ girls such that it consists of at least $2$ girls and at most $2$ boys.
Since the committee size is $5$,the possible combinations of (girls,boys) satisfying the conditions are:
$1$. $5$ girls and $0$ boys: $\binom{5}{5} \times \binom{8}{0} = 1 \times 1 = 1$
$2$. $4$ girls and $1$ boy: $\binom{5}{4} \times \binom{8}{1} = 5 \times 8 = 40$
$3$. $3$ girls and $2$ boys: $\binom{5}{3} \times \binom{8}{2} = 10 \times 28 = 280$
Total number of ways = $1 + 40 + 280 = 321$.

(D) Let $A, B, C$ be the events that the first,second,and third critics favor the book respectively.
The probabilities are:
$P(A) = \frac{2}{2+5} = \frac{2}{7}, P(A') = \frac{5}{7}$
$P(B) = \frac{3}{3+4} = \frac{3}{7}, P(B') = \frac{4}{7}$
$P(C) = \frac{4}{4+3} = \frac{4}{7}, P(C') = \frac{3}{7}$
The majority is in favor if at least two critics favor the book.
The probability is $P(A \cap B \cap C') + P(A \cap B' \cap C) + P(A' \cap B \cap C) + P(A \cap B \cap C)$.
$= (\frac{2}{7} \times \frac{3}{7} \times \frac{3}{7}) + (\frac{2}{7} \times \frac{4}{7} \times \frac{4}{7}) + (\frac{5}{7} \times \frac{3}{7} \times \frac{4}{7}) + (\frac{2}{7} \times \frac{3}{7} \times \frac{4}{7})$
$= \frac{18}{343} + \frac{32}{343} + \frac{60}{343} + \frac{24}{343} = \frac{134}{343}$.

(A) The probability that ship $A$ reaches safely is $P(A) = \frac{2}{2+5} = \frac{2}{7}$.
The probability that ship $B$ reaches safely is $P(B) = \frac{3}{3+7} = \frac{3}{10}$.
The probability that ship $C$ reaches safely is $P(C) = \frac{6}{6+11} = \frac{6}{17}$.
Since the events are independent,the probability that all of them arrive safely is $P(A \cap B \cap C) = P(A) \times P(B) \times P(C)$.
$P(A \cap B \cap C) = \frac{2}{7} \times \frac{3}{10} \times \frac{6}{17} = \frac{36}{1190} = \frac{18}{595}$.

(C) The total number of ways to choose $3$ vertices out of $6$ is given by $n(S) = {}^{6}C_{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
An equilateral triangle is formed if we select alternate vertices of the regular hexagon.
In a regular hexagon with vertices labeled $1, 2, 3, 4, 5, 6$,the possible sets of vertices forming an equilateral triangle are $\{1, 3, 5\}$ and $\{2, 4, 6\}$.
Thus,the number of favorable outcomes is $n(A) = 2$.
The probability $P(A)$ is given by $P(A) = \frac{n(A)}{n(S)} = \frac{2}{20} = \frac{1}{10}$.

(A) Let $X$ be the event that the number on the ticket is a perfect square.
$\therefore X = \{1, 4, 9, 16, 25, 36, 49, 64, 81, 100\}$
$\therefore n(X) = 10$
$\text{Also, } n(S) = 100$
$\therefore \text{Required probability} = \frac{n(X)}{n(S)} = \frac{10}{100} = \frac{1}{10}$

(D) Total number of cards $n(S) = 52$.
Let event $A$ be drawing a black card and event $B$ be drawing a face card.
Number of black cards $n(A) = 26$.
Number of face cards $n(B) = 12$.
Number of black face cards $n(A \cap B) = 6$ (since there are $3$ black face cards in each of the two black suits).
The probability of drawing a black card or a face card is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = \frac{26}{52} + \frac{12}{52} - \frac{6}{52} = \frac{32}{52} = \frac{8}{13}$.

(A) The sample space $S$ consists of all possible pairs $(x, y)$ where $x, y \in \{1, 2, 3, 5, 7, 11\}$.
Since there are $6$ faces on each die,the total number of outcomes is $n(S) = 6 \times 6 = 36$.
Let $A$ be the event that the sum of the numbers is a prime number.
The possible sums are:
$1+1=2$ (prime),$1+2=3$ (prime),$1+3=4$,$1+5=6$,$1+7=8$,$1+11=12$
$2+1=3$ (prime),$2+2=4$,$2+3=5$ (prime),$2+5=7$ (prime),$2+7=9$,$2+11=13$ (prime)
$3+1=4$,$3+2=5$ (prime),$3+3=6$,$3+5=8$,$3+7=10$,$3+11=14$
$5+1=6$,$5+2=7$ (prime),$5+3=8$,$5+5=10$,$5+7=12$,$5+11=16$
$7+1=8$,$7+2=9$,$7+3=10$,$7+5=12$,$7+7=14$,$7+11=18$
$11+1=12$,$11+2=13$ (prime),$11+3=14$,$11+5=16$,$11+7=18$,$11+11=22$
The favorable outcomes are: $(1,1), (1,2), (2,1), (2,3), (2,5), (2,11), (3,2), (5,2), (11,2)$.
Thus,$n(A) = 9$.
The probability $P(A) = \frac{n(A)}{n(S)} = \frac{9}{36} = \frac{1}{4}$.

(A) Total number of numbers $= 6 + 8 = 14$.
Number of ways to choose $4$ numbers out of $14$ is $^{14}C_4 = \frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1} = 1001$.
The product of $4$ numbers is negative if:
$i$. One number is negative and three are positive: $^{8}C_1 \times ^{6}C_3 = 8 \times 20 = 160$.
$ii$. Three numbers are negative and one is positive: $^{8}C_3 \times ^{6}C_1 = 56 \times 6 = 336$.
Total favorable ways $= 160 + 336 = 496$.
Required probability $= \frac{496}{1001}$.

(A) Given that $A$,$B$,and $C$ are mutually exclusive and exhaustive events,we have $P(A) + P(B) + P(C) = 1$.
Odds in favor of $A$ are $4 : 6$,so $P(A) = \frac{4}{4+6} = \frac{4}{10}$.
Odds against $B$ are $7 : 3$,which means odds in favor of $B$ are $3 : 7$,so $P(B) = \frac{3}{3+7} = \frac{3}{10}$.
Substituting these into the sum: $\frac{4}{10} + \frac{3}{10} + P(C) = 1$.
$\frac{7}{10} + P(C) = 1 \implies P(C) = 1 - \frac{7}{10} = \frac{3}{10}$.
The probability of the complement event $C'$ is $P(C') = 1 - P(C) = 1 - \frac{3}{10} = \frac{7}{10}$.
The odds against $C$ are defined as $P(C') : P(C) = \frac{7}{10} : \frac{3}{10} = 7 : 3$.

(D) The probability that the first critic favours the book is $P(A) = \frac{2}{2+5} = \frac{2}{7}$.
$\therefore P(A') = 1 - \frac{2}{7} = \frac{5}{7}$.
The probability that the second critic favours the book is $P(B) = \frac{3}{3+4} = \frac{3}{7}$.
$\therefore P(B') = 1 - \frac{3}{7} = \frac{4}{7}$.
The probability that the third critic favours the book is $P(C) = \frac{4}{4+3} = \frac{4}{7}$.
$\therefore P(C') = 1 - \frac{4}{7} = \frac{3}{7}$.
Majority will be in favour of the book if at least two critics favour the book.
Hence,the probability is $P(A \cap B \cap C') + P(A \cap B' \cap C) + P(A' \cap B \cap C) + P(A \cap B \cap C)$.
$= P(A) \cdot P(B) \cdot P(C') + P(A) \cdot P(B') \cdot P(C) + P(A') \cdot P(B) \cdot P(C) + P(A) \cdot P(B) \cdot P(C)$.
$= \left(\frac{2}{7} \times \frac{3}{7} \times \frac{3}{7}\right) + \left(\frac{2}{7} \times \frac{4}{7} \times \frac{4}{7}\right) + \left(\frac{5}{7} \times \frac{3}{7} \times \frac{4}{7}\right) + \left(\frac{2}{7} \times \frac{3}{7} \times \frac{4}{7}\right)$.
$= \frac{18}{343} + \frac{32}{343} + \frac{60}{343} + \frac{24}{343} = \frac{134}{343}$.

