WBJEE 2017 Mathematics Question Paper with Answer and Solution

(A) Given equation: $(\log _{5} x)(\log _{x} 3x)(\log _{3x} y) = \log _{x} x^{3}$
Using the change of base formula $\log _{b} a = \frac{\log a}{\log b}$,we get:
$\frac{\log x}{\log 5} \times \frac{\log 3x}{\log x} \times \frac{\log y}{\log 3x} = 3 \log _{x} x$
Simplifying the left side by cancelling common terms:
$\frac{\log y}{\log 5} = 3(1)$
$\log _{5} y = 3$
Converting to exponential form:
$y = 5^{3} = 125$

(A) The given quadratic equation is $2px^{2} + (2p + q)x + q = 0$.
The discriminant $D$ is given by $D = b^{2} - 4ac$.
Substituting the coefficients $a = 2p$,$b = (2p + q)$,and $c = q$:
$D = (2p + q)^{2} - 4(2p)(q)$
$D = 4p^{2} + q^{2} + 4pq - 8pq$
$D = 4p^{2} + q^{2} - 4pq$
$D = (2p - q)^{2}$
Since $p$ and $q$ are integers,$D$ is a perfect square of an integer.
For a quadratic equation with rational coefficients,if the discriminant is a perfect square,the roots are rational.

(C) The given quadratic equation is $ax^{2} + bx + 1 = 0$.
For the equation to have real roots,the discriminant $D$ must be greater than or equal to $0$.
$D = b^{2} - 4ac \geq 0$.
Substituting $c = 1$,we get $b^{2} - 4a \geq 0$,which implies $b^{2} \geq 4a$.
Given $a, b \in \{1, 2, 3\}$,we test the possible values:
If $a = 1$,$b^{2} \geq 4 \implies b \in \{2, 3\}$. Pairs: $(1, 2), (1, 3)$.
If $a = 2$,$b^{2} \geq 8 \implies b = 3$. Pair: $(2, 3)$.
If $a = 3$,$b^{2} \geq 12 \implies$ no value of $b$ satisfies this.
The possible ordered pairs $(a, b)$ are $(1, 2), (1, 3), (2, 3)$.
Thus,the number of possible ordered pairs is $3$.

(C) Let $y = \frac{x^{2}+2 x+4}{2 x^{2}+4 x+9}$.
$y(2 x^{2}+4 x+9) = x^{2}+2 x+4$.
$2 y x^{2} + 4 y x + 9 y = x^{2} + 2 x + 4$.
$(2 y-1) x^{2} + (4 y-2) x + (9 y-4) = 0$.
Since $x$ is a real number,the discriminant $D \geq 0$.
$D = (4 y-2)^{2} - 4(2 y-1)(9 y-4) \geq 0$.
$4(2 y-1)^{2} - 4(2 y-1)(9 y-4) \geq 0$.
$4(2 y-1) [ (2 y-1) - (9 y-4) ] \geq 0$.
$4(2 y-1) (-7 y + 3) \geq 0$.
$(2 y-1) (7 y - 3) \leq 0$.
This inequality holds for $\frac{3}{7} \leq y \leq \frac{1}{2}$.
Therefore,the maximum value of $y$ is $\frac{1}{2}$.

(C) Given expression: $\frac{(1+i)^{n}}{(1-i)^{n-2}}$
$= \frac{(1+i)^{n}}{(1-i)^{n} \cdot (1-i)^{-2}}$
$= \left(\frac{1+i}{1-i}\right)^{n} \cdot (1-i)^{2}$
$= \left(\frac{(1+i)(1+i)}{(1-i)(1+i)}\right)^{n} \cdot (1 + i^{2} - 2i)$
$= \left(\frac{1 + 2i + i^{2}}{1 - i^{2}}\right)^{n} \cdot (1 - 1 - 2i)$
$= \left(\frac{1 + 2i - 1}{1 + 1}\right)^{n} \cdot (-2i)$
$= \left(\frac{2i}{2}\right)^{n} \cdot (-2i)$
$= i^{n} \cdot (-2i)$
$= -2i^{n+1}$

(D) Given $z = x + iy$. Then $\frac{z+i}{z-i} = \frac{x + iy + i}{x + iy - i} = \frac{x + i(y+1)}{x + i(y-1)}$.
Multiplying the numerator and denominator by the conjugate of the denominator,$x - i(y-1)$:
$\frac{z+i}{z-i} = \frac{[x + i(y+1)][x - i(y-1)]}{x^2 + (y-1)^2} = \frac{x^2 - ix(y-1) + ix(y+1) - i^2(y+1)(y-1)}{x^2 + (y-1)^2}$.
Since $i^2 = -1$,this simplifies to:
$\frac{x^2 + y^2 - 1 + i[x(y+1) - x(y-1)]}{x^2 + (y-1)^2} = \frac{x^2 + y^2 - 1 + 2xi}{x^2 + (y-1)^2}$.
For the expression to be purely imaginary,the real part must be zero:
$\text{Re}\left(\frac{z+i}{z-i}\right) = \frac{x^2 + y^2 - 1}{x^2 + (y-1)^2} = 0$.
This implies $x^2 + y^2 - 1 = 0$,or $x^2 + y^2 = 1$.
This is the equation of a circle with center $(0, 0)$ and radius $1$ (excluding the point $(0, 1)$ where the denominator is zero).

(A) Given the quadratic equation $x^{2}+x+1=0$.
Using the quadratic formula,the roots are $x = \frac{-1 \pm \sqrt{1^{2}-4(1)(1)}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}$.
Let $\alpha = \frac{-1+i\sqrt{3}}{2} = e^{i(2\pi/3)}$ and $\beta = \frac{-1-i\sqrt{3}}{2} = e^{-i(2\pi/3)}$.
Then,$\alpha^{n}+\beta^{n} = (e^{i(2\pi/3)})^{n} + (e^{-i(2\pi/3)})^{n} = e^{i(2n\pi/3)} + e^{-i(2n\pi/3)}$.
Using Euler's formula,$e^{i\theta} + e^{-i\theta} = 2\cos(\theta)$.
Therefore,$\alpha^{n}+\beta^{n} = 2\cos\left(\frac{2n\pi}{3}\right)$.

(C) We have,$|z-i|=|z+1|=1$.
Let $z=x+iy$.
Then $|z-i|=1$ $\Rightarrow |x+i(y-1)|=1$ $\Rightarrow x^2+(y-1)^2=1$ $(i)$.
Also,$|z+1|=1$ $\Rightarrow |(x+1)+iy|=1$ $\Rightarrow (x+1)^2+y^2=1$ (ii).
Expanding $(i)$: $x^2+y^2-2y+1=1 \Rightarrow x^2+y^2=2y$.
Expanding (ii): $x^2+2x+1+y^2=1 \Rightarrow x^2+y^2=-2x$.
Equating the two: $2y=-2x \Rightarrow y=-x$.
Substituting $y=-x$ into $(i)$: $x^2+(-x-1)^2=1$ $\Rightarrow x^2+x^2+2x+1=1$ $\Rightarrow 2x^2+2x=0$ $\Rightarrow 2x(x+1)=0$.
Thus,$x=0$ or $x=-1$.
If $x=0$,then $y=0$,so $z=0$.
If $x=-1$,then $y=1$,so $z=-1+i$.
Therefore,the complex numbers are $0$ and $-1+i$.

(B) The expression $(p+1)(p+2)(p+3) \ldots (p+q)$ represents the product of $q$ consecutive natural numbers starting from $(p+1)$.
We know that the product of $q$ consecutive natural numbers is always divisible by $q!$.
This is because the expression can be written as $\frac{(p+q)!}{p!} = q! \times \binom{p+q}{q}$.
Since $\binom{p+q}{q}$ is an integer for all $p, q \in N$,the expression is always divisible by $q!$.
Thus,the greatest integer that divides this product for all $p \in N$ is $q!$.

(B) $5$-digit number cannot have $0$ at the first position (ten-thousands place).
For the first position,we have $9$ choices (digits $1$ to $9$).
For the remaining $4$ positions,we need to choose $4$ digits from the remaining $9$ available digits (including $0$ and excluding the digit used in the first position).
The number of ways to arrange these $4$ digits is ${}^{9}P_{4}$.
Therefore,the total number of $5$-digit numbers with distinct digits is $9 \times {}^{9}P_{4}$.

(B) The number of ways to select $3$ consonants out of $7$ is ${}^{7}C_{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.
The number of ways to select $2$ vowels out of $4$ is ${}^{4}C_{2} = \frac{4 \times 3}{2 \times 1} = 6$.
The total number of letters selected is $3 + 2 = 5$.
These $5$ letters can be arranged among themselves in $5!$ ways,where $5! = 120$.
Therefore,the total number of words formed is ${}^{7}C_{3} \times {}^{4}C_{2} \times 5! = 35 \times 6 \times 120 = 25200$.

