IIT JEE 1985 Mathematics Question Paper with Answer and Solution

(B) Let $z_1, z_2, z_3$ be three complex numbers in $A.P.$
Then $2z_2 = z_1 + z_3$.
This implies $z_2 = \frac{z_1 + z_3}{2}$.
Thus,the complex number $z_2$ is the midpoint of the line segment joining the points $z_1$ and $z_3$.
Therefore,the three points $z_1, z_2,$ and $z_3$ are collinear and lie on a straight line in the complex plane.

(D) Given $|{z_1}| = |{z_2}| = 1$,we have ${z_1} = \cos {\theta _1} + i\sin {\theta _1}$ and ${z_2} = \cos {\theta _2} + i\sin {\theta _2}$,where ${\theta _1} = \arg({z_1})$ and ${\theta _2} = \arg({z_2})$.
Since ${z_1} = a + ib$ and ${z_2} = c + id$,we have $a = \cos {\theta _1}, b = \sin {\theta _1}, c = \cos {\theta _2}, d = \sin {\theta _2}$.
Given $R({z_1}\overline {{z_2}} ) = 0$,we have $R[(\cos {\theta _1} + i\sin {\theta _1})(\cos {\theta _2} - i\sin {\theta _2})] = \cos({\theta _1} - {\theta _2}) = 0$.
Thus,${\theta _1} - {\theta _2} = \pm \frac{\pi }{2}$,which implies ${\theta _1} = {\theta _2} \pm \frac{\pi }{2}$.
Now,${w_1} = a + ic = \cos {\theta _1} + i\cos {\theta _2}$. Since $\cos {\theta _2} = \sin {\theta _1}$ (or $-\sin {\theta _1}$),we have $|{w_1}|^2 = \cos^2 {\theta _1} + \cos^2 {\theta _2} = \cos^2 {\theta _1} + \sin^2 {\theta _1} = 1$,so $|{w_1}| = 1$.
Similarly,$|{w_2}|^2 = b^2 + d^2 = \sin^2 {\theta _1} + \sin^2 {\theta _2} = \sin^2 {\theta _1} + \cos^2 {\theta _1} = 1$,so $|{w_2}| = 1$.
Finally,$R({w_1}\overline {{w_2}} ) = R((a + ic)(b - id)) = R(ab - iad + ibc + cd) = ab + cd = \cos {\theta _1}\sin {\theta _1} + \cos {\theta _2}\sin {\theta _2} = \frac{1}{2}(\sin 2{\theta _1} + \sin 2{\theta _2})$.
Since ${\theta _1} = {\theta _2} + \frac{\pi }{2}$,$\sin 2{\theta _1} = \sin(2{\theta _2} + \pi) = -\sin 2{\theta _2}$,so $R({w_1}\overline {{w_2}} ) = 0$.
Thus,all the above are satisfied.

(D) Let $\alpha$ and $\beta$ be the first and $(2n - 1)^{th}$ terms of the $A.P.$,$G.P.$,and $H.P.$ respectively.
For $A.P.$: The $n^{th}$ term is $a = \frac{\alpha + \beta}{2}$ $(i)$
For $G.P.$: The $n^{th}$ term is $b = \sqrt{\alpha \beta}$ (ii)
For $H.P.$: The $n^{th}$ term is $c = \frac{2\alpha \beta}{\alpha + \beta}$ (iii)
From $(i)$,(ii),and (iii),we observe that $a, b, c$ are the Arithmetic Mean,Geometric Mean,and Harmonic Mean of $\alpha$ and $\beta$ respectively.
We know that for any two positive numbers $\alpha$ and $\beta$,$A.M. \ge G.M. \ge H.M.$
Thus,$a \ge b \ge c$,which matches option $(a)$.
Also,$ac = \left(\frac{\alpha + \beta}{2}\right) \left(\frac{2\alpha \beta}{\alpha + \beta}\right) = \alpha \beta = b^2$.
Therefore,$ac - b^2 = 0$,which matches option $(c)$.
Hence,both $(a)$ and $(c)$ are correct.

