IIT JEE 1984 Mathematics Question Paper with Answer and Solution

(C) Since $1, \omega, \omega^2, \dots, \omega^{n-1}$ are the $n^{th}$ roots of unity,they are the roots of the equation $x^n - 1 = 0$.
Therefore,we can write the polynomial as:
$x^n - 1 = (x - 1)(x - \omega)(x - \omega^2) \dots (x - \omega^{n-1})$
Dividing both sides by $(x - 1)$,we get:
$\frac{x^n - 1}{x - 1} = (x - \omega)(x - \omega^2) \dots (x - \omega^{n-1})$
Using the geometric series formula,the left side simplifies to:
$x^{n-1} + x^{n-2} + \dots + x + 1 = (x - \omega)(x - \omega^2) \dots (x - \omega^{n-1})$
Now,taking the limit as $x \to 1$ or simply substituting $x = 1$ on both sides:
$1^{n-1} + 1^{n-2} + \dots + 1 + 1 = (1 - \omega)(1 - \omega^2) \dots (1 - \omega^{n-1})$
Since there are $n$ terms on the left side,the sum is $n$.
Thus,$(1 - \omega)(1 - \omega^2) \dots (1 - \omega^{n-1}) = n$.

(A) The given series is $3 \cdot 8 + 6 \cdot 11 + 9 \cdot 14 + 12 \cdot 17 + \dots$
The first factors of the terms are $3, 6, 9, 12, \dots$,which form an arithmetic progression with the first term $a = 3$ and common difference $d = 3$. The ${n^{th}}$ term is $a_n = 3 + (n - 1)3 = 3n$.
The second factors of the terms are $8, 11, 14, 17, \dots$,which form an arithmetic progression with the first term $a = 8$ and common difference $d = 3$. The ${n^{th}}$ term is $b_n = 8 + (n - 1)3 = 3n + 5$.
Therefore,the ${n^{th}}$ term of the given series is $T_n = a_n \cdot b_n = 3n(3n + 5)$.

(B) Let $S_2$ be the sum of integers divisible by $2$,$S_5$ be the sum of integers divisible by $5$,and $S_{10}$ be the sum of integers divisible by both $2$ and $5$ (i.e.,divisible by $10$).
$S_2 = 2 + 4 + \dots + 100 = \frac{50}{2}(2 + 100) = 25 \times 102 = 2550$.
$S_5 = 5 + 10 + \dots + 100 = \frac{20}{2}(5 + 100) = 10 \times 105 = 1050$.
$S_{10} = 10 + 20 + \dots + 100 = \frac{10}{2}(10 + 100) = 5 \times 110 = 550$.
By the Principle of Inclusion-Exclusion,the required sum is $S_2 + S_5 - S_{10} = 2550 + 1050 - 550 = 3050$.

(D) Given the equation: $x - \frac{2}{x - 1} = 1 - \frac{2}{x - 1}$.
For the expression to be defined,the denominator must not be zero,so $x - 1 \neq 0$,which implies $x \neq 1$.
Adding $\frac{2}{x - 1}$ to both sides of the equation,we get $x = 1$.
However,we have the condition $x \neq 1$ from the original equation.
Since the only potential solution $x = 1$ is excluded by the domain of the equation,there are no roots.

(D) Let $y = \frac{(x - a)(x - b)}{(x - c)}$.
Then $y(x - c) = x^2 - (a + b)x + ab$,which rearranges to $x^2 - (a + b + y)x + (ab + cy) = 0$.
For $x$ to be real,the discriminant $D$ must be $\ge 0$:
$D = (a + b + y)^2 - 4(ab + cy) \ge 0$.
Expanding this,we get $y^2 + 2y(a + b - 2c) + (a - b)^2 \ge 0$.
For the function to assume all real values $y$,this quadratic in $y$ must be non-negative for all $y$. However,since the coefficient of $y^2$ is positive,this quadratic will take all real values if its discriminant $D_y < 0$.
$D_y = [2(a + b - 2c)]^2 - 4(a - b)^2 < 0$.
$4(a + b - 2c)^2 - 4(a - b)^2 < 0$.
$(a + b - 2c)^2 - (a - b)^2 < 0$.
Using $A^2 - B^2 = (A-B)(A+B)$:
$(a + b - 2c - a + b)(a + b - 2c + a - b) < 0$.
$(2b - 2c)(2a - 2c) < 0$.
$4(b - c)(a - c) < 0$.
$(c - b)(c - a) < 0$.
This inequality holds if $c$ lies between $a$ and $b$,i.e.,$a < c < b$ or $b < c < a$.