(B) Let $a_{n}$ be the general term of a $GP$ whose first term is $a$ and common ratio is $r$.
According to the problem,each term is equal to the sum of the next two terms:
$a_{n} = a_{n+1} + a_{n+2}$
$a r^{n-1} = a r^{n} + a r^{n+1}$
Since the terms are positive,$a \neq 0$ and $r > 0$. Dividing by $a r^{n-1}$:
$1 = r + r^{2}$
$r^{2} + r - 1 = 0$
Using the quadratic formula $r = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$:
$r = \frac{-1 \pm \sqrt{1^{2} - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$
Since the $GP$ consists of positive terms,the common ratio $r$ must be positive.
Therefore,$r = \frac{\sqrt{5}-1}{2}$.

(B) Given $(1+x+x^2)^9 = a_0 + a_1 x + a_2 x^2 + \ldots + a_{18} x^{18}$.
Put $x = 1$,we get $(1+1+1)^9 = a_0 + a_1 + a_2 + \ldots + a_{18} \Rightarrow 3^9 = a_0 + a_1 + a_2 + \ldots + a_{18} \quad (i)$.
Put $x = -1$,we get $(1-1+1)^9 = a_0 - a_1 + a_2 - a_3 + \ldots + a_{18} \Rightarrow 1 = a_0 - a_1 + a_2 - a_3 + \ldots + a_{18} \quad (ii)$.
Adding $(i)$ and $(ii)$,we get $3^9 + 1 = 2(a_0 + a_2 + a_4 + \ldots + a_{18})$.
Therefore,$a_0 + a_2 + a_4 + \ldots + a_{18} = \frac{3^9 + 1}{2} = \frac{19683 + 1}{2} = \frac{19684}{2} = 9842$.
Since $9842$ is an even number,option $B$ is correct.

(D) Let $f(x) = \sin x(\sin x + \cos x) = \sin^2 x + \sin x \cos x$.
Using the identities $\sin^2 x = \frac{1-\cos 2x}{2}$ and $\sin x \cos x = \frac{\sin 2x}{2}$,we get:
$f(x) = \frac{1-\cos 2x}{2} + \frac{\sin 2x}{2} = \frac{1}{2} + \frac{1}{2}(\sin 2x - \cos 2x)$.
We know that for any real $\theta$,$-\sqrt{a^2+b^2} \leq a \sin \theta + b \cos \theta \leq \sqrt{a^2+b^2}$.
For $\sin 2x - \cos 2x$,we have $a=1, b=-1$,so $-\sqrt{1^2+(-1)^2} \leq \sin 2x - \cos 2x \leq \sqrt{1^2+(-1)^2}$,which simplifies to $-\sqrt{2} \leq \sin 2x - \cos 2x \leq \sqrt{2}$.
Dividing by $2$,we get $-\frac{\sqrt{2}}{2} \leq \frac{\sin 2x - \cos 2x}{2} \leq \frac{\sqrt{2}}{2}$.
Adding $\frac{1}{2}$ to all parts:
$\frac{1}{2} - \frac{\sqrt{2}}{2} \leq \frac{1}{2} + \frac{\sin 2x - \cos 2x}{2} \leq \frac{1}{2} + \frac{\sqrt{2}}{2}$.
Thus,$\frac{1-\sqrt{2}}{2} \leq f(x) \leq \frac{1+\sqrt{2}}{2}$.
Since $f(x) = k$,the range of $k$ is $\frac{1-\sqrt{2}}{2} \leq k \leq \frac{1+\sqrt{2}}{2}$.

(B) Let the origin be shifted to $(p, q)$ such that the new axes are parallel to the original axes. The transformation equations are $x = x' + p$ and $y = y' + q$.
Substituting these into the given equation $2x^2 + 3xy + 4y^2 + x + 18y + 25 = 0$:
$2(x' + p)^2 + 3(x' + p)(y' + q) + 4(y' + q)^2 + (x' + p) + 18(y' + q) + 25 = 0$
Expanding the terms:
$2(x'^2 + 2px' + p^2) + 3(x'y' + qx' + py' + pq) + 4(y'^2 + 2qy' + q^2) + x' + p + 18y' + 18q + 25 = 0$
Grouping the terms by $x'$,$y'$,and constants:
$2x'^2 + 3x'y' + 4y'^2 + (4p + 3q + 1)x' + (3p + 8q + 18)y' + (2p^2 + 3pq + 4q^2 + p + 18q + 25) = 0$
Comparing this with the given transformed equation $2x^2 + 3xy + 4y^2 = 1$ (or $2x'^2 + 3x'y' + 4y'^2 - 1 = 0$),the coefficients of $x'$ and $y'$ must be zero:
$4p + 3q + 1 = 0$ ... $(i)$
$3p + 8q + 18 = 0$ ... (ii)
Solving these linear equations:
From $(i)$,$4p = -3q - 1 \implies p = \frac{-3q - 1}{4}$.
Substituting into (ii): $3(\frac{-3q - 1}{4}) + 8q + 18 = 0$
$-9q - 3 + 32q + 72 = 0 \implies 23q + 69 = 0 \implies q = -3$.
Substituting $q = -3$ into $(i)$: $4p + 3(-3) + 1 = 0 \implies 4p - 8 = 0 \implies p = 2$.
Thus,$p = 2, q = -3$.

(A) Let the coordinates of vertex $C$ be $(x, y)$.
Since $A = (2, -3)$ and $B = (-2, 1)$,the centroid $G$ of $\Delta ABC$ is given by:
$G = \left( \frac{2 - 2 + x}{3}, \frac{-3 + 1 + y}{3} \right) = \left( \frac{x}{3}, \frac{y - 2}{3} \right)$.
Given that the centroid lies on the line $2x + 3y = 1$,we substitute the coordinates of $G$ into the equation:
$2\left( \frac{x}{3} \right) + 3\left( \frac{y - 2}{3} \right) = 1$.
Multiplying by $3$ to clear the denominators:
$2x + 3(y - 2) = 3$.
$2x + 3y - 6 = 3$.
$2x + 3y = 9$.
Thus,the locus of point $C$ is $2x + 3y = 9$.

(D) The reflection of point $P(3,6)$ on the line $y=x$ gives the point $Q(6,3)$.
The reflection of point $Q(6,3)$ on the line $y=-x$ gives the point $Q^{\prime}(-3,-6)$.
Now,the slope of $PQ = \frac{3-6}{6-3} = \frac{-3}{3} = -1$.
The slope of $QQ^{\prime} = \frac{-6-3}{-3-6} = \frac{-9}{-9} = 1$.
Since the product of the slopes is $(-1) \times (1) = -1$,the lines $PQ$ and $QQ^{\prime}$ are perpendicular.
Therefore,$\Delta PQQ^{\prime}$ is a right-angled triangle with the right angle at $Q$.
The circumcentre of a right-angled triangle is the midpoint of its hypotenuse.
The hypotenuse is $PQ^{\prime}$,with endpoints $P(3,6)$ and $Q^{\prime}(-3,-6)$.
Midpoint of $PQ^{\prime} = \left(\frac{3+(-3)}{2}, \frac{6+(-6)}{2}\right) = (0,0)$.

(B) Let $(h, k)$ be any point on the line $7x - 9y + 10 = 0$. Then,$7h - 9k + 10 = 0$,which implies $h = \frac{9k - 10}{7}$.
Perpendicular distance $d_{1}$ from $(h, k)$ to $3x + 4y - 5 = 0$ is given by $d_{1} = \frac{|3h + 4k - 5|}{\sqrt{3^{2} + 4^{2}}} = \frac{|3h + 4k - 5|}{5}$.
Perpendicular distance $d_{2}$ from $(h, k)$ to $12x + 5y - 7 = 0$ is given by $d_{2} = \frac{|12h + 5k - 7|}{\sqrt{12^{2} + 5^{2}}} = \frac{|12h + 5k - 7|}{13}$.
Substituting $h = \frac{9k - 10}{7}$ into the expressions:
$3h + 4k - 5 = 3(\frac{9k - 10}{7}) + 4k - 5 = \frac{27k - 30 + 28k - 35}{7} = \frac{55k - 65}{7} = \frac{5(11k - 13)}{7}$.
So,$d_{1} = \frac{|5(11k - 13)|}{5 \times 7} = \frac{|11k - 13|}{7}$.
$12h + 5k - 7 = 12(\frac{9k - 10}{7}) + 5k - 7 = \frac{108k - 120 + 35k - 49}{7} = \frac{143k - 169}{7} = \frac{13(11k - 13)}{7}$.
So,$d_{2} = \frac{|13(11k - 13)|}{13 \times 7} = \frac{|11k - 13|}{7}$.
Since $d_{1} = \frac{|11k - 13|}{7}$ and $d_{2} = \frac{|11k - 13|}{7}$,we have $d_{1} = d_{2}$.