(B) Let the discriminants of $P(x) = 0$ and $Q(x) = 0$ be $D_1$ and $D_2$ respectively.
$D_1 = b^2 - 4ac$ and $D_2 = d^2 - 4(-a)c = d^2 + 4ac$.
Consider the sum of the discriminants: $D_1 + D_2 = (b^2 - 4ac) + (d^2 + 4ac) = b^2 + d^2$.
Since $b^2 \ge 0$ and $d^2 \ge 0$,we have $D_1 + D_2 \ge 0$.
If both $D_1 < 0$ and $D_2 < 0$,then $D_1 + D_2 < 0$,which contradicts $D_1 + D_2 \ge 0$.
Thus,at least one of $D_1$ or $D_2$ must be greater than or equal to $0$.
If $D_1 \ge 0$,$P(x) = 0$ has at least two real roots.
If $D_2 \ge 0$,$Q(x) = 0$ has at least two real roots.
Therefore,the equation $P(x) \cdot Q(x) = 0$ must have at least two real roots.

(A) Since $a, b, c$ are in $G.P.$,we have $b^2 = ac$.
The equation $ax^2 + 2bx + c = 0$ becomes $ax^2 + 2\sqrt{ac}x + c = 0$.
This can be written as $(\sqrt{a}x + \sqrt{c})^2 = 0$,which gives the repeated root $x = -\sqrt{\frac{c}{a}}$.
Since this is a common root,it must satisfy $dx^2 + 2ex + f = 0$.
Substituting $x = -\sqrt{\frac{c}{a}}$,we get $d(\frac{c}{a}) - 2e\sqrt{\frac{c}{a}} + f = 0$.
Dividing by $c$,we get $\frac{d}{a} - 2e\frac{1}{\sqrt{ac}} + \frac{f}{c} = 0$.
Since $b = \sqrt{ac}$,this becomes $\frac{d}{a} + \frac{f}{c} = \frac{2e}{b}$.
This condition implies that $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in $A.P.$

(A) Let the $r$ consecutive natural numbers be $n, n+1, n+2, \dots, n+r-1$.
Their product is given by $P = n(n+1)(n+2)\dots(n+r-1)$.
We know that the number of ways to choose $r$ objects from $n+r-1$ objects is given by the binomial coefficient $\binom{n+r-1}{r} = \frac{(n+r-1)!}{r!(n-1)!}$.
This can be rewritten as $\frac{(n+r-1)(n+r-2)\dots(n)}{r!} = \binom{n+r-1}{r}$.
Since $\binom{n+r-1}{r}$ is always an integer,it follows that the product $n(n+1)\dots(n+r-1)$ must be divisible by $r!$.

(D) We use the formula $\prod_{k=0}^{n-1} \cos(2^k \theta) = \frac{\sin(2^n \theta)}{2^n \sin \theta}$.
Here,$\theta = \frac{2\pi}{15}$ and $n = 4$.
$\cos \frac{2\pi}{15} \cos \frac{4\pi}{15} \cos \frac{8\pi}{15} \cos \frac{16\pi}{15} = \frac{\sin(2^4 \cdot \frac{2\pi}{15})}{2^4 \sin \frac{2\pi}{15}}$
$= \frac{\sin \frac{32\pi}{15}}{16 \sin \frac{2\pi}{15}}$
Since $\frac{32\pi}{15} = 2\pi + \frac{2\pi}{15}$,we have $\sin \frac{32\pi}{15} = \sin \frac{2\pi}{15}$.
$= \frac{\sin \frac{2\pi}{15}}{16 \sin \frac{2\pi}{15}} = \frac{1}{16}$.

(A) Let $l$ be the length of the ladder.
In the initial position,the ladder makes an angle $\alpha$ with the horizontal. The horizontal distance from the wall is $PA = l \cos \alpha$ and the vertical height is $AB = l \sin \alpha$.
In the final position,the ladder makes an angle $\beta$ with the horizontal. The horizontal distance from the wall is $QA = l \cos \beta$ and the vertical height is $AC = l \sin \beta$.
The foot of the ladder is pulled away by distance $x = QA - PA = l(\cos \beta - \cos \alpha)$.
The ladder slides down by distance $y = AB - AC = l(\sin \alpha - \sin \beta)$.
Now,taking the ratio $\frac{y}{x}$:
$\frac{y}{x} = \frac{l(\sin \alpha - \sin \beta)}{l(\cos \beta - \cos \alpha)} = \frac{2 \sin \frac{\alpha - \beta}{2} \cos \frac{\alpha + \beta}{2}}{2 \sin \frac{\alpha + \beta}{2} \sin \frac{\alpha - \beta}{2}}$
$\frac{y}{x} = \cot \frac{\alpha + \beta}{2}$
Therefore,$x = y \tan \frac{\alpha + \beta}{2}$.

