AIEEE 2007 Mathematics Question Paper with Answer and Solution

(A) Let $O$ be the centre of the circular park and $P$ be the top of the tower. Thus,$OP$ is the height of the tower.
Since $O$ is the centre,$OA = OB = R$ (radius of the park).
Given that $\angle AOB = 60^{\circ}$ and $OA = OB$,$\Delta AOB$ is an equilateral triangle.
Therefore,$OA = OB = AB = a$.
In the right-angled triangle $\Delta AOP$,where $\angle OAP = 30^{\circ}$ is the angle of elevation:
$\tan 30^{\circ} = \frac{OP}{OA}$
$\frac{1}{\sqrt{3}} = \frac{OP}{a}$
$OP = \frac{a}{\sqrt{3}}$
Thus,the height of the tower is $\frac{a}{\sqrt{3}}$.

(A) The inequality $|z - (-4)| \le 3$ represents the set of all points $z$ that lie on or inside a circle with center $C(-4, 0)$ and radius $r = 3$.
We want to find the maximum value of $|z - (-1)|$,which represents the distance between $z$ and the point $A(-1, 0)$.
The distance between the center $C(-4, 0)$ and the point $A(-1, 0)$ is $d = \sqrt{(-1 - (-4))^2 + (0 - 0)^2} = \sqrt{3^2} = 3$.
The maximum distance from a point $A$ to any point $z$ inside or on the circle is given by $d + r$.
Therefore,the maximum value is $3 + 3 = 6$.

(C) The set $S$ contains $12$ elements.
We need to partition $S$ into three disjoint sets $A, B, C$ of equal size.
Since $|S| = 12$,each set must contain $12 / 3 = 4$ elements.
The number of ways to distribute $12$ distinct items into $3$ unlabeled groups of size $4$ is given by the formula $\frac{1}{3!} \binom{12}{4, 4, 4}$.
This is calculated as:
$\frac{1}{3!} \times \binom{12}{4} \times \binom{8}{4} \times \binom{4}{4} = \frac{1}{3!} \times \frac{12!}{4!8!} \times \frac{8!}{4!4!} \times \frac{4!}{4!0!} = \frac{12!}{3!(4!)^3}$.

(B) The general term in the expansion of $(a - b)^n$ is given by $T_{r+1} = ^nC_r a^{n-r} (-b)^r$.
For the $5^{th}$ term $(r=4)$: $T_5 = ^nC_4 a^{n-4} (-b)^4 = ^nC_4 a^{n-4} b^4$.
For the $6^{th}$ term $(r=5)$: $T_6 = ^nC_5 a^{n-5} (-b)^5 = -^nC_5 a^{n-5} b^5$.
Given $T_5 + T_6 = 0$,we have:
$^nC_4 a^{n-4} b^4 - ^nC_5 a^{n-5} b^5 = 0$.
Rearranging the terms:
$^nC_4 a^{n-4} b^4 = ^nC_5 a^{n-5} b^5$.
Dividing both sides by $a^{n-5} b^4$:
$\frac{a}{b} = \frac{^nC_5}{^nC_4}$.
Using the formula $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$:
$\frac{a}{b} = \frac{n-5+1}{5} = \frac{n-4}{5}$.

(D) Let $S = \binom{20}{0} - \binom{20}{1} + \binom{20}{2} - \dots + \binom{20}{10}$.
We know that $\binom{n}{r} = \binom{n}{n-r}$.
Thus,$\binom{20}{0} = \binom{20}{20}$,$\binom{20}{1} = \binom{20}{19}$,...,$\binom{20}{9} = \binom{20}{11}$.
Consider the expansion $(1-1)^{20} = \sum_{r=0}^{20} (-1)^r \binom{20}{r} = 0$.
This gives $\binom{20}{0} - \binom{20}{1} + \dots - \binom{20}{9} + \binom{20}{10} - \binom{20}{11} + \dots + \binom{20}{20} = 0$.
Using symmetry,$\binom{20}{0} - \binom{20}{1} + \dots - \binom{20}{9} + \binom{20}{10} - \binom{20}{9} + \dots + \binom{20}{0} = 0$.
$2[\binom{20}{0} - \binom{20}{1} + \dots - \binom{20}{9}] + \binom{20}{10} = 0$.
Let $X = \binom{20}{0} - \binom{20}{1} + \dots - \binom{20}{9}$. Then $2X + \binom{20}{10} = 0$,so $X = -\frac{1}{2} \binom{20}{10}$.
The required sum is $S = X + \binom{20}{10} = -\frac{1}{2} \binom{20}{10} + \binom{20}{10} = \frac{1}{2} \binom{20}{10}$.