(B) Let $P(A), P(B),$ and $P(C)$ be the probabilities of students $A, B,$ and $C$ solving the problem respectively.
$P(A) = \frac{1}{2}, P(B) = \frac{1}{3}, P(C) = \frac{1}{4}$
The probability that the problem is not solved by any of them is the probability that all three fail to solve it.
$P(A') = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2}$
$P(B') = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3}$
$P(C') = 1 - P(C) = 1 - \frac{1}{4} = \frac{3}{4}$
Since the students try independently,the probability that none of them solve the problem is:
$P(\text{none solve}) = P(A') \times P(B') \times P(C') = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{1}{4}$
The probability that the problem is solved is the complement of the probability that it is not solved:
$P(\text{solved}) = 1 - P(\text{none solve}) = 1 - \frac{1}{4} = \frac{3}{4}$

(D) Let $a = \sqrt{3}+1$,$b = \sqrt{3}-1$,and $C = 60^{\circ}$.
Using the Law of Cosines:
$c^2 = a^2 + b^2 - 2ab \cos C$
$c^2 = (\sqrt{3}+1)^2 + (\sqrt{3}-1)^2 - 2(\sqrt{3}+1)(\sqrt{3}-1) \cos 60^{\circ}$
$c^2 = (3+1+2\sqrt{3}) + (3+1-2\sqrt{3}) - 2(3-1) \times \frac{1}{2}$
$c^2 = 8 - 2 = 6 \implies c = \sqrt{6}$.
Using the Law of Tangents:
$\tan\left(\frac{A-B}{2}\right) = \frac{a-b}{a+b} \cot\left(\frac{C}{2}\right)$
$\tan\left(\frac{A-B}{2}\right) = \frac{(\sqrt{3}+1) - (\sqrt{3}-1)}{(\sqrt{3}+1) + (\sqrt{3}-1)} \cot(30^{\circ})$
$\tan\left(\frac{A-B}{2}\right) = \frac{2}{2\sqrt{3}} \times \sqrt{3} = 1$.
Therefore,$\frac{A-B}{2} = 45^{\circ} \implies A-B = 90^{\circ}$.

(B) Using the sine rule,we have:
$\frac{\sin A}{a} = \frac{\sin B}{b}$
Substituting the given values:
$\frac{\sin 60^{\circ}}{6} = \frac{\sin B}{8}$
Since $B = \sin^{-1} x$,we have $\sin B = x$.
$\frac{\sqrt{3}/2}{6} = \frac{x}{8}$
$\frac{\sqrt{3}}{12} = \frac{x}{8}$
$x = \frac{8\sqrt{3}}{12} = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}}$

(D) By the cosine rule,we have:
$a^2 = b^2 + c^2 - 2bc \cos A$
$a^2 = (\sqrt{3})^2 + (1)^2 - 2(\sqrt{3})(1) \cos(30^{\circ})$
$a^2 = 3 + 1 - 2\sqrt{3} \left(\frac{\sqrt{3}}{2}\right)$
$a^2 = 4 - 3 = 1$
$\therefore a = 1$
Since $b = \sqrt{3} \approx 1.732$ is the largest side,the largest angle is $\angle B$.
Using the cosine rule for $\angle B$:
$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{1^2 + 1^2 - (\sqrt{3})^2}{2(1)(1)} = \frac{1 + 1 - 3}{2} = -\frac{1}{2}$
$\therefore B = 120^{\circ}$

(C) Given the equation: $a \cdot \cos^2 \frac{C}{2} + c \cdot \cos^2 \frac{A}{2} = \frac{3b}{2}$
Using the identity $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$,we get:
$a \left( \frac{1 + \cos C}{2} \right) + c \left( \frac{1 + \cos A}{2} \right) = \frac{3b}{2}$
Multiplying by $2$:
$a(1 + \cos C) + c(1 + \cos A) = 3b$
$a + a \cos C + c + c \cos A = 3b$
Using the projection formula $b = a \cos C + c \cos A$:
$a + c + b = 3b$
$a + c = 2b$
This condition implies that $a, b, c$ are in $A$.$P$.

(A) In $\triangle ABC$,we have $\frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{9}=k$.
This gives:
$a+b=7k$ $(i)$
$b+c=8k$ $(ii)$
$c+a=9k$ $(iii)$
Adding these equations,we get $2(a+b+c)=24k$,so $a+b+c=12k$ $(iv)$.
Subtracting $(i), (ii), (iii)$ from $(iv)$ respectively:
$c = (a+b+c) - (a+b) = 12k - 7k = 5k$
$a = (a+b+c) - (b+c) = 12k - 8k = 4k$
$b = (a+b+c) - (c+a) = 12k - 9k = 3k$
Since $a^2+b^2 = (4k)^2 + (3k)^2 = 16k^2 + 9k^2 = 25k^2 = (5k)^2 = c^2$,the triangle is a right-angled triangle with hypotenuse $c$.
Thus,$\angle C = 90^{\circ}$.
Area of $\triangle ABC = \frac{1}{2} \times a \times b = \frac{1}{2} \times 4k \times 3k = 6k^2$.
Therefore,$\frac{(A(\triangle ABC))^2}{k^4} = \frac{(6k^2)^2}{k^4} = \frac{36k^4}{k^4} = 36$.

(C) In $\triangle ABD$,by the Law of Sines:
$\frac{\sin(\angle BAD)}{BD} = \frac{\sin(\angle B)}{AD}$
$\Rightarrow \frac{\sin(\angle BAD)}{x} = \frac{\sin(\frac{\pi}{3})}{AD} = \frac{\sqrt{3}/2}{AD}$
$\Rightarrow AD = \frac{\sqrt{3}x}{2 \sin(\angle BAD)} \quad \dots (i)$
In $\triangle ADC$,by the Law of Sines:
$\frac{\sin(\angle CAD)}{DC} = \frac{\sin(\angle C)}{AD}$
$\Rightarrow \frac{\sin(\angle CAD)}{3x} = \frac{\sin(\frac{\pi}{4})}{AD} = \frac{1/\sqrt{2}}{AD}$
$\Rightarrow AD = \frac{3x}{\sqrt{2} \sin(\angle CAD)} \quad \dots (ii)$
Equating $(i)$ and $(ii)$:
$\frac{\sqrt{3}x}{2 \sin(\angle BAD)} = \frac{3x}{\sqrt{2} \sin(\angle CAD)}$
$\Rightarrow \frac{\sin(\angle BAD)}{\sin(\angle CAD)} = \frac{\sqrt{3} \cdot \sqrt{2}}{2 \cdot 3} = \frac{\sqrt{6}}{6} = \frac{1}{\sqrt{6}}$

(A) Let the angles of the triangle be $A = \frac{\pi}{4}$,$B = \frac{\pi}{3}$,and $C = \pi - (\frac{\pi}{4} + \frac{\pi}{3}) = \pi - \frac{7\pi}{12} = \frac{5\pi}{12}$.
Since $\frac{\pi}{4} < \frac{\pi}{3} < \frac{5\pi}{12}$,the smallest angle is $A = \frac{\pi}{4}$ and the greatest angle is $C = \frac{5\pi}{12}$.
By the Law of Sines,the ratio of the sides is proportional to the sine of their opposite angles: $\frac{a}{c} = \frac{\sin A}{\sin C}$.
$\frac{a}{c} = \frac{\sin(\frac{\pi}{4})}{\sin(\frac{5\pi}{12})} = \frac{\sin(\frac{\pi}{4})}{\sin(\frac{\pi}{4} + \frac{\pi}{6})}$.
Using the formula $\sin(x+y) = \sin x \cos y + \cos x \sin y$:
$\sin(\frac{5\pi}{12}) = \sin(\frac{\pi}{4})\cos(\frac{\pi}{6}) + \cos(\frac{\pi}{4})\sin(\frac{\pi}{6}) = (\frac{1}{\sqrt{2}})(\frac{\sqrt{3}}{2}) + (\frac{1}{\sqrt{2}})(\frac{1}{2}) = \frac{\sqrt{3}+1}{2\sqrt{2}}$.
Thus,$\frac{a}{c} = \frac{1/\sqrt{2}}{(\sqrt{3}+1)/(2\sqrt{2})} = \frac{2}{\sqrt{3}+1}$.
Rationalizing the denominator: $\frac{2(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{2(\sqrt{3}-1)}{3-1} = \sqrt{3}-1$.
Therefore,the ratio is $(\sqrt{3}-1) : 1$.

(D) Given the region $A = \{(x, y) / \frac{y^2}{2} \leq x \leq y+4\}$.
To find the points of intersection,we set $x = \frac{y^2}{2}$ and $x = y+4$.
Equating the two expressions for $x$:
$\frac{y^2}{2} = y+4$
$y^2 = 2y + 8$
$y^2 - 2y - 8 = 0$
$(y - 4)(y + 2) = 0$
Thus,$y = 4$ or $y = -2$.
For $y = 4$,$x = 4+4 = 8$. For $y = -2$,$x = -2+4 = 2$.
The points of intersection are $(8, 4)$ and $(2, -2)$.
The area $A$ is given by the integral with respect to $y$:
$A = \int_{-2}^{4} (y + 4 - \frac{y^2}{2}) dy$
$A = [\frac{y^2}{2} + 4y - \frac{y^3}{6}]_{-2}^{4}$
$A = (\frac{16}{2} + 4(4) - \frac{64}{6}) - (\frac{4}{2} + 4(-2) - \frac{-8}{6})$
$A = (8 + 16 - \frac{32}{3}) - (2 - 8 + \frac{4}{3})$
$A = (24 - \frac{32}{3}) - (-6 + \frac{4}{3})$
$A = \frac{72 - 32}{3} - \frac{-18 + 4}{3}$
$A = \frac{40}{3} - (-\frac{14}{3})$
$A = \frac{40 + 14}{3} = \frac{54}{3} = 18$ sq. units.

(B) The given curves are $y=x^2$ and $y=|x|$.
Since both curves are symmetric about the $y$-axis,the total area is twice the area in the first quadrant.
In the first quadrant,$y=|x|$ becomes $y=x$.
The intersection points are found by setting $x^2 = x$,which gives $x(x-1)=0$,so $x=0$ and $x=1$.
The required area is given by:
$\text{Area} = 2 \int_0^1 (x - x^2) dx$
$= 2 \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1$
$= 2 \left( \frac{1}{2} - \frac{1}{3} \right)$
$= 2 \left( \frac{3-2}{6} \right) = 2 \left( \frac{1}{6} \right) = \frac{1}{3} \text{ sq. units}$.