(A) Let the lines be $L_1: x + y - 1 = 0$,$L_2: 2x + 3y - 6 = 0$,and $L_3: 4x - y + 4 = 0$.
To find the vertices,we solve the equations in pairs:
$1$. $L_1$ and $L_2$: $x + y = 1$ and $2x + 3y = 6$. Solving gives $x = -3, y = 4$. Vertex $A = (-3, 4)$.
$2$. $L_2$ and $L_3$: $2x + 3y = 6$ and $4x - y = -4$. Solving gives $x = -\frac{3}{7}, y = \frac{16}{7}$. Vertex $B = (-\frac{3}{7}, \frac{16}{7})$.
$3$. $L_1$ and $L_3$: $x + y = 1$ and $4x - y = -4$. Solving gives $x = -\frac{3}{5}, y = \frac{8}{5}$. Vertex $C = (-\frac{3}{5}, \frac{8}{5})$.
To find the orthocentre,we find the equations of altitudes:
Altitude from $A$ to $BC$: Slope of $BC = \frac{8/5 - 16/7}{-3/5 - (-3/7)} = \frac{56-80}{35} / \frac{-21+15}{35} = \frac{-24}{-6} = 4$. The altitude is perpendicular to $BC$,so its slope is $-\frac{1}{4}$. Equation: $y - 4 = -\frac{1}{4}(x + 3) \implies x + 4y = 13$.
Altitude from $B$ to $AC$: Slope of $AC = \frac{8/5 - 4}{-3/5 - (-3)} = \frac{-12/5}{12/5} = -1$. The altitude is perpendicular to $AC$,so its slope is $1$. Equation: $y - \frac{16}{7} = 1(x + \frac{3}{7}) \implies x - y = -\frac{19}{7} \implies 7x - 7y = -19$.
Solving $x + 4y = 13$ and $7x - 7y = -19$: From the first,$x = 13 - 4y$. Substituting into the second: $7(13 - 4y) - 7y = -19 \implies 91 - 28y - 7y = -19 \implies -35y = -110 \implies y = \frac{22}{7}$. Then $x = 13 - 4(\frac{22}{7}) = \frac{91 - 88}{7} = \frac{3}{7}$.
The orthocentre is $(\frac{3}{7}, \frac{22}{7})$,which lies in the First quadrant.

(C) The given lines are:
$ax + by + c = 0$ $(i)$
$bx + cy + a = 0$ $(ii)$
$cx + ay + b = 0$ $(iii)$
Three lines $a_1x + b_1y + c_1 = 0$,$a_2x + b_2y + c_2 = 0$,and $a_3x + b_3y + c_3 = 0$ are concurrent if the determinant of their coefficients is zero:
$\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0$
Expanding the determinant:
$a(cb - a^2) - b(b^2 - ac) + c(ab - c^2) = 0$
$abc - a^3 - b^3 + abc + abc - c^3 = 0$
$3abc - (a^3 + b^3 + c^3) = 0$
Therefore,$a^3 + b^3 + c^3 - 3abc = 0$.

(C) The given circle is $(x - 1)^2 + y^2 = 1$,which expands to $x^2 + y^2 - 2x = 0$.
Let $(h, k)$ be the midpoint of a chord.
The equation of a chord with midpoint $(h, k)$ is given by $T = S_1$,where $T = xh + yk - (x + h)$ and $S_1 = h^2 + k^2 - 2h$.
So,the equation of the chord is $xh + yk - (x + h) = h^2 + k^2 - 2h$.
Since the chord passes through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$ into the equation:
$0(h) + 0(k) - (0 + h) = h^2 + k^2 - 2h$
$-h = h^2 + k^2 - 2h$
$h^2 + k^2 - h = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 - x = 0$.