(C) Let $y = \frac{x^2 - 3x + 4}{x^2 + 3x + 4}$.
Multiplying both sides by $(x^2 + 3x + 4)$,we get $y(x^2 + 3x + 4) = x^2 - 3x + 4$.
Rearranging the terms,we get $(y - 1)x^2 + 3(y + 1)x + 4(y - 1) = 0$.
Since $x$ is real,the discriminant $D$ must be greater than or equal to $0$.
$D = [3(y + 1)]^2 - 4(y - 1)(4(y - 1)) \ge 0$.
$9(y + 1)^2 - 16(y - 1)^2 \ge 0$.
$(3(y + 1))^2 - (4(y - 1))^2 \ge 0$.
Using the identity $a^2 - b^2 = (a - b)(a + b)$,we get $(3y + 3 - 4y + 4)(3y + 3 + 4y - 4) \ge 0$.
$(-y + 7)(7y - 1) \ge 0$.
Multiplying by $-1$,we get $(y - 7)(7y - 1) \le 0$.
This inequality holds when $\frac{1}{7} \le y \le 7$.
Thus,the maximum value is $7$ and the minimum value is $\frac{1}{7}$.

(A) Let $f(x) = (x - a)(x - c) + 2(x - b)(x - d)$.
Since $f(x)$ is a quadratic polynomial with a positive leading coefficient,we evaluate $f(x)$ at $x = a, b, c, d$:
$f(a) = (a - a)(a - c) + 2(a - b)(a - d) = 2(a - b)(a - d)$. Since $a < b$ and $a < d$,$(a - b) < 0$ and $(a - d) < 0$,so $f(a) > 0$.
$f(b) = (b - a)(b - c) + 2(b - b)(b - d) = (b - a)(b - c)$. Since $b > a$ and $b < c$,$(b - a) > 0$ and $(b - c) < 0$,so $f(b) < 0$.
$f(c) = (c - a)(c - c) + 2(c - b)(c - d) = 2(c - b)(c - d)$. Since $c > b$ and $c < d$,$(c - b) > 0$ and $(c - d) < 0$,so $f(c) < 0$.
$f(d) = (d - a)(d - c) + 2(d - b)(d - d) = (d - a)(d - c)$. Since $d > a$ and $d > c$,$(d - a) > 0$ and $(d - c) > 0$,so $f(d) > 0$.
Since $f(a) > 0$ and $f(b) < 0$,there exists a root in $(a, b)$.
Since $f(c) < 0$ and $f(d) > 0$,there exists a root in $(c, d)$.
Thus,the equation has two real and distinct roots.

(B) The given equation is ${x^2} - 3kx + 2{e^{2\log k}} - 1 = 0$.
Since ${e^{2\log k}} = {e^{\log {k^2}}} = {k^2}$,the equation becomes ${x^2} - 3kx + 2{k^2} - 1 = 0$.
The product of the roots of a quadratic equation $ax^2 + bx + c = 0$ is $\frac{c}{a}$.
Here,the product of roots is $2{k^2} - 1$.
Given that the product of roots is $7$,we have $2{k^2} - 1 = 7$,which implies $2{k^2} = 8$,so ${k^2} = 4$,giving $k = \pm 2$.
Since the term $\log k$ exists in the original equation,$k$ must be positive,so $k = 2$.
For the roots to be real,the discriminant $D = b^2 - 4ac$ must be $\ge 0$.
$D = (-3k)^2 - 4(1)(2{k^2} - 1) = 9{k^2} - 8{k^2} + 4 = {k^2} + 4$.
Since ${k^2} + 4 > 0$ for all real $k$,the roots are always real for any real $k$.
However,given the options and the condition $k=2$,the roots are real when $k=2$.

