AP EAMCET 2021 Chemistry Question Paper with Answer and Solution

(D) The given equation of the pair of lines is $x^2+2xy-35y^2-4x+44y-12=0$.
Factoring the quadratic expression,we get $(x+7y-6)(x-5y+2)=0$.
This represents two lines: $L_1: x+7y-6=0$ and $L_2: x-5y+2=0$.
The third line is $L_3: 5x+\lambda y-8=0$.
For the three lines to be concurrent,the determinant of their coefficients must be zero:
$\begin{vmatrix} 1 & 7 & -6 \\ 1 & -5 & 2 \\ 5 & \lambda & -8 \end{vmatrix} = 0$
Expanding the determinant:
$1(40-2\lambda) - 7(-8-10) - 6(\lambda+25) = 0$
$40 - 2\lambda + 126 - 6\lambda - 150 = 0$
$16 - 8\lambda = 0$
$8\lambda = 16$
$\lambda = 2$

(C) For three lines to be concurrent,the determinant of their coefficients must be zero:
$\begin{vmatrix} 4 & 3 & -1 \\ 1 & -1 & 5 \\ k & 5 & -3 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$4((-1)(-3) - (5)(5)) - 3((1)(-3) - (5)(k)) - 1((1)(5) - (-1)(k)) = 0$
$4(3 - 25) - 3(-3 - 5k) - 1(5 + k) = 0$
$4(-22) + 9 + 15k - 5 - k = 0$
$-88 + 4 + 14k = 0$
$14k = 84$
$k = 6$

(C) Since the given three lines are concurrent,the determinant of their coefficients must be zero:
$\left|\begin{array}{ccc} 4 & 3 & -1 \\ 1 & -1 & 5 \\ k & 5 & -3 \end{array}\right| = 0$
Expanding the determinant along the first row:
$4((-1)(-3) - (5)(5)) - 3((1)(-3) - (5)(k)) - 1((1)(5) - (-1)(k)) = 0$
$4(3 - 25) - 3(-3 - 5k) - 1(5 + k) = 0$
$4(-22) + 9 + 15k - 5 - k = 0$
$-88 + 4 + 14k = 0$
$-84 + 14k = 0$
$14k = 84$
$k = 6$

(C) The given equation is $5x^2 - 8xy + 3y^2 = 0$.
Dividing by $x^2$,we get $3(\frac{y}{x})^2 - 8(\frac{y}{x}) + 5 = 0$.
Let $m = \frac{y}{x}$ be the slope of the lines. Then $3m^2 - 8m + 5 = 0$.
Solving the quadratic equation $3m^2 - 8m + 5 = 0$:
$3m^2 - 3m - 5m + 5 = 0$
$3m(m - 1) - 5(m - 1) = 0$
$(3m - 5)(m - 1) = 0$
Thus,the slopes are $m_1 = \frac{5}{3}$ and $m_2 = 1$ (since $m_1 > m_2$).
The ratio $m_1 : m_2 = \frac{5}{3} : 1 = 5 : 3$.

(C) The given equation is $4x^2 - 5xy + y^2 = 0$.
Dividing by $x^2$,we get the quadratic equation in terms of $m = \frac{y}{x}$:
$m^2 - 5m + 4 = 0$.
Here,$m_1$ and $m_2$ are the roots of this equation.
Thus,$m_1 + m_2 = 5$ and $m_1 m_2 = 4$.
We know that $|m_1 - m_2| = \sqrt{(m_1 + m_2)^2 - 4m_1 m_2}$.
Substituting the values:
$|m_1 - m_2| = \sqrt{(5)^2 - 4(4)} = \sqrt{25 - 16} = \sqrt{9} = 3$.

(D) Given the equation:
$y^3 - 4x^2y = 0$
Factoring the expression:
$y(y^2 - 4x^2) = 0$
$y(y - 2x)(y + 2x) = 0$
This gives the three lines:
$L_1: y = 0$
$L_2: y = 2x$
$L_3: y = -2x$
All three lines pass through the origin $(0, 0)$.
Since all three lines intersect at a single common point $(0, 0)$,they are concurrent lines.

(B) Given the equation of the pair of lines: $x^2+2 h x y+2 y^2=0$ $(i)$.
Comparing this with the general equation $a x^2+2 h x y+b y^2=0$,we get $a=1$ and $b=2$.
Let the slopes of the lines be $m_1$ and $m_2$.
Then,$m_1+m_2 = -\frac{2h}{b} = -\frac{2h}{2} = -h$ $(ii)$ and $m_1 m_2 = \frac{a}{b} = \frac{1}{2}$ $(iii)$.
Given the ratio of slopes is $m_1:m_2 = 1:2$,so $m_2 = 2m_1$.
Substituting $m_2 = 2m_1$ into $(iii)$,we get $m_1(2m_1) = \frac{1}{2}$ $\Rightarrow 2m_1^2 = \frac{1}{2}$ $\Rightarrow m_1^2 = \frac{1}{4}$ $\Rightarrow m_1 = \pm \frac{1}{2}$.
If $m_1 = \frac{1}{2}$,then $m_2 = 1$,so $m_1+m_2 = \frac{3}{2}$. From $(ii)$,$-h = \frac{3}{2} \Rightarrow h = -\frac{3}{2}$.
If $m_1 = -\frac{1}{2}$,then $m_2 = -1$,so $m_1+m_2 = -\frac{3}{2}$. From $(ii)$,$-h = -\frac{3}{2} \Rightarrow h = \frac{3}{2}$.
Thus,$h = \pm \frac{3}{2}$.

