IIT JEE 1993 Mathematics Question Paper with Answer and Solution

(B) The total number of points is $m + n + k$. The number of ways to select $3$ points from these is $^{m+n+k}C_3$.
However,selecting $3$ points from the same line does not form a triangle. The number of such collinear sets is $^mC_3 + ^nC_3 + ^kC_3$.
Therefore,the required number of triangles is $^{m+n+k}C_3 - ^mC_3 - ^nC_3 - ^kC_3$.

(C) To find the number of divisors of $9600$,we first find its prime factorization.
$9600 = 96 \times 100 = (32 \times 3) \times (4 \times 25) = (2^5 \times 3) \times (2^2 \times 5^2) = 2^7 \times 3^1 \times 5^2$.
If a number $N$ is expressed as $N = p_1^{a} \times p_2^{b} \times p_3^{c}$,then the total number of divisors is given by $(a + 1)(b + 1)(c + 1)$.
Here,$a = 7$,$b = 1$,and $c = 2$.
Therefore,the number of divisors $= (7 + 1)(1 + 1)(2 + 1) = 8 \times 2 \times 3 = 48$.

(C) Given expression is $(3 + 2x)^{50} = 3^{50} (1 + \frac{2x}{3})^{50}$.
Substituting $x = \frac{1}{5}$,we get $3^{50} (1 + \frac{2}{15})^{50} = 3^{50} (1 + \frac{2}{15})^{50}$.
Let $T_{r+1}$ be the $(r+1)^{th}$ term.
We know that $\frac{T_{r+1}}{T_r} = \frac{n-r+1}{r} \cdot |\frac{a_2}{a_1}|$.
Here $n=50$,$a_1=1$,$a_2=\frac{2}{15}$.
So,$\frac{T_{r+1}}{T_r} = \frac{50-r+1}{r} \cdot \frac{2}{15} = \frac{51-r}{r} \cdot \frac{2}{15}$.
For the term to be the largest,we set $\frac{T_{r+1}}{T_r} \ge 1$.
$\frac{2(51-r)}{15r} \ge 1$ $\Rightarrow 102 - 2r \ge 15r$ $\Rightarrow 102 \ge 17r$ $\Rightarrow r \le 6$.
Since $r=6$ satisfies the condition,$T_{6+1} = T_7$ is the largest term.

(C) Given the equation: $\frac{2\cos A}{a} + \frac{\cos B}{b} + \frac{2\cos C}{c} = \frac{a}{bc} + \frac{b}{ca}$
Using the cosine rule $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$,$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$,and $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$:
$\frac{2(b^2 + c^2 - a^2)}{2abc} + \frac{a^2 + c^2 - b^2}{2abc} + \frac{2(a^2 + b^2 - c^2)}{2abc} = \frac{a^2 + b^2}{abc}$
Multiplying by $2abc$ on both sides:
$2(b^2 + c^2 - a^2) + (a^2 + c^2 - b^2) + 2(a^2 + b^2 - c^2) = 2(a^2 + b^2)$
$2b^2 + 2c^2 - 2a^2 + a^2 + c^2 - b^2 + 2a^2 + 2b^2 - 2c^2 = 2a^2 + 2b^2$
$3b^2 + c^2 + a^2 = 2a^2 + 2b^2$
$b^2 - a^2 + c^2 = 0$
$a^2 = b^2 + c^2$
Since $a^2 = b^2 + c^2$,by the converse of the Pythagorean theorem,$\angle A = 90^o$.

(D) Given $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$.
Using the Sine Rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Substituting these,we get $\frac{\cos A}{2R \sin A} = \frac{\cos B}{2R \sin B} = \frac{\cos C}{2R \sin C}$.
This simplifies to $\cot A = \cot B = \cot C$.
Since $A, B, C$ are angles of a triangle,$A = B = C = 60^\circ$.
Thus,$\Delta ABC$ is an equilateral triangle.
The area of an equilateral triangle is given by $\text{Area} = \frac{\sqrt{3}}{4} a^2$.
Given $a = 2$,$\text{Area} = \frac{\sqrt{3}}{4} (2)^2 = \sqrt{3}$.