(C) We know that $\cos(\pi - \theta) = -\cos \theta$.
Therefore,$\cos \frac{{7\pi }}{8} = \cos(\pi - \frac{\pi }{8}) = -\cos \frac{\pi }{8}$ and $\cos \frac{{5\pi }}{8} = \cos(\pi - \frac{{3\pi }}{8}) = -\cos \frac{{3\pi }}{8}$.
Substituting these into the expression:
$E = \left( {1 + \cos \frac{\pi }{8}} \right)\left( {1 + \cos \frac{{3\pi }}{8}} \right)\left( {1 - \cos \frac{{3\pi }}{8}} \right)\left( {1 - \cos \frac{\pi }{8}} \right)$
Grouping the terms:
$E = \left( {1 + \cos \frac{\pi }{8}} \right)\left( {1 - \cos \frac{\pi }{8}} \right)\left( {1 + \cos \frac{{3\pi }}{8}} \right)\left( {1 - \cos \frac{{3\pi }}{8}} \right)$
Using the identity $(a+b)(a-b) = a^2 - b^2$:
$E = \left( {1 - \cos^2 \frac{\pi }{8}} \right)\left( {1 - \cos^2 \frac{{3\pi }}{8}} \right)$
$E = \sin^2 \frac{\pi }{8} \sin^2 \frac{{3\pi }}{8}$
Using the identity $2\sin A \sin B = \cos(A-B) - \cos(A+B)$:
$E = \left( \sin \frac{\pi }{8} \sin \frac{{3\pi }}{8} \right)^2 = \left( \frac{1}{2} \left( \cos \frac{{2\pi }}{8} - \cos \frac{{4\pi }}{8} \right) \right)^2$
$E = \left( \frac{1}{2} \left( \cos \frac{\pi }{4} - \cos \frac{\pi }{2} \right) \right)^2$
Since $\cos \frac{\pi }{4} = \frac{1}{\sqrt{2}}$ and $\cos \frac{\pi }{2} = 0$:
$E = \left( \frac{1}{2} \left( \frac{1}{\sqrt{2}} - 0 \right) \right)^2 = \left( \frac{1}{2\sqrt{2}} \right)^2 = \frac{1}{8}$.

(B) We know that for any triangle $ABC$,the maximum value of $\cos A + \cos B + \cos C$ is $\frac{3}{2}$,which occurs if and only if $A = B = C = 60^{\circ}$.
Alternatively,using the identity $\cos A + \cos B + \cos C = 1 + 4 \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2}) = \frac{3}{2}$,we get $\sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2}) = \frac{1}{8}$.
Since the maximum value of $\sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$ is $\frac{1}{8}$ (by Jensen's inequality for concave functions),this equality holds only when $\frac{A}{2} = \frac{B}{2} = \frac{C}{2} = 30^{\circ}$,i.e.,$A = B = C = 60^{\circ}$.
Thus,the triangle is equilateral.

(C) Let the slope of the third side be $m$. The equation of the line passing through $(1, -10)$ is $y + 10 = m(x - 1)$,which simplifies to $mx - y - (m + 10) = 0$.
Since the triangle is isosceles,the third side makes equal angles with the two given sides. The slopes of the given lines are $m_1 = 7$ and $m_2 = -1$.
Using the formula for the angle between two lines,$\tan \theta = |\frac{m - m_1}{1 + m \cdot m_1}| = |\frac{m - m_2}{1 + m \cdot m_2}|$,we get:
$|\frac{m - 7}{1 + 7m}| = |\frac{m - (-1)}{1 + m(-1)}| = |\frac{m + 1}{1 - m}|$.
Solving $|\frac{m - 7}{1 + 7m}| = |\frac{m + 1}{1 - m}|$:
Case $1$: $\frac{m - 7}{1 + 7m} = \frac{m + 1}{1 - m}$ $\Rightarrow (m - 7)(1 - m) = (m + 1)(1 + 7m)$ $\Rightarrow m - m^2 - 7 + 7m = m + 7m^2 + 1 + 7m$ $\Rightarrow 8m^2 - m + 8 = 0$. This gives no real roots.
Case $2$: $\frac{m - 7}{1 + 7m} = -(\frac{m + 1}{1 - m})$ $\Rightarrow (m - 7)(1 - m) = -(m + 1)(1 + 7m)$ $\Rightarrow -m^2 + 8m - 7 = -(7m^2 + 8m + 1)$ $\Rightarrow 6m^2 + 16m - 6 = 0$ $\Rightarrow 3m^2 + 8m - 3 = 0$.
Factoring gives $(3m - 1)(m + 3) = 0$,so $m = \frac{1}{3}$ or $m = -3$.
For $m = \frac{1}{3}$,the equation is $y + 10 = \frac{1}{3}(x - 1)$ $\Rightarrow 3y + 30 = x - 1$ $\Rightarrow x - 3y - 31 = 0$.
For $m = -3$,the equation is $y + 10 = -3(x - 1)$ $\Rightarrow y + 10 = -3x + 3$ $\Rightarrow 3x + y + 7 = 0$.