(C) The given equation of the pair of straight lines is $x^2+4xy+y^2=0$.
Comparing this with the general equation $ax^2+2hxy+by^2=0$,we get $a=1$,$2h=4$ (so $h=2$),and $b=1$.
The angle $\theta$ between the pair of lines is given by the formula $\tan \theta = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$.
Substituting the values,we get $\tan \theta = \left| \frac{2\sqrt{2^2-(1)(1)}}{1+1} \right| = \left| \frac{2\sqrt{4-1}}{2} \right| = \sqrt{3}$.
Therefore,$\theta = \tan^{-1}(\sqrt{3}) = 60^{\circ}$.

(A) The given equation is of the form $Ax^2 + 2Hxy + By^2 = 0$,where $A = \cos \theta(\cos \theta + 1)$,$2H = -(2 \cos \theta + \sin^2 \theta)$,and $B = 1 - \cos \theta$.
The angle $\alpha$ between the lines is given by $\tan \alpha = \left| \frac{2\sqrt{H^2 - AB}}{A + B} \right|$.
First,calculate $A + B = \cos^2 \theta + \cos \theta + 1 - \cos \theta = \cos^2 \theta + 1$.
Next,calculate $H^2 - AB = \frac{(2 \cos \theta + \sin^2 \theta)^2}{4} - \cos \theta(\cos \theta + 1)(1 - \cos \theta)$.
$H^2 - AB = \frac{4 \cos^2 \theta + 4 \cos \theta \sin^2 \theta + \sin^4 \theta}{4} - \cos \theta(1 - \cos^2 \theta) = \cos^2 \theta + \cos \theta \sin^2 \theta + \frac{\sin^4 \theta}{4} - \cos \theta + \cos^3 \theta$.
Simplifying the expression inside the square root leads to $\frac{(1 + \cos^2 \theta)^2}{4}$.
Thus,$\tan \alpha = \frac{2 \sqrt{\frac{(1 + \cos^2 \theta)^2}{4}}}{\cos^2 \theta + 1} = \frac{1 + \cos^2 \theta}{1 + \cos^2 \theta} = 1$.
Therefore,$\alpha = \tan^{-1}(1) = \frac{\pi}{4}$.

(B) The acute angle $\theta$ between the pair of lines $ax^2 + 2hxy + by^2 = 0$ is given by $\tan \theta = \left|\frac{2\sqrt{h^2 - ab}}{a + b}\right|$.
Comparing the given equation $6x^2 + 11xy - 10y^2 = 0$ with $ax^2 + 2hxy + by^2 = 0$,we get $a = 6$,$b = -10$,and $2h = 11$,so $h = \frac{11}{2}$.
Substituting these values into the formula:
$\tan \theta = \left|\frac{2\sqrt{(\frac{11}{2})^2 - (6)(-10)}}{6 - 10}\right|$
$\tan \theta = \left|\frac{2\sqrt{\frac{121}{4} + 60}}{-4}\right|$
$\tan \theta = \left|\frac{2\sqrt{\frac{121 + 240}{4}}}{-4}\right| = \left|\frac{2 \cdot \frac{\sqrt{361}}{2}}{-4}\right| = \left|\frac{\sqrt{361}}{-4}\right| = \frac{\sqrt{361}}{4}$.
Therefore,$\theta = \tan^{-1}\left(\frac{\sqrt{361}}{4}\right)$.

(B) The given equation of the pair of lines is $ax^2+6xy+by^2-10x+10y-6=0$.
Comparing this with the general equation $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$,we get $A=a, H=3, B=b, G=-5, F=5, C=-6$.
For the equation to represent a pair of lines,the determinant must be zero:
$\begin{vmatrix} a & 3 & -5 \\ 3 & b & 5 \\ -5 & 5 & -6 \end{vmatrix} = 0$
$a(-6b-25) - 3(-18+25) - 5(15+5b) = 0$
$-6ab - 25a - 21 - 75 - 25b = 0$
$-6ab - 25(a+b) - 96 = 0$.
Since the lines are perpendicular,the coefficient of $x^2$ plus the coefficient of $y^2$ must be zero,so $a+b=0$,which implies $b=-a$.
Substituting $b=-a$ into the equation:
$-6a(-a) - 25(a-a) - 96 = 0$
$6a^2 - 96 = 0$
$6a^2 = 96$
$a^2 = 16$
$|a| = 4$.

(A) The angle $\theta$ between the pair of lines $ax^2 + 2hxy + by^2 = 0$ is given by $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a+b} \right|$.
For the first equation $x^2 + 2hxy + by^2 = 0$,the angle is $\theta$.
For the second equation $x^2 + 2xy \sec \theta + y^2 = 0$,we have $a=1$,$h=\sec \theta$,and $b=1$.
Let $\phi$ be the angle between these lines.
Then $\tan \phi = \left| \frac{2\sqrt{(\sec \theta)^2 - (1)(1)}}{1+1} \right|$.
$\tan \phi = \left| \frac{2\sqrt{\sec^2 \theta - 1}}{2} \right|$.
Since $\sec^2 \theta - 1 = \tan^2 \theta$,we have $\tan \phi = \sqrt{\tan^2 \theta} = |\tan \theta|$.
Therefore,$\phi = \theta$.

(C) The given equation is $(\sin^2 \alpha) y^2 - 2xy(\cos^2 \alpha) + (\cos^2 \alpha - 1) x^2 = 0$.
Comparing this with the standard form $ax^2 + 2hxy + by^2 = 0$,we get:
$a = \cos^2 \alpha - 1 = -\sin^2 \alpha$
$h = -\cos^2 \alpha$
$b = \sin^2 \alpha$
The angle $\theta$ between the pair of lines is given by $\tan \theta = \left| \frac{2 \sqrt{h^2 - ab}}{a + b} \right|$.
First,calculate $a + b$:
$a + b = -\sin^2 \alpha + \sin^2 \alpha = 0$.
Since the sum of the coefficients of $x^2$ and $y^2$ is zero $(a + b = 0)$,the lines are perpendicular to each other.
Therefore,$\theta = 90^{\circ}$.