(D) We know the expansion of $e^x$ is given by:
$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots$
Substitute $x = -1$ into the expansion:
$e^{-1} = 1 + (-1) + \frac{(-1)^2}{2!} + \frac{(-1)^3}{3!} + \frac{(-1)^4}{4!} + \dots$
$e^{-1} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$
Simplifying the expression:
$e^{-1} = 0 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$
$e^{-1} = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$
Thus,the sum of the series is $e^{-1}$.

(A) Let the terms of the geometric progression be $a, ar, ar^2, \ldots$ where $a > 0$ and $r > 0$.
Given that each term equals the sum of the next two terms,we have:
$a = ar + ar^2$
Since $a \neq 0$,we can divide by $a$:
$1 = r + r^2$
$r^2 + r - 1 = 0$
Using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$r = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)}$
$r = \frac{-1 \pm \sqrt{5}}{2}$
Since the terms are positive,the common ratio $r$ must be positive.
Therefore,$r = \frac{\sqrt{5} - 1}{2}$.

(C) The line $QP$ lies along the negative $x$-axis,so its angle with the positive $x$-axis is $\pi$. The line $QR$ passes through $(0, 0)$ and $(3, 3\sqrt{3})$,so its slope is $m = \frac{3\sqrt{3} - 0}{3 - 0} = \sqrt{3}$. The angle $\theta$ that $QR$ makes with the positive $x$-axis is $\tan \theta = \sqrt{3}$,which means $\theta = \frac{\pi}{3}$.
The angle $\angle PQR$ is the angle between the line $QP$ (angle $\pi$) and the line $QR$ (angle $\frac{\pi}{3}$).
The measure of $\angle PQR = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.
The bisector of $\angle PQR$ will make an angle $\alpha$ with the positive $x$-axis,where $\alpha = \frac{\pi + \frac{\pi}{3}}{2} = \frac{4\pi/3}{2} = \frac{2\pi}{3}$.
The slope of the bisector is $\tan(\frac{2\pi}{3}) = -\sqrt{3}$.
Since the bisector passes through the origin $Q(0, 0)$,its equation is $y - 0 = -\sqrt{3}(x - 0)$,which simplifies to $y = -\sqrt{3}x$,or $\sqrt{3}x + y = 0$.

(A) Given vertices are $A(h, k)$,$B(1, 1)$,and $C(2, 1)$.
Since $AC$ is the hypotenuse,the right angle is at $B$. Thus,$AB \perp BC$.
The slope of $AB$ is $m_1 = \frac{k-1}{h-1}$ and the slope of $BC$ is $m_2 = \frac{1-1}{2-1} = 0$.
Since $AB \perp BC$,the line $AB$ must be vertical,so $h = 1$.
Now,the area of the triangle is given by $\frac{1}{2} \times \text{base} \times \text{height} = 1$.
Here,base $BC = \sqrt{(2-1)^2 + (1-1)^2} = 1$.
Height $AB = \sqrt{(1-1)^2 + (k-1)^2} = |k-1|$.
So,$\frac{1}{2} \times 1 \times |k-1| = 1$.
$|k-1| = 2$.
$k-1 = 2$ or $k-1 = -2$.
$k = 3$ or $k = -1$.
Therefore,the set of values for $k$ is $\{-1, 3\}$.

(A) The given pair of lines is $my^2 + (1 - m^2)xy - mx^2 = 0$.
This can be rewritten as $-mx^2 + (1 - m^2)xy + my^2 = 0$,or $mx^2 - (1 - m^2)xy - my^2 = 0$.
The lines $xy = 0$ represent the coordinate axes,$x = 0$ and $y = 0$.
The angle bisectors of the lines $x = 0$ and $y = 0$ are given by $x^2 - y^2 = 0$,which implies $x = y$ and $x = -y$.
If one of the lines $my^2 + (1 - m^2)xy - mx^2 = 0$ is a bisector,it must be either $x - y = 0$ or $x + y = 0$.
For $x - y = 0$,the equation is $x^2 - xy - xy + y^2 = 0$ (not matching) or by substituting $y = x$ into the original equation:
$m(x^2) + (1 - m^2)x^2 - mx^2 = 0 \Rightarrow (1 - m^2)x^2 = 0$,which implies $m^2 = 1$,so $m = \pm 1$.
However,if $m = 1$,the equation becomes $y^2 - x^2 = 0$,i.e.,$(y - x)(y + x) = 0$,which are the bisectors of $xy = 0$.
If $m = -1$,the equation becomes $-y^2 + 2xy + x^2 = 0$,which are not the bisectors.
Thus,$m = \pm 1$. Given the options,$m = 1$ is the correct choice.