(B) Let the vertices be $A = \left( 2, \frac{\sqrt{3} - 1}{2} \right)$,$B = \left( \frac{1}{2}, -\frac{1}{2} \right)$,and $C = \left( 2, -\frac{1}{2} \right)$.
Observe that the side $AC$ is vertical (both $x$-coordinates are $2$) and the side $BC$ is horizontal (both $y$-coordinates are $-\frac{1}{2}$).
Since $AC$ is perpendicular to $BC$,the triangle is a right-angled triangle with the right angle at vertex $C = \left( 2, -\frac{1}{2} \right)$.
In a right-angled triangle,the orthocentre is the vertex where the right angle is located.
Therefore,the orthocentre is $\left( 2, -\frac{1}{2} \right)$.

(A) Let the line passing through $A(-5, -4)$ make an angle $\theta$ with the $x$-axis. The equation of the line is $\frac{x + 5}{\cos \theta} = \frac{y + 4}{\sin \theta} = r$.
For point $B$ on $x + 3y + 2 = 0$,we have $(r_1 \cos \theta - 5) + 3(r_1 \sin \theta - 4) + 2 = 0$,which gives $r_1(\cos \theta + 3 \sin \theta) = 15$,so $\frac{15}{AB} = \cos \theta + 3 \sin \theta$.
For point $C$ on $2x + y + 4 = 0$,we have $2(r_2 \cos \theta - 5) + (r_2 \sin \theta - 4) + 4 = 0$,which gives $r_2(2 \cos \theta + \sin \theta) = 10$,so $\frac{10}{AC} = 2 \cos \theta + \sin \theta$.
For point $D$ on $x - y - 5 = 0$,we have $(r_3 \cos \theta - 5) - (r_3 \sin \theta - 4) - 5 = 0$,which gives $r_3(\cos \theta - \sin \theta) = 6$,so $\frac{6}{AD} = \cos \theta - \sin \theta$.
Substituting these into the given relation $(\frac{15}{AB})^2 + (\frac{10}{AC})^2 = (\frac{6}{AD})^2$,we get $(\cos \theta + 3 \sin \theta)^2 + (2 \cos \theta + \sin \theta)^2 = (\cos \theta - \sin \theta)^2$.
Expanding this,$\cos^2 \theta + 9 \sin^2 \theta + 6 \sin \theta \cos \theta + 4 \cos^2 \theta + \sin^2 \theta + 4 \sin \theta \cos \theta = \cos^2 \theta + \sin^2 \theta - 2 \sin \theta \cos \theta$.
Simplifying,$4 \cos^2 \theta + 9 \sin^2 \theta + 12 \sin \theta \cos \theta = 0$,which is $(2 \cos \theta + 3 \sin \theta)^2 = 0$.
Thus,$2 \cos \theta + 3 \sin \theta = 0$,so $\tan \theta = -\frac{2}{3}$.
The equation of the line is $y + 4 = -\frac{2}{3}(x + 5)$,which simplifies to $2x + 3y + 22 = 0$.

(D) Let the centre of the circle be $(h, k)$ and its radius be $r$.
Since the circle touches the $y$-axis,the radius $r = |h|$. Assuming $h > 0$,we have $r = h$.
The given circle is $x^2 + y^2 - 6x - 6y + 14 = 0$. Its centre $C_1$ is $(3, 3)$ and its radius $R_1 = \sqrt{3^2 + 3^2 - 14} = \sqrt{18 - 14} = \sqrt{4} = 2$.
Since the circles touch externally,the distance between their centres $C(h, k)$ and $C_1(3, 3)$ is equal to the sum of their radii:
$CC_1 = R_1 + r$
$\sqrt{(h - 3)^2 + (k - 3)^2} = 2 + h$
Squaring both sides:
$(h - 3)^2 + (k - 3)^2 = (2 + h)^2$
$h^2 - 6h + 9 + k^2 - 6k + 9 = 4 + 4h + h^2$
$k^2 - 6k + 18 - 6h = 4 + 4h$
$k^2 - 10h - 6k + 14 = 0$
Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 - 10x - 6y + 14 = 0$.