(A) The given equation is $ab(x^2 - y^2) + (a^2 - b^2)xy = 0$.
Expanding this,we get $abx^2 + (a^2 - b^2)xy - aby^2 = 0$.
This is a homogeneous equation of the second degree in $x$ and $y$ of the form $Ax^2 + 2Hxy + By^2 = 0$,where $A = ab$,$2H = a^2 - b^2$,and $B = -ab$.
The sum of the coefficients of $x^2$ and $y^2$ is $A + B = ab + (-ab) = 0$.
Since the sum of the coefficients of $x^2$ and $y^2$ is zero,the lines represented by the equation are perpendicular to each other.
Therefore,the angle between the lines is $\frac{\pi}{2}$.

(A) Let the equation be represented as $(a y^2+p x y+e x^2)(y^2+q x y+k x^2) = 0$.
For two lines to be perpendicular,their slopes $m_1$ and $m_2$ must satisfy $m_1 m_2 = -1$.
By comparing the coefficients of the given homogeneous equation $a y^4+b x y^3+c x^2 y^2+d x^3 y+e x^4=0$ with the product of two quadratic factors,we derive the condition.
Specifically,if we assume the factors are $(y^2 - m_1 x y - k_1 x^2)(y^2 - m_2 x y - k_2 x^2) = 0$ where $m_1 m_2 = -1$,the expansion leads to the relation:
$(b+d)(a d+b e)+(e-a)^2(a+c+e)=0$.

(D) The equation of the angle bisectors for the pair of lines $a x^2+2 h x y+b y^2=0$ is given by $\frac{x^2-y^2}{a-b}=\frac{x y}{h}$.
For the first pair $x^2-2 p x y-y^2=0$,we have $a=1, b=-1, h=-p$.
The angle bisectors are $\frac{x^2-y^2}{1-(-1)}=\frac{x y}{-p}$,which simplifies to $\frac{x^2-y^2}{2}=\frac{x y}{-p}$,or $x^2-y^2+\frac{2 x y}{p}=0$.
Given that this pair of bisectors is $x^2-2 q x y-y^2=0$,we compare the coefficients:
Comparing $x^2+\frac{2}{p} x y-y^2=0$ with $x^2-2 q x y-y^2=0$,we get $\frac{2}{p}=-2 q$.
Therefore,$p q=-1$.

(B) The given equation is $3 x^2-5 x y+4 y^2=0$.
Comparing this with the general form $a x^2+2 h x y+b y^2=0$,we get $a=3$,$2 h=-5$,and $b=4$.
The equation of the bisectors of the angle between the pair of lines is given by $\frac{x^2-y^2}{a-b}=\frac{x y}{h}$.
Substituting the values,we get $\frac{x^2-y^2}{3-4}=\frac{x y}{-5/2}$.
This simplifies to $\frac{x^2-y^2}{-1}=\frac{2 x y}{-5}$.
Multiplying both sides by $-5$,we get $5(x^2-y^2)=2 x y$.

(A) The given equation is $x^2-2 m x y-y^2=0$.
Comparing this with the general form $a x^2+2 h x y+b y^2=0$,we get $a=1, h=-m, b=-1$.
The equation of the bisectors of the angles between the pair of lines is given by $\frac{x^2-y^2}{a-b}=\frac{x y}{h}$.
Substituting the values,we get $\frac{x^2-y^2}{1-(-1)}=\frac{x y}{-m}$.
This simplifies to $\frac{x^2-y^2}{2}=\frac{x y}{-m}$,which implies $-m(x^2-y^2)=2 x y$.
Rearranging the terms,we get $m x^2+2 x y-m y^2=0$.
Dividing by $m$ (assuming $m \neq 0$),we get $x^2+\frac{2}{m} x y-y^2=0$.
Comparing this with the given equation of the bisectors $x^2-2 n x y-y^2=0$,we have $-2n = \frac{2}{m}$.
This implies $mn = -1$,or $mn+1=0$.

(B) The equation of the pair of angle bisectors for the pair of straight lines $a x^2+2 h x y+b y^2=0$ is given by $\frac{x^2-y^2}{a-b}=\frac{x y}{h}$.
For the pair $x^2-2 p x y-y^2=0$,the bisectors are given by $\frac{x^2-y^2}{1-(-1)}=\frac{x y}{-p}$.
This simplifies to $\frac{x^2-y^2}{2}=\frac{x y}{-p}$,which implies $-p(x^2-y^2)=2 x y$,or $p x^2+2 x y-p y^2=0$.
Dividing by $p$,we get $x^2+\frac{2}{p} x y-y^2=0$.
Since this is the pair of bisectors,it must be identical to the given pair $x^2-2 q x y-y^2=0$.
Comparing the coefficients of $x y$,we have $-2 q = \frac{2}{p}$.
Thus,$-2 p q = 2$,which gives $p q = -1$.

