IIT JEE 1986 Mathematics Question Paper with Answer and Solution

(D) Let $z = (1 - \cos \theta ) + i(2\sin \theta )$. We want the real part of $z^{-1} = \frac{1}{z}$.
Using trigonometric identities,$1 - \cos \theta = 2\sin^2(\frac{\theta}{2})$ and $2\sin \theta = 4\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})$.
So,$z = 2\sin(\frac{\theta}{2}) [\sin(\frac{\theta}{2}) + 2i\cos(\frac{\theta}{2})]$.
Then,$z^{-1} = \frac{1}{2\sin(\frac{\theta}{2})} \cdot \frac{1}{\sin(\frac{\theta}{2}) + 2i\cos(\frac{\theta}{2})}$.
Multiply numerator and denominator by the conjugate $\sin(\frac{\theta}{2}) - 2i\cos(\frac{\theta}{2})$:
$z^{-1} = \frac{1}{2\sin(\frac{\theta}{2})} \cdot \frac{\sin(\frac{\theta}{2}) - 2i\cos(\frac{\theta}{2})}{\sin^2(\frac{\theta}{2}) + 4\cos^2(\frac{\theta}{2})}$.
The real part is $\frac{\sin(\frac{\theta}{2})}{2\sin(\frac{\theta}{2}) [\sin^2(\frac{\theta}{2}) + 4\cos^2(\frac{\theta}{2})]} = \frac{1}{2[\sin^2(\frac{\theta}{2}) + 4\cos^2(\frac{\theta}{2})]}$.
Using $\sin^2(\frac{\theta}{2}) = \frac{1 - \cos \theta}{2}$ and $\cos^2(\frac{\theta}{2}) = \frac{1 + \cos \theta}{2}$:
Real part $= \frac{1}{2[\frac{1 - \cos \theta}{2} + 4(\frac{1 + \cos \theta}{2})]} = \frac{1}{1 - \cos \theta + 4 + 4\cos \theta} = \frac{1}{5 + 3\cos \theta }$.

(A) Let ${z_1} = a + ib$ and ${z_2} = c - id$,where $a > 0$ and $d > 0$.
Given $|{z_1}| = |{z_2}|$,we have ${a^2} + {b^2} = {c^2} + {d^2}$.
Let $w = \frac{{z_1 + z_2}}{{z_1 - z_2}}$.
Then $\bar{w} = \frac{{\bar{z_1} + \bar{z_2}}}{{\bar{z_1} - \bar{z_2}}}$.
Since $|{z_1}| = |{z_2}| = r$,we have $\bar{z_1} = \frac{r^2}{z_1}$ and $\bar{z_2} = \frac{r^2}{z_2}$.
Substituting these,$\bar{w} = \frac{{\frac{r^2}{z_1} + \frac{r^2}{z_2}}}{{\frac{r^2}{z_1} - \frac{r^2}{z_2}}} = \frac{{z_2 + z_1}}{{z_2 - z_1}} = -\frac{{z_1 + z_2}}{{z_1 - z_2}} = -w$.
Since $\bar{w} = -w$,the complex number $w$ is purely imaginary.
For example,let ${z_1} = 2 + i$ and ${z_2} = 1 - 2i$.
Then $\frac{{z_1 + z_2}}{{z_1 - z_2}} = \frac{{3 - i}}{{1 + 3i}} = \frac{{(3 - i)(1 - 3i)}}{{1^2 + 3^2}} = \frac{{3 - 9i - i - 3}}{{10}} = \frac{{-10i}}{{10}} = -i$,which is purely imaginary.

(B) Let $z = x + iy$. The vertices of the triangle are $z = (x, y)$,$iz = (-y, x)$,and $z + iz = (x - y, x + y)$.
The area $A$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $A = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Substituting the coordinates:
$A = \frac{1}{2} |x(x - (x + y)) + (-y)((x + y) - y) + (x - y)(y - x)|$
$A = \frac{1}{2} |x(-y) - y(x) + (x - y)(-(x - y))|$
$A = \frac{1}{2} |-xy - xy - (x - y)^2|$
$A = \frac{1}{2} |-2xy - (x^2 - 2xy + y^2)|$
$A = \frac{1}{2} |-2xy - x^2 + 2xy - y^2|$
$A = \frac{1}{2} |-x^2 - y^2| = \frac{1}{2} (x^2 + y^2) = \frac{1}{2} |z|^2$.