(C) The equation of the circle is $x^2 + y^2 - 4x - 2y - 11 = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2$,$f = -1$,and $c = -11$.
The center of the circle is $(-g, -f) = (2, 1)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 1 + 11} = \sqrt{16} = 4$.
The length of the tangent $L$ from point $(x_1, y_1) = (4, 5)$ is given by $L = \sqrt{x_1^2 + y_1^2 - 4x_1 - 2y_1 - 11}$.
$L = \sqrt{4^2 + 5^2 - 4(4) - 2(5) - 11} = \sqrt{16 + 25 - 16 - 10 - 11} = \sqrt{4} = 2$.
The quadrilateral formed by the two tangents,the two radii,and the center is composed of two congruent right-angled triangles.
The area of the quadrilateral is $2 \times (\frac{1}{2} \times \text{base} \times \text{height}) = 2 \times (\frac{1}{2} \times L \times r) = L \times r$.
Area $= 2 \times 4 = 8 \text{ sq. units}$.

(A) Let $(x', y')$ be the midpoint of a chord.
The equation of the chord with midpoint $(x', y')$ for the circle $x^2 + y^2 - 2x = 0$ is given by $T = S_1$.
$x x' + y y' - (x + x') = x'^2 + y'^2 - 2x'$
Since the chord passes through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$ into the equation:
$0 + 0 - (0 + x') = x'^2 + y'^2 - 2x'$
$-x' = x'^2 + y'^2 - 2x'$
$x'^2 + y'^2 - x' = 0$
Replacing $(x', y')$ with $(x, y)$,the locus is $x^2 + y^2 - x = 0$,which represents a circle.

(D) To find $\lim_{x \to 0} f(x)$,we evaluate the left-hand limit $(LHL)$ and the right-hand limit $(RHL)$ at $x = 0$.
For the $LHL$: $\lim_{x \to 0^-} f(x)$. As $x \to 0^-$,$x$ is slightly less than $0$,so $[x] = -1$.
Thus,$f(x) = \frac{\sin(-1)}{-1} = \frac{-\sin(1)}{-1} = \sin(1)$.
So,$\lim_{x \to 0^-} f(x) = \sin(1)$.
For the $RHL$: $\lim_{x \to 0^+} f(x)$. As $x \to 0^+$,$x$ is in the interval $[0, 1)$,so $[x] = 0$.
According to the definition of $f(x)$,when $[x] = 0$,$f(x) = 0$.
So,$\lim_{x \to 0^+} f(x) = 0$.
Since $\lim_{x \to 0^-} f(x) = \sin(1)$ and $\lim_{x \to 0^+} f(x) = 0$,the left-hand limit is not equal to the right-hand limit.
Therefore,$\lim_{x \to 0} f(x)$ does not exist.

(C) We know that the addition theorem of probability states that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Given the condition $P(A \cup B) = P(A \cap B)$,we substitute this into the formula:
$P(A \cap B) = P(A) + P(B) - P(A \cap B)$.
Rearranging the terms,we get $2P(A \cap B) = P(A) + P(B)$.
However,for the specific case where $P(A \cup B) = P(A \cap B)$,it implies that $P(A \setminus B) = 0$ and $P(B \setminus A) = 0$,which means $P(A) = P(A \cap B) = P(B)$.
Therefore,the correct relation is $P(A) = P(B)$.

(A) We are given the determinant $\Delta = \left| \begin{array}{ccc} a & a^2 & a^3 + 1 \\ b & b^2 & b^3 + 1 \\ c & c^2 & c^3 + 1 \end{array} \right|$.
By the property of determinants,we can split the third column into two parts:
$\Delta = \left| \begin{array}{ccc} a & a^2 & a^3 \\ b & b^2 & b^3 \\ c & c^2 & c^3 \end{array} \right| + \left| \begin{array}{ccc} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{array} \right|$.
Taking $abc$ common from the first determinant,we get:
$\Delta = abc \left| \begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array} \right| + \left| \begin{array}{ccc} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{array} \right|$.
In the second determinant,perform column swaps to match the first: $C_3 \leftrightarrow C_2$ then $C_2 \leftrightarrow C_1$. This results in two sign changes,so the sign remains positive:
$\Delta = abc \left| \begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array} \right| + \left| \begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array} \right| = (1 + abc) \left| \begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array} \right|$.
The value of the determinant $\left| \begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array} \right|$ is $(a - b)(b - c)(c - a)$.
Thus,$\Delta = (1 + abc)(a - b)(b - c)(c - a) = 0$.
Since $a, b, c$ are unequal,$(a - b)(b - c)(c - a) \neq 0$.
Therefore,the condition is $1 + abc = 0$.