(C) The equation of the circle is $x^2+y^2=a^2$. The line is $x=\frac{a}{\sqrt{2}}$.
Substituting $x=\frac{a}{\sqrt{2}}$ in the circle equation,we get $\frac{a^2}{2}+y^2=a^2$,which implies $y^2=\frac{a^2}{2}$,so $y=\pm\frac{a}{\sqrt{2}}$.
The required area is the area of the region bounded by the circle and the line $x=\frac{a}{\sqrt{2}}$ on the right side.
Area $= 2 \int_{\frac{a}{\sqrt{2}}}^a \sqrt{a^2-x^2} dx$
Using the formula $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$:
Area $= 2 \left[ \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a}) \right]_{\frac{a}{\sqrt{2}}}^a$
$= 2 \left[ (0 + \frac{a^2}{2}\sin^{-1}(1)) - (\frac{a}{2\sqrt{2}}\sqrt{a^2-\frac{a^2}{2}} + \frac{a^2}{2}\sin^{-1}(\frac{1}{\sqrt{2}})) \right]$
$= 2 \left[ \frac{a^2}{2}(\frac{\pi}{2}) - (\frac{a}{2\sqrt{2}}\cdot\frac{a}{\sqrt{2}} + \frac{a^2}{2}(\frac{\pi}{4})) \right]$
$= 2 \left[ \frac{a^2\pi}{4} - \frac{a^2}{4} - \frac{a^2\pi}{8} \right]$
$= 2 \left[ \frac{a^2\pi}{8} - \frac{a^2}{4} \right] = \frac{a^2\pi}{4} - \frac{a^2}{2} = \frac{a^2}{2}(\frac{\pi}{2}-1)$.

(C) The curves $y=3x+1$ and $y=4x+1$ intersect at the point where $3x+1 = 4x+1$,which gives $x=0$.
Given the boundary $x=3$,the region is bounded between $x=0$ and $x=3$.
In this interval,$4x+1 \geq 3x+1$.
Therefore,the required area is given by the integral:
$\text{Area} = \int_{0}^{3} [(4x+1) - (3x+1)] \, dx$
$= \int_{0}^{3} x \, dx$
$= \left[ \frac{x^2}{2} \right]_{0}^{3}$
$= \frac{3^2}{2} - \frac{0^2}{2} = \frac{9}{2} \text{ sq. units}$.

(A) The given curve is $y = a\sqrt{x} + bx$. Since it passes through $(1, 2)$,we have $2 = a(1) + b(1)$,which gives $a + b = 2$ ...$(i)$.
The area bounded by the curve,$x = 4$,and the $X$-axis is given by $\int_0^4 (a\sqrt{x} + bx) dx = 8$.
Evaluating the integral:
$\int_0^4 (ax^{1/2} + bx) dx = \left[ a \cdot \frac{x^{3/2}}{3/2} + b \cdot \frac{x^2}{2} \right]_0^4 = 8$.
Substituting the limits:
$\left( \frac{2a}{3} \cdot 4^{3/2} + \frac{b}{2} \cdot 4^2 \right) = 8$.
$\frac{2a}{3} \cdot 8 + \frac{b}{2} \cdot 16 = 8$.
$\frac{16a}{3} + 8b = 8$.
Dividing by $8$,we get $\frac{2a}{3} + b = 1$,which simplifies to $2a + 3b = 3$ ...(ii).
Solving equations $(i)$ and (ii):
From $(i)$,$b = 2 - a$. Substituting into (ii):
$2a + 3(2 - a) = 3$.
$2a + 6 - 3a = 3$.
$-a = -3 \Rightarrow a = 3$.
Substituting $a = 3$ into $(i)$:
$3 + b = 2 \Rightarrow b = -1$.
Thus,$a = 3$ and $b = -1$.

(A) The curves are $y=(x-1)^2$,$y=(x+1)^2$,and the line $y=\frac{1}{4}$.
By symmetry,the area is twice the area in the first quadrant bounded by $y=(x-1)^2$ and $y=\frac{1}{4}$ from $x=0$ to $x=\frac{1}{2}$.
$\text{Required Area} = 2 \int_0^{\frac{1}{2}} \left[ \frac{1}{4} - (x-1)^2 \right] dx$ is incorrect based on the graph; the region is bounded above by the curves and below by $y=\frac{1}{4}$.
Correct setup: The region is bounded by $y=(x+1)^2$ and $y=(x-1)^2$ meeting at $x=0$ and the line $y=\frac{1}{4}$.
Area $= 2 \int_0^{1/2} [y_{upper} - y_{lower}] dx = 2 \int_0^{1/2} [(x+1)^2 - (x-1)^2] dx$ is not correct. Looking at the graph,the area is bounded by $y=(x-1)^2$ and $y=(x+1)^2$ and $y=1/4$.
Area $= 2 \int_0^{1/2} [y_{upper} - y_{lower}] dx = 2 \int_0^{1/2} [\text{curve} - 1/4] dx$.
Specifically,for $x \in [0, 1/2]$,the curve is $y=(x-1)^2$.
Area $= 2 \int_0^{1/2} [(x-1)^2 - 1/4] dx = 2 \left[ \frac{(x-1)^3}{3} - \frac{x}{4} \right]_0^{1/2} = 2 \left[ (\frac{(-1/2)^3}{3} - \frac{1/2}{4}) - (\frac{(-1)^3}{3} - 0) \right] = 2 \left[ -\frac{1}{24} - \frac{1}{8} + \frac{1}{3} \right] = 2 \left[ \frac{-1-3+8}{24} \right] = 2 \left[ \frac{4}{24} \right] = \frac{1}{3} \text{ sq. units.}$

(C) The required area is given by the integral of the upper curve minus the lower curve between the limits $x=1$ and $x=2$.
$\text{Area} = \int_1^2 (e^x - \log x) dx$
$= \int_1^2 e^x dx - \int_1^2 \log x dx$
$= [e^x]_1^2 - [x \log x - x]_1^2$
$= (e^2 - e^1) - [(2 \log 2 - 2) - (1 \log 1 - 1)]$
Since $\log 1 = 0$,we have:
$= e^2 - e - [2 \log 2 - 2 - 0 + 1]$
$= e^2 - e - [2 \log 2 - 1]$
$= e^2 - e + 1 - 2 \log 2 \text{ sq. units}$

(C) Let $x$ be the mass of the material at time $t$.
The rate of decay is given by $\frac{dx}{dt} = -kx$.
Integrating this,we get $\ln(x) = -kt + C$.
At $t=0, x=27$,so $C = \ln(27)$.
Thus,$\ln(x) = -kt + \ln(27)$.
At $t=3, x=8$,so $\ln(8) = -3k + \ln(27)$,which gives $3k = \ln(27) - \ln(8) = \ln(\frac{27}{8})$.
Therefore,$k = \frac{1}{3} \ln(\frac{27}{8}) = \ln((\frac{27}{8})^{1/3}) = \ln(\frac{3}{2})$.
We need to find the mass after one more hour,i.e.,at $t=4$.
$\ln(x) = -4 \ln(\frac{3}{2}) + \ln(27) = \ln((\frac{3}{2})^{-4}) + \ln(27) = \ln(\frac{16}{81} \times 27) = \ln(\frac{16}{3})$.
Therefore,$x = \frac{16}{3} \text{ grams}$.

(C) According to Newton's law of cooling,$\frac{d\theta}{dt} = -k(\theta - \theta_0)$,where $\theta_0 = 25^{\circ} C$.
Integrating,we get $\ln(\theta - 25) = -kt + C$.
At $t = 0$,$\theta = 80^{\circ} C$,so $C = \ln(55)$.
Thus,$\ln(\theta - 25) = -kt + \ln(55)$,or $\ln\left(\frac{\theta - 25}{55}\right) = -kt$.
At $t = 30$ min,$\theta = 50^{\circ} C$,so $\ln\left(\frac{50 - 25}{55}\right) = -30k$ $\Rightarrow \ln\left(\frac{25}{55}\right) = -30k$ $\Rightarrow \ln\left(\frac{5}{11}\right) = -30k$.
For $t = 60$ min,$\ln\left(\frac{\theta - 25}{55}\right) = -60k = 2(-30k) = 2 \ln\left(\frac{5}{11}\right)$.
Therefore,$\frac{\theta - 25}{55} = \left(\frac{5}{11}\right)^2 = \frac{25}{121}$.
$\theta - 25 = 55 \times \frac{25}{121} = 5 \times \frac{25}{11} = \frac{125}{11} \approx 11.36$.
$\theta = 25 + 11.36 = 36.36^{\circ} C$.