(A) Given circle equation: $x^{2}+y^{2}-4x-6y+9=0$.
Comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=-2, f=-3, c=9$.
Centre $B = (-g, -f) = (2, 3)$.
Radius $r = \sqrt{g^{2}+f^{2}-c} = \sqrt{(-2)^{2}+(-3)^{2}-9} = \sqrt{4+9-9} = 2$.
Let the centre of the required circle be $A(1, 1)$.
The distance $AB$ between the centres is $AB = \sqrt{(2-1)^{2}+(3-1)^{2}} = \sqrt{1^{2}+2^{2}} = \sqrt{5}$.
Since the diameter of the first circle is a chord of the second circle,the radius $R$ of the second circle is the hypotenuse of the right-angled triangle formed by the distance between centres and the radius of the first circle.
$R = \sqrt{AB^{2}+r^{2}} = \sqrt{(\sqrt{5})^{2}+2^{2}} = \sqrt{5+4} = \sqrt{9} = 3$.

(D) The given equations of the circles are $x^{2}+y^{2}-4x-4y=0$ and $2x^{2}+2y^{2}=32$.
Simplifying the second equation,we get $x^{2}+y^{2}=16$.
The equation of the common chord is obtained by subtracting the two equations: $(x^{2}+y^{2}-4x-4y) - (x^{2}+y^{2}-16) = 0$.
This simplifies to $-4x-4y+16=0$,or $x+y=4$.
Let the origin be $O(0,0)$. The circle $x^{2}+y^{2}-4x-4y=0$ passes through the origin. The common chord $x+y=4$ intersects the circle at points $A$ and $B$.
Since the circle $x^{2}+y^{2}-4x-4y=0$ can be written as $(x-2)^{2}+(y-2)^{2}=8$,its center is $C(2,2)$ and radius is $r = \sqrt{8} = 2\sqrt{2}$.
The distance from the center $C(2,2)$ to the line $x+y-4=0$ is $d = \frac{|2+2-4|}{\sqrt{1^{2}+1^{2}}} = 0$.
Since the distance is $0$,the common chord is a diameter of the first circle.
Any diameter subtends a right angle at any point on the circumference. Since the origin $(0,0)$ lies on the circle,the angle subtended by the diameter at the origin is $\frac{\pi}{2}$.

(C) Given,the equation of the circle is $x^{2}+y^{2}+2x-2y-2=0$.
This can be rewritten as $(x+1)^{2}+(y-1)^{2}=4$.
Thus,the centre is $(-1, 1)$ and the radius $r = 2$.
Let $P(h, k)$ be the mid-point of a chord $AB$.
The distance $OP$ from the centre $O(-1, 1)$ to the mid-point $P(h, k)$ is given by $OP = \sqrt{(h+1)^{2}+(k-1)^{2}}$.
Since the chord $AB$ subtends an angle of $90^{\circ}$ at the centre,$\triangle OAP$ is a right-angled triangle with $\angle OAP = 45^{\circ}$.
In $\triangle OAP$,$\sin 45^{\circ} = \frac{OP}{OA}$.
$\frac{1}{\sqrt{2}} = \frac{\sqrt{(h+1)^{2}+(k-1)^{2}}}{2}$.
Squaring both sides,we get $\frac{1}{2} = \frac{(h+1)^{2}+(k-1)^{2}}{4}$.
$(h+1)^{2}+(k-1)^{2} = 2$.
Expanding this,$h^{2}+2h+1+k^{2}-2k+1 = 2$.
$h^{2}+k^{2}+2h-2k = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^{2}+y^{2}+2x-2y=0$.

(A) The given equation is $x^{2}+2 x y+y^{2}-5 x+5 y-5=0$.
This can be rewritten as $(x+y)^{2} = 5x - 5y + 5$.
Let $X = x+y$ and $Y = x-y+1$.
The equation is of the form $X^{2} = 5Y$,which represents a parabola.
The axis of a parabola of the form $(ax+by+c)^2 = k(bx-ay+d)$ is given by $ax+by+c=0$.
Here,$x+y=0$ is the axis of the parabola.

(C) Given equation of the conic is $x^{2}-6x+4y+1=0$.
Completing the square for the $x$ terms:
$(x^{2}-6x+9)-9+4y+1=0$
$(x-3)^{2}+4y-8=0$
$(x-3)^{2}=-4(y-2)$.
This is a parabola of the form $(x-h)^{2}=-4a(y-k)$,where $(h,k)=(3,2)$ and $4a=4$,so $a=1$.
The focus of this parabola is given by $(h, k-a)$.
Substituting the values,we get $(3, 2-1) = (3,1)$.

(A) The equation of the tangent to the parabola $y^{2}=4ax$ at the point $(at^{2}, 2at)$ is given by $ty = x + at^{2}$,which can be rewritten as $x - ty + at^{2} = 0$.
The equation of the normal to the hyperbola $x^{2}-y^{2}=a^{2}$ at the point $(a \sec \theta, a \tan \theta)$ is given by $\frac{ax}{\sec \theta} + \frac{ay}{\tan \theta} = a^{2} + a^{2}$,which simplifies to $x \cos \theta + y \cot \theta = 2a$.
Comparing the two equations $x - ty + at^{2} = 0$ and $x \cos \theta + y \cot \theta - 2a = 0$,we have the ratios of coefficients:
$\frac{1}{\cos \theta} = \frac{-t}{\cot \theta} = \frac{at^{2}}{-2a}$.
From $\frac{1}{\cos \theta} = \frac{-t}{\cot \theta}$,we get $t = -\frac{\cot \theta}{\cos \theta} = -\operatorname{cosec} \theta$.
From $\frac{1}{\cos \theta} = \frac{at^{2}}{-2a}$,we get $t^{2} = -2 \sec \theta$.

(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a > b$.
The coordinates of the foci are $S(ae, 0)$ and $S^{\prime}(-ae, 0)$.
The extremity of the minor axis is $B(0, b)$.
The slope of $SB$ is $m_1 = \frac{b - 0}{0 - ae} = -\frac{b}{ae}$.
The slope of $S^{\prime}B$ is $m_2 = \frac{b - 0}{0 - (-ae)} = \frac{b}{ae}$.
Since $\angle SBS^{\prime} = 90^{\circ}$,the product of the slopes is $-1$,so $m_1 \times m_2 = -1$.
$(-\frac{b}{ae}) \times (\frac{b}{ae}) = -1$.
$\frac{b^2}{a^2 e^2} = 1 \Rightarrow b^2 = a^2 e^2$.
Using the relation $b^2 = a^2(1 - e^2)$,we get $a^2(1 - e^2) = a^2 e^2$.
$1 - e^2 = e^2$ $\Rightarrow 2e^2 = 1$ $\Rightarrow e^2 = \frac{1}{2}$.
Since eccentricity $e > 0$,we have $e = \frac{1}{\sqrt{2}}$.

(A) Given the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$,we have $a^{2}=9$ and $b^{2}=5$,so $a=3$ and $b=\sqrt{5}$.
The eccentricity $e$ is given by $e = \sqrt{1-\frac{b^{2}}{a^{2}}} = \sqrt{1-\frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.
The foci are at $(\pm ae, 0) = (\pm 3 \times \frac{2}{3}, 0) = (\pm 2, 0)$.
The ends of the latus recta are at $(\pm 2, \pm \frac{b^{2}}{a}) = (\pm 2, \pm \frac{5}{3})$.
Consider the point $P(2, \frac{5}{3})$ in the first quadrant. The equation of the tangent at $P$ is $\frac{x(2)}{9} + \frac{y(5/3)}{5} = 1$,which simplifies to $\frac{2x}{9} + \frac{y}{3} = 1$,or $2x + 3y = 9$.
This line intersects the $x$-axis at $A(\frac{9}{2}, 0)$ and the $y$-axis at $B(0, 3)$.
The quadrilateral formed by the four tangents is symmetric about both axes. The area of the quadrilateral is $4 \times \text{Area}(\Delta OAB) = 4 \times (\frac{1}{2} \times OA \times OB) = 4 \times \frac{1}{2} \times \frac{9}{2} \times 3 = 27 \text{ sq units}$.

(D) Let the coordinates of point $M$ be $(h, k)$. Let $\angle MAB = \theta$,then $\angle MBA = 2\theta$.
From the coordinates of $A(-1, 0)$ and $B(2, 0)$,we have:
$\tan \theta = \frac{k}{h - (-1)} = \frac{k}{h+1}$
$\tan(\pi - 2\theta) = \frac{k}{h-2} \implies -\tan 2\theta = \frac{k}{h-2} \implies \tan 2\theta = \frac{k}{2-h}$
Using the identity $\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}$,we get:
$\frac{k}{2-h} = \frac{2(\frac{k}{h+1})}{1 - (\frac{k}{h+1})^2}$
$\frac{k}{2-h} = \frac{2k(h+1)}{(h+1)^2 - k^2}$
Assuming $k \neq 0$,we divide by $k$:
$\frac{1}{2-h} = \frac{2(h+1)}{(h+1)^2 - k^2}$
$(h+1)^2 - k^2 = 2(h+1)(2-h)$
$h^2 + 2h + 1 - k^2 = 2(2h - h^2 + 2 - h) = 2(-h^2 + h + 2) = -2h^2 + 2h + 4$
$3h^2 - k^2 = 3$
This is the equation of a hyperbola.