(C) Let $L = \mathop {\lim }\limits_{x \to 0} {\left\{ {\tan \left( {\frac{\pi }{4} + x} \right)} \right\}^{1/x}}$.
Using the formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we get $\tan(\frac{\pi}{4} + x) = \frac{1 + \tan x}{1 - \tan x}$.
So,$L = \mathop {\lim }\limits_{x \to 0} {\left( {\frac{{1 + \tan x}}{{1 - \tan x}}} \right)^{1/x}}$.
This is of the form $1^\infty$,so $L = e^{\mathop {\lim }\limits_{x \to 0} \frac{1}{x} (\frac{1 + \tan x}{1 - \tan x} - 1)}$.
$L = e^{\mathop {\lim }\limits_{x \to 0} \frac{1}{x} (\frac{1 + \tan x - 1 + \tan x}{1 - \tan x})} = e^{\mathop {\lim }\limits_{x \to 0} \frac{2 \tan x}{x(1 - \tan x)}}$.
Since $\mathop {\lim }\limits_{x \to 0} \frac{\tan x}{x} = 1$,we have $L = e^{2 \times 1 \times \frac{1}{1 - 0}} = e^2$.

(A) Let $X_1, X_2, X_3, X_4$ be the outcomes of the four rolls.
We want the probability that $2 \le X_i \le 5$ for all $i = 1, 2, 3, 4$.
For each roll,the possible outcomes are $\{1, 2, 3, 4, 5, 6\}$,so there are $6$ total outcomes.
The favorable outcomes for each roll are $\{2, 3, 4, 5\}$,which gives $4$ favorable outcomes.
The probability of a single roll resulting in a value between $2$ and $5$ inclusive is $p = \frac{4}{6} = \frac{2}{3}$.
Since the four rolls are independent,the probability that all four rolls result in values between $2$ and $5$ is $p^4 = (\frac{2}{3})^4 = \frac{16}{81}$.

(B) Given $x = 1 + \cos^2\phi + \cos^4\phi + \dots = \frac{1}{1 - \cos^2\phi} = \frac{1}{\sin^2\phi}$.
Given $y = 1 + \sin^2\phi + \sin^4\phi + \dots = \frac{1}{1 - \sin^2\phi} = \frac{1}{\cos^2\phi}$.
Given $z = 1 + \cos^2\phi \sin^2\phi + \cos^4\phi \sin^4\phi + \dots = \frac{1}{1 - \cos^2\phi \sin^2\phi}$.
Now,$xy = \frac{1}{\sin^2\phi \cos^2\phi}$.
Then $xy + z = \frac{1}{\sin^2\phi \cos^2\phi} + \frac{1}{1 - \cos^2\phi \sin^2\phi} = \frac{1 - \cos^2\phi \sin^2\phi + \sin^2\phi \cos^2\phi}{\sin^2\phi \cos^2\phi (1 - \cos^2\phi \sin^2\phi)} = \frac{1}{\sin^2\phi \cos^2\phi (1 - \cos^2\phi \sin^2\phi)}$.
Also,$xyz = \left(\frac{1}{\sin^2\phi}\right) \left(\frac{1}{\cos^2\phi}\right) \left(\frac{1}{1 - \cos^2\phi \sin^2\phi}\right) = \frac{1}{\sin^2\phi \cos^2\phi (1 - \cos^2\phi \sin^2\phi)}$.
Thus,$xyz = xy + z$.

(B) Given $k = \sin \frac{\pi}{18} \sin \frac{5\pi}{18} \sin \frac{7\pi}{18}$.
Using the identity $\sin \theta = \cos \left( \frac{\pi}{2} - \theta \right)$:
$k = \cos \left( \frac{\pi}{2} - \frac{\pi}{18} \right) \cos \left( \frac{\pi}{2} - \frac{5\pi}{18} \right) \cos \left( \frac{\pi}{2} - \frac{7\pi}{18} \right)$
$k = \cos \frac{8\pi}{18} \cos \frac{4\pi}{18} \cos \frac{2\pi}{18} = \cos \frac{4\pi}{9} \cos \frac{2\pi}{9} \cos \frac{\pi}{9}$.
Using the formula $\cos \theta \cos 2\theta \cos 4\theta = \frac{\sin 8\theta}{8 \sin \theta}$ where $\theta = \frac{\pi}{9}$:
$k = \frac{\sin \left( 8 \cdot \frac{\pi}{9} \right)}{8 \sin \frac{\pi}{9}} = \frac{\sin \frac{8\pi}{9}}{8 \sin \frac{\pi}{9}}$.
Since $\sin \frac{8\pi}{9} = \sin \left( \pi - \frac{\pi}{9} \right) = \sin \frac{\pi}{9}$:
$k = \frac{\sin \frac{\pi}{9}}{8 \sin \frac{\pi}{9}} = \frac{1}{8}$.