(B) The given lines are $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$.
Dividing the second equation by $2$,we get $3x - 4y - 3.5 = 0$.
Since the lines are parallel,the distance between them is the diameter of the circle.
The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.
Here,$a = 3, b = -4, c_1 = 4, c_2 = -3.5$.
$d = \frac{|4 - (-3.5)|}{\sqrt{3^2 + (-4)^2}} = \frac{|7.5|}{\sqrt{9 + 16}} = \frac{7.5}{5} = 1.5$.
Since the diameter is $1.5$,the radius $r = \frac{d}{2} = \frac{1.5}{2} = 0.75 = \frac{3}{4}$.

(C) Let the mid-point of the chord be $C(h, k)$.
Since the chord subtends a right angle at the origin $O(0, 0)$,the triangle formed by the origin and the endpoints of the chord is an isosceles right-angled triangle.
The distance from the origin to the midpoint $C(h, k)$ is $d = \sqrt{h^2 + k^2}$.
In the right-angled triangle formed by the origin,the midpoint,and one endpoint of the chord,the angle at the origin is $45^\circ$.
Using trigonometry,$\cos(45^\circ) = \frac{d}{r}$,where $r$ is the radius of the circle.
Given $r = 2$,we have $\frac{1}{\sqrt{2}} = \frac{\sqrt{h^2 + k^2}}{2}$.
Squaring both sides,$\frac{1}{2} = \frac{h^2 + k^2}{4}$,which simplifies to $h^2 + k^2 = 2$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 = 2$.

(A) Let the coordinates of $A$ be $(x_1, y_1)$ and $B$ be $(x_2, y_2)$.
From the given equations,the roots for abscissae are $x_1, x_2$,so $x_1 + x_2 = -2a$ and $x_1x_2 = -b^2$.
Similarly,the roots for ordinates are $y_1, y_2$,so $y_1 + y_2 = -2p$ and $y_1y_2 = -q^2$.
The equation of a circle with diameter $AB$ is given by $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
Expanding this,we get $x^2 - (x_1 + x_2)x + x_1x_2 + y^2 - (y_1 + y_2)y + y_1y_2 = 0$.
Substituting the values,we get $x^2 - (-2a)x + (-b^2) + y^2 - (-2p)y + (-q^2) = 0$.
This simplifies to $x^2 + y^2 + 2ax + 2py - b^2 - q^2 = 0$.

(B) The given expression is $\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1 + 2 + 3 + \dots + n}{1 - {n^2}}} \right]$.
Using the formula for the sum of the first $n$ natural numbers,$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$,the expression becomes:
$\mathop {\lim }\limits_{n \to \infty } \frac{n(n+1)}{2(1 - {n^2})}$
$= \mathop {\lim }\limits_{n \to \infty } \frac{n^2 + n}{2 - 2n^2}$
Dividing the numerator and denominator by $n^2$:
$= \mathop {\lim }\limits_{n \to \infty } \frac{1 + \frac{1}{n}}{\frac{2}{n^2} - 2} = \frac{1 + 0}{0 - 2} = -\frac{1}{2}$.

