IIT JEE 1983 Mathematics Question Paper with Answer and Solution

(B) In a parallelogram $ABCD$,the diagonals $AC$ and $BD$ bisect each other at the same midpoint.
The midpoint of diagonal $AC$ is given by $\frac{z_1 + z_3}{2}$.
The midpoint of diagonal $BD$ is given by $\frac{z_2 + z_4}{2}$.
Since the diagonals bisect each other,we have $\frac{z_1 + z_3}{2} = \frac{z_2 + z_4}{2}$.
Therefore,$z_1 + z_3 = z_2 + z_4$.

(B) Given $\omega = \frac{1 - iz}{z - i}$ and $|\omega| = 1$.
$|\frac{1 - iz}{z - i}| = 1$
$|1 - iz| = |z - i|$
Substitute $z = x + iy$:
$|1 - i(x + iy)| = |x + iy - i|$
$|1 - ix + y| = |x + i(y - 1)|$
$|(1 + y) - ix| = |x + i(y - 1)|$
Squaring both sides:
$(1 + y)^2 + (-x)^2 = x^2 + (y - 1)^2$
$1 + y^2 + 2y + x^2 = x^2 + y^2 + 1 - 2y$
$2y = -2y$
$4y = 0 \implies y = 0$.
Since $z = x + iy$ and $y = 0$,$z = x$,which means $z$ lies on the real axis.

(A) Let the vertices of the triangle be $O(0), A(z_1),$ and $B(z_2)$. Since $\Delta OAB$ is an equilateral triangle,the rotation of vector $OA$ by $60^\circ$ ($\pi/3$ radians) about the origin $O$ must coincide with the vector $OB$ (or vice versa).
Thus,we have $z_2 = z_1 e^{\pm i\pi/3}$.
This implies $\frac{z_2}{z_1} = \cos(\pi/3) \pm i \sin(\pi/3) = \frac{1}{2} \pm i \frac{\sqrt{3}}{2}$.
Rearranging gives $2z_2 = z_1(1 \pm i\sqrt{3})$,or $2z_2 - z_1 = \pm i\sqrt{3} z_1$.
Squaring both sides: $(2z_2 - z_1)^2 = -3z_1^2$.
$4z_2^2 - 4z_1 z_2 z_1^2 = -3z_1^2$.
$4z_1^2 4z_2^2 = 4z_1 z_2$.
Dividing by $4$,we get $z_1^2 z_2^2 = z_1 z_2$.

(C) Let $x = 2.\overline{357} = 2.357357357...$
This can be written as $x = 2 + 0.357357357...$
Let $y = 0.357357357... = \frac{357}{1000} + \frac{357}{1000^2} + \frac{357}{1000^3} + ...$
This is an infinite geometric series with first term $a = \frac{357}{1000}$ and common ratio $r = \frac{1}{1000}$.
The sum $S_{\infty} = \frac{a}{1-r} = \frac{\frac{357}{1000}}{1 - \frac{1}{1000}} = \frac{\frac{357}{1000}}{\frac{999}{1000}} = \frac{357}{999}$.
Therefore,$x = 2 + \frac{357}{999} = \frac{2 \times 999 + 357}{999} = \frac{1998 + 357}{999} = \frac{2355}{999}$.

(B) Given the quadratic equation $2x^2 + 3x + 1 = 0$.
Comparing this with $ax^2 + bx + c = 0$,we get $a = 2, b = 3, c = 1$.
The discriminant $D = b^2 - 4ac = (3)^2 - 4(2)(1) = 9 - 8 = 1$.
Since $D > 0$ and $D$ is a perfect square,the roots are real and rational.
Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$,we get $x = \frac{-3 \pm \sqrt{1}}{2(2)} = \frac{-3 \pm 1}{4}$.
Thus,the roots are $x = \frac{-2}{4} = -\frac{1}{2}$ and $x = \frac{-4}{4} = -1$.
Both roots are rational numbers.

(B) Let the roots be $\alpha$ and $\alpha^n$.
From the relation between roots and coefficients,we have:
$\alpha + \alpha^n = -\frac{b}{a}$ and $\alpha \cdot \alpha^n = \alpha^{n+1} = \frac{c}{a}$.
From the second equation,$\alpha = (\frac{c}{a})^{\frac{1}{n+1}}$.
Substituting this into the first equation:
$(\frac{c}{a})^{\frac{1}{n+1}} + ((\frac{c}{a})^{\frac{1}{n+1}})^n = -\frac{b}{a}$
$(\frac{c}{a})^{\frac{1}{n+1}} + (\frac{c}{a})^{\frac{n}{n+1}} = -\frac{b}{a}$
Multiply both sides by $a$:
$a \cdot \frac{c^{\frac{1}{n+1}}}{a^{\frac{1}{n+1}}} + a \cdot \frac{c^{\frac{n}{n+1}}}{a^{\frac{n}{n+1}}} = -b$
$a^{1 - \frac{1}{n+1}} c^{\frac{1}{n+1}} + a^{1 - \frac{n}{n+1}} c^{\frac{n}{n+1}} = -b$
$a^{\frac{n}{n+1}} c^{\frac{1}{n+1}} + a^{\frac{1}{n+1}} c^{\frac{n}{n+1}} = -b$
$(a^n c)^{\frac{1}{n+1}} + (a c^n)^{\frac{1}{n+1}} = -b$.

(A) First,arrange $m$ men in a row in $m!$ ways.
Since $n < m$ and no two women can sit together,we consider the gaps between the men.
In any one of the $m!$ arrangements,there are $(m + 1)$ available spaces (including the ends) where $n$ women can be placed.
The number of ways to arrange $n$ women in these $(m + 1)$ spaces is given by $^{m+1}P_n$.
Therefore,by the fundamental principle of counting,the total number of arrangements is $m! \times {}^{m+1}P_n$.
$= m! \times \frac{(m + 1)!}{(m + 1 - n)!} = \frac{m! (m + 1)!}{(m - n + 1)!}$.