(B) The equation of a circle with center $(h, k)$ tangent to the $x$-axis is $(x-h)^{2} + (y-k)^{2} = k^{2}$.
Since the circle passes through the point $(-1, 1)$,we substitute these coordinates into the equation:
$(-1-h)^{2} + (1-k)^{2} = k^{2}$
$1 + 2h + h^{2} + 1 - 2k + k^{2} = k^{2}$
$h^{2} + 2h + 2 - 2k = 0$
For $h$ to be a real coordinate,the discriminant $D$ of this quadratic equation in $h$ must be greater than or equal to $0$:
$D = (2)^{2} - 4(1)(2 - 2k) \ge 0$
$4 - 8 + 8k \ge 0$
$8k - 4 \ge 0$
$8k \ge 4$
$k \ge \frac{1}{2}$

(D) The equation of the parabola is $y^2 = 8x$,which is of the form $y^2 = 4ax$,where $4a = 8$,so $a = 2$.
The directrix of the parabola $y^2 = 4ax$ is given by $x = -a$,which is $x = -2$.
It is a standard property of parabolas that the locus of the intersection of two perpendicular tangents is the directrix of the parabola.
Since the point lies on the given tangent line $y = x + 2$ and also on the directrix $x = -2$,we substitute $x = -2$ into the line equation:
$y = (-2) + 2 = 0$.
Thus,the required point is $(-2, 0)$.

(A) The given equation is $x^2 \sec^2 \theta - y^2 \csc^2 \theta = 1$,which can be written as $\frac{x^2}{\cos^2 \theta} - \frac{y^2}{\sin^2 \theta} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we get $a^2 = \cos^2 \theta$ and $b^2 = \sin^2 \theta$.
The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{\sin^2 \theta}{\cos^2 \theta}} = \sqrt{1 + \tan^2 \theta} = \sqrt{\sec^2 \theta} = \sec \theta$.
The coordinates of the foci are $(\pm ae, 0)$.
$ae = \sqrt{a^2} \cdot e = \sqrt{\cos^2 \theta} \cdot \sec \theta = \cos \theta \cdot \frac{1}{\cos \theta} = 1$.
Since the distance of the foci from the center is $ae = 1$,which is independent of $\theta$,the foci remain constant.

(A) Let the number of boys be $x$ and the number of girls be $y$.
The total marks of boys is $52x$ and the total marks of girls is $42y$.
The combined average marks is given by $\frac{52x + 42y}{x + y} = 50$.
Multiplying both sides by $(x + y)$,we get $52x + 42y = 50(x + y)$.
Expanding the equation: $52x + 42y = 50x + 50y$.
Rearranging the terms: $52x - 50x = 50y - 42y$.
$2x = 8y$,which simplifies to $x = 4y$.
The total number of students is $x + y = 4y + y = 5y$.
The percentage of boys is $\frac{x}{x + y} \times 100 = \frac{4y}{5y} \times 100 = 80\%$.

(C) Let $\alpha$ and $\beta$ be the roots of the equation $x^2 + ax + 1 = 0$.
Then,$\alpha + \beta = -a$ and $\alpha \beta = 1$.
The difference between the roots is given by $|\alpha - \beta| = \sqrt{(\alpha + \beta)^2 - 4\alpha \beta}$.
Substituting the values,we get $|\alpha - \beta| = \sqrt{a^2 - 4}$.
According to the given condition,$|\alpha - \beta| < \sqrt{5}$.
Therefore,$\sqrt{a^2 - 4} < \sqrt{5}$.
Squaring both sides,we get $a^2 - 4 < 5$,which implies $a^2 < 9$.
This inequality holds when $|a| < 3$,which means $a \in (-3, 3)$.

