IIT JEE 1973 Mathematics Question Paper with Answer and Solution

(A) Let $x = 0.4232323...$ (Equation $1$)
Multiply by $10$ on both sides: $10x = 4.232323...$ (Equation $2$)
Multiply Equation $1$ by $1000$: $1000x = 423.232323...$ (Equation $3$)
Subtract Equation $2$ from Equation $3$:
$1000x - 10x = 423.232323... - 4.232323...$
$990x = 419$
$x = \frac{419}{990}$.

(A) The $n^{th}$ term of the series is given by $T_n = n(2n + 1)^2$.
Expanding this,we get $T_n = n(4n^2 + 4n + 1) = 4n^3 + 4n^2 + n$.
To find the sum of $20$ terms,we calculate $S_{20} = \sum_{n=1}^{20} (4n^3 + 4n^2 + n) = 4\sum_{n=1}^{20} n^3 + 4\sum_{n=1}^{20} n^2 + \sum_{n=1}^{20} n$.
Using the standard summation formulas:
$\sum_{n=1}^{N} n^3 = [\frac{N(N+1)}{2}]^2 = [\frac{20 \cdot 21}{2}]^2 = 210^2 = 44100$.
$\sum_{n=1}^{N} n^2 = \frac{N(N+1)(2N+1)}{6} = \frac{20 \cdot 21 \cdot 41}{6} = 2870$.
$\sum_{n=1}^{N} n = \frac{N(N+1)}{2} = \frac{20 \cdot 21}{2} = 210$.
Substituting these values: $S_{20} = 4(44100) + 4(2870) + 210 = 176400 + 11480 + 210 = 188090$.

(A) The vertices of the quadrilateral are formed by the intersection of the lines:
$1$. $x=0$ and $y=0$ intersect at $B(0,0)$.
$2$. $x=0$ and $6x+y=3$ intersect at $A(0,3)$.
$3$. $y=0$ and $x+y=1$ intersect at $C(1,0)$.
$4$. $x+y=1$ and $6x+y=3$ intersect at $D(\frac{2}{5}, \frac{3}{5})$.
The diagonal passing through the origin $B(0,0)$ must also pass through the vertex $D(\frac{2}{5}, \frac{3}{5})$.
The equation of a line passing through $(0,0)$ and $(x_1, y_1)$ is $y = \frac{y_1}{x_1}x$.
Substituting $x_1 = \frac{2}{5}$ and $y_1 = \frac{3}{5}$:
$y = \frac{3/5}{2/5}x$
$y = \frac{3}{2}x$
$2y = 3x$
$3x - 2y = 0$.

(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 $a$: $a = \frac{2}{\sqrt{2} \times \sqrt{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{2}{3}}$.

(B) The given lines are $ax + by + c = 0$,$ax + by - c = 0$,$ax - by + c = 0$,and $ax - by - c = 0$.
These can be rewritten in intercept form as $\frac{x}{\pm c/a} + \frac{y}{\pm c/b} = 1$.
The vertices of the parallelogram are the intersection points of these lines:
$A(\frac{c}{a}, 0)$,$B(0, \frac{c}{b})$,$C(-\frac{c}{a}, 0)$,and $D(0, -\frac{c}{b})$.
The diagonals of this quadrilateral are $AC$ and $BD$,which lie along the $x$-axis and $y$-axis respectively.
Since the axes are perpendicular,the diagonals are perpendicular,making the parallelogram a rhombus.
The length of diagonal $AC = |\frac{c}{a} - (-\frac{c}{a})| = \frac{2c}{a}$.
The length of diagonal $BD = |\frac{c}{b} - (-\frac{c}{b})| = \frac{2c}{b}$.
The area of the rhombus is $\frac{1}{2} \times AC \times BD = \frac{1}{2} \times \frac{2c}{a} \times \frac{2c}{b} = \frac{2c^2}{ab}$.

(A) For the first circle $x^2 + y^2 - 2x - 4y = 0$,the center $C_1 = (1, 2)$ and radius $R_1 = \sqrt{1^2 + 2^2 - 0} = \sqrt{5}$.
For the second circle $x^2 + y^2 - 8y - 4 = 0$,the center $C_2 = (0, 4)$ and radius $R_2 = \sqrt{0^2 + 4^2 - (-4)} = \sqrt{20} = 2\sqrt{5}$.
The distance between the centers $C_1$ and $C_2$ is $C_1C_2 = \sqrt{(1-0)^2 + (2-4)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1+4} = \sqrt{5}$.
We observe that $|R_2 - R_1| = |2\sqrt{5} - \sqrt{5}| = \sqrt{5}$.
Since $C_1C_2 = |R_2 - R_1|$,the circles touch each other internally.

(C) Given limit: $L = \mathop {\lim }\limits_{x \to \pi /2} \frac{2x - \pi}{\cos x}$.
Since the form is $\frac{0}{0}$,we apply $L$-Hospital's rule by differentiating the numerator and denominator with respect to $x$:
$L = \mathop {\lim }\limits_{x \to \pi /2} \frac{\frac{d}{dx}(2x - \pi)}{\frac{d}{dx}(\cos x)}$
$L = \mathop {\lim }\limits_{x \to \pi /2} \frac{2}{-\sin x}$
Substituting $x = \pi /2$:
$L = \frac{2}{-\sin(\pi /2)} = \frac{2}{-1} = -2$.
Thus,the correct option is $C$.

(C) Since the $5$ balls are identical and the $10$ boxes are identical,distributing the balls such that no box contains more than one ball is equivalent to selecting $5$ boxes out of $10$ to place the balls in.
Because the boxes are identical,the arrangement does not depend on the specific labels of the boxes,but rather on the selection of the positions.
However,in standard combinatorial problems of this type where boxes are considered distinct positions,the number of ways is given by the combination formula $C(n, r) = \binom{n}{r}$.
If the boxes are truly identical (indistinguishable),there is only $1$ way to place the balls. Given the options provided,the question implies the boxes are distinct positions,leading to the calculation $\binom{10}{5} = \frac{10!}{5!5!}$.