(A) Let $A = \begin{bmatrix} 1 & a & b \\ \omega & 1 & c \\ \omega^2 & \omega & 1 \end{bmatrix}$. For the matrix to be non-singular,$|A| \neq 0$.
$|A| = 1(1 - \omega c) - a(\omega - \omega^2 c) + b(\omega^2 - \omega^2) = 1 - \omega c - a\omega + a\omega^2 c = (1 - \omega c) - a\omega(1 - \omega c) = (1 - \omega c)(1 - a\omega)$.
For $|A| \neq 0$,we must have $(1 - \omega c) \neq 0$ and $(1 - a\omega) \neq 0$.
This implies $c \neq \frac{1}{\omega} = \omega^2$ and $a \neq \frac{1}{\omega} = \omega^2$.
Since $a, b, c \in \{\omega, \omega^2\}$,the condition $a \neq \omega^2$ implies $a = \omega$,and $c \neq \omega^2$ implies $c = \omega$.
The variable $b$ can be either $\omega$ or $\omega^2$.
Thus,the possible matrices are $\begin{bmatrix} 1 & \omega & \omega \\ \omega & 1 & \omega \\ \omega^2 & \omega & 1 \end{bmatrix}$ and $\begin{bmatrix} 1 & \omega & \omega^2 \\ \omega & 1 & \omega \\ \omega^2 & \omega & 1 \end{bmatrix}$.
Therefore,the number of distinct matrices in the set $S$ is $2$.

(D) Since the function is continuous at $x = \frac{\pi}{4}$ and $x = \frac{\pi}{2}$,we have:
At $x = \frac{\pi}{4}$:
$\lim_{x \to \frac{\pi}{4}^-} (x + a \sqrt{2} \sin x) = \lim_{x \to \frac{\pi}{4}^+} (2x \cot x + b)$
$\frac{\pi}{4} + a \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 2 \cdot \frac{\pi}{4} \cdot 1 + b$
$\frac{\pi}{4} + a = \frac{\pi}{2} + b \implies a - b = \frac{\pi}{4}$ . . . $(i)$
At $x = \frac{\pi}{2}$:
$\lim_{x \to \frac{\pi}{2}^-} (2x \cot x + b) = \lim_{x \to \frac{\pi}{2}^+} (a \cos 2x - b \sin x)$
$2 \cdot \frac{\pi}{2} \cdot 0 + b = a \cos \pi - b \sin \frac{\pi}{2}$
$b = -a - b \implies a + 2b = 0$ . . . $(ii)$
Solving $(i)$ and $(ii)$:
From $(ii)$,$a = -2b$. Substituting in $(i)$:
$-2b - b = \frac{\pi}{4} \implies -3b = \frac{\pi}{4} \implies b = -\frac{\pi}{12}$
Then $a = -2(-\frac{\pi}{12}) = \frac{\pi}{6}$.
Thus,$a = \frac{\pi}{6}$ and $b = -\frac{\pi}{12}$.

(D) The greatest integer function $f(x) = [x]$ is discontinuous at every integer value of $x$.
Given the interval $x \in \left(-\frac{7}{2}, 100\right)$,we can write this as $x \in (-3.5, 100)$.
The integers contained in this interval are $\{-3, -2, -1, 0, 1, 2, \dots, 99\}$.
To find the total number of integers,we use the formula: $\text{Number of terms} = (\text{Last term} - \text{First term}) + 1$.
Here,the first term is $-3$ and the last term is $99$.
Total number of discontinuities = $(99 - (-3)) + 1 = 99 + 3 + 1 = 103$.
Thus,there are $103$ points of discontinuity.

(A) For $f(x)$ to be continuous at $x=0$,we must have $\text{L.H.L.} = \text{R.H.L.} = f(0)$.
First,calculate the $\text{L.H.L.}$ at $x=0$:
$\text{L.H.L.} = \lim_{x \to 0^-} \frac{1-\cos kx}{x^2} = \lim_{x \to 0^-} \frac{2\sin^2(\frac{kx}{2})}{x^2} = 2 \lim_{x \to 0^-} \left( \frac{\sin(\frac{kx}{2})}{\frac{kx}{2}} \cdot \frac{k}{2} \right)^2 = 2 \cdot \frac{k^2}{4} = \frac{k^2}{2}$.
Next,calculate the $\text{R.H.L.}$ at $x=0$:
$\text{R.H.L.} = \lim_{x \to 0^+} \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}$.
Rationalizing the denominator:
$\text{R.H.L.} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{(\sqrt{16+\sqrt{x}}-4)(\sqrt{16+\sqrt{x}}+4)} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{16+\sqrt{x}-16} = \lim_{x \to 0^+} (\sqrt{16+\sqrt{x}}+4) = \sqrt{16}+4 = 8$.
Since the function is continuous at $x=0$,$\text{L.H.L.} = \text{R.H.L.}$:
$\frac{k^2}{2} = 8 \implies k^2 = 16 \implies k = \pm 4$.
Given the options,the correct value is $4$.

(A) $f(x)$ is continuous in $0 \leq x \leq \pi$,therefore it is continuous at $x = \frac{\pi}{4}$ and $x = \frac{\pi}{2}$.
At $x = \frac{\pi}{4}$:
$\lim_{x \rightarrow \frac{\pi}{4}^-} (x + a \sqrt{2} \sin x) = \lim_{x \rightarrow \frac{\pi}{4}^+} (2x \cot x + b)$
$\frac{\pi}{4} + a \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 2 \cdot \frac{\pi}{4} \cdot 1 + b$
$\frac{\pi}{4} + a = \frac{\pi}{2} + b \Rightarrow a - b = \frac{\pi}{4} \quad (i)$
At $x = \frac{\pi}{2}$:
$\lim_{x \rightarrow \frac{\pi}{2}^-} (2x \cot x + b) = \lim_{x \rightarrow \frac{\pi}{2}^+} (a \cos 2x - b \sin x)$
$2 \cdot \frac{\pi}{2} \cdot 0 + b = a \cos \pi - b \sin \frac{\pi}{2}$
$b = -a - b \Rightarrow a = -2b \quad (ii)$
Substituting $(ii)$ into $(i)$:
$-2b - b = \frac{\pi}{4} \Rightarrow -3b = \frac{\pi}{4} \Rightarrow b = -\frac{\pi}{12}$
Then $a = -2(-\frac{\pi}{12}) = \frac{\pi}{6}$.
Finally,$2a + 3b = 2(\frac{\pi}{6}) + 3(-\frac{\pi}{12}) = \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12}$.

(D) Given $f(x) = \frac{4}{x^4} \left[ 1 - \cos \frac{x}{2} - \cos \frac{x}{4} + \cos \frac{x}{2} \cdot \cos \frac{x}{4} \right]$.
Factoring the expression inside the brackets:
$f(x) = \frac{4}{x^4} \left[ (1 - \cos \frac{x}{2}) - \cos \frac{x}{4} (1 - \cos \frac{x}{2}) \right] = \frac{4}{x^4} (1 - \cos \frac{x}{2}) (1 - \cos \frac{x}{4})$.
Since $f(x)$ is continuous at $x = 0$,$f(0) = \lim_{x \to 0} f(x)$.
Using the identity $1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$:
$f(0) = \lim_{x \to 0} \frac{4}{x^4} \left( 2 \sin^2 \frac{x}{4} \right) \left( 2 \sin^2 \frac{x}{8} \right) = 16 \lim_{x \to 0} \frac{\sin^2 \frac{x}{4}}{x^2} \cdot \frac{\sin^2 \frac{x}{8}}{x^2}$.
Multiply and divide by $(\frac{1}{4})^2$ and $(\frac{1}{8})^2$:
$f(0) = 16 \lim_{x \to 0} \left( \frac{\sin \frac{x}{4}}{\frac{x}{4}} \right)^2 \cdot \frac{1}{16} \cdot \left( \frac{\sin \frac{x}{8}}{\frac{x}{8}} \right)^2 \cdot \frac{1}{64} = 16 \cdot \frac{1}{16} \cdot \frac{1}{64} \cdot (1)^2 \cdot (1)^2 = \frac{1}{64}$.

(B) Since $f(x)$ is continuous in $[-1, 1]$,it must be continuous at $x = 0$.
Therefore,$\lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{+}} f(x) = f(0)$.
First,calculate the right-hand limit: $\lim_{x \rightarrow 0^{+}} f(x) = \lim_{x \rightarrow 0^{+}} \frac{2x+1}{x-2} = \frac{2(0)+1}{0-2} = -\frac{1}{2}$.
Next,calculate the left-hand limit: $\lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{-}} \frac{\sqrt{1+mx} - \sqrt{1-mx}}{x}$.
Rationalizing the numerator: $\lim_{x \rightarrow 0^{-}} \frac{(\sqrt{1+mx} - \sqrt{1-mx})(\sqrt{1+mx} + \sqrt{1-mx})}{x(\sqrt{1+mx} + \sqrt{1-mx})} = \lim_{x \rightarrow 0^{-}} \frac{(1+mx) - (1-mx)}{x(\sqrt{1+mx} + \sqrt{1-mx})} = \lim_{x \rightarrow 0^{-}} \frac{2mx}{x(\sqrt{1+mx} + \sqrt{1-mx})} = \frac{2m}{1+1} = m$.
Equating the limits: $m = -\frac{1}{2}$.

(B) Given $f(x) = f(1-x)$ and $R_1 = \int_{-1}^2 x f(x) dx$.
Using the property $\int_{a}^b g(x) dx = \int_{a}^b g(a+b-x) dx$,where $a = -1$ and $b = 2$,we have $a+b = 1$.
Thus,$R_1 = \int_{-1}^2 (1-x) f(1-x) dx$.
Since $f(1-x) = f(x)$,we get $R_1 = \int_{-1}^2 (1-x) f(x) dx$.
$R_1 = \int_{-1}^2 f(x) dx - \int_{-1}^2 x f(x) dx$.
$R_1 = \int_{-1}^2 f(x) dx - R_1$.
$2 R_1 = \int_{-1}^2 f(x) dx$.
Since $R_2$ is the area bounded by $y = f(x)$ from $x = -1$ to $x = 2$,$R_2 = \int_{-1}^2 f(x) dx$.
Therefore,$R_2 = 2 R_1$.