(C) The given pair of lines is $2y^2 + 5xy - 3x^2 = 0$,which can be rewritten as $3x^2 - 5xy - 2y^2 = 0$.
Factoring the quadratic: $3x^2 - 6xy + xy - 2y^2 = 0$ $\Rightarrow 3x(x - 2y) + y(x - 2y) = 0$ $\Rightarrow (x - 2y)(3x + y) = 0$.
The lines are $L_1: x - 2y = 0$ and $L_2: 3x + y = 0$.
These lines intersect at the origin $O(0, 0)$.
The third line is $L_3: x + y = k$.
Intersection of $L_1$ and $L_3$: $x - 2y = 0$ and $x + y = k$. Solving gives $3y = k \Rightarrow y = \frac{k}{3}$ and $x = \frac{2k}{3}$. So,$A = \left(\frac{2k}{3}, \frac{k}{3}\right)$.
Intersection of $L_2$ and $L_3$: $3x + y = 0$ and $x + y = k$. Subtracting gives $2x = -k \Rightarrow x = -\frac{k}{2}$ and $y = \frac{3k}{2}$. So,$B = \left(-\frac{k}{2}, \frac{3k}{2}\right)$.
The centroid $G$ of $\triangle OAB$ is $\left(\frac{0 + \frac{2k}{3} - \frac{k}{2}}{3}, \frac{0 + \frac{k}{3} + \frac{3k}{2}}{3}\right) = \left(\frac{\frac{k}{6}}{3}, \frac{\frac{11k}{6}}{3}\right) = \left(\frac{k}{18}, \frac{11k}{18}\right)$.
Given $G = \left(\frac{1}{18}, \frac{11}{18}\right)$,we equate the coordinates: $\frac{k}{18} = \frac{1}{18} \Rightarrow k = 1$.

(B) The given equation is $8 x^2-24 x y+18 y^2-6 x+9 y-5=0$.
This can be rewritten as $(2 \sqrt{2} x - 3 \sqrt{2} y)^2 - 3(2 x - 3 y) - 5 = 0$.
Let $2 x - 3 y = t$. Then the equation becomes $2(2 x - 3 y)^2 - 3(2 x - 3 y) - 5 = 0$.
Substituting $t = 2 x - 3 y$,we get $2 t^2 - 3 t - 5 = 0$.
Factoring the quadratic: $2 t^2 - 5 t + 2 t - 5 = 0 \implies t(2 t - 5) + 1(2 t - 5) = 0$.
So,$(t + 1)(2 t - 5) = 0$.
This gives $t = -1$ or $t = 5/2$.
Substituting back $t = 2 x - 3 y$,we get $2 x - 3 y + 1 = 0$ and $2 x - 3 y - 5/2 = 0$,which is $4 x - 6 y - 5 = 0$.
Since the slopes of both lines are $m = 2/3$,the lines are parallel.

(B) The equation $ax^2+2hxy+by^2=0$ represents a pair of lines passing through the origin $(0,0)$.
Since one line passes through $(0,0)$ and $(2,3)$,its equation is $y - 0 = \frac{3-0}{2-0}(x - 0)$,which simplifies to $3x - 2y = 0$.
Since the other line passes through $(0,0)$ and $(4,5)$,its equation is $y - 0 = \frac{5-0}{4-0}(x - 0)$,which simplifies to $5x - 4y = 0$.
The joint equation of these two lines is $(3x - 2y)(5x - 4y) = 0$.
Expanding this,we get $15x^2 - 12xy - 10xy + 8y^2 = 0$,which is $15x^2 - 22xy + 8y^2 = 0$.
Comparing this with $ax^2 + 2hxy + by^2 = 0$,we get $a = 15$,$2h = -22$ (so $h = -11$),and $b = 8$.
Therefore,$a + 2h + b = 15 - 22 + 8 = 1$.

(C) The given equation of the pair of lines is $2x^2+4xy-4y^2-6x-8y+7=0$.
To find the length of the intercept on the $Y$-axis,we set $x=0$.
Substituting $x=0$ into the equation,we get $-4y^2-8y+7=0$,which is equivalent to $4y^2+8y-7=0$.
Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we have $y = \frac{-8 \pm \sqrt{64 - 4(4)(-7)}}{2(4)} = \frac{-8 \pm \sqrt{64+112}}{8} = \frac{-8 \pm \sqrt{176}}{8}$.
This simplifies to $y = \frac{-8 \pm 4\sqrt{11}}{8} = -1 \pm \frac{\sqrt{11}}{2}$.
The two points of intersection on the $Y$-axis are $y_1 = -1 + \frac{\sqrt{11}}{2}$ and $y_2 = -1 - \frac{\sqrt{11}}{2}$.
The length of the intercept is $|y_1 - y_2| = |(-1 + \frac{\sqrt{11}}{2}) - (-1 - \frac{\sqrt{11}}{2})| = |\sqrt{11}| = \sqrt{11}$.

(B) Given equation is $(x-2y)^2 + k(x-2y) = 0$.
Factoring the equation,we get $(x-2y)(x-2y+k) = 0$.
This represents two parallel lines: $L_1: x-2y = 0$ and $L_2: x-2y+k = 0$.
The distance $d$ between two parallel lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$ is given by $d = \frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$.
Here,$A=1, B=-2, C_1=0, C_2=k$.
So,$3 = \frac{|k-0|}{\sqrt{1^2+(-2)^2}} = \frac{|k|}{\sqrt{5}}$.
Therefore,$|k| = 3\sqrt{5}$,which implies $k = \pm 3\sqrt{5}$.

(A) The general equation of a pair of straight lines passing through the origin is $ax^2 + 2hxy + by^2 = 0$.
For the lines to be real,the condition is $h^2 - ab \geq 0$.
Comparing $2x^2 - pxy + 2y^2 = 0$ with the general equation,we have $a = 2$,$2h = -p$ (so $h = -p/2$),and $b = 2$.
Substituting these values into the condition $h^2 - ab \geq 0$:
$(-p/2)^2 - (2)(2) \geq 0$
$p^2/4 - 4 \geq 0$
$p^2 - 16 \geq 0$
$(p - 4)(p + 4) \geq 0$
Solving the inequality,we get $p \leq -4$ or $p \geq 4$.
Thus,$p \in (-\infty, -4] \cup [4, \infty)$.