(B) Let $\alpha$ be the common root of the equations $x^2 + ax + b = 0$ and $x^2 + bx + a = 0$.
Then,$\alpha^2 + a\alpha + b = 0$ and $\alpha^2 + b\alpha + a = 0$.
Subtracting the two equations,we get $(\alpha^2 + a\alpha + b) - (\alpha^2 + b\alpha + a) = 0$.
$(a - b)\alpha + (b - a) = 0$.
$(a - b)\alpha = (a - b)$.
Assuming $a \neq b$,we get $\alpha = 1$.
Substituting $\alpha = 1$ into the first equation: $(1)^2 + a(1) + b = 0$.
$1 + a + b = 0$.
Therefore,$a + b = -1$.

(A) The total number of balls in the box is $2 + 3 + 4 = 9$.
We need to select $3$ balls such that at least one is black.
This can be calculated by subtracting the number of ways to select $3$ balls with no black balls from the total number of ways to select $3$ balls.
Total ways to select $3$ balls from $9$ is $^{9}C_{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
Number of ways to select $3$ balls such that no black ball is chosen (i.e.,selecting from the $2$ white and $4$ red balls,total $6$ non-black balls) is $^{6}C_{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
Therefore,the number of ways to select at least one black ball is $84 - 20 = 64$.

(D) Let $S = \sum_{r=0}^{n} (-1)^r (r+1) C_r^2$.
We know that $\sum_{r=0}^{n} (-1)^r C_r^2 = (-1)^{n/2} \frac{n!}{(n/2)!(n/2)!}$ for even $n$.
Also,$\sum_{r=0}^{n} (-1)^r r C_r^2 = (-1)^{n/2} \frac{n}{2} \frac{n!}{(n/2)!(n/2)!}$.
Thus,$S = \sum_{r=0}^{n} (-1)^r C_r^2 + \sum_{r=0}^{n} (-1)^r r C_r^2$.
$S = (-1)^{n/2} \frac{n!}{(n/2)!(n/2)!} + (-1)^{n/2} \frac{n}{2} \frac{n!}{(n/2)!(n/2)!}$.
$S = (-1)^{n/2} \frac{n!}{(n/2)!(n/2)!} (1 + n/2) = (-1)^{n/2} \frac{n!}{(n/2)!(n/2)!} \frac{n+2}{2}$.
Multiplying by the coefficient $\frac{2(n/2)!(n/2)!}{n!}$,we get:
$\frac{2(n/2)!(n/2)!}{n!} \times (-1)^{n/2} \frac{n!}{(n/2)!(n/2)!} \frac{n+2}{2} = (-1)^{n/2}(n+2)$.

(D) To determine which inequality is satisfied by the interior points of the triangle,we check the vertices $(1, 3)$,$(5, 0)$,and $(-1, 2)$ for each inequality.
For $3x + 2y \ge 0$:
$(1, 3) \implies 3(1) + 2(3) = 9 > 0$
$(5, 0) \implies 3(5) + 2(0) = 15 > 0$
$(-1, 2) \implies 3(-1) + 2(2) = 1 > 0$
Since all vertices satisfy the inequality,the interior points also satisfy it.
For $2x + y - 13 \le 0$:
$(1, 3) \implies 2(1) + 3 - 13 = -8 \le 0$
$(5, 0) \implies 2(5) + 0 - 13 = -3 \le 0$
$(-1, 2) \implies 2(-1) + 2 - 13 = -13 \le 0$
Since all vertices satisfy the inequality,the interior points also satisfy it.
For $2x - 3y - 12 \le 0$:
$(1, 3) \implies 2(1) - 3(3) - 12 = -19 \le 0$
$(5, 0) \implies 2(5) - 3(0) - 12 = -2 \le 0$
$(-1, 2) \implies 2(-1) - 3(2) - 12 = -20 \le 0$
Since all vertices satisfy the inequality,the interior points also satisfy it.
Therefore,all the given inequalities are satisfied.