(D) For $x < 3$,$|x-3| = -(x-3)$,so $f(x) = \frac{x-3}{-(x-3)} + a = -1 + a = a - 1$.
For $x > 3$,$|x-3| = (x-3)$,so $f(x) = \frac{x-3}{x-3} + b = 1 + b$.
Since $f(x)$ is continuous at $x=3$,the left-hand limit,right-hand limit,and the value of the function at $x=3$ must be equal.
$\lim_{x \to 3^-} f(x) = f(3) \implies a - 1 = a + b \implies b = -1$.
$\lim_{x \to 3^+} f(x) = f(3) \implies 1 + b = a + b \implies a = 1$.
Therefore,$a - b = 1 - (-1) = 1 + 1 = 2$.

(A) For $f(x)$ to be continuous at $x=0$,the condition $\lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x)$ must hold.
First,calculate the left-hand limit $(LHL)$:
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1-\cos 4x}{x^2} = \lim_{x \to 0^-} \frac{2\sin^2(2x)}{x^2} = 2 \lim_{x \to 0^-} \left(\frac{\sin 2x}{2x}\right)^2 \times 4 = 2 \times 1^2 \times 4 = 8$.
Next,calculate the right-hand limit $(RHL)$:
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}$.
Rationalizing the denominator:
$\lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{16+\sqrt{x}-16} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{\sqrt{x}} = \lim_{x \to 0^+} (\sqrt{16+\sqrt{x}}+4) = \sqrt{16}+4 = 8$.
Since $\text{LHL} = \text{RHL} = 8$,for continuity,$a = 8$.

(C) Since $f(x)$ is the probability density function (p.d.f.) of $X$,the total area under the curve must be $1$.
$\int_{-\infty}^{\infty} f(x) dx = 1$
$\int_0^1 ax dx + \int_1^2 a dx + \int_2^3 (3a - ax) dx = 1$
$a \left[ \frac{x^2}{2} \right]_0^1 + a [x]_1^2 + \left[ 3ax - \frac{ax^2}{2} \right]_2^3 = 1$
$a(\frac{1}{2}) + a(1) + [(9a - \frac{9a}{2}) - (6a - \frac{4a}{2})] = 1$
$\frac{a}{2} + a + [\frac{9a}{2} - 4a] = 1$
$\frac{a}{2} + a + \frac{a}{2} = 1$
$2a = 1$
$a = \frac{1}{2}$

(B) Given the function $f(t) = \frac{1}{t^2 + t - 2}$.
Factoring the denominator,we get $f(t) = \frac{1}{(t + 2)(t - 1)}$.
The function $f(t)$ is discontinuous where the denominator is zero,i.e.,at $t = -2$ and $t = 1$.
Now,substitute $t = \frac{1}{x - 1}$ to find the values of $x$:
For $t = -2$:
$\frac{1}{x - 1} = -2 \implies x - 1 = -\frac{1}{2} \implies x = 1 - \frac{1}{2} = \frac{1}{2}$.
For $t = 1$:
$\frac{1}{x - 1} = 1 \implies x - 1 = 1 \implies x = 2$.
Additionally,the expression $t = \frac{1}{x - 1}$ is discontinuous at $x = 1$.
However,checking the options provided,the points of discontinuity for the composite function $f(x)$ are $x = \frac{1}{2}$ and $x = 2$.

(B) The cumulative distribution function (c.d.f.) $F(x)$ is defined as the integral of the probability density function (p.d.f.) $f(x)$ from $-\infty$ to $x$.
For $0 < x < 1$,we have:
$F(x) = \int_0^x f(t) dt$
$F(x) = \int_0^x 12t^2(1-t) dt$
$F(x) = 12 \int_0^x (t^2 - t^3) dt$
$F(x) = 12 \left[ \frac{t^3}{3} - \frac{t^4}{4} \right]_0^x$
$F(x) = 12 \left( \frac{x^3}{3} - \frac{x^4}{4} \right)$
$F(x) = 4x^3 - 3x^4$
Thus,the correct option is $B$.

(B) Since $f(x)$ is continuous on $[-2,2]$,it must be continuous at $x=0$ and $x=1$.
For continuity at $x=0$,$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$.
$\lim_{x \to 0^-} (\frac{\sin ax}{x} + 3) = a + 3$.
$\lim_{x \to 0^+} (2x + 7) = 7$.
Equating them,$a + 3 = 7 \implies a = 4$.
For continuity at $x=1$,$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$.
$\lim_{x \to 1^-} (2x + 7) = 2(1) + 7 = 9$.
$\lim_{x \to 1^+} (\sqrt{x^2+8} - b) = \sqrt{1+8} - b = 3 - b$.
Equating them,$9 = 3 - b \implies b = -6$.
Therefore,$2a + 3b = 2(4) + 3(-6) = 8 - 18 = -10$.

(B) The function is given by $f(x) = [x] \cdot \cos \left( \frac{2x - 1}{2} \pi \right)$.
We know that $[x]$ is the greatest integer function,which is discontinuous at all integer values of $x$.
Let $x = n$,where $n \in \mathbb{Z}$.
At $x = n$,the term $\cos \left( \frac{2n - 1}{2} \pi \right) = \cos \left( n\pi - \frac{\pi}{2} \right) = \cos \left( n\pi \right) \cos \left( \frac{\pi}{2} \right) + \sin \left( n\pi \right) \sin \left( \frac{\pi}{2} \right) = 0 + 0 = 0$.
Since the product of a discontinuous function and a function that is zero at those points can be continuous,we check the limit at $x = n$.
For $x = n + h$ (where $h \to 0^+$),$f(n+h) = [n+h] \cdot \cos \left( \frac{2(n+h) - 1}{2} \pi \right) = n \cdot \cos \left( n\pi - \frac{\pi}{2} + h\pi \right) = n \cdot \sin(h\pi) \approx n \cdot h\pi \to 0$.
For $x = n - h$ (where $h \to 0^+$),$f(n-h) = [n-h] \cdot \cos \left( \frac{2(n-h) - 1}{2} \pi \right) = (n-1) \cdot \cos \left( n\pi - \frac{\pi}{2} - h\pi \right) = (n-1) \cdot \sin(-h\pi) \approx -(n-1) \cdot h\pi \to 0$.
Since the left-hand limit and right-hand limit are both $0$,and $f(n) = n \cdot 0 = 0$,the function is continuous at all integers.
However,the question implies a standard property check. Re-evaluating: $[x]$ is discontinuous at all integers,but $f(x)$ is continuous everywhere because the product with the zero-valued cosine term compensates for the jump discontinuity of $[x]$ at every integer $n$.

(C) Since the function $f$ is continuous at $x=0$,we must have $f(0) = \lim_{x \to 0} f(x)$.
Given $f(0) = k$,we calculate the limit:
$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{1}{x} \log \left(\frac{1+3x}{1-2x}\right)$
$= \lim_{x \to 0} \frac{1}{x} [\log(1+3x) - \log(1-2x)]$
$= \lim_{x \to 0} \left[ \frac{\log(1+3x)}{x} - \frac{\log(1-2x)}{x} \right]$
Using the standard limit $\lim_{u \to 0} \frac{\log(1+u)}{u} = 1$:
$= \lim_{x \to 0} \left[ 3 \cdot \frac{\log(1+3x)}{3x} - (-2) \cdot \frac{\log(1-2x)}{-2x} \right]$
$= 3(1) + 2(1) = 5$.
Therefore,$k = 5$.

(A) To check the differentiability of $f(x) = |x - \pi|(e^{|x|} - 1) \sin |x|$ at $x = \pi$ and $x = 0$:
$1$. Differentiability at $x = \pi$:
$f(\pi) = 0$.
$f'(x) = \frac{d}{dx} [|x - \pi|(e^{|x|} - 1) \sin |x|]$.
Since $|x - \pi|$ is differentiable at $x = \pi$ only if the other factors are zero,note that at $x = \pi$,$(e^{|\pi|} - 1) \sin |\pi| = (e^{\pi} - 1) \cdot 0 = 0$.
Using the definition of the derivative,$\lim_{h \to 0} \frac{f(\pi + h) - f(\pi)}{h} = \lim_{h \to 0} \frac{|h|(e^{|\pi + h|} - 1) \sin |\pi + h|}{h} = \lim_{h \to 0} \frac{|h|}{h} (e^{\pi} - 1) \sin \pi = 0$.
Thus,$f(x)$ is differentiable at $x = \pi$.
$2$. Differentiability at $x = 0$:
$f(0) = 0$.
$\lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0} \frac{|h - \pi|(e^{|h|} - 1) \sin |h|}{h}$.
As $h \to 0$,$e^{|h|} - 1 \approx |h|$ and $\sin |h| \approx |h|$.
So,the limit becomes $\lim_{h \to 0} \frac{|-\pi| \cdot |h| \cdot |h|}{h} = \lim_{h \to 0} \pi \cdot \frac{h^2}{h} = \lim_{h \to 0} \pi h = 0$.
Thus,$f(x)$ is differentiable at $x = 0$.
Since the function is differentiable everywhere,the set $S$ is an empty set $\phi$.