(A) The general term in the expansion of $\left( \frac{x}{2} - \frac{3}{x^2} \right)^{10}$ is given by $T_{r+1} = {}^{10}C_r \left( \frac{x}{2} \right)^{10-r} \left( -\frac{3}{x^2} \right)^r$.
Simplifying this,we get $T_{r+1} = {}^{10}C_r (-1)^r \frac{3^r}{2^{10-r}} x^{10-r-2r} = {}^{10}C_r (-1)^r \frac{3^r}{2^{10-r}} x^{10-3r}$.
To find the coefficient of $x^4$,we set the exponent of $x$ equal to $4$:
$10 - 3r = 4$ $\Rightarrow 3r = 6$ $\Rightarrow r = 2$.
Substituting $r = 2$ into the expression for the term:
$T_{2+1} = {}^{10}C_2 (-1)^2 \frac{3^2}{2^{10-2}} x^4 = \frac{10 \times 9}{2 \times 1} \times 1 \times \frac{9}{2^8} x^4$.
$T_3 = 45 \times \frac{9}{256} x^4 = \frac{405}{256} x^4$.
Thus,the coefficient of $x^4$ is $\frac{405}{256}$.

(A) Given the binomial expansion: $(1 + ax)^n = 1 + n(ax) + \frac{n(n - 1)}{2}(ax)^2 + .... = 1 + 8x + 24x^2 + ....$
Comparing the coefficients of $x$ and $x^2$:
$na = 8$ --- $(1)$
$\frac{n(n - 1)}{2}a^2 = 24$ --- $(2)$
From $(1)$,$n = \frac{8}{a}$. Substituting this into $(2)$:
$\frac{(\frac{8}{a})(\frac{8}{a} - 1)}{2}a^2 = 24$
$\frac{8}{2a} \times (8 - a) \times a = 24$
$4(8 - a) = 24$
$8 - a = 6$
$a = 2$
Substituting $a = 2$ into $(1)$:
$n(2) = 8 \Rightarrow n = 4$.
Thus,the values are $a = 2$ and $n = 4$.

(A) Given $\tan A = \frac{1 - \cos B}{\sin B}$.
Using the half-angle formulas $1 - \cos B = 2\sin^2(B/2)$ and $\sin B = 2\sin(B/2)\cos(B/2)$:
$\tan A = \frac{2\sin^2(B/2)}{2\sin(B/2)\cos(B/2)} = \frac{\sin(B/2)}{\cos(B/2)} = \tan(B/2)$.
Thus,$A = B/2$,which implies $2A = B$.
Therefore,$\tan 2A = \tan B$.

(B) Let the height of the lighthouse be $h = 60 \ m$ and the distance of the boat from the foot of the lighthouse be $x \ m$.
In the right-angled triangle formed by the lighthouse and the sea level,the angle of elevation of the top from the boat is equal to the angle of depression,which is $15^\circ$.
Thus,$\tan(15^\circ) = \frac{\text{height}}{\text{distance}} = \frac{60}{x}$.
Therefore,$x = \frac{60}{\tan(15^\circ)} = 60 \cot(15^\circ)$.
Using the value $\cot(15^\circ) = \frac{\cos(15^\circ)}{\sin(15^\circ)} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$.
So,$x = 60 \left( \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \right) \ m$.
Thus,the correct option is $B$.

(D) Let the points be $A(0, 8/3)$,$B(1, 3)$,and $C(82, 30)$.
To check if these points form a triangle,we calculate the area of the triangle using the formula:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Substituting the values:
$\text{Area} = \frac{1}{2} |0(3 - 30) + 1(30 - 8/3) + 82(8/3 - 3)|$
$\text{Area} = \frac{1}{2} |0 + (90 - 8)/3 + 82(-1/3)|$
$\text{Area} = \frac{1}{2} |82/3 - 82/3| = 0$
Since the area is $0$,the points are collinear and do not form a triangle. Therefore,the correct option is $D$.

(A) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Area of $\Delta ABC = \frac{1}{2} |6(5 - (-2)) + (-3)(-2 - 3) + 4(3 - 5)| = \frac{1}{2} |6(7) - 3(-5) + 4(-2)| = \frac{1}{2} |42 + 15 - 8| = \frac{1}{2} |49| = 24.5$.
Area of $\Delta PBC = \frac{1}{2} |x(5 - (-2)) + (-3)(-2 - y) + 4(y - 5)| = \frac{1}{2} |7x + 6 + 3y + 4y - 20| = \frac{1}{2} |7x + 7y - 14| = \frac{7}{2} |x + y - 2|$.
Ratio = $\frac{\text{Area}(\Delta PBC)}{\text{Area}(\Delta ABC)} = \frac{\frac{7}{2} |x + y - 2|}{\frac{49}{2}} = \left| \frac{x + y - 2}{7} \right|$.

(B) Let the vertices be $A(at_1t_2, a(t_1 + t_2))$,$B(at_2t_3, a(t_2 + t_3))$,and $C(at_3t_1, a(t_3 + t_1))$.
The slope of $BC$ is $m_{BC} = \frac{a(t_2 + t_3) - a(t_3 + t_1)}{at_2t_3 - at_3t_1} = \frac{a(t_2 - t_1)}{at_3(t_2 - t_1)} = \frac{1}{t_3}$.
The slope of the altitude from $A$ to $BC$ is $-t_3$. The equation of this altitude is $y - a(t_1 + t_2) = -t_3(x - at_1t_2)$,which simplifies to $t_3x + y = a(t_1 + t_2 + t_1t_2t_3)$.
Similarly,the equation of the altitude from $B$ to $AC$ is $t_1x + y = a(t_2 + t_3 + t_1t_2t_3)$.
Subtracting the two equations: $(t_3 - t_1)x = a(t_1 - t_3)$,so $x = -a$.
Substituting $x = -a$ into the first altitude equation: $-at_3 + y = a(t_1 + t_2 + t_1t_2t_3)$,which gives $y = a(t_1 + t_2 + t_3 + t_1t_2t_3)$.
Thus,the orthocentre is $(-a, a(t_1 + t_2 + t_3 + t_1t_2t_3))$.