(D) The perpendicular bisectors of the sides of a triangle intersect at the circumcenter $O$.
Since the perpendicular bisectors of $AB$ and $AC$ are $x - y + 5 = 0$ and $x + 2y = 0$,their intersection point $O$ is found by solving these equations:
$x - y = -5$ and $x + 2y = 0$.
Subtracting the equations,we get $-3y = -5$,so $y = \frac{5}{3}$.
Then $x = -2y = -\frac{10}{3}$.
Thus,the circumcenter $O$ is $(-\frac{10}{3}, \frac{5}{3})$.
Since $O$ is the circumcenter,$OA = OB = OC$.
Let $B = (x_1, y_1)$. The reflection of $A(1, -2)$ about the line $x - y + 5 = 0$ gives $B$.
The formula for the reflection of $(x_0, y_0)$ about $ax + by + c = 0$ is $\frac{x - x_0}{a} = \frac{y - y_0}{b} = -2 \frac{ax_0 + by_0 + c}{a^2 + b^2}$.
For $A(1, -2)$ and $x - y + 5 = 0$:
$\frac{x_1 - 1}{1} = \frac{y_1 + 2}{-1} = -2 \frac{1 - (-2) + 5}{1^2 + (-1)^2} = -2 \frac{8}{2} = -8$.
So,$x_1 - 1 = -8 \implies x_1 = -7$ and $y_1 + 2 = 8 \implies y_1 = 6$. Thus,$B = (-7, 6)$.
Similarly,for $C(x_2, y_2)$,the reflection of $A(1, -2)$ about $x + 2y = 0$:
$\frac{x_2 - 1}{1} = \frac{y_2 + 2}{2} = -2 \frac{1 + 2(-2)}{1^2 + 2^2} = -2 \frac{-3}{5} = \frac{6}{5}$.
So,$x_2 - 1 = \frac{6}{5} \implies x_2 = \frac{11}{5}$ and $y_2 + 2 = \frac{12}{5} \implies y_2 = \frac{2}{5}$. Thus,$C = (\frac{11}{5}, \frac{2}{5})$.
The equation of line $BC$ passing through $(-7, 6)$ and $(\frac{11}{5}, \frac{2}{5})$ is:
$y - 6 = \frac{\frac{2}{5} - 6}{\frac{11}{5} - (-7)} (x - (-7)) = \frac{-\frac{28}{5}}{\frac{46}{5}} (x + 7) = -\frac{14}{23} (x + 7)$.
$23(y - 6) = -14(x + 7) \implies 23y - 138 = -14x - 98 \implies 14x + 23y - 40 = 0$.

(B) Let the coordinates of the point $P(h, k)$ divide the line segment joining $(1, 0)$ and $(2\cos \theta, 2\sin \theta)$ in the ratio $2 : 3$.
Using the section formula,we have:
$h = \frac{2(2\cos \theta) + 3(1)}{2 + 3} = \frac{4\cos \theta + 3}{5}$
$k = \frac{2(2\sin \theta) + 3(0)}{2 + 3} = \frac{4\sin \theta}{5}$
From these,we get:
$\cos \theta = \frac{5h - 3}{4}$ and $\sin \theta = \frac{5k}{4}$
Using the identity $\cos^2 \theta + \sin^2 \theta = 1$,we substitute the values:
$\left(\frac{5h - 3}{4}\right)^2 + \left(\frac{5k}{4}\right)^2 = 1$
$(5h - 3)^2 + (5k)^2 = 16$
Replacing $(h, k)$ with $(x, y)$,the locus is $(5x - 3)^2 + (5y)^2 = 16$,which represents a circle.

(A) The equation of the common chord of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$.
First,write the equations in the form $x^2 + y^2 + 2gx + 2fy + c = 0$.
The first circle is $3x^2 + 3y^2 - 2x + 12y - 9 = 0$. Dividing by $3$,we get $x^2 + y^2 - \frac{2}{3}x + 4y - 3 = 0$ $(S_1)$.
The second circle is $x^2 + y^2 + 6x + 2y - 15 = 0$ $(S_2)$.
The common chord is $S_1 - S_2 = 0$:
$(x^2 + y^2 - \frac{2}{3}x + 4y - 3) - (x^2 + y^2 + 6x + 2y - 15) = 0$
$(-\frac{2}{3} - 6)x + (4 - 2)y + (-3 + 15) = 0$
$-\frac{20}{3}x + 2y + 12 = 0$
Multiplying by $-3$,we get $20x - 6y - 36 = 0$,which simplifies to $10x - 3y - 18 = 0$.