(C) Given function $f(x) = \log x$ on the interval $[1, e]$.
First,we find the derivative of the function:
$f'(x) = \frac{1}{x}$.
According to Lagrange's Mean Value Theorem $(LMVT)$,there exists at least one point $c \in (1, e)$ such that:
$f'(c) = \frac{f(e) - f(1)}{e - 1}$.
Substituting the values:
$f(e) = \log e = 1$ and $f(1) = \log 1 = 0$.
So,$\frac{1}{c} = \frac{1 - 0}{e - 1}$.
$\frac{1}{c} = \frac{1}{e - 1}$.
Therefore,$c = e - 1$.

(C) Let $I = \int_0^1 (\cos^{-1} x)(1) \, dx$.
Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = \cos^{-1} x$ and $dv = dx$.
Then $du = -\frac{1}{\sqrt{1-x^2}} \, dx$ and $v = x$.
$I = [x \cos^{-1} x]_0^1 - \int_0^1 x \left(-\frac{1}{\sqrt{1-x^2}}\right) \, dx$.
$I = [x \cos^{-1} x]_0^1 + \int_0^1 \frac{x}{\sqrt{1-x^2}} \, dx$.
For the integral $\int \frac{x}{\sqrt{1-x^2}} \, dx$,let $t = 1-x^2$,then $dt = -2x \, dx$,so $x \, dx = -\frac{1}{2} \, dt$.
$\int \frac{x}{\sqrt{1-x^2}} \, dx = -\sqrt{1-x^2}$.
Thus,$I = [x \cos^{-1} x - \sqrt{1-x^2}]_0^1$.
Evaluating at the limits:
$I = [1 \cdot \cos^{-1}(1) - \sqrt{1-1^2}] - [0 \cdot \cos^{-1}(0) - \sqrt{1-0^2}]$.
$I = [1 \cdot 0 - 0] - [0 - 1] = 0 - (-1) = 1$.

(A) Let $I = \int_1^2 \frac{dx}{(x^2-2x+4)^{\frac{3}{2}}} = \int_1^2 \frac{dx}{((x-1)^2+3)^{\frac{3}{2}}}$.
Substitute $x-1 = \sqrt{3} \tan \theta$,so $dx = \sqrt{3} \sec^2 \theta \ d\theta$.
When $x=1$,$\theta=0$. When $x=2$,$\tan \theta = \frac{1}{\sqrt{3}}$,so $\theta = \frac{\pi}{6}$.
$I = \int_0^{\frac{\pi}{6}} \frac{\sqrt{3} \sec^2 \theta}{(3 \sec^2 \theta)^{\frac{3}{2}}} \ d\theta = \int_0^{\frac{\pi}{6}} \frac{\sqrt{3} \sec^2 \theta}{3\sqrt{3} \sec^3 \theta} \ d\theta = \frac{1}{3} \int_0^{\frac{\pi}{6}} \cos \theta \ d\theta$.
$I = \frac{1}{3} [\sin \theta]_0^{\frac{\pi}{6}} = \frac{1}{3} (\sin \frac{\pi}{6} - \sin 0) = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$.
Given $\frac{k}{k+5} = \frac{1}{6}$,we have $6k = k+5$,which implies $5k = 5$,so $k = 1$.

$(A)$ Let $I = \int_0^{\frac{1}{2}} \frac{x^2}{\left(1-x^2\right)^{\frac{3}{2}}} \,d x$.
Substitute $x = \sin \theta$, then $dx = \cos \theta d\theta$.
When $x = 0, \theta = 0$. When $x = \frac{1}{2}, \theta = \frac{\pi}{6}$.
The denominator becomes $(1 - \sin^2 \theta)^{\frac{3}{2}} = (\cos^2 \theta)^{\frac{3}{2}} = \cos^3 \theta$.
Thus, $I = \int_0^{\frac{\pi}{6}} \frac{\sin^2 \theta \cdot \cos \theta}{\cos^3 \theta} d\theta = \int_0^{\frac{\pi}{6}} \tan^2 \theta d\theta$.
Using the identity $\tan^2 \theta = \sec^2 \theta - 1$, we get:
$I = \int_0^{\frac{\pi}{6}} (\sec^2 \theta - 1) d\theta = [\tan \theta - \theta]_0^{\frac{\pi}{6}}$.
$I = (\tan \frac{\pi}{6} - \frac{\pi}{6}) - (\tan 0 - 0) = \frac{1}{\sqrt{3}} - \frac{\pi}{6} = \frac{\sqrt{3}}{3} - \frac{\pi}{6} = \frac{2\sqrt{3} - \pi}{6}$.
Given that $I = \frac{k}{6}$, comparing the expressions gives $k = 2\sqrt{3} - \pi$.

(C) Let $I = \int_0^\pi \frac{x \tan x}{\sec x + \cos x} \,dx$ $\quad \dots (i)$
Using the property $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$, we have:
$I = \int_0^\pi \frac{(\pi - x) \tan(\pi - x)}{\sec(\pi - x) + \cos(\pi - x)} \,dx$
Since $\tan(\pi - x) = -\tan x$, $\sec(\pi - x) = -\sec x$, and $\cos(\pi - x) = -\cos x$:
$I = \int_0^\pi \frac{(\pi - x)(-\tan x)}{-(\sec x + \cos x)} \,dx = \int_0^\pi \frac{(\pi - x) \tan x}{\sec x + \cos x} \,dx$ $\quad \dots (ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^\pi \frac{\pi \tan x}{\sec x + \cos x} \,dx = \pi \int_0^\pi \frac{\sin x / \cos x}{1/\cos x + \cos x} \,dx = \pi \int_0^\pi \frac{\sin x}{1 + \cos^2 x} \,dx$
Let $t = \cos x$, then $dt = -\sin x \,dx$. When $x=0, t=1$; when $x=\pi, t=-1$.
$2I = \pi \int_1^{-1} \frac{-dt}{1 + t^2} = \pi \int_{-1}^1 \frac{dt}{1 + t^2} = \pi [\tan^{-1} t]_{-1}^1$
$2I = \pi (\tan^{-1}(1) - \tan^{-1}(-1)) = \pi (\frac{\pi}{4} - (-\frac{\pi}{4})) = \pi (\frac{\pi}{2}) = \frac{\pi^2}{2}$
Therefore, $I = \frac{\pi^2}{4}$.

(B) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec^{\frac{2}{3}} x \operatorname{cosec}^{\frac{4}{3}} x \, dx$
$= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{\cos^{\frac{2}{3}} x \sin^{\frac{4}{3}} x} \, dx$
$= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{\left(\frac{\sin x}{\cos x}\right)^{\frac{4}{3}} \cos^2 x} \, dx$
$= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sec^2 x}{\tan^{\frac{4}{3}} x} \, dx$
Let $\tan x = t$,then $\sec^2 x \, dx = dt$.
When $x = \frac{\pi}{6}$,$t = \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}} = 3^{-\frac{1}{2}}$.
When $x = \frac{\pi}{3}$,$t = \tan(\frac{\pi}{3}) = \sqrt{3} = 3^{\frac{1}{2}}$.
$I = \int_{3^{-\frac{1}{2}}}^{3^{\frac{1}{2}}} t^{-\frac{4}{3}} \, dt = \left[ \frac{t^{-\frac{1}{3}}}{-\frac{1}{3}} \right]_{3^{-\frac{1}{2}}}^{3^{\frac{1}{2}}} = -3 \left[ t^{-\frac{1}{3}} \right]_{3^{-\frac{1}{2}}}^{3^{\frac{1}{2}}}$
$= -3 \left( (3^{\frac{1}{2}})^{-\frac{1}{3}} - (3^{-\frac{1}{2}})^{-\frac{1}{3}} \right) = -3 \left( 3^{-\frac{1}{6}} - 3^{\frac{1}{6}} \right)$
$= 3 \cdot 3^{\frac{1}{6}} - 3 \cdot 3^{-\frac{1}{6}} = 3^{\frac{7}{6}} - 3^{\frac{5}{6}}$

(D) Let $h(x) = [f(x)+f(-x)][g(x)-g(-x)]$.
We evaluate $h(-x)$:
$h(-x) = [f(-x)+f(x)][g(-x)-g(x)]$.
This can be rewritten as:
$h(-x) = [f(x)+f(-x)] \cdot -[g(x)-g(-x)]$.
$h(-x) = -[f(x)+f(-x)][g(x)-g(-x)] = -h(x)$.
Since $h(-x) = -h(x)$,the function $h(x)$ is an odd function.
According to the property of definite integrals,if $h(x)$ is an odd function,then $\int_{-a}^{a} h(x) \, dx = 0$.
Therefore,$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [f(x)+f(-x)][g(x)-g(-x)] \, dx = 0$.