(B) To find the point of intersection of the lines $x + 2y - 10 = 0$ and $2x + y + 5 = 0$,we solve them simultaneously.
Multiply the first equation by $2$: $2x + 4y - 20 = 0$.
Subtract the second equation from this: $(2x + 4y - 20) - (2x + y + 5) = 0$,which simplifies to $3y - 25 = 0$,so $y = 25/3$.
Substitute $y = 25/3$ into $x + 2y - 10 = 0$: $x + 2(25/3) - 10 = 0 \implies x + 50/3 - 30/3 = 0 \implies x = -20/3$.
The point of intersection is $(-20/3, 25/3)$.
Testing the options,for $5x + 4y = 0$:
$5(-20/3) + 4(25/3) = -100/3 + 100/3 = 0$.
Thus,the line $5x + 4y = 0$ passes through the point of intersection.

(B) The equations of the angle bisectors are given by $\frac{3x - 4y + 7}{\sqrt{3^2 + (-4)^2}} = \pm \frac{12x + 5y - 2}{\sqrt{12^2 + 5^2}}$
$\frac{3x - 4y + 7}{5} = \pm \frac{12x + 5y - 2}{13}$
Case $1$: $13(3x - 4y + 7) = 5(12x + 5y - 2)$ $\Rightarrow 39x - 52y + 91 = 60x + 25y - 10
$ $\Rightarrow 21x + 77y - 101 = 0$
Case $2$: $13(3x - 4y + 7) = -5(12x + 5y - 2)$ $\Rightarrow 39x - 52y + 91 = -60x - 25y + 10
$ $\Rightarrow 99x - 27y + 81 = 0$ $\Rightarrow 11x - 3y + 9 = 0$
To identify the acute angle bisector,check the sign of $a_1a_2 + b_1b_2$. Here $a_1=3, b_1=-4, a_2=12, b_2=5$.
$a_1a_2 + b_1b_2 = (3)(12) + (-4)(5) = 36 - 20 = 16 > 0$.
Since the expression is positive,the equation $\frac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = -\frac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}}$ gives the acute angle bisector.
This corresponds to $13(3x - 4y + 7) = -5(12x + 5y - 2)$,which simplifies to $11x - 3y + 9 = 0$.

(C) Let $p$ be the length of the perpendicular from the vertex $(2, -1)$ to the base $x + y - 2 = 0$.
The formula for the perpendicular distance from a point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is $p = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
Substituting the values,we get $p = \frac{|1(2) + 1(-1) - 2|}{\sqrt{1^2 + 1^2}} = \frac{|2 - 1 - 2|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.
For an equilateral triangle with side length $a$,the altitude $p$ is given by $p = a \sin(60^{\circ}) = \frac{a\sqrt{3}}{2}$.
Equating the two expressions for $p$: $\frac{1}{\sqrt{2}} = \frac{a\sqrt{3}}{2}$.
Solving for $a$: $a = \frac{2}{\sqrt{2} \times \sqrt{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{2}{3}}$.

(A) The vertices of the triangle are the intersection points of the given lines.
Solving the pairs of equations:
$1$) $x + y = 4$ and $3x + y = 4$: Subtracting gives $2x = 0 \implies x = 0, y = 4$. Vertex $A(0, 4)$.
$2$) $3x + y = 4$ and $x + 3y = 4$: Solving gives $x = 1, y = 1$. Vertex $B(1, 1)$.
$3$) $x + y = 4$ and $x + 3y = 4$: Subtracting gives $2y = 0 \implies y = 0, x = 4$. Vertex $C(4, 0)$.
Now,calculate the lengths of the sides:
$AB = \sqrt{(1-0)^2 + (1-4)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{10}$
$BC = \sqrt{(4-1)^2 + (0-1)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{10}$
$AC = \sqrt{(4-0)^2 + (0-4)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{32} = 4\sqrt{2}$
Since $AB = BC$,the triangle is isosceles.

(A) Given line: $4x - 3y - 10 = 0 \implies x = \frac{3y + 10}{4}$.
Substituting this into the circle equation $x^2 + y^2 - 2x + 4y - 20 = 0$:
$(\frac{3y + 10}{4})^2 + y^2 - 2(\frac{3y + 10}{4}) + 4y - 20 = 0$.
Multiplying by $16$ to clear the denominator:
$(3y + 10)^2 + 16y^2 - 8(3y + 10) + 64y - 320 = 0$.
$9y^2 + 60y + 100 + 16y^2 - 24y - 80 + 64y - 320 = 0$.
$25y^2 + 100y - 300 = 0$.
$y^2 + 4y - 12 = 0$.
$(y + 6)(y - 2) = 0$.
So,$y = -6$ or $y = 2$.
For $y = -6$,$x = \frac{3(-6) + 10}{4} = \frac{-8}{4} = -2$.
For $y = 2$,$x = \frac{3(2) + 10}{4} = \frac{16}{4} = 4$.
The points are $(-2, -6)$ and $(4, 2)$.

(C) The equation of a chord of a circle $S = 0$ with mid-point $(x_1, y_1)$ is given by $T = S_1$,where $T = x x_1 + y y_1 - a^2$ and $S_1 = x_1^2 + y_1^2 - a^2$.
Substituting these into the formula:
$x x_1 + y y_1 - a^2 = x_1^2 + y_1^2 - a^2$
Adding $a^2$ to both sides,we get:
$x x_1 + y y_1 = x_1^2 + y_1^2$.