(B) Let the coordinates of point $M$ be $(h, k)$.
Given $A = (0, 3)$ and $M = (h, k)$. Since $M$ lies on the line segment $AB$ such that $AM = 2AB$,point $B$ is the midpoint of $AM$.
Thus,the coordinates of $B$ are $\left( \frac{h+0}{2}, \frac{k+3}{2} \right) = \left( \frac{h}{2}, \frac{k+3}{2} \right)$.
Since $B$ lies on the circle $x^2 + 4x + (y - 3)^2 = 0$,we substitute the coordinates of $B$ into the circle equation:
$\left( \frac{h}{2} \right)^2 + 4\left( \frac{h}{2} \right) + \left( \frac{k+3}{2} - 3 \right)^2 = 0$
$\Rightarrow \frac{h^2}{4} + 2h + \left( \frac{k-3}{2} \right)^2 = 0$
$\Rightarrow \frac{h^2}{4} + 2h + \frac{(k-3)^2}{4} = 0$
Multiplying by $4$,we get $h^2 + 8h + (k-3)^2 = 0$,which simplifies to $h^2 + k^2 + 8h - 6k + 9 = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 + 8x - 6y + 9 = 0$,which represents a circle.

(A) Since $\frac{1 + 3p}{3}, \frac{1 - p}{4}$ and $\frac{1 - 2p}{2}$ are the probabilities of three events,each probability must lie in the interval $[0, 1]$.
$1$) $0 \le \frac{1 + 3p}{3} \le 1 \Rightarrow 0 \le 1 + 3p \le 3 \Rightarrow -1 \le 3p \le 2 \Rightarrow -\frac{1}{3} \le p \le \frac{2}{3}$
$2$) $0 \le \frac{1 - p}{4} \le 1 \Rightarrow 0 \le 1 - p \le 4 \Rightarrow -1 \le -p \le 3 \Rightarrow -3 \le p \le 1$
$3$) $0 \le \frac{1 - 2p}{2} \le 1 \Rightarrow 0 \le 1 - 2p \le 2 \Rightarrow -1 \le -2p \le 1 \Rightarrow -\frac{1}{2} \le p \le \frac{1}{2}$
For mutually exclusive events,the sum of probabilities must be $\le 1$:
$\frac{1 + 3p}{3} + \frac{1 - p}{4} + \frac{1 - 2p}{2} \le 1$
Multiply by $12$: $4(1 + 3p) + 3(1 - p) + 6(1 - 2p) \le 12$
$4 + 12p + 3 - 3p + 6 - 12p \le 12$
$13 - 3p \le 12 \Rightarrow -3p \le -1 \Rightarrow p \ge \frac{1}{3}$
Taking the intersection of all conditions:
$p \in [-\frac{1}{3}, \frac{2}{3}] \cap [-3, 1] \cap [-\frac{1}{2}, \frac{1}{2}] \cap [\frac{1}{3}, \infty)$
Intersection is $\frac{1}{3} \le p \le \frac{1}{2}$.

(D) Let $z = \cos \frac{{2\pi }}{7} + i\sin \frac{{2\pi }}{7}$. By De Moivre's theorem,${z^k} = \cos \frac{{2\pi k}}{7} + i\sin \frac{{2\pi k}}{7}$.
The given sum is $S = \sum\limits_{k = 1}^6 {\left( {\sin \frac{{2\pi k}}{7} - i\cos \frac{{2\pi k}}{7}} \right)}$.
We can rewrite the term inside the sum as: $\sin \frac{{2\pi k}}{7} - i\cos \frac{{2\pi k}}{7} = -i \left( \cos \frac{{2\pi k}}{7} + i\sin \frac{{2\pi k}}{7} \right) = -i z^k$.
Thus,$S = -i \sum\limits_{k = 1}^6 z^k = -i (z + z^2 + z^3 + z^4 + z^5 + z^6)$.
Since $z$ is a $7^{th}$ root of unity ($z^7 = 1$ and $z \neq 1$),the sum of all $7^{th}$ roots of unity is $1 + z + z^2 + z^3 + z^4 + z^5 + z^6 = 0$.
Therefore,$\sum\limits_{k = 1}^6 z^k = -1$.
Substituting this into the expression for $S$,we get $S = -i(-1) = i$.