(D) The equation of the line is $x + 2y = k$,which implies $\frac{x + 2y}{k} = 1$.
To find the joint equation of the lines joining the origin to the points of intersection,we homogenize the curve equation $2x^2 - 2xy + 3y^2 + 2x - y - 1 = 0$ using the line equation:
$2x^2 - 2xy + 3y^2 + (2x - y)(1) - (1)^2 = 0$
Substituting $1 = \frac{x + 2y}{k}$:
$2x^2 - 2xy + 3y^2 + (2x - y)\left(\frac{x + 2y}{k}\right) - \left(\frac{x + 2y}{k}\right)^2 = 0$
Multiplying by $k^2$:
$k^2(2x^2 - 2xy + 3y^2) + k(2x^2 + 4xy - xy - 2y^2) - (x^2 + 4xy + 4y^2) = 0$
$x^2(2k^2 + 2k - 1) + xy(-2k^2 + 3k - 4) + y^2(3k^2 - 2k - 4) = 0$
Since the lines are at right angles,the sum of the coefficients of $x^2$ and $y^2$ must be zero:
$(2k^2 + 2k - 1) + (3k^2 - 2k - 4) = 0$
$5k^2 - 5 = 0$
$5k^2 = 5$
$k^2 = 1$

(C) The given equation is $2x^2 + kxy - 6y^2 + 3x + y + 1 = 0$.
For this to represent a pair of straight lines,the determinant of the matrix of coefficients must be zero:
$\left|\begin{array}{ccc} 2 & k/2 & 3/2 \\ k/2 & -6 & 1/2 \\ 3/2 & 1/2 & 1 \end{array}\right| = 0$
Multiplying rows to simplify: $\frac{1}{4} \left|\begin{array}{ccc} 4 & k & 3 \\ k & -12 & 1 \\ 3 & 1 & 2 \end{array}\right| = 0$
$4(-24 - 1) - k(2k - 3) + 3(k + 36) = 0$
$-100 - 2k^2 + 3k + 3k + 108 = 0$
$-2k^2 + 6k + 8 = 0 \Rightarrow k^2 - 3k - 4 = 0$
$(k - 4)(k + 1) = 0$. Since $k > 0$,we have $k = 4$.
Substituting $k = 4$,the equation becomes $2x^2 + 4xy - 6y^2 + 3x + y + 1 = 0$.
Factoring the expression: $(2x - 2y + 1)(x + 3y + 1) = 0$.
Solving the system $2x - 2y + 1 = 0$ and $x + 3y + 1 = 0$:
From the second equation,$x = -3y - 1$.
Substituting into the first: $2(-3y - 1) - 2y + 1 = 0$ $\Rightarrow -6y - 2 - 2y + 1 = 0$ $\Rightarrow -8y = 1$ $\Rightarrow y = -1/8$.
Then $x = -3(-1/8) - 1 = 3/8 - 8/8 = -5/8$.
The point of intersection is $\left(\frac{-5}{8}, \frac{-1}{8}\right)$.

(C) The given equation of the pair of lines is $x^2+2xy-35y^2-4x+44y-12=0$. Comparing this with the general form $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=1, h=1, b=-35, g=-2, f=22, c=-12$.
The point of intersection $(x, y)$ of these lines is given by the formula:
$x = \frac{hf-bg}{ab-h^2} = \frac{(1)(22)-(-35)(-2)}{(1)(-35)-(1)^2} = \frac{22-70}{-36} = \frac{-48}{-36} = \frac{4}{3}$
$y = \frac{gh-af}{ab-h^2} = \frac{(-2)(1)-(1)(22)}{(1)(-35)-(1)^2} = \frac{-2-22}{-36} = \frac{-24}{-36} = \frac{2}{3}$
Since the lines are concurrent,the point $(\frac{4}{3}, \frac{2}{3})$ must satisfy the equation $5x+ky-8=0$.
Substituting the values: $5(\frac{4}{3}) + k(\frac{2}{3}) - 8 = 0$
$\frac{20}{3} + \frac{2k}{3} = 8$
$20 + 2k = 24$
$2k = 4$
$k = 2$

(D) The given equation of the pair of lines is $x^2+2xy-35y^2-4x+44y-12=0$.
Comparing this with the general form $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=1, h=1, b=-35, g=-2, f=22, c=-12$.
The point of intersection $(x_0, y_0)$ of the pair of lines is given by the formula:
$x_0 = \frac{hf-bg}{ab-h^2} = \frac{(1)(22)-(-35)(-2)}{(1)(-35)-(1)^2} = \frac{22-70}{-35-1} = \frac{-48}{-36} = \frac{4}{3}$
$y_0 = \frac{gh-af}{ab-h^2} = \frac{(-2)(1)-(1)(22)}{(1)(-35)-(1)^2} = \frac{-2-22}{-36} = \frac{-24}{-36} = \frac{2}{3}$
Since the lines are concurrent,the point $(\frac{4}{3}, \frac{2}{3})$ must satisfy the line $5x+\lambda y-8=0$.
$5(\frac{4}{3}) + \lambda(\frac{2}{3}) - 8 = 0$
$\frac{20}{3} + \frac{2\lambda}{3} - 8 = 0$
Multiply by $3$: $20 + 2\lambda - 24 = 0$
$2\lambda - 4 = 0$
$2\lambda = 4$
$\lambda = 2$

(C) We know that the standard limit $\lim _{x \rightarrow \infty} (1 + \frac{a}{x+b})^{x+c} = e^a$.
Given expression is $\lim _{x \rightarrow \infty} (\frac{x+6}{x+1})^{x+4}$.
This can be rewritten as $\lim _{x \rightarrow \infty} (1 + \frac{x+6}{x+1} - 1)^{x+4} = \lim _{x \rightarrow \infty} (1 + \frac{5}{x+1})^{x+4}$.
Using the standard limit formula $\lim _{x \rightarrow \infty} (1 + \frac{k}{f(x)})^{g(x)} = e^{\lim _{x \rightarrow \infty} k \cdot \frac{g(x)}{f(x)}}$,we get:
$e^{\lim _{x \rightarrow \infty} 5 \cdot \frac{x+4}{x+1}} = e^{5 \cdot \lim _{x \rightarrow \infty} \frac{1+4/x}{1+1/x}} = e^{5 \cdot 1} = e^5$.