(D) Let $I = \int_{-1}^3 \left(\cot^{-1}\left(\frac{x}{x^2+1}\right) + \cot^{-1}\left(\frac{x^2+1}{x}\right)\right) dx$.
Since $\cot^{-1}(u) = \tan^{-1}(\frac{1}{u})$ for $u > 0$,we observe the term $\cot^{-1}(\frac{x}{x^2+1})$.
For $x > 0$,$\cot^{-1}(\frac{x}{x^2+1}) = \tan^{-1}(\frac{x^2+1}{x})$.
However,the identity $\tan^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$ holds for all $u \in \mathbb{R}$.
Thus,the integrand simplifies to $\frac{\pi}{2}$ for all $x \neq 0$.
$I = \int_{-1}^3 \frac{\pi}{2} dx = \frac{\pi}{2} [x]_{-1}^3$.
$I = \frac{\pi}{2} (3 - (-1)) = \frac{\pi}{2} (4) = 2\pi$.

(B) Let $I = \int_0^{\pi} \frac{dx}{4+3 \cos x}$.
Using the substitution $\tan \frac{x}{2} = t$,we have $dx = \frac{2 dt}{1+t^2}$ and $\cos x = \frac{1-t^2}{1+t^2}$.
As $x$ goes from $0$ to $\pi$,$t$ goes from $\tan(0) = 0$ to $\tan(\frac{\pi}{2}) = \infty$.
Substituting these into the integral:
$I = \int_0^{\infty} \frac{\frac{2 dt}{1+t^2}}{4 + 3(\frac{1-t^2}{1+t^2})} = \int_0^{\infty} \frac{2 dt}{4(1+t^2) + 3(1-t^2)} = \int_0^{\infty} \frac{2 dt}{4 + 4t^2 + 3 - 3t^2} = \int_0^{\infty} \frac{2 dt}{7 + t^2}$.
Using the standard integral formula $\int \frac{dx}{a^2+x^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$:
$I = 2 \left[ \frac{1}{\sqrt{7}} \tan^{-1}(\frac{t}{\sqrt{7}}) \right]_0^{\infty} = \frac{2}{\sqrt{7}} [\tan^{-1}(\infty) - \tan^{-1}(0)]$.
Since $\tan^{-1}(\infty) = \frac{\pi}{2}$ and $\tan^{-1}(0) = 0$:
$I = \frac{2}{\sqrt{7}} \times \frac{\pi}{2} = \frac{\pi}{\sqrt{7}}$.

(D) To evaluate the integral $\int_0^4 |2x - 5| \, dx$,we first find the point where the expression inside the absolute value changes sign.
Set $2x - 5 = 0$,which gives $x = \frac{5}{2}$.
Since $\frac{5}{2}$ lies between $0$ and $4$,we split the integral at $x = \frac{5}{2}$:
$\int_0^4 |2x - 5| \, dx = \int_0^{\frac{5}{2}} -(2x - 5) \, dx + \int_{\frac{5}{2}}^4 (2x - 5) \, dx$
$= \int_0^{\frac{5}{2}} (5 - 2x) \, dx + \int_{\frac{5}{2}}^4 (2x - 5) \, dx$
$= [5x - x^2]_0^{\frac{5}{2}} + [x^2 - 5x]_{\frac{5}{2}}^4$
$= (5(\frac{5}{2}) - (\frac{5}{2})^2) - (0) + ((4)^2 - 5(4)) - ((\frac{5}{2})^2 - 5(\frac{5}{2}))$
$= (\frac{25}{2} - \frac{25}{4}) + ((16 - 20) - (\frac{25}{4} - \frac{25}{2}))$
$= \frac{25}{4} + (-4 - (-\frac{25}{4}))$
$= \frac{25}{4} + (-4 + \frac{25}{4}) = \frac{25}{4} + \frac{9}{4} = \frac{34}{4} = \frac{17}{2}$.

(B) Let $I = \int_0^\pi \left| \sin x - \frac{2x}{\pi} \right| dx$.
Consider the function $f(x) = \sin x - \frac{2x}{\pi}$.
At $x = \frac{\pi}{2}$,$f(\frac{\pi}{2}) = \sin(\frac{\pi}{2}) - \frac{2(\pi/2)}{\pi} = 1 - 1 = 0$.
For $0 \le x \le \frac{\pi}{2}$,$\sin x \ge \frac{2x}{\pi}$,so $|\sin x - \frac{2x}{\pi}| = \sin x - \frac{2x}{\pi}$.
For $\frac{\pi}{2} \le x \le \pi$,$\sin x \le \frac{2x}{\pi}$,so $|\sin x - \frac{2x}{\pi}| = \frac{2x}{\pi} - \sin x$.
Thus,$I = \int_0^{\pi/2} (\sin x - \frac{2x}{\pi}) dx + \int_{\pi/2}^{\pi} (\frac{2x}{\pi} - \sin x) dx$.
Evaluating the integrals:
$I = [-\cos x - \frac{x^2}{\pi}]_0^{\pi/2} + [\frac{x^2}{\pi} + \cos x]_{\pi/2}^{\pi}$.
$I = [(-\cos(\frac{\pi}{2}) - \frac{(\pi/2)^2}{\pi}) - (-\cos(0) - 0)] + [(\frac{\pi^2}{\pi} + \cos(\pi)) - (\frac{(\pi/2)^2}{\pi} + \cos(\frac{\pi}{2}))]$.
$I = [(0 - \frac{\pi}{4}) - (-1)] + [(\pi - 1) - (\frac{\pi}{4} + 0)]$.
$I = 1 - \frac{\pi}{4} + \pi - 1 - \frac{\pi}{4} = \pi - \frac{\pi}{2} = \frac{\pi}{2}$.

(B) We need to evaluate the integral $\int_{-2}^{3} f(x) dx$.
Split the integral based on the definition of $f(x)$:
$\int_{-2}^{3} f(x) dx = \int_{-2}^{2} f(x) dx + \int_{2}^{3} f(x) dx$
For $|x| \leq 2$,$f(x) = e^{\cos x} \sin x$. Note that $g(x) = e^{\cos x} \sin x$ is an odd function because $g(-x) = e^{\cos(-x)} \sin(-x) = e^{\cos x} (-\sin x) = -g(x)$.
The integral of an odd function over a symmetric interval $[-a, a]$ is $0$. Thus,$\int_{-2}^{2} e^{\cos x} \sin x dx = 0$.
For $x > 2$,$f(x) = 2$. Therefore,$\int_{2}^{3} 2 dx = 2[x]_{2}^{3} = 2(3 - 2) = 2(1) = 2$.
Adding these results,we get $0 + 2 = 2$.

(C) We are given $I_n = \int_0^{\frac{\pi}{4}} \tan^n \theta \, d\theta$.
Consider the sum $I_n + I_{n-2} = \int_0^{\frac{\pi}{4}} (\tan^n \theta + \tan^{n-2} \theta) \, d\theta$.
Factor out $\tan^{n-2} \theta$:
$I_n + I_{n-2} = \int_0^{\frac{\pi}{4}} \tan^{n-2} \theta (\tan^2 \theta + 1) \, d\theta$.
Since $1 + \tan^2 \theta = \sec^2 \theta$,we have:
$I_n + I_{n-2} = \int_0^{\frac{\pi}{4}} \tan^{n-2} \theta \sec^2 \theta \, d\theta$.
Let $u = \tan \theta$,then $du = \sec^2 \theta \, d\theta$.
When $\theta = 0$,$u = 0$. When $\theta = \frac{\pi}{4}$,$u = 1$.
Thus,$I_n + I_{n-2} = \int_0^1 u^{n-2} \, du = \left[ \frac{u^{n-1}}{n-1} \right]_0^1 = \frac{1}{n-1}$.
For $n = 12$,we have $I_{12} + I_{10} = \frac{1}{12-1} = \frac{1}{11}$.

(C) Given $f^{\prime}(x)=f(x)$.
Dividing by $f(x)$,we get $\frac{f^{\prime}(x)}{f(x)}=1$.
Integrating both sides with respect to $x$,we get $\ln|f(x)| = x + C$.
Since $f(0)=1$,we have $\ln(1) = 0 + C$,which implies $C=0$.
Thus,$f(x) = e^x$.
Given $f(x)+g(x)=x^2$,we have $g(x) = x^2 - e^x$.
Now,we need to evaluate the integral $I = \int_0^1 f(x)g(x) dx = \int_0^1 e^x(x^2 - e^x) dx$.
$I = \int_0^1 (x^2 e^x - e^{2x}) dx = \int_0^1 x^2 e^x dx - \int_0^1 e^{2x} dx$.
Using integration by parts for $\int x^2 e^x dx = x^2 e^x - \int 2x e^x dx = x^2 e^x - 2(x e^x - e^x) = e^x(x^2 - 2x + 2)$.
Evaluating the definite integral: $[e^x(x^2 - 2x + 2)]_0^1 = (e(1-2+2)) - (e^0(0-0+2)) = e - 2$.
Evaluating the second part: $\int_0^1 e^{2x} dx = [\frac{1}{2} e^{2x}]_0^1 = \frac{1}{2}(e^2 - 1)$.
Therefore,$I = (e - 2) - \frac{1}{2}(e^2 - 1) = e - 2 - \frac{e^2}{2} + \frac{1}{2} = e - \frac{e^2}{2} - \frac{3}{2}$.

(C) Given $I_1 = \int_{1-h}^{h} x f(x(1-x)) dx$ and $I_2 = \int_{1-h}^{h} f(x(1-x)) dx$.
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we have $a+b = (1-h) + h = 1$.
Applying this to $I_1$:
$I_1 = \int_{1-h}^{h} (1-x) f((1-x)(1-(1-x))) dx$
$I_1 = \int_{1-h}^{h} (1-x) f((1-x)x) dx$
$I_1 = \int_{1-h}^{h} f(x(1-x)) dx - \int_{1-h}^{h} x f(x(1-x)) dx$
$I_1 = I_2 - I_1$
$2I_1 = I_2$
Therefore,$\frac{I_1}{I_2} = \frac{1}{2}$.