(D) Given $p^{2} + q^{2} = 1$ where $p, q > 0$.
We know the identity $(p+q)^{2} = p^{2} + q^{2} + 2pq$.
Substituting the given value: $(p+q)^{2} = 1 + 2pq$.
By the $AM \geq GM$ inequality,for positive real numbers $p^{2}$ and $q^{2}$,we have $\frac{p^{2} + q^{2}}{2} \geq \sqrt{p^{2}q^{2}} = pq$.
Since $p^{2} + q^{2} = 1$,we get $\frac{1}{2} \geq pq$,which implies $2pq \leq 1$.
Substituting this into the identity: $(p+q)^{2} = 1 + 2pq \leq 1 + 1 = 2$.
Taking the square root on both sides,we get $p+q \leq \sqrt{2}$.
Thus,the maximum value of $(p+q)$ is $\sqrt{2}$.

(B) Let the angles made by the line with the positive directions of the $x$,$y$,and $z$-axes be $\alpha$,$\beta$,and $\gamma$ respectively.
Given $\alpha = \frac{\pi}{4}$ and $\beta = \frac{\pi}{4}$.
The direction cosines of the line are $l = \cos \alpha$,$m = \cos \beta$,and $n = \cos \gamma$.
We know that $l^2 + m^2 + n^2 = 1$.
Substituting the values,we get $\cos^2(\frac{\pi}{4}) + \cos^2(\frac{\pi}{4}) + \cos^2 \gamma = 1$.
Since $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$,we have $(\frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{2}})^2 + \cos^2 \gamma = 1$.
$\frac{1}{2} + \frac{1}{2} + \cos^2 \gamma = 1$.
$1 + \cos^2 \gamma = 1$.
$\cos^2 \gamma = 0$,which implies $\cos \gamma = 0$.
Therefore,$\gamma = \frac{\pi}{2}$.

(B) Let the direction ratios of the line $L$ be $(a, b, c)$.
Since the line $L$ lies in both planes,it is perpendicular to the normals of both planes.
The normal vectors are $\vec{n_1} = (2, 3, 1)$ and $\vec{n_2} = (1, 3, 2)$.
The direction vector $\vec{v}$ of the line $L$ is given by the cross product $\vec{n_1} \times \vec{n_2}$:
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & 3 & 2 \end{vmatrix} = \hat{i}(6-3) - \hat{j}(4-1) + \hat{k}(6-3) = 3\hat{i} - 3\hat{j} + 3\hat{k}$.
We can simplify the direction vector to $(1, -1, 1)$.
The direction cosines $(\ell, m, n)$ are obtained by normalizing the vector $(1, -1, 1)$:
Magnitude $= \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$.
Thus,$\ell = \frac{1}{\sqrt{3}}$,$m = -\frac{1}{\sqrt{3}}$,$n = \frac{1}{\sqrt{3}}$.
The angle $\alpha$ with the positive $x$-axis satisfies $\cos \alpha = \ell = \frac{1}{\sqrt{3}}$.

(D) The equation of the given sphere is $x^2 + y^2 + z^2 - 6x - 12y - 2z + 20 = 0$.
Comparing this with the general equation $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$,we get $2u = -6, 2v = -12, 2w = -2$.
Thus,$u = -3, v = -6, w = -1$.
The center of the sphere is $(-u, -v, -w) = (3, 6, 1)$.
Let the given end of the diameter be $A = (2, 3, 5)$ and the other end be $B = (\alpha, \beta, \gamma)$.
Since the center of the sphere is the midpoint of the diameter,we have:
$\frac{\alpha + 2}{2} = 3 \Rightarrow \alpha + 2 = 6 \Rightarrow \alpha = 4$
$\frac{\beta + 3}{2} = 6 \Rightarrow \beta + 3 = 12 \Rightarrow \beta = 9$
$\frac{\gamma + 5}{2} = 1 \Rightarrow \gamma + 5 = 2 \Rightarrow \gamma = -3$
Therefore,the coordinates of the other end are $(4, 9, -3)$.