(C) Let $L = \mathop {\lim }\limits_{x \to 1} (1 - x)\tan \left( {\frac{{\pi x}}{2}} \right)$.
Substitute $y = 1 - x$,so as $x \to 1$,$y \to 0$ and $x = 1 - y$.
Then $L = \mathop {\lim }\limits_{y \to 0} y \tan \left( {\frac{{\pi (1 - y)}}{2}} \right) = \mathop {\lim }\limits_{y \to 0} y \tan \left( {\frac{\pi }{2} - \frac{{\pi y}}{2}} \right)$.
Using the identity $\tan \left( {\frac{\pi }{2} - \theta } \right) = \cot \theta$,we get:
$L = \mathop {\lim }\limits_{y \to 0} y \cot \left( {\frac{{\pi y}}{2}} \right) = \mathop {\lim }\limits_{y \to 0} \frac{y}{\tan \left( {\frac{{\pi y}}{2}} \right)}$.
Multiply and divide by $\frac{\pi }{2}$:
$L = \mathop {\lim }\limits_{y \to 0} \frac{1}{\frac{\pi }{2}} \cdot \frac{{\frac{{\pi y}}{2}}}{{\tan \left( {\frac{{\pi y}}{2}} \right)}} = \frac{2}{\pi } \cdot 1 = \frac{2}{\pi }$.

(B) When three dice are rolled,the total number of possible outcomes is $6 \times 6 \times 6 = 216$.
The favorable outcomes where the same number appears on each die are $(1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), \text{ and } (6, 6, 6)$.
There are $6$ such favorable outcomes.
Therefore,the required probability is $\frac{6}{216} = \frac{1}{36}$.

(B) The probability that exactly one of the events $A$ or $B$ occurs is given by the probability of the symmetric difference of $A$ and $B$,denoted as $P(A \Delta B)$.
This is equivalent to the probability that $A$ occurs and $B$ does not,or $B$ occurs and $A$ does not:
$P(A \cap \bar{B}) + P(\bar{A} \cap B)$
Using the property $P(A \cap \bar{B}) = P(A) - P(A \cap B)$ and $P(\bar{A} \cap B) = P(B) - P(A \cap B)$:
$= (P(A) - P(A \cap B)) + (P(B) - P(A \cap B))$
$= P(A) + P(B) - 2P(A \cap B)$.

(A) Let the complex numbers $z_1, z_2, z_3$ denote the vertices $A, B, C$ of an equilateral triangle $ABC$.
Since $|z_1| = |z_2| = |z_3|$,the origin $O$ is equidistant from all vertices $A, B, C$.
This implies that the origin $O$ is the circumcenter of the equilateral triangle $ABC$.
For an equilateral triangle,the circumcenter,centroid,and orthocenter coincide.
If $G$ is the centroid,then $G = \frac{z_1 + z_2 + z_3}{3}$.
Since the centroid $G$ coincides with the circumcenter $O$ (which is the origin $0$),we have $\frac{z_1 + z_2 + z_3}{3} = 0$.
Therefore,$z_1 + z_2 + z_3 = 0$.

(B) Given that $a, b, c$ are in $A.P.$,we have the relation $2b = a + c$,which can be rewritten as $a - 2b + c = 0$.
The equation of the straight line is $ax + by + c = 0$.
Comparing this with the condition $a(1) + b(-2) + c = 0$,we can see that the line satisfies the equation for the point $(x, y) = (1, -2)$.
Therefore,the line $ax + by + c = 0$ always passes through the point $(1, -2)$.

(A) Total number of points = $3 + 4 + 5 = 12$.
The total number of ways to select $3$ points out of $12$ is $^{12}C_3 = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$.
However,points lying on the same side are collinear and cannot form a triangle. We must subtract the cases where $3$ points are selected from the same side:
$1$. Points on side $AB$: $^{3}C_3 = 1$ way.
$2$. Points on side $BC$: $^{4}C_3 = 4$ ways.
$3$. Points on side $CA$: $^{5}C_3 = 10$ ways.
Total collinear sets = $1 + 4 + 10 = 15$.
Number of triangles = $220 - 15 = 205$.

(A) For a system of homogeneous linear equations to have a non-zero solution,the determinant of the coefficient matrix must be equal to zero.
The coefficient matrix is given by:
$A = \begin{bmatrix} \lambda & 1 & 1 \\ -1 & \lambda & 1 \\ -1 & -1 & \lambda \end{bmatrix}$
Setting the determinant to zero:
$\begin{vmatrix} \lambda & 1 & 1 \\ -1 & \lambda & 1 \\ -1 & -1 & \lambda \end{vmatrix} = 0$
Expanding along the first row:
$\lambda(\lambda^2 - (-1)) - 1(-\lambda - (-1)) + 1(1 - (-\lambda)) = 0$
$\lambda(\lambda^2 + 1) - 1(-\lambda + 1) + 1(1 + \lambda) = 0$
$\lambda^3 + \lambda + \lambda - 1 + 1 + \lambda = 0$
$\lambda^3 + 3\lambda = 0$
$\lambda(\lambda^2 + 3) = 0$
This gives $\lambda = 0$ or $\lambda^2 = -3$. Since $\lambda$ must be a real value,$\lambda^2 = -3$ has no real solutions.
Therefore,the only real value for $\lambda$ is $0$.