(B) The equation of a family of circles passing through the intersection of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 + \lambda S_2 = 0$.
Given $S_1: x^2 + y^2 + 13x - 3y = 0$ and $S_2: 2x^2 + 2y^2 + 4x - 7y - 25 = 0$.
The required equation is $(x^2 + y^2 + 13x - 3y) + \lambda (2x^2 + 2y^2 + 4x - 7y - 25) = 0$.
Since the circle passes through $(1, 1)$,we substitute $x = 1$ and $y = 1$:
$(1^2 + 1^2 + 13(1) - 3(1)) + \lambda (2(1^2) + 2(1^2) + 4(1) - 7(1) - 25) = 0$
$(1 + 1 + 13 - 3) + \lambda (2 + 2 + 4 - 7 - 25) = 0$
$12 + \lambda (-24) = 0$
$12 = 24\lambda \implies \lambda = \frac{1}{2}$.
Substituting $\lambda = \frac{1}{2}$ into the family equation:
$(x^2 + y^2 + 13x - 3y) + \frac{1}{2}(2x^2 + 2y^2 + 4x - 7y - 25) = 0$
Multiply by $2$:
$2x^2 + 2y^2 + 26x - 6y + 2x^2 + 2y^2 + 4x - 7y - 25 = 0$
$4x^2 + 4y^2 + 30x - 13y - 25 = 0$.

(C) The tangent to the parabola $y = x^2$ at $(2, 4)$ is given by $\frac{1}{2}(y + 4) = x(2)$,which simplifies to $4x - y - 4 = 0$.
Since the circle touches the parabola at $(2, 4)$,the centre $(h, k)$ of the circle must lie on the normal to the parabola at $(2, 4)$.
The slope of the tangent is $4$,so the slope of the normal is $-\frac{1}{4}$.
The equation of the normal at $(2, 4)$ is $y - 4 = -\frac{1}{4}(x - 2)$,which simplifies to $x + 4y = 18$.
Thus,$h + 4k = 18$ $(i)$.
Since the circle passes through $(0, 1)$ and $(2, 4)$,the distance from the centre $(h, k)$ to these points must be equal:
$(h - 2)^2 + (k - 4)^2 = (h - 0)^2 + (k - 1)^2$.
Expanding this,$h^2 - 4h + 4 + k^2 - 8k + 16 = h^2 + k^2 - 2k + 1$,which simplifies to $4h + 6k = 19$ $(ii)$.
Solving equations $(i)$ and $(ii)$ simultaneously: from $(i)$,$h = 18 - 4k$. Substituting into $(ii)$ gives $4(18 - 4k) + 6k = 19$,so $72 - 16k + 6k = 19$,which means $10k = 53$,so $k = \frac{53}{10}$.
Then $h = 18 - 4(\frac{53}{10}) = 18 - \frac{106}{5} = \frac{90 - 106}{5} = -\frac{16}{5}$.
The centre is $\left( -\frac{16}{5}, \frac{53}{10} \right)$.

(B) Using $L$'$H$ôpital's rule,we differentiate the numerator and denominator with respect to $x$:
$\mathop {\lim }\limits_{x \to 0} \frac{{{2^x} - 1}}{{{{(1 + x)}^{1/2}} - 1}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}({2^x} - 1)}}{{\frac{d}{{dx}}({{(1 + x)}^{1/2}} - 1)}}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{{2^x}\ln 2}}{{\frac{1}{2}{{(1 + x)}^{ - 1/2}}}}$
Substituting $x = 0$:
$= \frac{{{2^0}\ln 2}}{{\frac{1}{2}{{(1 + 0)}^{ - 1/2}}}} = \frac{{1 \cdot \ln 2}}{{\frac{1}{2}}} = 2\ln 2 = \ln({2^2}) = \ln 4$.

(B) The word $ASSASSIN$ contains $8$ letters: $A(2), S(4), I(1), N(1)$.
Total number of arrangements $ = \frac{8!}{2!4!1!1!} = \frac{40320}{2 \times 24} = 840$.
To ensure no two $S$ occur together,we first arrange the remaining letters $A, A, I, N$. The number of ways to arrange these $4$ letters is $\frac{4!}{2!} = 12$.
These $4$ letters create $5$ gaps (including ends): $\_ L_1 \_ L_2 \_ L_3 \_ L_4 \_$.
We need to place $4$ $S$'s in these $5$ gaps. The number of ways to choose $4$ gaps out of $5$ is $\binom{5}{4} = 5$.
Favorable arrangements $ = 12 \times 5 = 60$.
Required probability $ = \frac{60}{840} = \frac{1}{14}$.

(D) The inclusion-exclusion principle states that $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(AB) - P(BC) - P(AC) + P(ABC)$.
Substituting the given values: $P(A \cup B \cup C) = 0.3 + 0.4 + 0.8 - 0.08 - x - 0.28 + 0.09 = 1.23 - x$.
Given $P(A \cup B \cup C) \ge 0.75$,we have $1.23 - x \ge 0.75$,which implies $x \le 0.48$.
Also,for any events $B$ and $C$,$P(BC) \ge P(ABC) = 0.09$.
Furthermore,$P(B \cup C) = P(B) + P(C) - P(BC) = 0.4 + 0.8 - x = 1.2 - x \le 1$,so $x \ge 0.2$.
Additionally,$P(BC) \ge P(ABC) + P(AB) - P(A) = 0.09 + 0.08 - 0.3 = -0.13$ (always true).
However,the condition $P(BC) \ge P(ABC) + P(AC) - P(A) = 0.09 + 0.28 - 0.3 = 0.07$ must hold.
Also,$P(B \cap C \cap A^c) = P(BC) - P(ABC) = x - 0.09 \ge 0$,so $x \ge 0.09$.
Since $P(B \cup C) \le 1$,$x \ge 0.2$. Combining these,$0.2 \le x \le 0.48$. None of the options match this range.