(B) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\Delta = \frac{1}{2} \left| \begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array} \right|$.
Given that the determinants are equal,it implies that the areas of the two triangles are equal.
Equality of area does not imply that the triangles are congruent or similar.
Therefore,the correct option is $B$.

(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:
$\frac{a \cdot (b \times c)}{c \times a \cdot b} + \frac{b \cdot (a \times c)}{c \cdot (a \times b)} = \frac{[a, b, c]}{[c, a, b]} + \frac{[b, a, c]}{[c, a, b]}$
$= \frac{[a, b, c]}{[a, b, c]} + \frac{-[a, b, c]}{[a, b, c]}$
$= 1 - 1 = 0$.

(C) The function $f(x) = x(\sqrt{x} - \sqrt{x + 1})$ is defined for $x \ge 0$ because of the term $\sqrt{x}$.
For $x < 0$,the term $\sqrt{x}$ is not defined in the set of real numbers.
Since the function is not defined in the neighborhood $(-\delta, 0)$ for any $\delta > 0$,it cannot be continuous at $x = 0$.
By definition,a function must be continuous at a point to be differentiable at that point.
Therefore,$f(x)$ is not differentiable at $x = 0$.

(C) To find the derivative of $\log(\log x)$ with respect to $x$,we use the chain rule.
Let $y = \log(\log x)$.
Applying the chain rule,$\frac{dy}{dx} = \frac{d}{d(\log x)} \log(\log x) \cdot \frac{d}{dx} \log x$.
Since $\frac{d}{du} \log u = \frac{1}{u}$ and $\frac{d}{dx} \log x = \frac{1}{x}$,we get:
$\frac{dy}{dx} = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x}$.
This can be written as $(x \log x)^{-1}$.
Therefore,the correct option is $C$.

(B) Given $f(x) = \log_{x}(\log x)$.
Using the base change formula,we can write $f(x) = \frac{\log(\log x)}{\log x}$.
Now,differentiate $f(x)$ with respect to $x$ using the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$:
$f'(x) = \frac{\left(\frac{d}{dx}(\log(\log x))\right)(\log x) - (\log(\log x))(\frac{d}{dx}(\log x))}{(\log x)^2}$
$f'(x) = \frac{\left(\frac{1}{\log x} \cdot \frac{1}{x}\right)(\log x) - (\log(\log x))(\frac{1}{x})}{(\log x)^2}$
$f'(x) = \frac{\frac{1}{x} - \frac{\log(\log x)}{x}}{(\log x)^2} = \frac{1 - \log(\log x)}{x(\log x)^2}$.
Now,evaluate $f'(x)$ at $x = e$:
$f'(e) = \frac{1 - \log(\log e)}{e(\log e)^2}$.
Since $\log e = 1$ and $\log(1) = 0$,we have:
$f'(e) = \frac{1 - 0}{e(1)^2} = \frac{1}{e}$.

(C) Let $I = \int \frac{x^5}{\sqrt{1 + x^3}} dx$.
Put $1 + x^3 = t^2$,then $3x^2 dx = 2t dt$,which implies $x^2 dx = \frac{2}{3} t dt$.
Also,$x^3 = t^2 - 1$.
Substituting these into the integral:
$I = \int \frac{x^3 \cdot x^2 dx}{\sqrt{1 + x^3}} = \int \frac{(t^2 - 1) \cdot \frac{2}{3} t dt}{t}$
$I = \frac{2}{3} \int (t^2 - 1) dt$
$I = \frac{2}{3} \left( \frac{t^3}{3} - t \right) + c$
$I = \frac{2}{9} t^3 - \frac{2}{3} t + c$
Since $t = (1 + x^3)^{1/2}$,we have:
$I = \frac{2}{9} (1 + x^3)^{3/2} - \frac{2}{3} (1 + x^3)^{1/2} + c$.