(D) We use the formula $2\tan^{-1}(x) = \tan^{-1}\left( \frac{2x}{1-x^2} \right)$.
First,calculate $2\tan^{-1}\left( \frac{1}{5} \right)$:
$2\tan^{-1}\left( \frac{1}{5} \right) = \tan^{-1}\left( \frac{2(1/5)}{1-(1/5)^2} \right) = \tan^{-1}\left( \frac{2/5}{1-1/25} \right) = \tan^{-1}\left( \frac{2/5}{24/25} \right) = \tan^{-1}\left( \frac{2}{5} \times \frac{25}{24} \right) = \tan^{-1}\left( \frac{5}{12} \right)$.
Now,substitute this into the expression:
$\tan \left[ \tan^{-1}\left( \frac{5}{12} \right) - \frac{\pi}{4} \right] = \tan \left[ \tan^{-1}\left( \frac{5}{12} \right) - \tan^{-1}(1) \right]$.
Using the formula $\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left( \frac{x-y}{1+xy} \right)$:
$\tan \left[ \tan^{-1}\left( \frac{5/12 - 1}{1 + (5/12)(1)} \right) \right] = \frac{5/12 - 1}{1 + 5/12} = \frac{-7/12}{17/12} = -\frac{7}{17}$.

(B) Since the forces $a, b$ and $c$ are mutually perpendicular,they can be represented along the $x, y$ and $z$ axes respectively.
Let the vectors be $\vec{a} = 2\hat{i}$,$\vec{b} = 10\hat{j}$,and $\vec{c} = 11\hat{k}$.
The resultant vector is $\vec{R} = \vec{a} + \vec{b} + \vec{c} = 2\hat{i} + 10\hat{j} + 11\hat{k}$.
The magnitude of the resultant is given by $|\vec{R}| = \sqrt{(2)^2 + (10)^2 + (11)^2}$.
$|\vec{R}| = \sqrt{4 + 100 + 121} = \sqrt{225} = 15$.

(D) Let the points be $A$,$B$,and $C$ with position vectors $\vec{OA} = a + b$,$\vec{OB} = a - b$,and $\vec{OC} = a + kb$.
For the points to be collinear,the vectors $\vec{AB}$ and $\vec{BC}$ must be parallel,meaning $\vec{AB} = \lambda \vec{BC}$ for some scalar $\lambda$.
Calculate $\vec{AB} = \vec{OB} - \vec{OA} = (a - b) - (a + b) = -2b$.
Calculate $\vec{BC} = \vec{OC} - \vec{OB} = (a + kb) - (a - b) = (k + 1)b$.
For collinearity,$-2b = \lambda(k + 1)b$.
This implies $-2 = \lambda(k + 1)$. Since $\lambda$ can be any non-zero real number,$k$ can take any real value except for the case where the points coincide. However,in the context of vector collinearity,the condition holds for any $k \in \mathbb{R}$ as long as the vectors are linearly dependent.
Thus,the points are collinear for every real number $k$.

(B) The magnitude of a vector remains invariant under rotation of the coordinate axes.
Given the original components are $(2p, 1)$ and the new components are $(p+1, 1)$.
The square of the magnitude is given by $x^2 + y^2$.
Therefore,$(2p)^2 + 1^2 = (p+1)^2 + 1^2$.
$4p^2 + 1 = p^2 + 2p + 1 + 1$.
$3p^2 - 2p - 1 = 0$.
Solving the quadratic equation: $3p^2 - 3p + p - 1 = 0$.
$3p(p-1) + 1(p-1) = 0$.
$(3p+1)(p-1) = 0$.
Thus,$p = 1$ or $p = -\frac{1}{3}$.

