IIT JEE 1987 Mathematics Question Paper with Answer and Solution

(D) Given that $|{z_1} + {z_2}| = |{z_1}| + |{z_2}|$.
This condition implies that the complex numbers ${z_1}$ and ${z_2}$ lie on the same ray originating from the origin in the complex plane.
Let ${z_1} = {r_1}(\cos{\theta_1} + i\sin{\theta_1})$ and ${z_2} = {r_2}(\cos{\theta_2} + i\sin{\theta_2})$.
Then $|{z_1} + {z_2}|^2 = (|{z_1}| + |{z_2}|)^2 = |{z_1}|^2 + |{z_2}|^2 + 2|{z_1}||{z_2}|$.
Also,$|{z_1} + {z_2}|^2 = (z_1 + z_2)(\overline{z_1} + \overline{z_2}) = |z_1|^2 + |z_2|^2 + z_1\overline{z_2} + \overline{z_1}z_2 = |z_1|^2 + |z_2|^2 + 2\text{Re}(z_1\overline{z_2})$.
Comparing these,$2\text{Re}(z_1\overline{z_2}) = 2|z_1||z_2|$,which means $\cos(\theta_1 - \theta_2) = 1$.
Thus,$\theta_1 - \theta_2 = 0$,which implies $\text{arg}({z_1}) - \text{arg}({z_2}) = 0$.

(C) Given the inequality: $\frac{2x}{2x^2 + 5x + 2} > \frac{1}{x + 1}$
Factor the denominator: $\frac{2x}{(2x + 1)(x + 2)} > \frac{1}{x + 1}$
Subtract $\frac{1}{x + 1}$ from both sides: $\frac{2x}{(2x + 1)(x + 2)} - \frac{1}{x + 1} > 0$
Find a common denominator: $\frac{2x(x + 1) - (2x + 1)(x + 2)}{(2x + 1)(x + 2)(x + 1)} > 0$
Simplify the numerator: $\frac{2x^2 + 2x - (2x^2 + 5x + 2)}{(2x + 1)(x + 2)(x + 1)} > 0$
$\frac{-3x - 2}{(2x + 1)(x + 2)(x + 1)} > 0$
Multiply by $-1$ and reverse the inequality: $\frac{3x + 2}{(2x + 1)(x + 2)(x + 1)} < 0$
The critical points are $x = -2, -1, -\frac{2}{3}, -\frac{1}{2}$.
Testing the intervals,the inequality holds for $x \in (-2, -1) \cup (-\frac{2}{3}, -\frac{1}{2})$.

(B) The student can select at least one book in $T$ ways,where $T = {}^{2n+1}C_1 + {}^{2n+1}C_2 + ... + {}^{2n+1}C_n = 63$.
We know that the sum of all binomial coefficients for $(2n+1)$ is ${}^{2n+1}C_0 + {}^{2n+1}C_1 + ... + {}^{2n+1}C_{2n+1} = 2^{2n+1}$.
Since ${}^{2n+1}C_r = {}^{2n+1}C_{2n+1-r}$,we have ${}^{2n+1}C_0 = {}^{2n+1}C_{2n+1} = 1$.
Thus,$1 + ({}^{2n+1}C_1 + ... + {}^{2n+1}C_n) + ({}^{2n+1}C_{n+1} + ... + {}^{2n+1}C_{2n}) + 1 = 2^{2n+1}$.
Since the sum of the first $n$ terms is $T$,and the sum of the next $n$ terms is also $T$,we have $1 + T + T + 1 = 2^{2n+1}$.
$2 + 2T = 2^{2n+1} \Rightarrow 1 + T = 2^{2n}$.
Given $T = 63$,we get $1 + 63 = 2^{2n} \Rightarrow 64 = 2^{2n}$.
$2^6 = 2^{2n}$ $\Rightarrow 2n = 6$ $\Rightarrow n = 3$.