(A) The total number of ways to draw $n$ cards from $52$ is $52 \times 51 \times \dots \times (52 - n + 1)$.
For the $n^{th}$ card to be the second ace,we must have exactly one ace in the first $(n-1)$ cards and an ace on the $n^{th}$ draw.
The number of ways to choose the position of the first ace in the first $(n-1)$ draws is $(n-1)$.
The number of ways to choose the first ace is $4$,and the remaining $(n-2)$ cards are chosen from the $48$ non-ace cards in $P(48, n-2)$ ways.
The number of ways to choose the second ace on the $n^{th}$ draw is $3$.
The total number of favorable outcomes is $(n-1) \times 4 \times P(48, n-2) \times 3$.
The total number of outcomes for $n$ draws is $P(52, n)$.
$P(N=n) = \frac{(n-1) \times 4 \times 3 \times \frac{48!}{(48-(n-2))!}}{\frac{52!}{(52-n)!}} = \frac{12(n-1) \times 48! \times (52-n)!}{52! \times (50-n)!} = \frac{12(n-1)(52-n)(51-n)}{52 \times 51 \times 50 \times 49} = \frac{(n-1)(52-n)(51-n)}{50 \times 49 \times 17 \times 13}$.

(C) In the expansion of $(1 + x)^{2n}$,the coefficient of $x^k$ is given by $^{2n}C_k$,where $0 \le k \le 2n$.
Given that the coefficients of $x^{3r}$ and $x^{r+2}$ are equal,we have $^{2n}C_{3r} = ^{2n}C_{r+2}$.
Using the property $^{n}C_a = ^{n}C_b$,which implies $a = b$ or $a + b = n$,we get two cases:
Case $1$: $3r = r + 2$ $\Rightarrow 2r = 2$ $\Rightarrow r = 1$. However,it is given that $r > 1$,so this case is rejected.
Case $2$: $3r + (r + 2) = 2n$ $\Rightarrow 4r + 2 = 2n$ $\Rightarrow 2n = 4r + 2$ $\Rightarrow n = 2r + 1$.
Thus,the correct relation is $n = 2r + 1$.

(D) The locus of point $P(x,y)$ is defined by the condition $|AP - BP| = 6$.
Here,$A = (0,4)$ and $B = (0,-4)$.
$AP = \sqrt{x^2 + (y-4)^2}$ and $BP = \sqrt{x^2 + (y+4)^2}$.
$|\sqrt{x^2 + (y-4)^2} - \sqrt{x^2 + (y+4)^2}| = 6$.
This is the definition of a hyperbola with foci at $(0,4)$ and $(0,-4)$.
The distance between foci $2ae = 8$,so $ae = 4$.
The constant difference $2a = 6$,so $a = 3$.
Since $b^2 = a^2(e^2 - 1) = (ae)^2 - a^2 = 4^2 - 3^2 = 16 - 9 = 7$.
The center is at $(0,0)$ and the transverse axis is along the $y$-axis.
The equation is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$,which is $\frac{y^2}{9} - \frac{x^2}{7} = 1$.

(D) Let the vertices be $A(5, -1)$,$B(-2, 3)$,and $C(h, k)$. The orthocentre $H$ is $(0, 0)$.
Since $CH \perp AB$,the slope of $AB$ is $m_{AB} = \frac{3 - (-1)}{-2 - 5} = \frac{4}{-7} = -\frac{4}{7}$.
The slope of the altitude $CH$ is $m_{CH} = -\frac{1}{m_{AB}} = \frac{7}{4}$.
Since $CH$ passes through $(0, 0)$ and $(h, k)$,its slope is $\frac{k}{h} = \frac{7}{4}$,which implies $7h - 4k = 0$ ---$(1)$.
Similarly,since $AH \perp BC$,the slope of $BC$ is $m_{BC} = \frac{k - 3}{h - (-2)} = \frac{k - 3}{h + 2}$.
The slope of the altitude $AH$ is $m_{AH} = \frac{0 - (-1)}{0 - 5} = -\frac{1}{5}$.
Since $AH \perp BC$,$m_{BC} \times m_{AH} = -1$,so $\left(\frac{k - 3}{h + 2}\right) \times \left(-\frac{1}{5}\right) = -1$.
$\frac{k - 3}{h + 2} = 5$ $\Rightarrow k - 3 = 5h + 10$ $\Rightarrow 5h - k + 13 = 0$ ---$(2)$.
Solving equations $(1)$ and $(2)$: From $(1)$,$k = \frac{7h}{4}$. Substituting into $(2)$: $5h - \frac{7h}{4} + 13 = 0$.
$\frac{20h - 7h}{4} + 13 = 0$ $\Rightarrow 13h = -52$ $\Rightarrow h = -4$.
Then $k = \frac{7(-4)}{4} = -7$.
Thus,the third vertex is $(-4, -7)$.

(A) Given the determinant equation: $\left| \begin{array}{ccc} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{array} \right| = 0$.
Applying the row operation $R_1 \to R_1 + R_2 + R_3$,we get:
$\left| \begin{array}{ccc} x+9 & x+9 & x+9 \\ 2 & x & 2 \\ 7 & 6 & x \end{array} \right| = 0$.
Taking $(x+9)$ as a common factor from $R_1$:
$(x+9) \left| \begin{array}{ccc} 1 & 1 & 1 \\ 2 & x & 2 \\ 7 & 6 & x \end{array} \right| = 0$.
Expanding the determinant:
$(x+9) [1(x^2 - 12) - 1(2x - 14) + 1(12 - 7x)] = 0$.
$(x+9) [x^2 - 12 - 2x + 14 + 12 - 7x] = 0$.
$(x+9) (x^2 - 9x + 14) = 0$.
Factoring the quadratic expression:
$(x+9) (x-2) (x-7) = 0$.
Thus,the roots are $x = -9, 2, 7$.
The other two roots are $2$ and $7$.