(A) Given the function $y = \frac{x + 2}{x - 1}$.
To find $x$ in terms of $y$,we perform the following steps:
$y(x - 1) = x + 2$
$yx - y = x + 2$
$yx - x = y + 2$
$x(y - 1) = y + 2$
$x = \frac{y + 2}{y - 1}$
Since $f(y) = \frac{y + 2}{y - 1}$,we conclude that $x = f(y)$.

(C) Given $f(x) = \begin{cases} 0, & x < 0 \\ x^2, & x \ge 0 \end{cases}$.
First,we check the continuity at $x = 0$:
$\lim_{x \to 0^-} f(x) = \lim_{h \to 0} f(0 - h) = 0$.
$\lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(0 + h) = \lim_{h \to 0} (0 + h)^2 = 0$.
Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 0$,$f(x)$ is continuous at $x = 0$.
Next,we check the differentiability at $x = 0$:
$Lf'(0) = \lim_{h \to 0} \frac{f(0 - h) - f(0)}{-h} = \lim_{h \to 0} \frac{0 - 0}{-h} = 0$.
$Rf'(0) = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2 - 0}{h} = \lim_{h \to 0} h = 0$.
Since $Lf'(0) = Rf'(0) = 0$,$f(x)$ is differentiable at $x = 0$ and $f'(0) = 0$.
The derivative is $f'(x) = \begin{cases} 0, & x < 0 \\ 2x, & x \ge 0 \end{cases}$.
Now,check the continuity of $f'(x)$ at $x = 0$:
$\lim_{x \to 0^-} f'(x) = 0$ and $\lim_{x \to 0^+} f'(x) = \lim_{x \to 0^+} 2x = 0$.
Since $\lim_{x \to 0^-} f'(x) = \lim_{x \to 0^+} f'(x) = f'(0) = 0$,$f'(x)$ is continuous at $x = 0$.
Finally,check the differentiability of $f'(x)$ at $x = 0$:
$Lf''(0) = \lim_{h \to 0} \frac{f'(0 - h) - f'(0)}{-h} = \lim_{h \to 0} \frac{0 - 0}{-h} = 0$.
$Rf''(0) = \lim_{h \to 0} \frac{f'(0 + h) - f'(0)}{h} = \lim_{h \to 0} \frac{2(0 + h) - 0}{h} = \lim_{h \to 0} \frac{2h}{h} = 2$.
Since $Lf''(0) \neq Rf''(0)$,$f'(x)$ is not differentiable at $x = 0$.

(A) Let $I = \int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \, dx$.
Substitute $t = \sin^{-1} x$,then $x = \sin t$ and $dt = \frac{1}{\sqrt{1 - x^2}} \, dx$.
The integral becomes $I = \int t \sin t \, dt$.
Using integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = t$ and $dv = \sin t \, dt$. Then $du = dt$ and $v = -\cos t$.
$I = -t \cos t - \int (-\cos t) \, dt = -t \cos t + \sin t + c$.
Since $t = \sin^{-1} x$,we have $\sin t = x$ and $\cos t = \sqrt{1 - \sin^2 t} = \sqrt{1 - x^2}$.
Substituting these back,we get $I = -(\sin^{-1} x) \sqrt{1 - x^2} + x + c = x - \sqrt{1 - x^2} \sin^{-1} x + c$.

(A) Let $I = \int \frac{dx}{\sin x + \sin 2x} = \int \frac{dx}{\sin x(1 + 2\cos x)}$.
Multiply numerator and denominator by $\sin x$:
$I = \int \frac{\sin x \, dx}{\sin^2 x(1 + 2\cos x)} = \int \frac{\sin x \, dx}{(1 - \cos^2 x)(1 + 2\cos x)} = \int \frac{\sin x \, dx}{(1 - \cos x)(1 + \cos x)(1 + 2\cos x)}$.
Let $\cos x = t$,then $-\sin x \, dx = dt$,so $\sin x \, dx = -dt$.
$I = - \int \frac{dt}{(1 - t)(1 + t)(1 + 2t)}$.
Using partial fractions:
$\frac{1}{(1 - t)(1 + t)(1 + 2t)} = \frac{A}{1 - t} + \frac{B}{1 + t} + \frac{C}{1 + 2t}$.
Solving for coefficients,we get $A = \frac{1}{6}$,$B = -\frac{1}{2}$,$C = \frac{4}{3}$.
$I = - \int \left( \frac{1/6}{1 - t} - \frac{1/2}{1 + t} + \frac{4/3}{1 + 2t} \right) dt$.
$I = - \left[ \frac{1}{6} \frac{\log |1 - t|}{-1} - \frac{1}{2} \log |1 + t| + \frac{4}{3} \cdot \frac{1}{2} \log |1 + 2t| \right] + C$.
$I = \frac{1}{6} \log |1 - t| + \frac{1}{2} \log |1 + t| - \frac{2}{3} \log |1 + 2t| + C$.
Substituting $t = \cos x$ back:
$I = \frac{1}{6} \log |1 - \cos x| + \frac{1}{2} \log |1 + \cos x| - \frac{2}{3} \log |1 + 2\cos x| + C$.