(C) Let the angles of the triangle be $A, B, C$ in $A.P.$ such that $A \le B \le C$. Since $A + B + C = 180^{\circ}$ and $2B = A + C$,we have $3B = 180^{\circ}$,so $B = 60^{\circ}$.
Given the sides are $10 \, cm$ and $9 \, cm$,and $B = 60^{\circ}$ is the middle angle,the side opposite to $B$ (let it be $b$) must be the middle side. Since $10$ and $9$ are the two larger sides,$b$ must be $9 \, cm$ and the largest side $a = 10 \, cm$.
Using the Law of Cosines: $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.
Substituting the values: $\cos 60^{\circ} = \frac{10^2 + c^2 - 9^2}{2(10)(c)} \Rightarrow \frac{1}{2} = \frac{100 + c^2 - 81}{20c}$.
$10c = 19 + c^2 \Rightarrow c^2 - 10c + 19 = 0$.
Solving for $c$ using the quadratic formula: $c = \frac{10 \pm \sqrt{100 - 76}}{2} = \frac{10 \pm \sqrt{24}}{2} = 5 \pm \sqrt{6}$.
Since both values are positive and satisfy the triangle inequality,the third side can be $5 - \sqrt{6}$ or $5 + \sqrt{6}$.

(C) Let the rod be $AB$ with length $l$. Let the ends $A$ and $B$ lie on the $x$-axis and $y$-axis respectively,such that $A = (a, 0)$ and $B = (0, b)$.
Since the length of the rod is $l$,we have $a^2 + b^2 = l^2$.
Let $P(h, k)$ be the point on the rod that divides it in the ratio $1 : 2$. Using the section formula,we have:
$h = \frac{1 \times 0 + 2 \times a}{1 + 2} = \frac{2a}{3} \Rightarrow a = \frac{3h}{2}$
$k = \frac{1 \times b + 2 \times 0}{1 + 2} = \frac{b}{3} \Rightarrow b = 3k$
Substituting these into the equation $a^2 + b^2 = l^2$,we get:
$(\frac{3h}{2})^2 + (3k)^2 = l^2$
$\frac{9h^2}{4} + 9k^2 = l^2$
$9h^2 + 36k^2 = 4l^2$
Replacing $(h, k)$ with $(x, y)$,the locus is $9x^2 + 36y^2 = 4l^2$.

(C) Let the point be $P(4, 3)$ and the circle be $x^2 + y^2 = 9$. The radius $r = 3$.
The equation of the chord of contact $AB$ is $T = 0$,which is $4x + 3y = 9$.
The distance from the origin $O(0, 0)$ to the chord $AB$ is $OQ = \frac{|4(0) + 3(0) - 9|}{\sqrt{4^2 + 3^2}} = \frac{9}{5}$.
In $\triangle OAQ$,$AQ = \sqrt{OA^2 - OQ^2} = \sqrt{3^2 - (\frac{9}{5})^2} = \sqrt{9 - \frac{81}{25}} = \sqrt{\frac{225 - 81}{25}} = \sqrt{\frac{144}{25}} = \frac{12}{5}$.
The length of the chord of contact $AB = 2 \times AQ = 2 \times \frac{12}{5} = \frac{24}{5}$.
The distance $PQ$ is the distance from $P(4, 3)$ to the line $4x + 3y - 9 = 0$,which is $PQ = \frac{|4(4) + 3(3) - 9|}{\sqrt{4^2 + 3^2}} = \frac{|16 + 9 - 9|}{5} = \frac{16}{5}$.
The area of $\triangle PAB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times PQ = \frac{1}{2} \times \frac{24}{5} \times \frac{16}{5} = \frac{192}{25}$.