(B) Let $\theta = \cos^{-1} \frac{4}{5}$. Then $\cos \theta = \frac{4}{5}$.
Since $\tan \theta = \frac{\sqrt{1 - \cos^2 \theta}}{\cos \theta} = \frac{\sqrt{1 - (16/25)}}{4/5} = \frac{3/5}{4/5} = \frac{3}{4}$,we have $\theta = \tan^{-1} \frac{3}{4}$.
Substituting this into the expression,we get $\tan \left[ \tan^{-1} \frac{3}{4} + \tan^{-1} \frac{2}{3} \right]$.
Using the formula $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right)$,we have:
$\tan \left[ \tan^{-1} \left( \frac{3/4 + 2/3}{1 - (3/4)(2/3)} \right) \right]$
$= \tan \left[ \tan^{-1} \left( \frac{9/12 + 8/12}{1 - 6/12} \right) \right]$
$= \tan \left[ \tan^{-1} \left( \frac{17/12}{6/12} \right) \right]$
$= \tan \left[ \tan^{-1} \left( \frac{17}{6} \right) \right] = \frac{17}{6}$.

(B) Given vectors are $a = i + 2j + 2k$ and $b = 3i + 6j + 2k.$
First,calculate the magnitude of vector $b$:
$|b| = \sqrt{3^2 + 6^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7.$
Next,find the unit vector in the direction of $a$,denoted as $\hat{a}$:
$|a| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3.$
$\hat{a} = \frac{a}{|a|} = \frac{i + 2j + 2k}{3}.$
The required vector has magnitude $|b|$ and direction $\hat{a}$,so it is $|b|\hat{a}$:
$7 \times \left( \frac{i + 2j + 2k}{3} \right) = \frac{7}{3}(i + 2j + 2k).$
Thus,the correct option is $B$.

(C) Given the equation: $\overrightarrow {AO} + \overrightarrow {OB} = \overrightarrow {BO} + \overrightarrow {OC}$.
By the triangle law of vector addition,$\overrightarrow {AO} + \overrightarrow {OB} = \overrightarrow {AB}$.
Similarly,$\overrightarrow {BO} + \overrightarrow {OC} = \overrightarrow {BC}$.
Therefore,$\overrightarrow {AB} = \overrightarrow {BC}$.
This implies that the magnitude of vector $\overrightarrow {AB}$ is equal to the magnitude of vector $\overrightarrow {BC}$,i.e.,$|\overrightarrow {AB}| = |\overrightarrow {BC}|$.
Since the lengths of the two sides $AB$ and $BC$ are equal,the points $A, B, C$ form an isosceles triangle.

(A) Let the points be $A(60, 3)$,$B(40, -8)$,and $C(a, -52)$.
Since the points are collinear,the vector $\overrightarrow{AB}$ must be parallel to the vector $\overrightarrow{BC}$.
$\overrightarrow{AB} = (40 - 60)i + (-8 - 3)j = -20i - 11j$.
$\overrightarrow{BC} = (a - 40)i + (-52 - (-8))j = (a - 40)i - 44j$.
For collinearity,$\overrightarrow{AB} = k\overrightarrow{BC}$ for some scalar $k$.
$-20i - 11j = k((a - 40)i - 44j)$.
Comparing the coefficients of $j$: $-11 = -44k$,which gives $k = \frac{11}{44} = \frac{1}{4}$.
Comparing the coefficients of $i$: $-20 = k(a - 40)$.
Substituting $k = \frac{1}{4}$: $-20 = \frac{1}{4}(a - 40)$.
$-80 = a - 40$.
$a = -80 + 40 = -40$.

(C) Let the horizontal force be $\vec{F_1} = P_1 \hat{i}$ and the inclined force be $\vec{F_2}$.
The resultant force $\vec{R}$ is in the vertical direction,so $\vec{R} = P \hat{j}$.
From the parallelogram law of vector addition,$\vec{R} = \vec{F_1} + \vec{F_2}$,so $\vec{F_2} = \vec{R} - \vec{F_1} = -P_1 \hat{i} + P \hat{j}$.
The angle between $\vec{F_2}$ and the vertical axis (y-axis) is $60^\circ$.
Using the dot product formula: $\cos 60^\circ = \frac{\vec{F_2} \cdot \hat{j}}{|\vec{F_2}| |\hat{j}|}$.
$\frac{1}{2} = \frac{(-P_1 \hat{i} + P \hat{j}) \cdot \hat{j}}{\sqrt{(-P_1)^2 + P^2}} = \frac{P}{\sqrt{P_1^2 + P^2}}$.
Squaring both sides: $\frac{1}{4} = \frac{P^2}{P_1^2 + P^2} \Rightarrow P_1^2 + P^2 = 4P^2 \Rightarrow P_1^2 = 3P^2 \Rightarrow P_1 = P\sqrt{3}$.
The magnitude of the inclined force is $|\vec{F_2}| = \sqrt{P_1^2 + P^2} = \sqrt{3P^2 + P^2} = \sqrt{4P^2} = 2P$.
Thus,the two forces are $P\sqrt{3}$ and $2P$.