(B) Let $t = \sin^{-1} x$.
Then $dt = \frac{1}{\sqrt{1 - x^2}} dx$.
When $x = 0$,$t = 0$.
When $x = 1/2$,$t = \sin^{-1}(1/2) = \pi/6$.
Also,$x = \sin t$.
Substituting these into the integral:
$\int_0^{\pi/6} t \sin t \, dt$.
Using integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = t$,$dv = \sin t \, dt$.
Then $du = dt$,$v = -\cos t$.
$\int t \sin t \, dt = -t \cos t - \int (-\cos t) \, dt = -t \cos t + \sin t$.
Evaluating the definite integral:
$[-t \cos t + \sin t]_0^{\pi/6} = (-\frac{\pi}{6} \cos(\frac{\pi}{6}) + \sin(\frac{\pi}{6})) - (0 + 0)$.
$= -\frac{\pi}{6} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{1}{2} - \frac{\sqrt{3} \pi}{12}$.

(C) Given the conditions:
$(\vec{a} - \vec{d}) \cdot (\vec{b} - \vec{c}) = 0$
$(\vec{b} - \vec{d}) \cdot (\vec{c} - \vec{a}) = 0$
These can be written in terms of vectors as:
$\vec{DA} \cdot \vec{CB} = 0 \Rightarrow \vec{DA} \perp \vec{CB}$
$\vec{DB} \cdot \vec{AC} = 0 \Rightarrow \vec{DB} \perp \vec{AC}$
Since $\vec{DA}$ is perpendicular to the side $BC$ and $\vec{DB}$ is perpendicular to the side $AC,$ the point $D$ is the intersection of the altitudes of $\Delta ABC.$
Therefore,the point $D$ is the orthocentre of $\Delta ABC.$

(B) The given lines can be written in symmetric form as follows:
For the first line $x = ay + b$ and $z = cy + d$,we have $\frac{x - b}{a} = y = \frac{z - d}{c}$.
Thus,the direction ratios of the first line are $(a, 1, c)$.
For the second line $x = a'y + b'$ and $z = c'y + d'$,we have $\frac{x - b'}{a'} = y = \frac{z - d'}{c'}$.
Thus,the direction ratios of the second line are $(a', 1, c')$.
Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.
Substituting the direction ratios,we get $(a)(a') + (1)(1) + (c)(c') = 0$.
Therefore,$aa' + 1 + cc' = 0$,which implies $aa' + cc' = -1$.

(B) We have the integral $I = \int \frac{1}{x^2(x^4 + 1)^{3/4}} dx$.
Taking $x^4$ common from the bracket,we get:
$I = \int \frac{1}{x^2 [x^4(1 + \frac{1}{x^4})]^{3/4}} dx = \int \frac{1}{x^2 \cdot x^3 (1 + \frac{1}{x^4})^{3/4}} dx = \int \frac{1}{x^5 (1 + \frac{1}{x^4})^{3/4}} dx$.
Let $t = 1 + \frac{1}{x^4}$.
Then $dt = -\frac{4}{x^5} dx$,which implies $\frac{dx}{x^5} = -\frac{1}{4} dt$.
Substituting these into the integral:
$I = \int -\frac{1}{4} t^{-3/4} dt = -\frac{1}{4} \cdot \frac{t^{1/4}}{1/4} + c = -t^{1/4} + c$.
Substituting $t$ back:
$I = -(1 + \frac{1}{x^4})^{1/4} + c = -(\frac{x^4 + 1}{x^4})^{1/4} + c = -\frac{(x^4 + 1)^{1/4}}{x} + c$.