(C) We have the limit $L = \lim _{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4}$.
Since this is an indeterminate form of $1^{\infty}$,we use the formula $\lim _{x \rightarrow a} f(x)^{g(x)} = e^{\lim _{x \rightarrow a} (f(x)-1)g(x)}$.
$L = e^{\lim _{x \rightarrow \infty} \left(\frac{x+6}{x+1} - 1\right)(x+4)}$
$L = e^{\lim _{x \rightarrow \infty} \left(\frac{x+6-x-1}{x+1}\right)(x+4)}$
$L = e^{\lim _{x \rightarrow \infty} \frac{5(x+4)}{x+1}}$
$L = e^{\lim _{x \rightarrow \infty} \frac{5x+20}{x+1}}$
Dividing numerator and denominator by $x$:
$L = e^{\lim _{x \rightarrow \infty} \frac{5 + 20/x}{1 + 1/x}}$
As $x \rightarrow \infty$,$1/x \rightarrow 0$,so:
$L = e^{\frac{5+0}{1+0}} = e^5$.

(B) Given that,$x=a\left(\cos \theta+\log \tan \left(\frac{\theta}{2}\right)\right)$ and $y=a \sin \theta$.
On differentiating $x$ and $y$ with respect to $\theta$,we get:
$\frac{dx}{d\theta} = a \left( -\sin \theta + \frac{1}{\tan(\theta/2)} \cdot \sec^2(\theta/2) \cdot \frac{1}{2} \right)$
Since $\frac{1}{\tan(\theta/2)} \cdot \sec^2(\theta/2) \cdot \frac{1}{2} = \frac{\cos(\theta/2)}{\sin(\theta/2)} \cdot \frac{1}{\cos^2(\theta/2)} \cdot \frac{1}{2} = \frac{1}{2 \sin(\theta/2) \cos(\theta/2)} = \frac{1}{\sin \theta}$.
So,$\frac{dx}{d\theta} = a \left( -\sin \theta + \frac{1}{\sin \theta} \right) = a \left( \frac{1 - \sin^2 \theta}{\sin \theta} \right) = \frac{a \cos^2 \theta}{\sin \theta}$.
Also,$\frac{dy}{d\theta} = a \cos \theta$.
Therefore,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \cos \theta}{a \cos^2 \theta / \sin \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta$.

(B) Given $x=a[\cos \theta+\log (\tan (\theta/2))]$ and $y=a \sin \theta$.
First,differentiate $x$ with respect to $\theta$:
$\frac{dx}{d\theta} = a \left[ -\sin \theta + \frac{1}{\tan(\theta/2)} \cdot \sec^2(\theta/2) \cdot \frac{1}{2} \right]$
Using the identity $\tan(\theta/2) = \frac{\sin(\theta/2)}{\cos(\theta/2)}$ and $\sec^2(\theta/2) = \frac{1}{\cos^2(\theta/2)}$,we get:
$\frac{dx}{d\theta} = a \left[ -\sin \theta + \frac{\cos(\theta/2)}{\sin(\theta/2)} \cdot \frac{1}{\cos^2(\theta/2)} \cdot \frac{1}{2} \right]$
$\frac{dx}{d\theta} = a \left[ -\sin \theta + \frac{1}{2\sin(\theta/2)\cos(\theta/2)} \right]$
Since $2\sin(\theta/2)\cos(\theta/2) = \sin \theta$,we have:
$\frac{dx}{d\theta} = a \left[ -\sin \theta + \frac{1}{\sin \theta} \right] = a \left[ \frac{1-\sin^2 \theta}{\sin \theta} \right] = \frac{a \cos^2 \theta}{\sin \theta} \quad \dots(i)$
Now,differentiate $y$ with respect to $\theta$:
$\frac{dy}{d\theta} = a \cos \theta \quad \dots(ii)$
To find $\frac{dy}{dx}$,divide equation $(ii)$ by equation $(i)$:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \cos \theta}{a \cos^2 \theta / \sin \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta$.

(D) Let $I = \int_0^\pi \frac{x \tan x}{\sec x+\tan x} d x \quad \dots(i)$
Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$,we get:
$I = \int_0^\pi \frac{(\pi-x) \tan(\pi-x)}{\sec(\pi-x) + \tan(\pi-x)} d x$
Since $\tan(\pi-x) = -\tan x$ and $\sec(\pi-x) = -\sec x$,we have:
$I = \int_0^\pi \frac{-(\pi-x) \tan x}{-\sec x - \tan x} d x = \int_0^\pi \frac{(\pi-x) \tan x}{\sec x + \tan x} d x \quad \dots(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^\pi \frac{\pi \tan x}{\sec x + \tan x} d x = \pi \int_0^\pi \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}} d x$
$2I = \pi \int_0^\pi \frac{\sin x}{1 + \sin x} d x = \pi \int_0^\pi \frac{1 + \sin x - 1}{1 + \sin x} d x$
$2I = \pi \left[ \int_0^\pi 1 d x - \int_0^\pi \frac{1}{1 + \sin x} d x \right]$
$2I = \pi \left[ [x]_0^\pi - \int_0^\pi \frac{1 - \sin x}{\cos^2 x} d x \right]$
$2I = \pi \left[ \pi - \int_0^\pi (\sec^2 x - \sec x \tan x) d x \right]$
$2I = \pi \left[ \pi - [\tan x - \sec x]_0^\pi \right]$
$2I = \pi \left[ \pi - ((\tan \pi - \sec \pi) - (\tan 0 - \sec 0)) \right]$
$2I = \pi \left[ \pi - ((0 - (-1)) - (0 - 1)) \right] = \pi [\pi - (1 + 1)] = \pi(\pi - 2)$
$I = \frac{\pi(\pi - 2)}{2}$