(C) Given $f(x) = \int_{1}^{x} \frac{\log_{e} t}{1+t} dt$.
Consider $f\left(\frac{1}{x}\right) = \int_{1}^{1/x} \frac{\log_{e} t}{1+t} dt$.
Let $t = \frac{1}{u}$,then $dt = -\frac{1}{u^2} du$.
When $t=1, u=1$. When $t=1/x, u=x$.
$f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\log_{e}(1/u)}{1+(1/u)} \left(-\frac{1}{u^2}\right) du = \int_{1}^{x} \frac{-\log_{e} u}{\frac{u+1}{u}} \left(\frac{1}{u^2}\right) du = \int_{1}^{x} \frac{-\log_{e} u}{u(u+1)} du$.
Now,$F(x) = f(x) + f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\log_{e} t}{1+t} dt - \int_{1}^{x} \frac{\log_{e} t}{t(1+t)} dt$.
$F(x) = \int_{1}^{x} \log_{e} t \left( \frac{1}{1+t} - \frac{1}{t(1+t)} \right) dt = \int_{1}^{x} \log_{e} t \left( \frac{t-1}{t(1+t)} \right) dt$.
Wait,let's re-evaluate: $F(x) = \int_{1}^{x} \frac{\log_{e} t}{1+t} \left( 1 - \frac{1}{t} \right) dt = \int_{1}^{x} \frac{\log_{e} t}{1+t} \left( \frac{t-1}{t} \right) dt$.
Actually,$f(x) + f(1/x) = \int_{1}^{x} \frac{\log_{e} t}{1+t} dt + \int_{1}^{1/x} \frac{\log_{e} t}{1+t} dt$.
Using the property $\int_{1}^{1/x} \frac{\log_{e} t}{1+t} dt = \int_{1}^{x} \frac{\log_{e}(1/u)}{1+(1/u)} (-du/u^2) = \int_{1}^{x} \frac{\log_{e} u}{u(u+1)} du$.
$F(x) = \int_{1}^{x} \frac{\log_{e} t}{1+t} (1 + 1/t) dt = \int_{1}^{x} \frac{\log_{e} t}{t} dt$.
$F(x) = \left[ \frac{(\log_{e} t)^2}{2} \right]_{1}^{x} = \frac{(\log_{e} x)^2}{2}$.
For $x=e$,$F(e) = \frac{(\log_{e} e)^2}{2} = \frac{1}{2} = 0.5$.

(D) We are given the equation $\int_{\sqrt{2}}^{x} \frac{dt}{t\sqrt{t^2-1}} = \frac{\pi}{2}$.
Using the standard integral formula $\int \frac{dt}{t\sqrt{t^2-1}} = \sec^{-1} t$,we evaluate the definite integral:
$\left[\sec^{-1} t\right]_{\sqrt{2}}^{x} = \frac{\pi}{2}$.
Substituting the limits,we get:
$\sec^{-1} x - \sec^{-1} \sqrt{2} = \frac{\pi}{2}$.
Since $\sec^{-1} \sqrt{2} = \frac{\pi}{4}$,the equation becomes:
$\sec^{-1} x - \frac{\pi}{4} = \frac{\pi}{2}$.
Solving for $\sec^{-1} x$:
$\sec^{-1} x = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$.
Therefore,$x = \sec\left(\frac{3\pi}{4}\right) = -\sqrt{2}$.
Since $-\sqrt{2}$ is not among the given options,the correct choice is $(D)$.

(A) The curves are $y^2 = x$ and $y = |x|$.
Since $y = |x|$,the curves are $y^2 = x$ and $y = x$ (for $x \ge 0$) and $y = -x$ (for $x < 0$).
However,$y^2 = x$ implies $x \ge 0$,so we only consider the region where $x \ge 0$ and $y = x$.
The intersection points are found by $x^2 = x$,which gives $x(x-1) = 0$,so $x = 0$ and $x = 1$.
In the interval $[0, 1]$,$\sqrt{x} \ge x$.
The required area is given by $\int_{0}^{1} (\sqrt{x} - x) dx$.
$= \left[ \frac{x^{3/2}}{3/2} - \frac{x^2}{2} \right]_{0}^{1}$.
$= \left( \frac{2}{3} - \frac{1}{2} \right) - (0 - 0) = \frac{4-3}{6} = \frac{1}{6}$.

(D) Given,$D = \left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{array} \right|$.
Apply row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$:
$D = \left| \begin{array}{ccc} 1 & 1 & 1 \\ 0 & x & 0 \\ 0 & 0 & y \end{array} \right|$.
Expanding along the first column:
$D = 1 \cdot (x \cdot y - 0 \cdot 0) - 0 + 0 = xy$.
Since $D = xy$,it is clearly divisible by both $x$ and $y$.