(A) Let the points be $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$.
Define two vectors in the plane: $\vec{AB} = (2-1)i + (0-(-1))j + (-1-2)k = i + j - 3k$ and $\vec{AC} = (0-1)i + (2-(-1))j + (1-2)k = -i + 3j - k$.
Alternatively,using the provided solution vectors: $\vec{u} = \vec{AB} = i + j - 3k$ and $\vec{v} = \vec{BC} = (0-2)i + (2-0)j + (1-(-1))k = -2i + 2j + 2k$.
The normal vector $\vec{n}$ to the plane is given by the cross product $\vec{u} \times \vec{v}$:
$\vec{n} = \begin{vmatrix} i & j & k \\ 1 & 1 & -3 \\ -2 & 2 & 2 \end{vmatrix} = i(2 - (-6)) - j(2 - 6) + k(2 - (-2)) = 8i + 4j + 4k$.
The unit vector is $\hat{n} = \pm \frac{\vec{n}}{|\vec{n}|} = \pm \frac{8i + 4j + 4k}{\sqrt{8^2 + 4^2 + 4^2}} = \pm \frac{8i + 4j + 4k}{\sqrt{64 + 16 + 16}} = \pm \frac{8i + 4j + 4k}{\sqrt{96}} = \pm \frac{8i + 4j + 4k}{4\sqrt{6}} = \pm \frac{2i + j + k}{\sqrt{6}}$.

(B) The vertices of the triangle are $A(1, -1, 2)$,$B(2, 1, -1)$,and $C(3, -1, 2)$.
First,we find the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$:
$\overrightarrow{AB} = (2-1)\hat{i} + (1-(-1))\hat{j} + (-1-2)\hat{k} = \hat{i} + 2\hat{j} - 3\hat{k}$
$\overrightarrow{AC} = (3-1)\hat{i} + (-1-(-1))\hat{j} + (2-2)\hat{k} = 2\hat{i} + 0\hat{j} + 0\hat{k} = 2\hat{i}$
The area of the triangle is given by $\frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}|$.
Calculating the cross product:
$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 0 & 0 \end{vmatrix} = \hat{i}(0 - 0) - \hat{j}(0 - (-6)) + \hat{k}(0 - 4) = -6\hat{j} - 4\hat{k}$.
Now,find the magnitude:
$|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{(-6)^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$.
Therefore,the area of the triangle is $\frac{1}{2} \times 2\sqrt{13} = \sqrt{13}$ square units.

(C) Let the vectors be $\vec{a} = 2i - 3j + 0k$,$\vec{b} = i + j - k$,and $\vec{c} = 3i + 0j - k$.
The volume of a parallelepiped with concurrent edges $\vec{a}, \vec{b}, \vec{c}$ is given by the scalar triple product $|[\vec{a} \, \vec{b} \, \vec{c}]|$.
This is calculated as the determinant of the matrix formed by the components of the vectors:
$V = |\det \begin{bmatrix} 2 & -3 & 0 \\ 1 & 1 & -1 \\ 3 & 0 & -1 \end{bmatrix}|$
Expanding the determinant along the first row:
$V = |2(1(-1) - (-1)(0)) - (-3)(1(-1) - (-1)(3)) + 0|$
$V = |2(-1) + 3(-1 + 3)|$
$V = |-2 + 3(2)|$
$V = |-2 + 6| = |4| = 4$.
Thus,the volume of the parallelepiped is $4$ cubic units.

(A) Given that $x$ is a non-zero vector such that $x \cdot a = 0, x \cdot b = 0$,and $x \cdot c = 0$.
This implies that the vector $x$ is perpendicular to each of the vectors $a, b$,and $c$.
If $x$ is a non-zero vector perpendicular to $a, b$,and $c$,then $a, b$,and $c$ must lie in a plane perpendicular to $x$.
Therefore,the vectors $a, b$,and $c$ are coplanar.
For any three coplanar vectors,their scalar triple product is zero,i.e.,$[a, b, c] = 0$.

(D) Given $f(x) = \cos (\log x)$.
Then $f(y) = \cos (\log y)$.
We need to evaluate $f(x)f(y) - \frac{1}{2}[f(x/y) + f(xy)]$.
Using the property $\log(a/b) = \log a - \log b$ and $\log(ab) = \log a + \log b$:
$f(x/y) = \cos(\log(x/y)) = \cos(\log x - \log y)$
$f(xy) = \cos(\log(xy)) = \cos(\log x + \log y)$
Now,substitute these into the expression:
$f(x)f(y) - \frac{1}{2}[\cos(\log x - \log y) + \cos(\log x + \log y)]$
Using the trigonometric identity $\cos(A - B) + \cos(A + B) = 2\cos A \cos B$,where $A = \log x$ and $B = \log y$:
$= \cos(\log x)\cos(\log y) - \frac{1}{2}[2\cos(\log x)\cos(\log y)]$
$= \cos(\log x)\cos(\log y) - \cos(\log x)\cos(\log y) = 0$.

(C) Given $f(x) = (a - x^n)^{1/n}$.
To find $f[f(x)]$,we substitute $f(x)$ into the function $f$:
$f[f(x)] = (a - [f(x)]^n)^{1/n}$
Substitute the expression for $f(x)$:
$f[f(x)] = (a - [(a - x^n)^{1/n}]^n)^{1/n}$
Simplify the inner term:
$f[f(x)] = (a - (a - x^n))^{1/n}$
$f[f(x)] = (a - a + x^n)^{1/n}$
$f[f(x)] = (x^n)^{1/n}$
$f[f(x)] = x$.

(B) For the function $f(x)$ to be continuous at $x = 0$,the value $f(0)$ must be equal to $\lim_{x \to 0} f(x)$.
Given $f(x) = \frac{\log(1 + ax) - \log(1 - bx)}{x}$.
Using the standard limit $\lim_{x \to 0} \frac{\log(1 + kx)}{x} = k$,we can rewrite the limit as:
$\lim_{x \to 0} f(x) = \lim_{x \to 0} \left( \frac{\log(1 + ax)}{x} - \frac{\log(1 - bx)}{x} \right)$
$= \lim_{x \to 0} \left( a \cdot \frac{\log(1 + ax)}{ax} - (-b) \cdot \frac{\log(1 - bx)}{-bx} \right)$
$= a(1) + b(1) = a + b$.
Therefore,$f(0) = a + b$.