(B) Given expression: $3\left[ \sin^4\left( \frac{3\pi}{2} - \alpha \right) + \sin^4(3\pi + \alpha) \right] - 2\left[ \sin^6\left( \frac{\pi}{2} + \alpha \right) + \sin^6(5\pi - \alpha) \right]$
Using trigonometric reduction formulas:
$\sin(\frac{3\pi}{2} - \alpha) = -\cos \alpha$,$\sin(3\pi + \alpha) = -\sin \alpha$,$\sin(\frac{\pi}{2} + \alpha) = \cos \alpha$,$\sin(5\pi - \alpha) = \sin \alpha$
Substituting these values:
$= 3\left[ (-\cos \alpha)^4 + (-\sin \alpha)^4 \right] - 2\left[ (\cos \alpha)^6 + (\sin \alpha)^6 \right]$
$= 3(\cos^4 \alpha + \sin^4 \alpha) - 2(\cos^6 \alpha + \sin^6 \alpha)$
Using identities $\sin^4 \alpha + \cos^4 \alpha = 1 - 2\sin^2 \alpha \cos^2 \alpha$ and $\sin^6 \alpha + \cos^6 \alpha = 1 - 3\sin^2 \alpha \cos^2 \alpha$:
$= 3(1 - 2\sin^2 \alpha \cos^2 \alpha) - 2(1 - 3\sin^2 \alpha \cos^2 \alpha)$
$= 3 - 6\sin^2 \alpha \cos^2 \alpha - 2 + 6\sin^2 \alpha \cos^2 \alpha$
$= 3 - 2 = 1$

(A) Using the Law of Sines,we have $\frac{\sin A}{a} = \frac{\sin B}{b}$,which implies $a \sin B = b \sin A$.
From option $(A)$,$b \sin A = a$,so $a \sin B = a$,which gives $\sin B = 1$,meaning $B = \frac{\pi}{2}$.
Since $A < \frac{\pi}{2}$,the sum of angles $A + B < \pi$,so the triangle $ABC$ is possible.
From option $(B)$,$b \sin A > a$,which implies $a \sin B > a$,or $\sin B > 1$,which is impossible.
Similarly,for option $(C)$,$b \sin A > a$ leads to $\sin B > 1$,which is also impossible.

(B) Let the intercepts of the line on the $x$-axis and $y$-axis be $a$ and $b$ respectively. The coordinates of the points are $A(a, 0)$ and $B(0, b)$.
Given that the point $P(-5, 4)$ divides the line segment $AB$ in the ratio $1 : 2$.
Using the section formula,the coordinates of $P$ are given by $\left( \frac{1 \cdot 0 + 2 \cdot a}{1 + 2}, \frac{1 \cdot b + 2 \cdot 0}{1 + 2} \right) = \left( \frac{2a}{3}, \frac{b}{3} \right)$.
Equating the coordinates,we get $\frac{2a}{3} = -5 \implies a = -\frac{15}{2}$ and $\frac{b}{3} = 4 \implies b = 12$.
The intercept form of the line is $\frac{x}{a} + \frac{y}{b} = 1$.
Substituting the values,we get $\frac{x}{-15/2} + \frac{y}{12} = 1 \implies -\frac{2x}{15} + \frac{y}{12} = 1$.
Multiplying by $60$,we get $-8x + 5y = 60$,which simplifies to $8x - 5y + 60 = 0$.