(D) Let $I = \int_0^{\pi /2} \frac{x \sin x \cos x}{\cos^4 x + \sin^4 x} \, dx$ .....$(i)$
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:
$I = \int_0^{\pi /2} \frac{(\frac{\pi}{2} - x) \cos x \sin x}{\sin^4 x + \cos^4 x} \, dx$ .....$(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^{\pi /2} \frac{\frac{\pi}{2} \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx$
$I = \frac{\pi}{4} \int_0^{\pi /2} \frac{\sin x \cos x}{\cos^4 x + \sin^4 x} \, dx$
Divide numerator and denominator by $\cos^4 x$:
$I = \frac{\pi}{4} \int_0^{\pi /2} \frac{\tan x \sec^2 x}{1 + \tan^4 x} \, dx$
Let $\tan^2 x = t$,then $2 \tan x \sec^2 x \, dx = dt$,so $\tan x \sec^2 x \, dx = \frac{1}{2} dt$. When $x=0, t=0$; when $x=\frac{\pi}{2}, t \to \infty$.
$I = \frac{\pi}{4} \int_0^{\infty} \frac{1}{1 + t^2} \cdot \frac{1}{2} \, dt = \frac{\pi}{8} [\tan^{-1} t]_0^{\infty} = \frac{\pi}{8} \cdot \frac{\pi}{2} = \frac{\pi^2}{16}$.

(C) Let $I = \int_0^\pi {e^{\cos^2 x} \cos^3((2n + 1)x)} \, dx$.
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx$,we have:
$I = \int_0^\pi {e^{\cos^2(\pi - x)} \cos^3((2n + 1)(\pi - x))} \, dx$.
Since $\cos(\pi - x) = -\cos x$,we have $\cos^2(\pi - x) = \cos^2 x$.
Also,$\cos((2n + 1)(\pi - x)) = \cos((2n + 1)\pi - (2n + 1)x) = \cos((2n + 1)\pi) \cos((2n + 1)x) + \sin((2n + 1)\pi) \sin((2n + 1)x)$.
Since $(2n + 1)$ is an odd integer,$\cos((2n + 1)\pi) = -1$ and $\sin((2n + 1)\pi) = 0$.
Thus,$\cos((2n + 1)(\pi - x)) = -\cos((2n + 1)x)$.
Therefore,$\cos^3((2n + 1)(\pi - x)) = (-\cos((2n + 1)x))^3 = -\cos^3((2n + 1)x)$.
Substituting this back into the integral,we get $I = \int_0^\pi e^{\cos^2 x} (-\cos^3((2n + 1)x)) \, dx = -I$.
$2I = 0 \implies I = 0$.

(D) Let $b = b_1 i + b_2 j + b_3 k$.
Given $a \cdot b = 3$,we have $b_1 + b_2 + b_3 = 3$ ......$(i)$.
Given $a \times b = c$,where $a = (1, 1, 1)$ and $c = (0, 1, -1)$,we calculate the cross product:
$a \times b = \begin{vmatrix} i & j & k \\ 1 & 1 & 1 \\ b_1 & b_2 & b_3 \end{vmatrix} = (b_3 - b_2)i + (b_1 - b_3)j + (b_2 - b_1)k$.
Comparing this with $c = (0, 1, -1)$,we get:
$b_3 - b_2 = 0$ ......$(ii)$
$b_1 - b_3 = 1$ ......$(iii)$
$b_2 - b_1 = -1$ ......$(iv)$
From $(ii)$,$b_2 = b_3$. Substituting into $(i)$,$b_1 + 2b_2 = 3$.
From $(iii)$,$b_1 = 1 + b_3 = 1 + b_2$.
Substituting $b_1$ into $b_1 + 2b_2 = 3$,we get $(1 + b_2) + 2b_2 = 3$,so $3b_2 = 2$,which means $b_2 = \frac{2}{3}$.
Then $b_3 = \frac{2}{3}$ and $b_1 = 1 + \frac{2}{3} = \frac{5}{3}$.
Thus,$b = \left( \frac{5}{3}, \frac{2}{3}, \frac{2}{3} \right)$.