(A) Given the matrix $A = \begin{bmatrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{bmatrix}$.
Since $A$ is an upper triangular matrix,its determinant $|A|$ is the product of its diagonal elements.
$|A| = 5 \times \alpha \times 5 = 25\alpha$.
We are given that $|A|^2 = 25$.
Substituting the value of $|A|$,we get $(25\alpha)^2 = 25$.
$625\alpha^2 = 25$.
$\alpha^2 = \frac{25}{625} = \frac{1}{25}$.
Taking the square root on both sides,$|\alpha| = \sqrt{\frac{1}{25}} = \frac{1}{5}$.

(D) For the function to be continuous at $x = 0$,we must define $f(0) = \lim_{x \to 0} f(x)$.
$\lim_{x \to 0} f(x) = \lim_{x \to 0} \left[ \frac{1}{x} - \frac{2}{e^{2x} - 1} \right]$
$= \lim_{x \to 0} \frac{e^{2x} - 1 - 2x}{x(e^{2x} - 1)}$
Using the Taylor series expansion $e^{2x} = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \dots$,we get:
$= \lim_{x \to 0} \frac{(1 + 2x + \frac{4x^2}{2} + \dots) - 1 - 2x}{x(1 + 2x + \dots - 1)}$
$= \lim_{x \to 0} \frac{\frac{4x^2}{2} + \dots}{x(2x + \dots)}$
$= \lim_{x \to 0} \frac{2x^2 + \dots}{2x^2 + \dots} = \frac{2}{2} = 1$.
Thus,$f(0) = 1$.

(D) Given $f(x) = \tan^{-1}(\sin x + \cos x)$.
To find where the function is increasing,we calculate the derivative $f'(x)$.
$f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot (\cos x - \sin x)$.
$f'(x) = \frac{\cos x - \sin x}{1 + (\sin^2 x + \cos^2 x + 2\sin x \cos x)} = \frac{\cos x - \sin x}{2 + \sin 2x}$.
For the function to be increasing,we require $f'(x) > 0$.
Since $2 + \sin 2x > 0$ for all $x$,we must have $\cos x - \sin x > 0$.
$\cos x > \sin x$.
Dividing by $\cos x$ (assuming $\cos x > 0$),we get $1 > \tan x$,which implies $x < \frac{\pi}{4}$.
Considering the domain of the function,the interval where $f'(x) > 0$ is $(-\frac{\pi}{2}, \frac{\pi}{4})$.

(A) According to the Mean Value Theorem,there exists a value $c \in (1, 3)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.
Given $f(x) = \log_e x$,we have $f'(x) = \frac{1}{x}$.
Here,$a = 1$ and $b = 3$.
So,$f'(c) = \frac{f(3) - f(1)}{3 - 1} = \frac{\log_e 3 - \log_e 1}{2}$.
Since $\log_e 1 = 0$,we get $f'(c) = \frac{\log_e 3}{2}$.
Substituting $f'(c) = \frac{1}{c}$,we have $\frac{1}{c} = \frac{\log_e 3}{2}$.
Therefore,$c = \frac{2}{\log_e 3} = 2 \log_3 e$.

(C) The function is defined as $f(x) = \text{Min}\{x + 1, |x| + 1\}$.
We can analyze this by considering two cases for $x$:
Case $1$: $x \ge 0$. Then $|x| = x$,so $f(x) = \text{Min}\{x + 1, x + 1\} = x + 1$.
Case $2$: $x < 0$. Then $|x| = -x$,so $f(x) = \text{Min}\{x + 1, -x + 1\}$.
For $x < 0$,$x + 1 < -x + 1$ is equivalent to $2x < 0$,which is true for all $x < 0$.
Thus,$f(x) = x + 1$ for all $x < 0$.
Combining these,$f(x) = x + 1$ for all $x \in R$.
Since $f(x) = x + 1$ is a linear polynomial,it is differentiable everywhere in $R$.
Therefore,option $C$ is correct.