(C) Let the vertices be $A(2, -1)$,$B(0, 2)$,$C(2, 3)$,and $D(4, 0)$.
The diagonals are $AC$ and $BD$.
The slope of diagonal $AC$ $(m_1)$ is $\frac{3 - (-1)}{2 - 2} = \frac{4}{0} = \infty$ (vertical line).
The slope of diagonal $BD$ $(m_2)$ is $\frac{0 - 2}{4 - 0} = \frac{-2}{4} = -\frac{1}{2}$.
Since one diagonal is vertical,the angle $\theta$ between the diagonals is given by $\tan \theta = |\frac{1}{m_2}| = |\frac{1}{-1/2}| = 2$.
Therefore,$\theta = \tan^{-1}(2)$.

(D) Let $A, B$ and $C$ be the events that the student passes tests $I, II$ and $III$ respectively. The student is successful if he passes in $(I \text{ and } II)$ or $(I \text{ and } III)$.
This is represented by the event $(A \cap B) \cup (A \cap C)$.
Using the inclusion-exclusion principle,$P((A \cap B) \cup (A \cap C)) = P(A \cap B) + P(A \cap C) - P(A \cap B \cap C)$.
Since the events are independent,$P(A \cap B) = P(A)P(B) = pq$,$P(A \cap C) = P(A)P(C) = p(\frac{1}{2})$,and $P(A \cap B \cap C) = P(A)P(B)P(C) = pq(\frac{1}{2})$.
Thus,the probability of success is $pq + \frac{p}{2} - \frac{pq}{2} = \frac{pq}{2} + \frac{p}{2} = \frac{p}{2}(q + 1)$.
Given the probability of success is $\frac{1}{2}$,we have $\frac{p}{2}(q + 1) = \frac{1}{2}$,which simplifies to $p(q + 1) = 1$.
If $p=1$,then $1+q=1 \Rightarrow q=0$. If $p=\frac{2}{3}$,then $\frac{2}{3}(q+1)=1 \Rightarrow q+1=\frac{3}{2} \Rightarrow q=\frac{1}{2}$.
Since there are multiple pairs $(p, q)$ satisfying $p(q+1)=1$,there are infinitely many values for $p$ and $q$. Thus,all options $A, B$ and $C$ are correct.

(D) Let $\Delta = \left| {\begin{array}{*{20}{c}}{a + b}&{a + 2b}&{a + 3b}\\{a + 2b}&{a + 3b}&{a + 4b}\\{a + 4b}&{a + 5b}&{a + 6b}\end{array}} \right|$.
Applying the row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_2$:
$R_2 - R_1 = (a+2b)-(a+b) = b, (a+3b)-(a+2b) = b, (a+4b)-(a+3b) = b$.
$R_3 - R_2 = (a+4b)-(a+2b) = 2b, (a+5b)-(a+3b) = 2b, (a+6b)-(a+4b) = 2b$.
Thus,$\Delta = \left| {\begin{array}{*{20}{c}}{a + b}&{a + 2b}&{a + 3b}\\b&b&b\\2b&2b&2b\end{array}} \right|$.
Since $R_2$ and $R_3$ are proportional (specifically,$R_3 = 2R_2$),the determinant is $0$.
Alternatively,by substituting $a = 1$ and $b = 1$,the determinant becomes $\left| {\begin{array}{*{20}{c}}2&3&4\\3&4&5\\5&6&7\end{array}} \right| = 2(28-30) - 3(21-25) + 4(18-20) = 2(-2) - 3(-4) + 4(-2) = -4 + 12 - 8 = 0$.

(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 $G.P.$

(D) The principal value branch of ${\sin^{-1}}(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Since $\frac{2\pi}{3}$ does not lie in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$,we simplify the expression.
We know that $\sin(\pi - \theta) = \sin(\theta)$.
Therefore,$\sin(\frac{2\pi}{3}) = \sin(\pi - \frac{\pi}{3}) = \sin(\frac{\pi}{3})$.
Now,${\sin^{-1}}[\sin(\frac{2\pi}{3})] = {\sin^{-1}}[\sin(\frac{\pi}{3})]$.
Since $\frac{\pi}{3} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,the value is $\frac{\pi}{3}$.