(C) Given $x = \log_3 5$ and $y = \log_{17} 25 = 2 \log_{17} 5$.
Taking reciprocals:
$\frac{1}{x} = \log_5 3 = \log_5 (9^{1/2}) = \frac{1}{2} \log_5 9$.
$\frac{1}{y} = \frac{1}{2} \log_5 17$.
Since $17 > 9$,it follows that $\log_5 17 > \log_5 9$.
Therefore,$\frac{1}{2} \log_5 17 > \frac{1}{2} \log_5 9$,which implies $\frac{1}{y} > \frac{1}{x}$.
Since $x$ and $y$ are both positive,$\frac{1}{y} > \frac{1}{x}$ implies $x > y$.

(A) Let the determinant be $\Delta = \left| \begin{array}{ccc} 1 & \log_x y & \log_x z \\ \log_y x & 1 & \log_y z \\ \log_z x & \log_z y & 1 \end{array} \right|$.
Using the property $\log_a b = \frac{\ln b}{\ln a}$,we can rewrite the elements as:
$\Delta = \left| \begin{array}{ccc} 1 & \frac{\ln y}{\ln x} & \frac{\ln z}{\ln x} \\ \frac{\ln x}{\ln y} & 1 & \frac{\ln z}{\ln y} \\ \frac{\ln x}{\ln z} & \frac{\ln y}{\ln z} & 1 \end{array} \right|$.
Taking $\frac{1}{\ln x}$ common from $R_1$,$\frac{1}{\ln y}$ from $R_2$,and $\frac{1}{\ln z}$ from $R_3$ is not directly helpful,so let's multiply $R_1$ by $\ln x$,$R_2$ by $\ln y$,and $R_3$ by $\ln z$ and divide the determinant by $(\ln x \ln y \ln z)$:
$\Delta = \frac{1}{\ln x \ln y \ln z} \left| \begin{array}{ccc} \ln x & \ln y & \ln z \\ \ln x & \ln y & \ln z \\ \ln x & \ln y & \ln z \end{array} \right|$.
Since all three rows are identical,the value of the determinant is $0$.

(A) Let $\cos ^{ - 1}x = \theta$. Then $x = \cos \theta$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - x^2}$.
Now,$\tan \theta = \frac{\sin \theta}{\cos \theta}$.
Substituting the values,we get $\tan \theta = \frac{\sqrt{1 - x^2}}{x}$.
Therefore,$\tan (\cos ^{ - 1}x) = \frac{\sqrt{1 - x^2}}{x}$.

(B) Since the vectors are coplanar,their scalar triple product must be zero.
$\begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} = 0$
Applying the column operation $C_2 \to C_2 - C_1$:
$\begin{vmatrix} a & 0 & c \\ 1 & -1 & 1 \\ c & 0 & b \end{vmatrix} = 0$
Expanding along the second column:
$-(-1) \begin{vmatrix} a & c \\ c & b \end{vmatrix} = 0$
$ab - c^2 = 0 \Rightarrow c^2 = ab$
Thus,$c = \sqrt{ab}$,which is the geometric mean of $a$ and $b$.

(A) Any vector $r$ in the plane of $b$ and $c$ can be written as $r = \lambda b + \mu c$. For simplicity,let $r = b + tc$.
Substituting the vectors,$r = (i + 2j - k) + t(i + j - 2k) = (1 + t)i + (2 + t)j - (1 + 2t)k$.
The projection of $r$ on $a$ is given by $\frac{|r \cdot a|}{|a|} = \sqrt{2/3}$.
First,calculate $r \cdot a = (1 + t)(2) + (2 + t)(-1) + (-1 - 2t)(1) = 2 + 2t - 2 - t - 1 - 2t = -t - 1$.
Also,$|a| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}$.
Thus,$\frac{|-t - 1|}{\sqrt{6}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{2} \cdot \sqrt{2}}{\sqrt{3} \cdot \sqrt{2}} = \frac{2}{\sqrt{6}}$.
So,$|-t - 1| = 2$,which implies $-t - 1 = 2$ or $-t - 1 = -2$.
If $-t - 1 = 2$,then $t = -3$. Substituting $t = -3$ into $r$,we get $r = (1 - 3)i + (2 - 3)j - (1 - 6)k = -2i - j + 5k$.
If $-t - 1 = -2$,then $t = 1$. Substituting $t = 1$ into $r$,we get $r = (1 + 1)i + (2 + 1)j - (1 + 2)k = 2i + 3j - 3k$.
Therefore,the vectors are $2i + 3j - 3k$ and $-2i - j + 5k$.