(C) Given,the equation of the hyperbola is $x^{2}-y^{2}+1=0$,which can be rewritten as $y^{2}-x^{2}=1$.
Comparing this with the standard form $\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1$,we get $a^{2}=1$ and $b^{2}=1$,so $a=1$ and $b=1$.
The eccentricity $e$ is given by $e=\sqrt{1+\frac{a^{2}}{b^{2}}}=\sqrt{1+\frac{1}{1}}=\sqrt{2}$.
The foci are $(0, \pm be) = (0, \pm \sqrt{2})$.
The line segment joining these foci is a diameter of the circle.
The center of the circle is the midpoint of the foci: $(\frac{0+0}{2}, \frac{\sqrt{2}+(-\sqrt{2})}{2}) = (0,0)$.
The length of the diameter is the distance between $(0, \sqrt{2})$ and $(0, -\sqrt{2})$,which is $2\sqrt{2}$.
Thus,the radius $r = \frac{2\sqrt{2}}{2} = \sqrt{2}$.
The equation of the circle with center $(0,0)$ and radius $r=\sqrt{2}$ is $x^{2}+y^{2}=r^{2}$,which gives $x^{2}+y^{2}=2$.

(B) The equation of the hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. The focus $S$ is $(ae, 0)$.
The line is $bx-ay=0$,which can be written as $bx-ay+0=0$.
The length of the perpendicular $SP$ from $S(ae, 0)$ to the line $bx-ay=0$ is:
$SP = \left| \frac{b(ae) - a(0)}{\sqrt{b^{2}+(-a)^{2}}} \right| = \frac{abe}{\sqrt{b^{2}+a^{2}}}$.
Since $b^{2} = a^{2}(e^{2}-1)$,we have $a^{2}+b^{2} = a^{2}e^{2}$,so $\sqrt{a^{2}+b^{2}} = ae$.
Thus,$SP = \frac{abe}{ae} = b$.
The distance $CS$ is $ae$. In the right-angled triangle $\Delta SPC$,$CP^{2} = CS^{2} - SP^{2}$.
$CP^{2} = (ae)^{2} - b^{2} = a^{2}e^{2} - b^{2} = a^{2}(1 + \frac{b^{2}}{a^{2}}) - b^{2} = a^{2} + b^{2} - b^{2} = a^{2}$.
Therefore,$CP = a$.
The area of the rectangle with sides $SP$ and $CP$ is $SP \times CP = b \times a = ab$.

(C) Given the limit $L = \lim _{x \rightarrow 0} \frac{2 f(x)-3 f(2 x)+f(4 x)}{x^{2}}$.
Since the limit is of the form $\frac{0}{0}$,we apply $L$'$H$ôpital's rule:
$L = \lim _{x \rightarrow 0} \frac{2 f^{\prime}(x)-6 f^{\prime}(2 x)+4 f^{\prime}(4 x)}{2 x}$
$L = \lim _{x \rightarrow 0} \frac{f^{\prime}(x)-3 f^{\prime}(2 x)+2 f^{\prime}(4 x)}{x}$
Applying $L$'$H$ôpital's rule again:
$L = \lim _{x}$ ${\rightarrow 0} \frac{f^{\prime \prime}(x)-3 f^{\prime \prime}(2 x) \cdot 2+2 f^{\prime \prime}(4 x) \cdot 4}{1}$
$L = f^{\prime \prime}(0)-6 f^{\prime \prime}(0)+8 f^{\prime \prime}(0)$
$L = k-6 k+8 k = 3 k$.

(B) Given $f(x) = \lim_{n \rightarrow \infty} n(x^{1/n} - 1)$.
Let $h = \frac{1}{n}$. As $n \rightarrow \infty$,$h \rightarrow 0$.
Then $f(x) = \lim_{h \rightarrow 0} \frac{x^h - 1}{h}$.
This is the standard limit definition of the derivative of $a^x$ at $x=0$,which is $\ln(x)$.
Thus,$f(x) = \ln(x)$.
Now,$f(xy) = \ln(xy) = \ln(x) + \ln(y) = f(x) + f(y)$.

(B) Let $y = \lim _{x \rightarrow 0} (\sin x)^{2 \tan x}$.
Taking the natural logarithm on both sides:
$\ln y = \lim _{x \rightarrow 0} 2 \tan x \ln(\sin x)$.
This is an indeterminate form of type $0 \times (-\infty)$. We rewrite it as:
$\ln y = 2 \lim _{x \rightarrow 0} \frac{\ln(\sin x)}{\cot x}$.
Applying $L'H\hat{o}pital's$ rule:
$\ln y = 2 \lim _{x \rightarrow 0} \frac{\frac{\cos x}{\sin x}}{-\csc^2 x} = 2 \lim _{x \rightarrow 0} \frac{\cos x / \sin x}{-1 / \sin^2 x} = 2 \lim _{x \rightarrow 0} (-\cos x \sin x)$.
$\ln y = 2 \times (-1 \times 0) = 0$.
Since $\ln y = 0$,we have $y = e^0 = 1$.

(D) Given that the mean of $n$ observations is $\bar{x}$.
Therefore,the sum of observations is $\sum_{i=1}^{n} x_{i} = n \bar{x}$.
When the observation $x_{q}$ is replaced by $x_{q}^{\prime}$,the new sum of observations becomes:
$\sum x_{new} = \sum x - x_{q} + x_{q}^{\prime} = n \bar{x} - x_{q} + x_{q}^{\prime}$.
The new mean $\bar{x}^{\prime}$ is given by:
$\bar{x}^{\prime} = \frac{\sum x_{new}}{n} = \frac{n \bar{x} - x_{q} + x_{q}^{\prime}}{n}$.

(B) In a non-leap year,the total number of days is $365$.
There are $52$ weeks and $1$ extra day in a non-leap year ($52 \times 7 = 364$ days).
Thus,a non-leap year always has $52$ Sundays.
The remaining $1$ day can be any of the following: Sunday,Monday,Tuesday,Wednesday,Thursday,Friday,or Saturday.
Out of these $7$ possible outcomes,only $1$ outcome is a Sunday.
$\therefore$ Total number of outcomes $= 7$.
Number of favourable outcomes $= 1$.
Hence,the required probability $= \frac{1}{7}$.

(B) Given,$f(x) = \int_{-1}^{x} |t| dt$.
Since $x \geq 0$,we can split the integral at $t = 0$:
$f(x) = \int_{-1}^{0} |t| dt + \int_{0}^{x} |t| dt$.
For $t \in [-1, 0]$,$|t| = -t$,and for $t \in [0, x]$,$|t| = t$.
So,$f(x) = \int_{-1}^{0} (-t) dt + \int_{0}^{x} t dt$.
Evaluating the integrals:
$f(x) = -\left[ \frac{t^{2}}{2} \right]_{-1}^{0} + \left[ \frac{t^{2}}{2} \right]_{0}^{x}$.
$f(x) = -\left( 0 - \frac{(-1)^{2}}{2} \right) + \left( \frac{x^{2}}{2} - 0 \right)$.
$f(x) = -\left( -\frac{1}{2} \right) + \frac{x^{2}}{2} = \frac{1}{2} + \frac{x^{2}}{2} = \frac{1}{2}(1 + x^{2})$.

(D) For every real number $x$,$x^2 \geq 0$.
$\therefore (x, x) \in P$.
Hence,$P$ is reflexive.
Now,let $(x, y) \in P$.
$\Rightarrow xy \geq 0$.
$\Rightarrow yx \geq 0$.
$\therefore (y, x) \in P$.
Hence,$P$ is symmetric.
Again,consider $(-1, 0) \in P$ because $(-1)(0) = 0 \geq 0$,and $(0, 2) \in P$ because $(0)(2) = 0 \geq 0$.
However,$(-1, 2) \notin P$ because $(-1)(2) = -2 < 0$.
Therefore,$P$ is not transitive.
Thus,the relation $P$ is reflexive and symmetric but not transitive.

(B) We have $x \rho y \iff x-y \in \{0\} \cup \mathbb{I}$,where $\mathbb{I}$ is the set of irrational numbers.
$1$. Reflexivity: For any $x \in R$,$x-x = 0$. Since $0$ is zero,$(x, x) \in \rho$. Thus,$\rho$ is reflexive.
$2$. Symmetry: If $(x, y) \in \rho$,then $x-y$ is $0$ or irrational. Then $y-x = -(x-y)$ is also $0$ or irrational. Thus,$(y, x) \in \rho$. Therefore,$\rho$ is symmetric.
$3$. Transitivity: Consider $x = 2, y = \sqrt{3}, z = 4$.
$(x, y) = (2, \sqrt{3}) \in \rho$ because $2-\sqrt{3}$ is irrational.
$(y, z) = (\sqrt{3}, 4) \in \rho$ because $\sqrt{3}-4$ is irrational.
However,$(x, z) = (2, 4) \notin \rho$ because $2-4 = -2$,which is a rational number (not zero or irrational).
Therefore,$\rho$ is not transitive.