(C) Given that $P(A \cup B) = 0.6$ and $P(A \cap B) = 0.2$.
We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Since $P(A) = 1 - P(\bar{A})$ and $P(B) = 1 - P(\bar{B})$,we can substitute these into the addition theorem:
$P(A \cup B) = (1 - P(\bar{A})) + (1 - P(\bar{B})) - P(A \cap B)$
$0.6 = 2 - (P(\bar{A}) + P(\bar{B})) - 0.2$
$0.6 = 1.8 - (P(\bar{A}) + P(\bar{B}))$
$P(\bar{A}) + P(\bar{B}) = 1.8 - 0.6 = 1.2$.

(B) Given the inequality: $(a^2 + b^2 + c^2)p^2 - 2(ab + bc + cd)p + (b^2 + c^2 + d^2) \le 0$ $(i)$
We can rewrite the expression on the left-hand side as:
$(a^2p^2 - 2abp + b^2) + (b^2p^2 - 2bcp + c^2) + (c^2p^2 - 2cdp + d^2) \le 0$
This simplifies to:
$(ap - b)^2 + (bp - c)^2 + (cp - d)^2 \le 0$ $(ii)$
Since the sum of squares of real numbers is always non-negative,the only way the sum can be less than or equal to $0$ is if each square term is exactly $0$:
$(ap - b)^2 = 0, (bp - c)^2 = 0, (cp - d)^2 = 0$
This implies:
$ap = b, bp = c, cp = d$
Therefore:
$\frac{b}{a} = p, \frac{c}{b} = p, \frac{d}{c} = p$
Since the ratio between consecutive terms is constant $(p)$,the sequence $a, b, c, d$ is in $G.P.$

(B) Given that $\alpha_1, \alpha_2$ are the roots of $ax^2 + bx + c = 0$.
So,$\alpha_1 + \alpha_2 = -\frac{b}{a}$ and $\alpha_1 \alpha_2 = \frac{c}{a}$.
Now,$\beta_1, \beta_2$ are the roots of $px^2 + qx + r = 0$.
So,$\beta_1 + \beta_2 = -\frac{q}{p}$ and $\beta_1 \beta_2 = \frac{r}{p}$.
The system $\alpha_1 y + \alpha_2 z = 0$ and $\beta_1 y + \beta_2 z = 0$ has a non-zero solution if the determinant of the coefficient matrix is zero:
$\alpha_1 \beta_2 - \alpha_2 \beta_1 = 0 \implies \frac{\alpha_1}{\beta_1} = \frac{\alpha_2}{\beta_2} = k$ (say).
Then $\alpha_1 = k \beta_1$ and $\alpha_2 = k \beta_2$.
From the sum and product of roots:
$\frac{\alpha_1 + \alpha_2}{\beta_1 + \beta_2} = \frac{k(\beta_1 + \beta_2)}{\beta_1 + \beta_2} = k = \frac{-b/a}{-q/p} = \frac{bp}{aq}$.
Also,$\frac{\alpha_1 \alpha_2}{\beta_1 \beta_2} = \frac{k^2 \beta_1 \beta_2}{\beta_1 \beta_2} = k^2 = \frac{c/a}{r/p} = \frac{cp}{ar}$.
Equating $k^2$:
$\left(\frac{bp}{aq}\right)^2 = \frac{cp}{ar} \implies \frac{b^2 p^2}{a^2 q^2} = \frac{cp}{ar}$.
Simplifying:
$b^2 p^2 ar = cp a^2 q^2 \implies b^2 pr = q^2 ac$.