(D) Let $I = \int_0^{\pi /2} \frac{dx}{1 + \tan^3 x}$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:
$I = \int_0^{\pi /2} \frac{dx}{1 + \tan^3(\pi/2 - x)} = \int_0^{\pi /2} \frac{dx}{1 + \cot^3 x}$.
$I = \int_0^{\pi /2} \frac{\tan^3 x}{1 + \tan^3 x} dx$.
Adding the two expressions for $I$:
$2I = \int_0^{\pi /2} \left( \frac{1}{1 + \tan^3 x} + \frac{\tan^3 x}{1 + \tan^3 x} \right) dx$.
$2I = \int_0^{\pi /2} \frac{1 + \tan^3 x}{1 + \tan^3 x} dx = \int_0^{\pi /2} 1 dx$.
$2I = [x]_0^{\pi /2} = \frac{\pi}{2}$.
Therefore,$I = \frac{\pi}{4}$.

(A) Let $I = \int_{\pi /4}^{3\pi /4} \frac{\phi}{1 + \sin \phi} \, d\phi$.
Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$,we have:
$I = \int_{\pi /4}^{3\pi /4} \frac{\pi - \phi}{1 + \sin(\pi - \phi)} \, d\phi = \int_{\pi /4}^{3\pi /4} \frac{\pi - \phi}{1 + \sin \phi} \, d\phi$.
Adding the two expressions for $I$:
$2I = \int_{\pi /4}^{3\pi /4} \frac{\phi + \pi - \phi}{1 + \sin \phi} \, d\phi = \pi \int_{\pi /4}^{3\pi /4} \frac{1}{1 + \sin \phi} \, d\phi$.
$2I = \pi \int_{\pi /4}^{3\pi /4} \frac{1 - \sin \phi}{\cos^2 \phi} \, d\phi = \pi \int_{\pi /4}^{3\pi /4} (\sec^2 \phi - \sec \phi \tan \phi) \, d\phi$.
$2I = \pi [\tan \phi - \sec \phi]_{\pi /4}^{3\pi /4}$.
$2I = \pi [(\tan(3\pi/4) - \sec(3\pi/4)) - (\tan(\pi/4) - \sec(\pi/4))]$.
$2I = \pi [(-1 - (-\sqrt{2})) - (1 - \sqrt{2})] = \pi [\sqrt{2} - 1 - 1 + \sqrt{2}] = \pi [2\sqrt{2} - 2] = 2\pi(\sqrt{2} - 1)$.
$I = \pi(\sqrt{2} - 1)$.
Since $\tan(\pi/8) = \sqrt{2} - 1$,the value is $\pi \tan(\pi/8)$.

(A) Given that $E$ and $F$ are independent events,$P(E \cap F) = P(E)P(F) = \frac{1}{12}$ $(i)$.
Also,$P(\bar{E} \cap \bar{F}) = P(\bar{E})P(\bar{F}) = \frac{1}{2}$.
Since $P(\bar{E}) = 1 - P(E)$ and $P(\bar{F}) = 1 - P(F)$,we have $(1 - P(E))(1 - P(F)) = \frac{1}{2}$.
$1 - (P(E) + P(F)) + P(E)P(F) = \frac{1}{2}$.
Substituting $P(E)P(F) = \frac{1}{12}$,we get $1 - (P(E) + P(F)) + \frac{1}{12} = \frac{1}{2}$.
$P(E) + P(F) = 1 + \frac{1}{12} - \frac{1}{2} = \frac{12 + 1 - 6}{12} = \frac{7}{12}$ $(ii)$.
Let $x = P(E)$ and $y = P(F)$. Then $x + y = \frac{7}{12}$ and $xy = \frac{1}{12}$.
The quadratic equation $t^2 - (x+y)t + xy = 0$ becomes $t^2 - \frac{7}{12}t + \frac{1}{12} = 0$.
$12t^2 - 7t + 1 = 0 \Rightarrow (4t - 1)(3t - 1) = 0$.
Thus,$t = \frac{1}{4}$ or $t = \frac{1}{3}$.
Therefore,the probabilities are $\frac{1}{3}$ and $\frac{1}{4}$.