(C) Given $R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$ and $S = \{(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)\}$.
$R \cup S = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)\}$.
$1$. Reflexivity: Since $(1, 1), (2, 2), (3, 3) \in R \cup S$,it is reflexive.
$2$. Symmetry: Since $(1, 2) \in R \cup S \implies (2, 1) \in R \cup S$ and $(1, 3) \in R \cup S \implies (3, 1) \in R \cup S$,it is symmetric.
$3$. Transitivity: We have $(2, 1) \in R \cup S$ and $(1, 3) \in R \cup S$. If it were transitive,$(2, 3)$ must be in $R \cup S$. However,$(2, 3) \notin R \cup S$. Thus,it is not transitive.
Therefore,$R \cup S$ is reflexive and symmetric but not transitive.

(C) For reflexivity: For any $a \in R$,we have $1 + a^2 > 0$. Thus,$(a, a) \in \rho$. So,$\rho$ is reflexive.
For symmetry: If $(a, b) \in \rho$,then $1 + ab > 0$. Since $ab = ba$,we have $1 + ba > 0$,which implies $(b, a) \in \rho$. So,$\rho$ is symmetric.
For transitivity: Consider $a = 1$,$b = -0.5$,and $c = -9$.
Check $(a, b)$: $1 + (1)(-0.5) = 0.5 > 0$,so $(1, -0.5) \in \rho$.
Check $(b, c)$: $1 + (-0.5)(-9) = 1 + 4.5 = 5.5 > 0$,so $(-0.5, -9) \in \rho$.
Check $(a, c)$: $1 + (1)(-9) = 1 - 9 = -8 < 0$,so $(1, -9) \notin \rho$.
Since $(1, -0.5) \in \rho$ and $(-0.5, -9) \in \rho$ but $(1, -9) \notin \rho$,the relation is not transitive.
Therefore,$\rho$ is reflexive and symmetric but not transitive.

(B) Given $A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$.
We calculate the powers of $A$:
$A^2 = A \cdot A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 & \frac{2(2+1)}{2} \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix}$.
$A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 & 6 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 & \frac{3(3+1)}{2} \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{bmatrix}$.
By observing the pattern,for any positive integer $n$,we have:
$A^n = \begin{bmatrix} 1 & n & \frac{n(n+1)}{2} \\ 0 & 1 & n \\ 0 & 0 & 1 \end{bmatrix}$.

(B) matrix $A$ is orthogonal if $A^T A = I$.
Taking the determinant on both sides,we get $|A^T A| = |I|$.
Since $|A^T| = |A|$,we have $|A|^2 = 1$,which implies $|A| = \pm 1$.
Since $|A| \neq 0$,every orthogonal matrix is non-singular.
Thus,$Q \subseteq P$.
Since there exist non-singular matrices that are not orthogonal (e.g.,any diagonal matrix with entries other than $1$ or $-1$),$Q$ is a proper subset of $P$.

(A) We have,$|A| = \begin{vmatrix} 1 & \cos \theta & 0 \\ -\cos \theta & 1 & \cos \theta \\ -1 & -\cos \theta & 1 \end{vmatrix}$
Expanding along the first row:
$|A| = 1[1 - (-\cos \theta)(\cos \theta)] - \cos \theta[-\cos \theta - (-\cos \theta)] + 0[\cos^2 \theta + 1]$
$|A| = 1[1 + \cos^2 \theta] - \cos \theta[0] + 0$
$|A| = 1 + \cos^2 \theta$
Now,we know that $-1 \leq \cos \theta \leq 1$.
Therefore,$0 \leq \cos^2 \theta \leq 1$.
Adding $1$ to all parts,we get $1 \leq 1 + \cos^2 \theta \leq 2$.
Thus,$1 \leq |A| \leq 2$.
Therefore,the value of $|A|$ lies in the closed interval $[1, 2]$.

(D) Let the given equation be $D_1 + D_2 = 0$.
We have $D_1 = \left|\begin{array}{ccc}a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1\end{array}\right|$.
Taking the transpose of $D_2$,we get $D_2 = \left|\begin{array}{ccc}a+1 & a-1 & (-1)^{n+2} a \\ b+1 & b-1 & (-1)^{n+1} b \\ c-1 & c+1 & (-1)^{n} c\end{array}\right|$.
Now,swapping the first and third columns of $D_2$ twice (or rearranging columns) to match the structure of $D_1$:
$D_2 = \left|\begin{array}{ccc}(-1)^{n+2} a & a+1 & a-1 \\ (-1)^{n+1} b & b+1 & b-1 \\ (-1)^{n} c & c-1 & c+1\end{array}\right|$.
Adding $D_1$ and $D_2$,we get:
$\left|\begin{array}{ccc}a(1 + (-1)^{n+2}) & a+1 & a-1 \\ b(-1 + (-1)^{n+1}) & b+1 & b-1 \\ c(1 + (-1)^{n}) & c-1 & c+1\end{array}\right| = 0$.
For the determinant to be zero for arbitrary $a, b, c$,the first column must be zero.
$1 + (-1)^{n+2} = 0 \Rightarrow (-1)^{n+2} = -1$,which implies $n+2$ is odd,so $n$ is odd.
$-1 + (-1)^{n+1} = 0 \Rightarrow (-1)^{n+1} = 1$,which implies $n+1$ is even,so $n$ is odd.
$1 + (-1)^{n} = 0 \Rightarrow (-1)^{n} = -1$,which implies $n$ is odd.
Thus,$n$ is any odd integer.

(B) We know that $\det(AB) = \det(A) \cdot \det(B)$.
Given $\det(AB) = 0$,we have $\det(A) \cdot \det(B) = 0$.
Calculating $\det(A) = (x+2)^2 - 9x = x^2 + 4x + 4 - 9x = x^2 - 5x + 4 = (x-1)(x-4)$.
Calculating $\det(B) = x(x+2) - 0 = x(x+2)$.
Thus,the equation becomes $(x-1)(x-4) \cdot x(x+2) = 0$.
Setting each factor to zero,we get $x-1=0, x-4=0, x=0, x+2=0$.
Therefore,the solutions are $x = 1, 4, 0, -2$.

(D) The given system of equations is a homogeneous system $AX = 0$,where $A = \begin{bmatrix} 8 & -3 & -5 \\ 5 & -8 & 3 \\ 3 & 5 & -8 \end{bmatrix}$.
To determine the nature of the solutions,we calculate the determinant of the coefficient matrix $D = |A|$.
$D = \begin{vmatrix} 8 & -3 & -5 \\ 5 & -8 & 3 \\ 3 & 5 & -8 \end{vmatrix}$
$D = 8((-8)(-8) - (3)(5)) - (-3)((5)(-8) - (3)(3)) + (-5)((5)(5) - (-8)(3))$
$D = 8(64 - 15) + 3(-40 - 9) - 5(25 + 24)$
$D = 8(49) + 3(-49) - 5(49)$
$D = 49(8 - 3 - 5) = 49(0) = 0$.
Since the determinant $D = 0$,the system of homogeneous equations has infinitely many non-zero solutions.

(A) Given the equation: $\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$
Using the formula $\tan ^{-1} A + \tan ^{-1} B = \tan ^{-1} \left( \frac{A+B}{1-AB} \right)$,we get:
$\tan ^{-1}\left[\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\left(\frac{x-1}{x-2}\right) \left(\frac{x+1}{x+2}\right)}\right]=\frac{\pi}{4}$
Taking $\tan$ on both sides:
$\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\frac{(x-1)(x+1)}{(x-2)(x+2)}} = \tan \frac{\pi}{4} = 1$
Simplifying the numerator and denominator:
$\frac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)-(x-1)(x+1)} = 1$
$\frac{(x^2+x-2)+(x^2-x-2)}{(x^2-4)-(x^2-1)} = 1$
$\frac{2x^2-4}{-3} = 1$
$2x^2-4 = -3$
$2x^2 = 1$
$x^2 = \frac{1}{2}$
$x = \pm \frac{1}{\sqrt{2}}$

(A) Given that $f: R \rightarrow R$ is an injective function satisfying $f(x) f(y) = f(x+y)$ for all $x, y \in R$.
This functional equation is satisfied by the exponential function $f(x) = a^x$ for some $a > 0, a \neq 1$.
Since $f(x), f(y), f(z)$ are in $G$.$P$.,we have $(f(y))^2 = f(x) \cdot f(z)$.
Substituting $f(x) = a^x$,we get $(a^y)^2 = a^x \cdot a^z$.
This simplifies to $a^{2y} = a^{x+z}$.
Equating the exponents,we get $2y = x+z$.
This condition implies that $x, y, z$ are in $A$.$P$.