(A) The scalar triple product is defined as $[a, b, c] = a \cdot (b \times c)$.
Given that $a, b, c$ are non-coplanar,$[a, b, c] \neq 0$.
We know the properties of scalar triple products: $[c, a, b] = [a, b, c]$ and $[b, a, c] = -[a, b, c]$.
Substituting these into the expression:
$\frac{a \cdot (b \times c)}{c \times a \cdot b} + \frac{b \cdot (a \times c)}{c \cdot (a \times b)} = \frac{[a, b, c]}{[c, a, b]} + \frac{[b, a, c]}{[c, a, b]}$
$= \frac{[a, b, c]}{[a, b, c]} + \frac{-[a, b, c]}{[a, b, c]}$
$= 1 - 1 = 0$.

(A) Let the position vectors of the points be $\vec{a} = 3i - 2j - k,$ $\vec{b} = 2i + 3j - 4k,$ $\vec{c} = -i + j + 2k,$ and $\vec{d} = 4i + 5j + \lambda k.$
Since the four points are coplanar,the scalar triple product of the vectors $(\vec{b}-\vec{a}),$ $(\vec{c}-\vec{a}),$ and $(\vec{d}-\vec{a})$ must be zero.
$\vec{b}-\vec{a} = (2-3)i + (3-(-2))j + (-4-(-1))k = -i + 5j - 3k$
$\vec{c}-\vec{a} = (-1-3)i + (1-(-2))j + (2-(-1))k = -4i + 3j + 3k$
$\vec{d}-\vec{a} = (4-3)i + (5-(-2))j + (\lambda-(-1))k = i + 7j + (\lambda+1)k$
For coplanarity,the determinant of these vectors is zero:
$\begin{vmatrix} -1 & 5 & -3 \\ -4 & 3 & 3 \\ 1 & 7 & \lambda+1 \end{vmatrix} = 0$
$-1(3(\lambda+1) - 21) - 5(-4(\lambda+1) - 3) - 3(-28 - 3) = 0$
$-1(3\lambda + 3 - 21) - 5(-4\lambda - 4 - 3) - 3(-31) = 0$
$-3\lambda + 18 + 20\lambda + 35 + 93 = 0$
$17\lambda + 146 = 0$
$\lambda = -\frac{146}{17}$

(D) Given that $c$ is a unit vector,$|c| = 1,$ so $|c|^2 = c_1^2 + c_2^2 + c_3^2 = 1$ .....$(i)$
Since $c \perp a$ and $c \perp b,$ we have $c \cdot a = 0$ and $c \cdot b = 0.$
This implies $a_1 c_1 + a_2 c_2 + a_3 c_3 = 0$ .....$(ii)$
and $b_1 c_1 + b_2 c_2 + b_3 c_3 = 0$ .....$(iii)$
The angle between $a$ and $b$ is $\frac{\pi}{6},$ so $a \cdot b = |a||b| \cos(\frac{\pi}{6}) = |a||b| \frac{\sqrt{3}}{2}.$
Squaring both sides,$(a \cdot b)^2 = |a|^2 |b|^2 \frac{3}{4} \Rightarrow (a_1 b_1 + a_2 b_2 + a_3 b_3)^2 = \frac{3}{4} (\Sigma a_1^2)(\Sigma b_1^2)$ .....$(iv)$
Let $D = \left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right|.$ Then $D^2 = \left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right| \left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right| = \left| \begin{array}{ccc} a \cdot a & a \cdot b & a \cdot c \\ b \cdot a & b \cdot b & b \cdot c \\ c \cdot a & c \cdot b & c \cdot c \end{array} \right|.$
Using the conditions $a \cdot c = 0,$ $b \cdot c = 0,$ and $c \cdot c = 1,$ we get $D^2 = \left| \begin{array}{ccc} |a|^2 & a \cdot b & 0 \\ a \cdot b & |b|^2 & 0 \\ 0 & 0 & 1 \end{array} \right| = |a|^2 |b|^2 - (a \cdot b)^2.$
Substituting $(a \cdot b)^2 = \frac{3}{4} |a|^2 |b|^2,$ we get $D^2 = |a|^2 |b|^2 - \frac{3}{4} |a|^2 |b|^2 = \frac{1}{4} |a|^2 |b|^2 = \frac{(\Sigma a_1^2)(\Sigma b_1^2)}{4}.$

(D) For the interval $-1 \le x \le 1$,consider the function $g(x) = x \sin \pi x$.
Since $\sin \pi x$ has the same sign as $x$ for $x \in (-1, 1)$,the product $x \sin \pi x$ is always non-negative.
Specifically,for $x \in (-1, 1)$,$0 \le x \sin \pi x < 1$.
Therefore,the greatest integer function $[x \sin \pi x] = 0$ for all $x \in [-1, 1]$.
Since $f(x) = 0$ is a constant function on the interval $[-1, 1]$,it is continuous everywhere in this interval,including at $x = 0$ and in the sub-interval $(-1, 0)$.
Furthermore,a constant function is differentiable everywhere,so $f(x)$ is differentiable in $(-1, 1)$.
Thus,all the given statements are correct.