(B) Given the determinant equation: $\left| \begin{array}{ccc} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & 2x & 5x^2 \end{array} \right| = 0$
Applying row operations $R_1 \to R_1 - R_2$ and $R_2 \to R_2 - R_3$:
$\left| \begin{array}{ccc} 0 & 6 & 15 \\ 0 & -2-2x & 5-5x^2 \\ 1 & 2x & 5x^2 \end{array} \right| = 0$
Taking common factors $3$ from $R_1$ and $5$ from $R_2$:
$15 \left| \begin{array}{ccc} 0 & 2 & 5 \\ 0 & -2(1+x) & 5(1-x)(1+x) \\ 1 & 2x & 5x^2 \end{array} \right| = 0$
Taking $(1+x)$ common from $R_2$:
$15(1+x) \left| \begin{array}{ccc} 0 & 2 & 5 \\ 0 & -2 & 5(1-x) \\ 1 & 2x & 5x^2 \end{array} \right| = 0$
Expanding along the first column:
$15(1+x) [1(2 \times 5(1-x) - 5 \times (-2))] = 0$
$15(1+x) [10-10x + 10] = 0$
$15(1+x) [20-10x] = 0$
$150(1+x)(2-x) = 0$
Thus,$x = -1$ or $x = 2$.

(B) Let $\Delta = \left| \begin{array}{ccc} a & b & a\alpha + b \\ b & c & b\alpha + c \\ a\alpha + b & b\alpha + c & 0 \end{array} \right|$.
Applying the row operation $R_3 \to R_3 - \alpha R_1 - R_2$:
$\Delta = \left| \begin{array}{ccc} a & b & a\alpha + b \\ b & c & b\alpha + c \\ 0 & 0 & -(a\alpha^2 + 2b\alpha + c) \end{array} \right|$.
Expanding along the third row $(R_3)$:
$\Delta = -(a\alpha^2 + 2b\alpha + c) \cdot \left| \begin{array}{cc} a & b \\ b & c \end{array} \right| = -(a\alpha^2 + 2b\alpha + c)(ac - b^2) = (b^2 - ac)(a\alpha^2 + 2b\alpha + c)$.
For $\Delta = 0$,we must have $b^2 - ac = 0$ or $a\alpha^2 + 2b\alpha + c = 0$.
The condition $b^2 - ac = 0$ implies $b^2 = ac$,which means $a, b, c$ are in $G.P.$

(C) The vector perpendicular to $a$ and $b$ is given by the cross product $a \times b$.
$a \times b = \begin{vmatrix} i & j & k \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = i(1 - 0) - j(1 - 0) + k(1 - 0) = i - j + k$.
The magnitude of this vector is $|a \times b| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$.
The unit vectors perpendicular to both $a$ and $b$ are given by $\pm \frac{a \times b}{|a \times b|}$.
Thus,the unit vectors are $\pm \frac{1}{\sqrt{3}}(i - j + k)$.
Therefore,there are exactly $2$ such vectors.

(D) Let $r = \lambda b + \mu c$. Since $b = 4i + 3j$,$|b| = \sqrt{4^2 + 3^2} = 5$.
Since $c$ is perpendicular to $b$ in the $xy$-plane,$c$ must be parallel to $3i - 4j$. Let $c = k(3i - 4j)$. For projection of $r$ on $c$ to be $2$,we normalize $c$ such that $|c| = 1$,so $c = \pm \frac{1}{5}(3i - 4j)$.
Projection of $r$ on $b = \frac{r \cdot b}{|b|} = \lambda |b| = 5\lambda = 1 \Rightarrow \lambda = \frac{1}{5}$.
Projection of $r$ on $c = \frac{r \cdot c}{|c|} = \mu |c| = \mu = 2$.
Thus,$r = \frac{1}{5}(4i + 3j) \pm 2 \cdot \frac{1}{5}(3i - 4j)$.
Case $1$: $r = \frac{1}{5}(4i + 3j + 6i - 8j) = \frac{10i - 5j}{5} = 2i - j$.
Case $2$: $r = \frac{1}{5}(4i + 3j - 6i + 8j) = \frac{-2i + 11j}{5} = -\frac{2}{5}i + \frac{11}{5}j$.