(A) Given the functional equation $\frac{f(x)}{f(y)}=f(x-y)$.
Setting $y=0$,we get $\frac{f(x)}{f(0)}=f(x)$,which implies $f(0)=1$.
Differentiating both sides with respect to $x$,we get $\frac{f^{\prime}(x)}{f(y)}=f^{\prime}(x-y)$.
Setting $x=0$,we have $\frac{f^{\prime}(0)}{f(y)}=f^{\prime}(-y)$.
Since $f^{\prime}(0)=p$,we get $f^{\prime}(-y) = \frac{p}{f(y)}$.
Also,differentiating the original equation with respect to $y$,we get $f(x) \cdot (-\frac{f^{\prime}(y)}{(f(y))^2}) = f^{\prime}(x-y) \cdot (-1)$.
This simplifies to $\frac{f(x) f^{\prime}(y)}{(f(y))^2} = f^{\prime}(x-y)$.
At $y=0$,$\frac{f(x) f^{\prime}(0)}{(f(0))^2} = f^{\prime}(x)$,so $f^{\prime}(x) = p f(x)$.
This is a linear differential equation with solution $f(x) = e^{px}$.
Then $f^{\prime}(x) = p e^{px}$.
Given $f^{\prime}(5) = q$,we have $p e^{5p} = q$,so $e^{5p} = \frac{q}{p}$.
We need $f^{\prime}(-5) = p e^{-5p} = \frac{p}{e^{5p}} = \frac{p}{q/p} = \frac{p^2}{q}$.

(C) Given functions are $F(x)=e^{x}$ and $G(x)=e^{-x}$.
We define $H(x) = G(F(x))$.
Substituting $F(x)$ into $G(x)$,we get $H(x) = G(e^{x}) = e^{-(e^{x})}$.
Now,we differentiate $H(x)$ with respect to $x$ using the chain rule:
$\frac{dH}{dx} = \frac{d}{dx}(e^{-e^{x}}) = e^{-e^{x}} \cdot \frac{d}{dx}(-e^{x}) = e^{-e^{x}} \cdot (-e^{x}) = -e^{x} \cdot e^{-e^{x}}$.
To find the value at $x=0$,we substitute $x=0$ into the derivative:
$\left. \frac{dH}{dx} \right|_{x=0} = -e^{0} \cdot e^{-e^{0}} = -1 \cdot e^{-1} = -\frac{1}{e}$.

(C) Given $f(x) = x^{13} + x^{11} + x^{9} + x^{7} + x^{5} + x^{3} + x + 19$.
Taking the derivative with respect to $x$:
$f'(x) = 13x^{12} + 11x^{10} + 9x^{8} + 7x^{6} + 5x^{4} + 3x^{2} + 1$.
Since all exponents of $x$ in $f'(x)$ are even and the coefficients are positive,$f'(x) \geq 1$ for all real $x$.
Thus,$f'(x) > 0$ for all $x \in \mathbb{R}$,which means $f(x)$ is a strictly increasing function.
$A$ strictly increasing function can intersect the $x$-axis at most once.
Therefore,$f(x) = 0$ has not more than one real root.

(C) Given $f(x) = \log_{5} \log_{3} x$.
Using the change of base formula $\log_{a} b = \frac{\ln b}{\ln a}$,we have:
$f(x) = \frac{\ln(\log_{3} x)}{\ln 5} = \frac{\ln(\frac{\ln x}{\ln 3})}{\ln 5} = \frac{\ln(\ln x) - \ln(\ln 3)}{\ln 5}$.
Differentiating with respect to $x$:
$f^{\prime}(x) = \frac{1}{\ln 5} \cdot \frac{d}{dx} [\ln(\ln x) - \ln(\ln 3)] = \frac{1}{\ln 5} \cdot \frac{1}{\ln x} \cdot \frac{1}{x}$.
Now,substituting $x = e$:
$f^{\prime}(e) = \frac{1}{\ln 5} \cdot \frac{1}{\ln e} \cdot \frac{1}{e} = \frac{1}{\ln 5 \cdot 1 \cdot e} = \frac{1}{e \ln 5}$.
Since $\ln 5 = \log_{e} 5$,we get $f^{\prime}(e) = \frac{1}{e \log_{e} 5}$.

(A) Given $y=e^{m \sin^{-1} x}$.
Differentiating with respect to $x$,we get $\frac{d y}{d x}=e^{m \sin^{-1} x} \cdot \frac{m}{\sqrt{1-x^{2}}}$.
This implies $\sqrt{1-x^{2}} \frac{d y}{d x}=m e^{m \sin^{-1} x} = m y$.
Squaring both sides or differentiating again with respect to $x$ using the product rule:
$\frac{d}{d x} \left( \sqrt{1-x^{2}} \frac{d y}{d x} \right) = \frac{d}{d x} (m y)$.
$\sqrt{1-x^{2}} \frac{d^{2} y}{d x^{2}} + \frac{1}{2 \sqrt{1-x^{2}}} (-2 x) \frac{d y}{d x} = m \frac{d y}{d x}$.
Multiplying throughout by $\sqrt{1-x^{2}}$:
$(1-x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = m \sqrt{1-x^{2}} \frac{d y}{d x}$.
Substituting $\sqrt{1-x^{2}} \frac{d y}{d x} = m y$ into the equation:
$(1-x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = m(m y) = m^{2} y$.
Rearranging gives $(1-x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - m^{2} y = 0$.
Comparing this with the given equation $(1-x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - k y = 0$,we find $k = m^{2}$.

(B) Given $f(0)=f(1)=f^{\prime}(0)=0$.
By Rolle's Theorem on the interval $[0, 1]$,since $f(0)=f(1)=0$,there exists at least one point $d \in (0, 1)$ such that $f^{\prime}(d)=0$.
Now,we have $f^{\prime}(0)=0$ and $f^{\prime}(d)=0$ where $d \in (0, 1)$.
Applying Rolle's Theorem to the function $f^{\prime}(x)$ on the interval $[0, d]$,since $f^{\prime}(0)=f^{\prime}(d)=0$,there must exist at least one point $c \in (0, d)$ such that $f^{\prime \prime}(c)=0$.
Thus,$f^{\prime \prime}(c)=0$ for some $c \in R$.

(B) Given $f(x) = x^{n}$.
Taking the derivative,we get $f^{\prime}(x) = n x^{n-1}$.
Substituting this into the given equation $f^{\prime}(\alpha+\beta) = f^{\prime}(\alpha) + f^{\prime}(\beta)$,we have:
$n(\alpha+\beta)^{n-1} = n\alpha^{n-1} + n\beta^{n-1}$.
For $n \neq 0$,we can divide by $n$:
$(\alpha+\beta)^{n-1} = \alpha^{n-1} + \beta^{n-1}$.
If $n=1$,then $(\alpha+\beta)^{0} = \alpha^{0} + \beta^{0} \Rightarrow 1 = 1 + 1$,which is $1 = 2$ (False).
If $n=2$,then $(\alpha+\beta)^{2-1} = \alpha^{2-1} + \beta^{2-1} \Rightarrow \alpha+\beta = \alpha+\beta$ (True).
If $n=0$,$f(x) = x^{0} = 1$,so $f^{\prime}(x) = 0$. Then $0 = 0 + 0$ (True).
However,checking the options provided,$n=2$ is the standard solution for this type of problem.

(D) Given the curve $y=x^{2}+2ax+b$.
At $x=\alpha$,the point is $P(\alpha, \alpha^{2}+2a\alpha+b)$.
At $x=\beta$,the point is $Q(\beta, \beta^{2}+2a\beta+b)$.
The slope of the chord $PQ$ is given by $m_{chord} = \frac{(\beta^{2}+2a\beta+b) - (\alpha^{2}+2a\alpha+b)}{\beta-\alpha}$.
$m_{chord} = \frac{(\beta^{2}-\alpha^{2}) + 2a(\beta-\alpha)}{\beta-\alpha} = \frac{(\beta-\alpha)(\beta+\alpha) + 2a(\beta-\alpha)}{\beta-\alpha} = \alpha+\beta+2a$.
The slope of the tangent to the curve at any point $x$ is $m_{tangent} = \frac{dy}{dx} = 2x+2a$.
Since the chord is parallel to the tangent,their slopes must be equal:
$2x+2a = \alpha+\beta+2a$.
Solving for $x$,we get $2x = \alpha+\beta$,which implies $x = \frac{\alpha+\beta}{2}$.

(B) Given the curve $xy = 1 - 2x$.
Rewriting the equation as $y = \frac{1 - 2x}{x} = \frac{1}{x} - 2$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\frac{1}{x^2}$.
Since $\frac{dy}{dx} < 0$ for all $x \neq 0$,the slope of the tangent is always negative.
The line $ax + by + c = 0$ can be written as $y = -\frac{a}{b}x - \frac{c}{b}$.
The slope of this line is $m = -\frac{a}{b}$.
Since the line is tangent to the curve,its slope must be equal to the derivative at the point of tangency,which is negative.
Therefore,$-\frac{a}{b} < 0$,which implies $\frac{a}{b} > 0$.
This condition $\frac{a}{b} > 0$ holds if both $a$ and $b$ have the same sign.
Thus,either $a > 0, b > 0$ or $a < 0, b < 0$.