(A) Given that the vectors $(1, a, a^2)$,$(1, b, b^2)$,and $(1, c, c^2)$ are non-coplanar,the determinant of the matrix formed by these vectors is non-zero:
$\Delta = \left| \begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array} \right| \neq 0$.
We are given the equation:
$\left| \begin{array}{ccc} a & a^2 & 1 + a^3 \\ b & b^2 & 1 + b^3 \\ c & c^2 & 1 + c^3 \end{array} \right| = 0$.
Using the property of determinants,we can split the third column:
$\left| \begin{array}{ccc} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{array} \right| + \left| \begin{array}{ccc} a & a^2 & a^3 \\ b & b^2 & b^3 \\ c & c^2 & c^3 \end{array} \right| = 0$.
In the second determinant,factor out $a, b, c$ from the rows:
$\left| \begin{array}{ccc} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{array} \right| + abc \left| \begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array} \right| = 0$.
Note that $\left| \begin{array}{ccc} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{array} \right| = \left| \begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array} \right| = \Delta$ (after two column swaps).
Thus,$\Delta + abc \Delta = 0 \Rightarrow \Delta(1 + abc) = 0$.
Since $\Delta \neq 0$,we have $1 + abc = 0$,which implies $abc = -1$.

(A) Let $x = \cos^2 \theta$. Then $dx = -2 \cos \theta \sin \theta \, d\theta$.
Substituting these into the integral:
$\int \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} (-2 \cos \theta \sin \theta) \, d\theta = \int \sqrt{\frac{2 \sin^2(\theta/2)}{2 \cos^2(\theta/2)}} (-2 \cos \theta \sin \theta) \, d\theta$
$= \int \tan(\theta/2) (-2 \cos \theta \sin \theta) \, d\theta = \int \frac{\sin(\theta/2)}{\cos(\theta/2)} (-4 \sin(\theta/2) \cos(\theta/2) \cos \theta) \, d\theta$
$= -4 \int \sin^2(\theta/2) \cos \theta \, d\theta = -2 \int (1 - \cos \theta) \cos \theta \, d\theta$
$= -2 \int (\cos \theta - \cos^2 \theta) \, d\theta = -2 \int \cos \theta \, d\theta + \int (1 + \cos 2\theta) \, d\theta$
$= -2 \sin \theta + \theta + \frac{1}{2} \sin 2\theta + c = \theta + \sin \theta \cos \theta - 2 \sin \theta + c$
Since $x = \cos^2 \theta$,we have $\cos \theta = \sqrt{x}$ and $\sin \theta = \sqrt{1 - x}$.
Substituting back: $\cos^{-1}\sqrt{x} + \sqrt{1 - x} \cdot \sqrt{x} - 2 \sqrt{1 - x} + c = \cos^{-1}\sqrt{x} + \sqrt{1 - x}(\sqrt{x} - 2) + c$.

(D) Let $A$ be the event that the maximum number on the two chosen tickets is $\le 10$. Let $B$ be the event that the minimum number on the two chosen tickets is $5$. We need to find the conditional probability $P(B|A) = \frac{n(A \cap B)}{n(A)}$.
First,we determine $n(A)$,the number of ways to choose two distinct tickets such that both are $\le 10$. The number of ways to choose $2$ tickets from $10$ is $\binom{10}{2} = \frac{10 \times 9}{2} = 45$.
Next,we determine $n(A \cap B)$,the number of ways to choose two tickets such that the maximum is $\le 10$ and the minimum is $5$. This means one ticket must be $5$,and the other ticket must be in the set $\{6, 7, 8, 9, 10\}$.
There are $5$ such pairs: $(5, 6), (5, 7), (5, 8), (5, 9), (5, 10)$.
Thus,$n(A \cap B) = 5$.
Therefore,the required probability is $P(B|A) = \frac{5}{45} = \frac{1}{9}$.
Since $\frac{1}{9}$ is not among the options,the correct answer is $(D)$ None of these.