(B) Let $I = \int {\frac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}\;dx}$.
We know that $1 = (\sin^2 x + \cos^2 x)^2 = \sin^4 x + \cos^4 x + 2\sin^2 x \cos^2 x$,so the denominator $1 - 2\sin^2 x \cos^2 x = \sin^4 x + \cos^4 x$.
Thus,$I = \int {\frac{{(\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4 x)}}{{\sin^4 x + \cos^4 x}}\;dx}$.
$I = \int {(\sin^4 x - \cos^4 x)\,dx}$.
Using the identity $a^2 - b^2 = (a-b)(a+b)$,we have $\sin^4 x - \cos^4 x = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x)$.
Since $\sin^2 x + \cos^2 x = 1$,the integral becomes $I = \int {(\sin^2 x - \cos^2 x)\,dx}$.
Using the identity $\cos 2x = \cos^2 x - \sin^2 x$,we have $\sin^2 x - \cos^2 x = -\cos 2x$.
Therefore,$I = \int {-\cos 2x\,dx} = -\frac{\sin 2x}{2} + c$.

(A) Given the differential equation $x \frac{dy}{dx} = y(\log y - \log x + 1)$.
Dividing by $x$,we get $\frac{dy}{dx} = \frac{y}{x} (\log(\frac{y}{x}) + 1)$.
This is a homogeneous differential equation.
Let $v = \frac{y}{x}$,so $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = v(\log v + 1)$.
$v + x \frac{dv}{dx} = v \log v + v$.
$x \frac{dv}{dx} = v \log v$.
Separating the variables: $\frac{dv}{v \log v} = \frac{dx}{x}$.
Integrating both sides: $\int \frac{dv}{v \log v} = \int \frac{dx}{x}$.
Let $u = \log v$,then $du = \frac{1}{v} dv$. The integral becomes $\int \frac{du}{u} = \log x + C$.
$\log(\log v) = \log x + \log c = \log(cx)$.
Taking the exponential of both sides: $\log v = cx$.
Since $v = \frac{y}{x}$,we have $\log(\frac{y}{x}) = cx$.
Therefore,$\frac{y}{x} = e^{cx}$,which implies $y = x e^{cx}$.

(B) Let ${y_1} = {\sec ^{ - 1}}\left( \frac{1}{{2{x^2} - 1}} \right)$ and ${y_2} = \sqrt {1 - {x^2}} $.
Using the substitution $x = \cos \theta$,we have ${y_1} = {\sec ^{ - 1}}\left( \frac{1}{{2\cos^2 \theta - 1}} \right) = {\sec ^{ - 1}}(\sec 2\theta) = 2\theta = 2\cos^{-1}x$.
Now,differentiate ${y_1}$ with respect to $x$: $\frac{{d{y_1}}}{{dx}} = 2 \times \left( -\frac{1}{\sqrt{1-x^2}} \right) = -\frac{2}{\sqrt{1-x^2}}$.
Differentiate ${y_2}$ with respect to $x$: $\frac{{d{y_2}}}{{dx}} = \frac{1}{2\sqrt{1-x^2}} \times (-2x) = -\frac{x}{\sqrt{1-x^2}}$.
Therefore,the differential coefficient is $\frac{{d{y_1}}}{{d{y_2}}} = \frac{d{y_1}/dx}{d{y_2}/dx} = \frac{-2/\sqrt{1-x^2}}{-x/\sqrt{1-x^2}} = \frac{2}{x}$.
At $x = \frac{1}{2}$,$\frac{{d{y_1}}}{{d{y_2}}} = \frac{2}{1/2} = 4$.