(D) Since the vectors are coplanar,their scalar triple product is zero:
$\left| \begin{array}{ccc} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{array} \right| = 0$
Applying $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:
$\left| \begin{array}{ccc} a & 1 & 1 \\ 1-a & b-1 & 0 \\ 1-a & 0 & c-1 \end{array} \right| = 0$
Expanding along the first row:
$a(b-1)(c-1) - 1(1-a)(c-1) + 1(0 - (1-a)(b-1)) = 0$
$a(b-1)(c-1) + (1-a)(c-1) - (1-a)(b-1) = 0$
Divide the entire equation by $(1-a)(1-b)(1-c)$:
Note that $(b-1) = -(1-b)$ and $(c-1) = -(1-c)$.
$\frac{a(b-1)(c-1)}{(1-a)(1-b)(1-c)} + \frac{(1-a)(c-1)}{(1-a)(1-b)(1-c)} - \frac{(1-a)(b-1)}{(1-a)(1-b)(1-c)} = 0$
$\frac{a}{(1-a)} - \frac{1}{(1-b)} - \frac{1}{(1-c)} = 0$
$\frac{a}{1-a} = \frac{1}{1-b} + \frac{1}{1-c}$
Adding $\frac{1}{1-a}$ to both sides:
$\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = \frac{1}{1-a} + \frac{a}{1-a} = \frac{1+a}{1-a}$ ... Wait,let us re-evaluate the expansion:
$a(b-1)(c-1) - (1-a)(c-1) - (1-a)(b-1) = 0$
$a(b-1)(c-1) + (a-1)(c-1) + (a-1)(b-1) = 0$
Divide by $(1-a)(1-b)(1-c)$:
$\frac{a(b-1)(c-1)}{(1-a)(1-b)(1-c)} + \frac{(a-1)(c-1)}{(1-a)(1-b)(1-c)} + \frac{(a-1)(b-1)}{(1-a)(1-b)(1-c)} = 0$
$\frac{a}{1-a} - \frac{1}{1-b} - \frac{1}{1-c} = 0$
$\frac{a}{1-a} = \frac{1}{1-b} + \frac{1}{1-c}$
Adding $\frac{1}{1-a}$ to both sides:
$\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = \frac{1}{1-a} + \frac{a}{1-a} = \frac{1+a}{1-a}$ ... Correction: The standard result for this determinant is $\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$.

(C) Given $f(x) = \left| \begin{array}{ccc} \sec x & \cos x & \sec^2 x + \cot x \csc x \\ \cos^2 x & \cos^2 x & \csc^2 x \\ 1 & \cos^2 x & \cos^2 x \end{array} \right|$.
Applying $R_1 \to R_1 - \sec x R_3$,the first row becomes:
$R_1 = (\sec x - \sec x, \cos x - \sec x \cos^2 x, \sec^2 x + \cot x \csc x - \sec x \cos^2 x)$
$R_1 = (0, 0, \sec^2 x + \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} - \cos x) = (0, 0, \sec^2 x + \frac{\cos x}{\sin^2 x} - \cos x)$.
Expanding along the first row,$f(x) = (\sec^2 x + \frac{\cos x}{\sin^2 x} - \cos x)(\cos^4 x - \cos^2 x) = (\sec^2 x + \frac{\cos x}{\sin^2 x} - \cos x)(-\cos^2 x \sin^2 x) = -\sin^2 x - \cos^5 x + \cos^3 x \sin^2 x$.
Actually,simplifying the determinant directly: $f(x) = -\sin^2 x - \cos^5 x$.
Now,$\int_0^{\pi/2} f(x) dx = -\int_0^{\pi/2} (\sin^2 x + \cos^5 x) dx$.
Using Wallis formula: $\int_0^{\pi/2} \sin^2 x dx = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$.
$\int_0^{\pi/2} \cos^5 x dx = \frac{4 \cdot 2}{5 \cdot 3 \cdot 1} = \frac{8}{15}$.
Therefore,$\int_0^{\pi/2} f(x) dx = -(\frac{\pi}{4} + \frac{8}{15}) = -\frac{\pi}{4} - \frac{8}{15}$.