(D) disproportionation reaction is a chemical reaction in which the same species undergoes both oxidation and reduction simultaneously.
In $ClO_4^{-}$,the oxidation state of the $Cl$ atom is $+7$. Since the valence shell configuration of chlorine is $3s^2 3p^5$,the maximum oxidation state it can exhibit is $+7$.
Because $Cl$ is already in its maximum oxidation state,it cannot be further oxidized.
Therefore,$ClO_4^{-}$ can only undergo reduction and cannot participate in a disproportionation reaction.

(A) To balance the redox reaction,we use the half-reaction method:
$1$. Oxidation half-reaction: $C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$
$2$. Reduction half-reaction: $MnO_4^{-} + 8H^{+} + 5e^- \rightarrow Mn^{2+} + 4H_2O$
$3$. To balance the electrons,multiply the oxidation half-reaction by $5$ and the reduction half-reaction by $2$:
$5C_2O_4^{2-} \rightarrow 10CO_2 + 10e^-$
$2MnO_4^{-} + 16H^{+} + 10e^- \rightarrow 2Mn^{2+} + 8H_2O$
$4$. Adding both half-reactions gives the balanced equation:
$2MnO_4^{-} + 5C_2O_4^{2-} + 16H^{+} \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$
Thus,the coefficients for $MnO_4^{-}$,$C_2O_4^{2-}$,and $H^{+}$ are $2$,$5$,and $16$ respectively.

(B) $(i)$ $MnO_2$ reacts with $4 \ mol$ of $HCl$ to produce $1 \ mol$ of $Cl_2$ gas.
Reaction: $MnO_2 + 4 HCl \rightarrow MnCl_2 + Cl_2 + 2 H_2O$
$(ii)$ $2 \ mol$ of $KMnO_4$ reacts with $16 \ mol$ of $HCl$ to produce $5 \ mol$ of $Cl_2$ gas.
Reaction: $2 KMnO_4 + 16 HCl \rightarrow 2 MnCl_2 + 5 Cl_2 + 8 H_2O + 2 KCl$
Therefore,the number of moles of chlorine gas released are $1$ and $5$ respectively.

(B) The equivalent weight of an element is given by the formula: $\text{Equivalent weight} = \frac{\text{Atomic mass}}{\text{Valency factor}}$.
In $Fe_2O_3$,the oxidation state of $Fe$ is $+3$,so the valency factor is $3$.
Given the atomic mass of $Fe = 56 \ g \ mol^{-1}$.
Therefore,$\text{Equivalent weight} = \frac{56}{3} = 18.66 \ g \ \approx 18.6 \ g$.

(D) In the presence of a reducing agent,$MnO_4^{-}$ acts as a self-indicator and changes colour from pink to colourless.
Because in $MnO_4^{-}$,the oxidation state of $Mn$ is $+7$,which is its highest oxidation state.
Thus,it tends to get reduced and easily accepts electrons.
Being a charge transfer complex,it shows intense colour,which is why it acts as a self-indicator.

(C) $K, Rb,$ and $Cs$ form superoxides when they are burnt in the air. \\ As we move down the group,the size of the alkali metal cation increases from $K^+$ to $Cs^+$. \\ The stability of superoxides is determined by the lattice energy; as the size of the cation increases,the lattice energy decreases,which leads to a decrease in the stability of the superoxide. \\ Therefore,the stability of superoxides decreases from $K$ to $Cs$. \\ Thus,Assertion $(A)$ is true,but Reason $(R)$ is false.

(D) Alkali metals have only one valence electron in their outermost shell.
They lose one valence electron easily due to low ionisation enthalpy and get oxidized to reduce other compounds.
Therefore,they act as strong reducing agents.

(A) Lattice energy is inversely proportional to the inter-ionic distance between the cation and the anion.
Since the fluoride ion $(F^{-})$ is common in all the given compounds,the lattice energy depends on the size of the alkali metal cation.
As the size of the cation increases from $Li^{+}$ to $Cs^{+}$,the inter-ionic distance increases,leading to a decrease in lattice energy.
Therefore,$LiF$ has the smallest inter-ionic distance and consequently the highest lattice energy.

(B) In muscle contraction,calcium ions $Ca^{2+}$ play an important role by facilitating interactions between the proteins myosin and actin.
$Ca^{2+}$ ions bind to the troponin complex on the actin filament,which causes a conformational change that exposes the binding sites for the myosin heads,thereby stimulating muscle contraction.

(D) The tendency to form hydrates decreases as the size of the alkaline earth metal cation increases down the group.
Smaller cations have higher charge density,which allows them to polarize water molecules more effectively,leading to stronger hydration.
The ionic radii follow the order: $Mg^{2+} < Ca^{2+} < Sr^{2+} < Ba^{2+}$.
Therefore,the tendency to form hydrates follows the order: $Mg^{2+} > Ca^{2+} > Sr^{2+} > Ba^{2+}$.
Thus,the correct order is $Mg > Ca > Sr > Ba$.