(B) We use the standard integral formula: $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + c$.
Rewrite the integrand as:
$\frac{x - 1}{(x + 1)^3} = \frac{(x + 1) - 2}{(x + 1)^3} = \frac{1}{(x + 1)^2} - \frac{2}{(x + 1)^3}$.
Let $f(x) = \frac{1}{(x + 1)^2}$.
Then $f'(x) = \frac{d}{dx} [(x + 1)^{-2}] = -2(x + 1)^{-3} = -\frac{2}{(x + 1)^3}$.
Thus,the integral becomes:
$\int e^x \left[ \frac{1}{(x + 1)^2} - \frac{2}{(x + 1)^3} \right] \, dx = \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + c$.
Substituting $f(x)$ back,we get:
$\frac{e^x}{(x + 1)^2} + c$.

(A) Let $I = \int_0^{\pi /4} {\frac{{\sin x + \cos x}}{{9 + 16\sin 2x}}\,dx}$.
Substitute $t = \sin x - \cos x$,then $dt = (\cos x + \sin x)dx$.
When $x = 0$,$t = \sin 0 - \cos 0 = -1$.
When $x = \pi /4$,$t = \sin(\pi /4) - \cos(\pi /4) = 0$.
Also,$t^2 = (\sin x - \cos x)^2 = \sin^2 x + \cos^2 x - 2\sin x \cos x = 1 - \sin 2x$,so $\sin 2x = 1 - t^2$.
Substituting these into the integral:
$I = \int_{-1}^0 \frac{dt}{9 + 16(1 - t^2)} = \int_{-1}^0 \frac{dt}{9 + 16 - 16t^2} = \int_{-1}^0 \frac{dt}{25 - 16t^2}$.
Using the formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right|$,we have:
$I = \frac{1}{16} \int_{-1}^0 \frac{dt}{(5/4)^2 - t^2} = \frac{1}{16} \cdot \frac{1}{2(5/4)} \left[ \log \left| \frac{5/4 + t}{5/4 - t} \right| \right]_{-1}^0$.
$I = \frac{1}{16} \cdot \frac{2}{5} \left[ \log \left| \frac{5 + 4t}{5 - 4t} \right| \right]_{-1}^0 = \frac{1}{40} \left[ \log(1) - \log \left| \frac{5 - 4}{5 + 4} \right| \right]$.
$I = \frac{1}{40} [0 - \log(1/9)] = \frac{1}{40} \log(9) = \frac{1}{40} \log(3^2) = \frac{2}{40} \log 3 = \frac{1}{20} \log 3$.

(C) Let $I = \int_0^{\pi /2} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}} \, dx$ ..... $(i)$
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:
$I = \int_0^{\pi /2} \frac{\sqrt{\cot(\pi/2 - x)}}{\sqrt{\cot(\pi/2 - x)} + \sqrt{\tan(\pi/2 - x)}} \, dx$
Since $\cot(\pi/2 - x) = \tan x$ and $\tan(\pi/2 - x) = \cot x$,we have:
$I = \int_0^{\pi /2} \frac{\sqrt{\tan x}}{\sqrt{\tan x} + \sqrt{\cot x}} \, dx$ ..... $(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^{\pi /2} \frac{\sqrt{\cot x} + \sqrt{\tan x}}{\sqrt{\cot x} + \sqrt{\tan x}} \, dx$
$2I = \int_0^{\pi /2} 1 \, dx$
$2I = [x]_0^{\pi /2} = \frac{\pi}{2} - 0 = \frac{\pi}{2}$
$I = \frac{\pi}{4}$

(B) Let the ordinate at $x = a$ divide the area into two equal parts.
The total area $A$ bounded by the curve $y = 1 + \frac{8}{x^2}$,the $x$-axis,and the lines $x = 2$ and $x = 4$ is given by:
$A = \int_{2}^{4} \left( 1 + \frac{8}{x^2} \right) dx$
$A = \left[ x - \frac{8}{x} \right]_{2}^{4}$
$A = \left( 4 - \frac{8}{4} \right) - \left( 2 - \frac{8}{2} \right) = (4 - 2) - (2 - 4) = 2 - (-2) = 4$ square units.
Since the ordinate $x = a$ divides this area into two equal parts,the area from $x = 2$ to $x = a$ must be half of the total area,which is $4 / 2 = 2$.
$\int_{2}^{a} \left( 1 + \frac{8}{x^2} \right) dx = 2$
$\left[ x - \frac{8}{x} \right]_{2}^{a} = 2$
$\left( a - \frac{8}{a} \right) - \left( 2 - \frac{8}{2} \right) = 2$
$a - \frac{8}{a} - (2 - 4) = 2$
$a - \frac{8}{a} + 2 = 2$
$a - \frac{8}{a} = 0$
$a^2 - 8 = 0$
$a^2 = 8$
$a = \pm 2\sqrt{2}$.
Since the region is in the first quadrant where $x > 0$,we have $a = 2\sqrt{2}$.

(D) Let $X$ be the random variable representing the number on a drawn ticket. Since the tickets are drawn with replacement,each draw is independent.
For each draw,the probability that the number on the ticket is less than or equal to $9$ is $P(X \le 9) = \frac{9}{15} = \frac{3}{5}$.
The probability that the greatest number on the $7$ drawn tickets is $9$ is given by $P(\text{max} = 9) = P(\text{max} \le 9) - P(\text{max} \le 8)$.
$P(\text{max} \le 9) = (\frac{9}{15})^7 = (\frac{3}{5})^7$.
$P(\text{max} \le 8) = (\frac{8}{15})^7$.
Therefore,the required probability is $(\frac{9}{15})^7 - (\frac{8}{15})^7 = (\frac{3}{5})^7 - (\frac{8}{15})^7$.
Since this result is not among the given options,the correct choice is $D$.