(A) All circles passing through the origin and having their center on the $x$-axis at $(a, 0)$ will have a radius of $a$.
The general equation of such circles is given by:
$(x - a)^2 + y^2 = a^2$
$x^2 - 2ax + a^2 + y^2 = a^2$
$x^2 + y^2 - 2ax = 0$ --- $(i)$
This equation has only one arbitrary constant,$a$. Therefore,we differentiate it once with respect to $x$:
$2x + 2y\frac{dy}{dx} - 2a = 0$
$x + y\frac{dy}{dx} = a$
Substituting the value of $a$ into equation $(i)$:
$x^2 + y^2 - 2x(x + y\frac{dy}{dx}) = 0$
$x^2 + y^2 - 2x^2 - 2xy\frac{dy}{dx} = 0$
$y^2 - x^2 - 2xy\frac{dy}{dx} = 0$
$y^2 = x^2 + 2xy\frac{dy}{dx}$

(C) Let $I = \int \frac{dx}{\cos x + \sqrt{3} \sin x}$.
Multiply and divide by $2$:
$I = \int \frac{dx}{2 \left( \frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x \right)}$.
Using $\sin \frac{\pi}{6} = \frac{1}{2}$ and $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$:
$I = \frac{1}{2} \int \frac{dx}{\sin \frac{\pi}{6} \cos x + \cos \frac{\pi}{6} \sin x}$.
Using the formula $\sin(A+B) = \sin A \cos B + \cos A \sin B$:
$I = \frac{1}{2} \int \frac{dx}{\sin(x + \frac{\pi}{6})} = \frac{1}{2} \int \csc(x + \frac{\pi}{6}) dx$.
Using the standard integral $\int \csc \theta d\theta = \log |\tan \frac{\theta}{2}| + C$:
$I = \frac{1}{2} \log |\tan(\frac{x}{2} + \frac{\pi}{12})| + C$.

(D) Given the equation: $\sin ^{-1}\left(\frac{x}{5}\right) + \csc ^{-1}\left(\frac{5}{4}\right) = \frac{\pi}{2}$
We know that $\csc ^{-1}(z) = \sin ^{-1}\left(\frac{1}{z}\right)$. Therefore,$\csc ^{-1}\left(\frac{5}{4}\right) = \sin ^{-1}\left(\frac{4}{5}\right)$.
Substituting this into the equation: $\sin ^{-1}\left(\frac{x}{5}\right) + \sin ^{-1}\left(\frac{4}{5}\right) = \frac{\pi}{2}$
Rearranging the terms: $\sin ^{-1}\left(\frac{x}{5}\right) = \frac{\pi}{2} - \sin ^{-1}\left(\frac{4}{5}\right)$
Using the identity $\sin ^{-1}(y) + \cos ^{-1}(y) = \frac{\pi}{2}$,we have $\frac{\pi}{2} - \sin ^{-1}\left(\frac{4}{5}\right) = \cos ^{-1}\left(\frac{4}{5}\right)$.
So,$\sin ^{-1}\left(\frac{x}{5}\right) = \cos ^{-1}\left(\frac{4}{5}\right)$.
To solve for $x$,convert $\cos ^{-1}\left(\frac{4}{5}\right)$ to $\sin ^{-1}$. Since $\cos \theta = \frac{4}{5}$,the opposite side is $\sqrt{5^2 - 4^2} = 3$,so $\sin \theta = \frac{3}{5}$.
Thus,$\sin ^{-1}\left(\frac{x}{5}\right) = \sin ^{-1}\left(\frac{3}{5}\right)$.
Comparing the arguments,$\frac{x}{5} = \frac{3}{5}$,which gives $x = 3$.

(B) The total number of outcomes when a pair of dice is thrown is $6 \times 6 = 36$.
To get a score of $9$,the possible outcomes are $(3, 6), (4, 5), (5, 4), (6, 3)$.
Thus,the number of favorable outcomes is $4$.
The probability of getting a score of $9$ in a single throw is $p = \frac{4}{36} = \frac{1}{9}$.
The probability of not getting a score of $9$ is $q = 1 - p = 1 - \frac{1}{9} = \frac{8}{9}$.
Using the binomial distribution formula $P(X = k) = nC_k \times p^k \times q^{n-k}$,where $n = 3$ and $k = 2$:
$P(X = 2) = 3C_2 \times (\frac{1}{9})^2 \times (\frac{8}{9})^{3-2}$
$= 3 \times \frac{1}{81} \times \frac{8}{9}$
$= \frac{24}{729} = \frac{8}{243}$.