(B) Given $f(x) = \sin x - \cos x - Kx + 5$.
Differentiating with respect to $x$,we get $f'(x) = \cos x + \sin x - K$.
For the function to be decreasing for all $x > 0$,we must have $f'(x) \leq 0$ for all $x$.
This implies $\cos x + \sin x - K \leq 0$,or $K \geq \cos x + \sin x$.
We know that the maximum value of $\cos x + \sin x$ is $\sqrt{1^2 + 1^2} = \sqrt{2}$.
Therefore,for $K \geq \cos x + \sin x$ to hold for all $x$,$K$ must be greater than or equal to the maximum value of the expression $\cos x + \sin x$.
Thus,$K \geq \sqrt{2}$.

(B, C) Let $S_1$ be the displacement of the first particle and $S_2$ be the displacement of the second particle at time $t$.
For the first particle: $S_1 = u t$.
For the second particle: $S_2 = \frac{1}{2} f t^2$.
The distance between them is $D(t) = |S_1 - S_2| = |u t - \frac{1}{2} f t^2|$.
To find the greatest distance,we differentiate $D(t)$ with respect to $t$ and set it to zero:
$\frac{dD}{dt} = u - f t = 0$.
This gives $t = \frac{u}{f}$.
At $t = \frac{u}{f}$,the distance is $D = u(\frac{u}{f}) - \frac{1}{2} f(\frac{u}{f})^2 = \frac{u^2}{f} - \frac{u^2}{2f} = \frac{u^2}{2f}$.
Thus,the particles are at the greatest distance at $t = \frac{u}{f}$ and the maximum distance is $\frac{u^2}{2f}$.
Therefore,options $B$ and $C$ are correct.

(C) Let $I = \int \cos (\log x) d x$.
Substitute $\log x = t$,which implies $x = e^t$.
Then,$dx = e^t dt$.
Substituting these into the integral,we get $I = \int e^t \cos t dt$.
Using the standard integration formula $\int e^{ax} \cos(bx) dx = \frac{e^{ax}}{a^2 + b^2} [a \cos(bx) + b \sin(bx)] + C$,where $a = 1$ and $b = 1$:
$I = \frac{e^t}{1^2 + 1^2} [1 \cdot \cos t + 1 \cdot \sin t] + C$.
$I = \frac{e^t}{2} [\cos t + \sin t] + C$.
Substituting $t = \log x$ and $e^t = x$ back into the expression:
$I = \frac{x}{2} [\cos(\log x) + \sin(\log x)] + C$.
Thus,$F(x) = \frac{x}{2} [\cos(\log x) + \sin(\log x)]$.

(A) Let $I = \int \frac{x^{2}-1}{x^{4}+3 x^{2}+1} d x$.
Divide the numerator and denominator by $x^{2}$:
$I = \int \frac{1 - 1/x^{2}}{x^{2} + 3 + 1/x^{2}} d x$.
Rewrite the denominator as $(x^{2} + 1/x^{2}) + 3$:
$I = \int \frac{1 - 1/x^{2}}{(x + 1/x)^{2} - 2 + 3} d x$.
$I = \int \frac{1 - 1/x^{2}}{(x + 1/x)^{2} + 1} d x$.
Let $t = x + 1/x$. Then $dt = (1 - 1/x^{2}) d x$.
Substituting these into the integral:
$I = \int \frac{dt}{t^{2} + 1}$.
Using the standard integral formula $\int \frac{1}{t^{2} + 1} dt = \tan^{-1}(t) + C$:
$I = \tan^{-1}(t) + C$.
Substituting back $t = x + 1/x$:
$I = \tan^{-1}(x + 1/x) + C$.

(D) We have $I_{1} = \int_{0}^{n} [x] dx = \sum_{k=0}^{n-1} \int_{k}^{k+1} k dx = \sum_{k=0}^{n-1} k(k+1-k) = \sum_{k=0}^{n-1} k = \frac{(n-1)n}{2}$.
Now,$I_{2} = \int_{0}^{n} \{x\} dx$. Since $\{x\} = x - [x]$,we have $I_{2} = \int_{0}^{n} x dx - \int_{0}^{n} [x] dx$.
$I_{2} = \left[ \frac{x^2}{2} \right]_{0}^{n} - I_{1} = \frac{n^2}{2} - \frac{n(n-1)}{2} = \frac{n^2 - n^2 + n}{2} = \frac{n}{2}$.
Therefore,$\frac{I_{1}}{I_{2}} = \frac{\frac{n(n-1)}{2}}{\frac{n}{2}} = n-1$.

(C) Let $I = \int_{0}^{100} e^{x-[x]} d x$.
Since $f(x) = x - [x]$ is a periodic function with period $T = 1$,we can use the property $\int_{0}^{nT} f(x) d x = n \int_{0}^{T} f(x) d x$.
Here,$n = 100$ and $T = 1$,so $I = 100 \int_{0}^{1} e^{x-[x]} d x$.
For $0 < x < 1$,the greatest integer function $[x] = 0$,so $x - [x] = x$.
Thus,$I = 100 \int_{0}^{1} e^{x} d x$.
Evaluating the integral,we get $I = 100 [e^{x}]_{0}^{1}$.
$I = 100 (e^{1} - e^{0}) = 100 (e - 1)$.

(B) Given $I = \int_{0}^{100 \pi} \sqrt{1 - \cos 2x} \, dx$.
Using the identity $1 - \cos 2x = 2 \sin^2 x$,we have $I = \int_{0}^{100 \pi} \sqrt{2 \sin^2 x} \, dx$.
$I = \sqrt{2} \int_{0}^{100 \pi} |\sin x| \, dx$.
Since $|\sin x|$ is a periodic function with period $\pi$,we can write $I = \sqrt{2} \times 100 \int_{0}^{\pi} |\sin x| \, dx$.
In the interval $[0, \pi]$,$\sin x \geq 0$,so $|\sin x| = \sin x$.
$I = 100 \sqrt{2} \int_{0}^{\pi} \sin x \, dx$.
$I = 100 \sqrt{2} [-\cos x]_{0}^{\pi}$.
$I = 100 \sqrt{2} [-\cos \pi - (-\cos 0)]$.
$I = 100 \sqrt{2} [-(-1) - (-1)]$.
$I = 100 \sqrt{2} [1 + 1] = 100 \sqrt{2} \times 2 = 200 \sqrt{2}$.

(C) For $x \in [10, 19]$,we have $|\sin x| \leq 1$ and $1+x^{6} > 10^{6}$.
Since $x \geq 10$,$1+x^{6} > 10^{6}$,which implies $\frac{1}{1+x^{6}} < 10^{-6}$.
Therefore,$|I| = \left| \int_{10}^{19} \frac{\sin x}{1+x^{6}} dx \right| \leq \int_{10}^{19} \frac{|\sin x|}{1+x^{6}} dx$.
Since $|\sin x| \leq 1$ and $\frac{1}{1+x^{6}} < 10^{-6}$,we have $|I| < \int_{10}^{19} 10^{-6} dx$.
$|I| < 10^{-6} \times (19 - 10) = 9 \times 10^{-6}$.
However,checking the options provided and the magnitude,$9 \times 10^{-6} < 10^{-5}$. Thus,$|I| < 10^{-5}$ is the correct bound.

(D) We know that for $x \in [0, 1]$,$0 \leq x^2 \leq 1$.
Since $e^x$ is an increasing function,we have $e^0 \leq e^{x^2} \leq e^1$,which implies $1 \leq e^{x^2} \leq e$.
Integrating the inequality over the interval $[0, 1]$:
$\int_{0}^{1} 1 \, dx \leq \int_{0}^{1} e^{x^2} \, dx \leq \int_{0}^{1} e \, dx$.
Calculating the integrals:
$[x]_{0}^{1} \leq \int_{0}^{1} e^{x^2} \, dx \leq [ex]_{0}^{1}$.
$1 \leq \int_{0}^{1} e^{x^2} \, dx \leq e$.
Thus,the value of the integral lies in the closed interval $[1, e]$.

(B) The given limit is $L = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{n}{n^{2}+r^{2}}$.
Dividing the numerator and denominator by $n^{2}$,we get:
$L = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{1/n}{1+(r/n)^{2}}$.
Using the definition of a definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}\right) = \int_{0}^{1} f(x) dx$,where $f(x) = \frac{1}{1+x^{2}}$.
Therefore,$L = \int_{0}^{1} \frac{1}{1+x^{2}} dx$.
Evaluating the integral,we get $L = [\tan ^{-1} x]_{0}^{1}$.
$L = \tan ^{-1}(1) - \tan ^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.