(B) Expanding the determinant along the first row:
$\cos(A+B) [\cos A \cos B - \sin A \sin B] - (-\sin(A+B)) [\sin A \cos B - (-\cos A \sin B)] + \cos(2B) [\sin A \sin A - (-\cos A \cos A)] = 0$
Simplify the terms inside the brackets:
$\cos(A+B) [\cos(A+B)] + \sin(A+B) [\sin(A+B)] + \cos(2B) [\sin^2 A + \cos^2 A] = 0$
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$:
$\cos^2(A+B) + \sin^2(A+B) + \cos(2B) = 1 = 0$
Since $\cos^2 \theta + \sin^2 \theta = 1$:
$1 + \cos(2B) = 0$
$\cos(2B) = -1$
$2B = (2n+1)\pi$
$B = (2n+1) \frac{\pi}{2}, n \in Z$

(D) Given equation: $y^2=2 c(x+\sqrt{c}) \dots (i)$
Differentiating with respect to $x$:
$2 y \frac{dy}{dx} = 2c \implies c = y \frac{dy}{dx} \dots (ii)$
Substituting $(ii)$ into $(i)$:
$y^2 = 2 \left(y \frac{dy}{dx}\right) \left(x + \sqrt{y \frac{dy}{dx}}\right)$
Dividing by $y$ (assuming $y \neq 0$):
$y = 2 \frac{dy}{dx} \left(x + \sqrt{y \frac{dy}{dx}}\right)$
$y - 2x \frac{dy}{dx} = 2 \frac{dy}{dx} \sqrt{y \frac{dy}{dx}}$
Squaring both sides:
$(y - 2x \frac{dy}{dx})^2 = 4 \left(\frac{dy}{dx}\right)^2 \left(y \frac{dy}{dx}\right)$
$(y - 2x \frac{dy}{dx})^2 = 4y \left(\frac{dy}{dx}\right)^3$
The highest order derivative is $\frac{dy}{dx}$,so the order is $1$. The power of the highest order derivative is $3$,so the degree is $3$.

(C) The general equation of a parabola with its axis parallel to the $Y$-axis is given by $(x-h)^2 = 4a(y-k)$,where $(h, k)$ is the vertex and $a$ is a constant.
This equation contains three arbitrary constants: $h$,$k$,and $a$.
To find the differential equation,we differentiate with respect to $x$ three times.
First differentiation: $2(x-h) = 4a \frac{dy}{dx} \implies (x-h) = 2a y_1$.
Second differentiation: $1 = 2a y_2$.
Third differentiation: $0 = 2a y_3$.
Since $2a \neq 0$,we must have $y_3 = 0$.

(B) Given equation: $x^k + y^k = a^k$
Differentiating with respect to $x$:
$k x^{k-1} + k y^{k-1} \frac{dy}{dx} = 0$
$\frac{dy}{dx} = -\frac{k x^{k-1}}{k y^{k-1}} = -\frac{x^{k-1}}{y^{k-1}} = -(\frac{x}{y})^{k-1}$
$\frac{dy}{dx} = -(\frac{y}{x})^{-(k-1)} = -(\frac{y}{x})^{1-k}$
So,$\frac{dy}{dx} + (\frac{y}{x})^{1-k} = 0$
Comparing this with the given equation $\frac{dy}{dx} + (\frac{y}{x})^{\frac{1}{3}} = 0$:
$1 - k = \frac{1}{3}$
$k = 1 - \frac{1}{3} = \frac{2}{3}$

(D) Given equation: $y = e^x(a \cos x + b \sin x)$
Differentiating with respect to $x$ using the product rule:
$\frac{dy}{dx} = e^x(a \cos x + b \sin x) + e^x(-a \sin x + b \cos x)$
Since $y = e^x(a \cos x + b \sin x)$,we can write:
$\frac{dy}{dx} = y + e^x(b \cos x - a \sin x)$
Differentiating again with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{dy}{dx} + [e^x(b \cos x - a \sin x) + e^x(-b \sin x - a \cos x)]$
Substitute $e^x(b \cos x - a \sin x) = \frac{dy}{dx} - y$:
$\frac{d^2y}{dx^2} = \frac{dy}{dx} + (\frac{dy}{dx} - y) - e^x(a \cos x + b \sin x)$
Since $e^x(a \cos x + b \sin x) = y$,we have:
$\frac{d^2y}{dx^2} = \frac{dy}{dx} + \frac{dy}{dx} - y - y$
$\frac{d^2y}{dx^2} = 2\frac{dy}{dx} - 2y$
Rearranging the terms:
$\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0$

(A) The general equation of a circle passing through the origin with its center on the $y$-axis is given by $x^{2} + (y-a)^{2} = a^{2}$,where $a$ is an arbitrary constant.
Expanding this,we get $x^{2} + y^{2} - 2ay + a^{2} = a^{2}$,which simplifies to $x^{2} + y^{2} - 2ay = 0$.
To eliminate the arbitrary constant $a$,we differentiate with respect to $x$:
$2x + 2y \frac{dy}{dx} - 2a \frac{dy}{dx} = 0$.
Dividing by $2$,we get $x + y \frac{dy}{dx} - a \frac{dy}{dx} = 0$,which implies $a = \frac{x + y \frac{dy}{dx}}{\frac{dy}{dx}} = x \frac{dx}{dy} + y$.
Substituting $a$ back into the original equation $x^{2} + y^{2} = 2ay$:
$x^{2} + y^{2} = 2(x \frac{dx}{dy} + y)y = 2xy \frac{dx}{dy} + 2y^{2}$.
Rearranging,$x^{2} - y^{2} = 2xy \frac{dx}{dy}$.
Since $\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$,we have $x^{2} - y^{2} = \frac{2xy}{\frac{dy}{dx}}$.
Thus,$(x^{2} - y^{2}) \frac{dy}{dx} = 2xy$,or $(x^{2} - y^{2}) \frac{dy}{dx} - 2xy = 0$.

(D) Given the differential equation: $(2+\sin x) \frac{dy}{dx} + (y+1) \cos x = 0$.
Rearranging the terms to separate the variables: $\frac{1}{y+1} dy = -\frac{\cos x}{2+\sin x} dx$.
Integrating both sides: $\int \frac{1}{y+1} dy = -\int \frac{\cos x}{2+\sin x} dx$.
This gives: $\ln(y+1) = -\ln(2+\sin x) + C$.
Using the initial condition $y(0)=1$: $\ln(1+1) = -\ln(2+\sin 0) + C \implies \ln 2 = -\ln 2 + C \implies C = 2\ln 2 = \ln 4$.
Substituting $C$ back into the equation: $\ln(y+1) = -\ln(2+\sin x) + \ln 4 = \ln\left(\frac{4}{2+\sin x}\right)$.
Taking the exponential of both sides: $y+1 = \frac{4}{2+\sin x} \implies y = \frac{4}{2+\sin x} - 1$.
Now,evaluating at $x = \frac{\pi}{2}$: $y\left(\frac{\pi}{2}\right) = \frac{4}{2+\sin(\pi/2)} - 1 = \frac{4}{2+1} - 1 = \frac{4}{3} - 1 = \frac{1}{3}$.

(C) Given differential equation is $(1+y^2) dx - xy dy = 0$.
Rearranging the terms,we get $(1+y^2) dx = xy dy$.
Separating the variables,we have $\frac{1}{x} dx = \frac{y}{1+y^2} dy$.
Integrating both sides: $\int \frac{1}{x} dx = \int \frac{y}{1+y^2} dy$.
This gives $\log |x| = \frac{1}{2} \log (1+y^2) + C$.
Given $x=1$ and $y=0$,we substitute these values: $\log(1) = \frac{1}{2} \log(1+0^2) + C$,which implies $0 = 0 + C$,so $C=0$.
The equation becomes $\log x = \frac{1}{2} \log (1+y^2)$.
Multiplying by $2$,we get $2 \log x = \log (1+y^2)$,which is $\log(x^2) = \log(1+y^2)$.
Taking the exponential of both sides,$x^2 = 1+y^2$,or $x^2 - y^2 = 1$.
This equation represents a rectangular hyperbola.

(C) Given differential equation is $\log \left(\frac{d y}{d x}\right)=a x+b y$.
Taking exponential on both sides,we get $\frac{d y}{d x}=e^{a x+b y} = e^{a x} \cdot e^{b y}$.
Separating the variables,we have $\frac{d y}{e^{b y}} = e^{a x} d x$,which is $e^{-b y} d y = e^{a x} d x$.
Integrating both sides,we get $\int e^{-b y} d y = \int e^{a x} d x$.
This results in $\frac{e^{-b y}}{-b} = \frac{e^{a x}}{a} + C$.
Rearranging the terms,we get $\frac{e^{a x}}{a} + \frac{e^{-b y}}{b} = -C$.
Multiplying by $ab$,we get $b e^{a x} + a e^{-b y} = -abC$.
Letting $c_1 = -abC$,we obtain $a e^{-b y} + b e^{a x} = c_1$.