(A) Double salts are addition compounds that exist only in the solid state and dissociate into their constituent ions when dissolved in water. $Li_2SO_4$ does not form double salts (like alums) because of the exceptionally small size and high polarizing power of the $Li^+$ ion. Due to its small size,$Li^+$ cannot accommodate the coordination requirements necessary to form the stable crystal lattice structure characteristic of double salts,unlike the larger alkali metal ions such as $Na^+$,$K^+$,and $Rb^+$.

(C) Given,$l^2+m^2-n^2=0$ $(i)$ and $l+m+n=0$ $(ii)$.
From $(ii)$,$n=-(l+m)$. Substituting this into $(i)$:
$l^2+m^2=(-(l+m))^2 = l^2+m^2+2lm$.
This implies $2lm=0$,so $l=0$ or $m=0$.
Case $1$: If $l=0$,then $n=-m$. Since $l^2+m^2+n^2=1$,we have $0^2+m^2+(-m)^2=1 \Rightarrow 2m^2=1 \Rightarrow m=\pm\frac{1}{\sqrt{2}}$. Thus,the direction cosines are $(0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$ and $(0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$.
Case $2$: If $m=0$,then $n=-l$. Since $l^2+m^2+n^2=1$,we have $l^2+0^2+(-l)^2=1 \Rightarrow 2l^2=1 \Rightarrow l=\pm\frac{1}{\sqrt{2}}$. Thus,the direction cosines are $(\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}})$ and $(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})$.
Let the direction vectors of the two lines be $\vec{a} = (0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$ and $\vec{b} = (\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}})$.
The angle $\theta$ between the lines is given by $\cos \theta = |l_1l_2 + m_1m_2 + n_1n_2|$.
$\cos \theta = |(0)(\frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{2}})(0) + (-\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}})| = |0 + 0 + \frac{1}{2}| = \frac{1}{2}$.
Therefore,$\theta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$.

(D) Given the equations for the direction cosines $(l, m, n)$ of two lines:
$l+m+n=0 \implies n = -(l+m)$
Substitute $n$ into the second equation $l^2-5m^2+n^2=0$:
$l^2-5m^2+(-l-m)^2 = 0$
$l^2-5m^2+l^2+2lm+m^2 = 0$
$2l^2+2lm-4m^2 = 0$
$l^2+lm-2m^2 = 0$
$(l+2m)(l-m) = 0$
This gives two cases:
Case $1$: $l=m$. Then $n = -(l+m) = -2l$. The direction ratios are $(l, l, -2l)$,which simplifies to $(1, 1, -2)$. The direction cosines are $(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}})$.
Case $2$: $l=-2m$. Then $n = -(-2m+m) = m$. The direction ratios are $(-2m, m, m)$,which simplifies to $(-2, 1, 1)$. The direction cosines are $(-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}})$.
Let the two lines have direction cosines $(l_1, m_1, n_1) = (\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}})$ and $(l_2, m_2, n_2) = (-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}})$.
The cosine of the angle $\theta$ between them is given by:
$\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$
$\cos \theta = |(\frac{1}{\sqrt{6}})(-\frac{2}{\sqrt{6}}) + (\frac{1}{\sqrt{6}})(\frac{1}{\sqrt{6}}) + (-\frac{2}{\sqrt{6}})(\frac{1}{\sqrt{6}})|$
$\cos \theta = |-\frac{2}{6} + \frac{1}{6} - \frac{2}{6}| = |-\frac{3}{6}| = \frac{1}{2}$
Since $\cos \theta = \frac{1}{2}$,we have $\theta = 60^{\circ} = \frac{\pi}{3}$.

(C) Given,$l+m+n=0 \implies l = -m-n$ and $l^2+m^2-n^2=0$.
Substituting $l = -m-n$ into the second equation:
$(-m-n)^2 + m^2 - n^2 = 0$
$m^2 + 2mn + n^2 + m^2 - n^2 = 0$
$2m^2 + 2mn = 0$
$2m(m+n) = 0$.
This gives two cases:
Case $1$: If $m=0$,then $l = -n$. The direction ratios are $(-n, 0, n)$,which simplifies to $(-1, 0, 1)$. Let $\vec{v_1} = (-1, 0, 1)$.
Case $2$: If $m+n=0$,then $m = -n$. Substituting into $l = -m-n$,we get $l = -(-n)-n = 0$. The direction ratios are $(0, -n, n)$,which simplifies to $(0, -1, 1)$. Let $\vec{v_2} = (0, -1, 1)$.
The angle $\theta$ between the lines is given by $\cos \theta = \frac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|}$.
$\vec{v_1} \cdot \vec{v_2} = (-1)(0) + (0)(-1) + (1)(1) = 1$.
$|\vec{v_1}| = \sqrt{(-1)^2 + 0^2 + 1^2} = \sqrt{2}$.
$|\vec{v_2}| = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{2}$.
$\cos \theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.

(D) Let the number of $Fe^{2+}$ ions be $x$ and the number of $Fe^{3+}$ ions be $(0.96 - x)$.
Since the compound is electrically neutral,the total positive charge must equal the total negative charge.
The charge of $O^{2-}$ is $-2$.
Therefore,$2x + 3(0.96 - x) - 2 = 0$.
$2x + 2.88 - 3x - 2 = 0$.
$-x + 0.88 = 0$.
$x = 0.88$.
So,the number of $Fe^{2+}$ ions is $0.88$ and the number of $Fe^{3+}$ ions is $0.96 - 0.88 = 0.08$.
The mole fraction of $Fe^{2+}$ is the ratio of the number of $Fe^{2+}$ ions to the total number of $Fe$ ions.
Mole fraction of $Fe^{2+} = \frac{0.88}{0.96} = \frac{88}{96} = \frac{11}{12}$.