(C) Let $I = \int_{0}^{a} \frac{dx}{1+f(x)} \quad \dots (i)$
Using the property $\int_{0}^{a} g(x) dx = \int_{0}^{a} g(a-x) dx$,we have:
$I = \int_{0}^{a} \frac{dx}{1+f(a-x)}$
Given $f(x) f(a-x) = 1$,we can write $f(a-x) = \frac{1}{f(x)}$.
Substituting this into the integral:
$I = \int_{0}^{a} \frac{dx}{1+\frac{1}{f(x)}} = \int_{0}^{a} \frac{f(x) dx}{f(x)+1} \quad \dots (ii)$
Adding equations $(i)$ and $(ii)$:
$2I = \int_{0}^{a} \frac{dx}{1+f(x)} + \int_{0}^{a} \frac{f(x) dx}{1+f(x)}$
$2I = \int_{0}^{a} \frac{1+f(x)}{1+f(x)} dx = \int_{0}^{a} 1 dx$
$2I = [x]_{0}^{a} = a$
$I = \frac{a}{2}$

(A) Given the parabolas are $x = -2y^{2}$ and $x = 1 - 3y^{2}$.
To find the points of intersection,set the equations equal:
$-2y^{2} = 1 - 3y^{2}$
$y^{2} = 1$
$y = \pm 1$
When $y = \pm 1$,$x = -2(1)^{2} = -2$.
The points of intersection are $(-2, 1)$ and $(-2, -1)$.
The area is bounded by $x = 1 - 3y^{2}$ (right curve) and $x = -2y^{2}$ (left curve) from $y = -1$ to $y = 1$.
Area $= \int_{-1}^{1} [(1 - 3y^{2}) - (-2y^{2})] dy$
$= \int_{-1}^{1} (1 - y^{2}) dy$
Since the function is even,Area $= 2 \int_{0}^{1} (1 - y^{2}) dy$
$= 2 [y - \frac{y^{3}}{3}]_{0}^{1}$
$= 2 [1 - \frac{1}{3}]$
$= 2 [\frac{2}{3}] = \frac{4}{3}$ sq units.

(A) Given the differential equation: $(x+y)^{2} \frac{dy}{dx} = a^{2}$.
Let $v = x+y$. Then,differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dv}{dx}$,which implies $\frac{dy}{dx} = \frac{dv}{dx} - 1$.
Substituting these into the original equation: $v^{2} (\frac{dv}{dx} - 1) = a^{2}$.
Rearranging the terms: $v^{2} \frac{dv}{dx} = v^{2} + a^{2}$,so $\frac{dv}{dx} = \frac{v^{2} + a^{2}}{v^{2}}$.
Separating the variables: $\frac{v^{2}}{v^{2} + a^{2}} dv = dx$.
Integrating both sides: $\int \frac{v^{2}}{v^{2} + a^{2}} dv = \int dx$.
This can be written as: $\int (1 - \frac{a^{2}}{v^{2} + a^{2}}) dv = x + C'$.
Integrating gives: $v - a \tan^{-1}(\frac{v}{a}) = x + C'$.
Substituting $v = x+y$: $(x+y) - a \tan^{-1}(\frac{x+y}{a}) = x + C'$.
Simplifying: $y - a \tan^{-1}(\frac{x+y}{a}) = C'$.
Rearranging: $\frac{y-C'}{a} = \tan^{-1}(\frac{x+y}{a})$.
Taking the tangent on both sides: $\tan(\frac{y-C'}{a}) = \frac{x+y}{a}$.
Letting $-C' = C$,we get $\frac{x+y}{a} = \tan(\frac{y+C}{a})$.

(B) Given the differential equation: $x^{2}(x^{2}-1) \frac{dy}{dx} + x(x^{2}+1)y = x^{2}-1$.
Divide throughout by $x^{2}(x^{2}-1)$ to get it in the standard form $\frac{dy}{dx} + Py = Q$:
$\frac{dy}{dx} + \frac{x^{2}+1}{x(x^{2}-1)}y = \frac{1}{x^{2}}$.
Here,$P = \frac{x^{2}+1}{x(x^{2}-1)}$.
The integrating factor $(IF)$ is given by $e^{\int P dx}$.
$\int P dx = \int \frac{x^{2}+1}{x(x-1)(x+1)} dx$.
Using partial fractions: $\frac{x^{2}+1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}$.
Solving for constants: $x^{2}+1 = A(x^{2}-1) + Bx(x+1) + Cx(x-1)$.
For $x=0$,$1 = -A \Rightarrow A = -1$.
For $x=1$,$2 = 2B \Rightarrow B = 1$.
For $x=-1$,$2 = 2C \Rightarrow C = 1$.
So,$\int P dx = \int (\frac{-1}{x} + \frac{1}{x-1} + \frac{1}{x+1}) dx = -\ln|x| + \ln|x-1| + \ln|x+1| = \ln|\frac{x^{2}-1}{x}|$.
$IF = e^{\ln|\frac{x^{2}-1}{x}|} = \frac{x^{2}-1}{x} = x - \frac{1}{x}$.

(C) Let $\vec{a}$ and $\vec{b}$ be two unit vectors,so $|\vec{a}| = 1$ and $|\vec{b}| = 1$.
Given that their sum is a unit vector,$|\vec{a} + \vec{b}| = 1$.
Squaring both sides,we get $|\vec{a} + \vec{b}|^2 = 1^2$.
Using the identity $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b})$,we have $1 + 1 + 2(\vec{a} \cdot \vec{b}) = 1$.
This simplifies to $2 + 2(\vec{a} \cdot \vec{b}) = 1$,which gives $2(\vec{a} \cdot \vec{b}) = -1$.
Now,we need to find the magnitude of their difference,$|\vec{a} - \vec{b}|$.
We know that $|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})$.
Substituting the known values,$|\vec{a} - \vec{b}|^2 = 1 + 1 - (-1) = 1 + 1 + 1 = 3$.
Therefore,$|\vec{a} - \vec{b}| = \sqrt{3}$ units.

(B) Let $x = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$.
Then,$x \times \hat{i} = (\alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}) \times \hat{i} = -\beta \hat{k} + \gamma \hat{j}$.
Similarly,$x \times \hat{j} = \alpha \hat{k} - \gamma \hat{i}$ and $x \times \hat{k} = -\alpha \hat{j} + \beta \hat{i}$.
Now,$(x \times \hat{i})^{2} = (x \times \hat{i}) \cdot (x \times \hat{i}) = (-\beta \hat{k} + \gamma \hat{j}) \cdot (-\beta \hat{k} + \gamma \hat{j}) = \beta^{2} + \gamma^{2}$.
Similarly,$(x \times \hat{j})^{2} = \alpha^{2} + \gamma^{2}$ and $(x \times \hat{k})^{2} = \alpha^{2} + \beta^{2}$.
Adding these,we get $(x \times \hat{i})^{2} + (x \times \hat{j})^{2} + (x \times \hat{k})^{2} = (\beta^{2} + \gamma^{2}) + (\alpha^{2} + \gamma^{2}) + (\alpha^{2} + \beta^{2}) = 2(\alpha^{2} + \beta^{2} + \gamma^{2})$.
Since $|x|^{2} = \alpha^{2} + \beta^{2} + \gamma^{2}$,the expression equals $2|x|^{2}$.

(B) Let the direction ratios of the three lines be $\vec{a} = (1, -1, 1)$,$\vec{b} = (2, -3, 0)$,and $\vec{c} = (1, 0, 3)$.
Since all three lines pass through the origin,they are coplanar if and only if the scalar triple product of their direction vectors is zero,i.e.,$[\vec{a}, \vec{b}, \vec{c}] = 0$.
We calculate the determinant of the matrix formed by these vectors:
$\Delta = \begin{vmatrix} 1 & -1 & 1 \\ 2 & -3 & 0 \\ 1 & 0 & 3 \end{vmatrix}$
$= 1((-3)(3) - (0)(0)) - (-1)((2)(3) - (0)(1)) + 1((2)(0) - (-3)(1))$
$= 1(-9 - 0) + 1(6 - 0) + 1(0 + 3)$
$= -9 + 6 + 3 = 0$
Since the determinant is $0$,the three vectors are linearly dependent,which means the three lines lie in the same plane.
Therefore,the lines are coplanar.

(D) The plane passes through $(1, 2, -3)$ and $(2, -2, 1)$. The vector connecting these two points is $\vec{v} = (2-1)\hat{i} + (-2-2)\hat{j} + (1-(-3))\hat{k} = \hat{i} - 4\hat{j} + 4\hat{k}$.
Since the plane is parallel to the $X$-axis,its normal is perpendicular to the unit vector $\hat{i} = (1, 0, 0)$.
The normal vector $\vec{n}$ to the plane is given by the cross product of $\vec{v}$ and $\hat{i}$:
$\vec{n} = \vec{v} \times \hat{i} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -4 & 4 \\ 1 & 0 & 0 \end{vmatrix} = \hat{i}(0) - \hat{j}(0-4) + \hat{k}(0-(-4)) = 4\hat{j} + 4\hat{k}$.
We can simplify the normal vector to $\vec{n}' = (0, 1, 1)$.
The equation of the plane passing through $(1, 2, -3)$ with normal $(0, 1, 1)$ is:
$0(x-1) + 1(y-2) + 1(z+3) = 0$
$y - 2 + z + 3 = 0$
$y + z + 1 = 0$.