(B) The function $f(x) = 4^{-x^2} + \cos^{-1}\left( \frac{x}{2} - 1 \right) + \log(\cos x)$ is defined if all its components are defined.
$1$. For $\cos^{-1}\left( \frac{x}{2} - 1 \right)$ to be defined,we must have $-1 \leq \frac{x}{2} - 1 \leq 1$.
Adding $1$ to all parts: $0 \leq \frac{x}{2} \leq 2$.
Multiplying by $2$: $0 \leq x \leq 4$.
$2$. For $\log(\cos x)$ to be defined,we must have $\cos x > 0$.
In the interval $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$,$\cos x > 0$ is always true.
$3$. Combining these conditions with the given interval $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$:
We need $x \in [0, 4]$ $AND$ $x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$.
Since $\frac{\pi}{2} \approx 1.57$,the intersection is $[0, \frac{\pi}{2})$.
Thus,the largest interval is $\left[ 0, \frac{\pi}{2} \right)$.

(D) Let the angles the line makes with the positive directions of the $x$,$y$,and $z$-axes be $\alpha, \beta$,and $\gamma$ respectively.
Given that $\alpha = 45^\circ$ and $\beta = 45^\circ$.
Let the direction cosines of the line be $l, m, n$.
Then $l = \cos \alpha = \cos 45^\circ = \frac{1}{\sqrt{2}}$,
$m = \cos \beta = \cos 45^\circ = \frac{1}{\sqrt{2}}$,
$n = \cos \gamma$.
We know that $l^2 + m^2 + n^2 = 1$.
Substituting the values,we get $(\frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{2}})^2 + \cos^2 \gamma = 1$.
$\frac{1}{2} + \frac{1}{2} + \cos^2 \gamma = 1$.
$1 + \cos^2 \gamma = 1$.
$\cos^2 \gamma = 0$.
$\cos \gamma = 0$.
Therefore,$\gamma = 90^\circ$.

(A) Given that $2\vec{u} \times 3\vec{v}$ is a unit vector,its magnitude must be $1$.
Thus,$|2\vec{u} \times 3\vec{v}| = 1$.
Since $\vec{u}$ and $\vec{v}$ are unit vectors,we have $|\vec{u}| = 1$ and $|\vec{v}| = 1$.
Using the property of the cross product,$|2\vec{u} \times 3\vec{v}| = 6|\vec{u}||\vec{v}|\sin\theta = 6(1)(1)\sin\theta = 6\sin\theta$.
According to the problem,$6\sin\theta = 1$,which implies $\sin\theta = \frac{1}{6}$.
Since $\theta$ is an acute angle,there exists exactly one value of $\theta$ such that $\theta = \arcsin(\frac{1}{6})$.

(D) Given vectors are $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$,and $\vec{c} = x\hat{i} + (x-2)\hat{j} - \hat{k}$.
Since $\vec{c}$ lies in the plane of $\vec{a}$ and $\vec{b}$,the scalar triple product $[\vec{a} \vec{b} \vec{c}]$ must be zero.
This implies the determinant of the components is zero:
$\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ x & x-2 & -1 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1((-1)(-1) - 2(x-2)) - 1((1)(-1) - 2(x)) + 1((1)(x-2) - (-1)(x)) = 0$
$1(1 - 2x + 4) - 1(-1 - 2x) + 1(x - 2 + x) = 0$
$(5 - 2x) + (1 + 2x) + (2x - 2) = 0$
$5 - 2x + 1 + 2x + 2x - 2 = 0$
$2x + 4 = 0$
$2x = -4$
$x = -2$

(D) Since $\vec{a}, \vec{b},$ and $\vec{c}$ are coplanar,their scalar triple product is zero:
$[\vec{a} \vec{b} \vec{c}] = 0$
$\begin{vmatrix} \alpha & 2 & \beta \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = 0$
$\alpha(1-0) - 2(1-0) + \beta(1-0) = 0 \Rightarrow \alpha - 2 + \beta = 0 \Rightarrow \alpha + \beta = 2 \dots (i)$
Since $\vec{a}$ bisects the angle between $\vec{b}$ and $\vec{c}$,$\vec{a}$ must be proportional to the sum of the unit vectors along $\vec{b}$ and $\vec{c}$:
$\hat{b} = \frac{\vec{b}}{|\vec{b}|} = \frac{(1, 1, 0)}{\sqrt{2}}, \hat{c} = \frac{\vec{c}}{|\vec{c}|} = \frac{(0, 1, 1)}{\sqrt{2}}$
$\vec{a} = k(\hat{b} + \hat{c}) = \frac{k}{\sqrt{2}}(1, 2, 1)$
Comparing this with $\vec{a} = (\alpha, 2, \beta)$,we get $\frac{k}{\sqrt{2}} = 1$,so $k = \sqrt{2}$.
Thus,$\alpha = 1$ and $\beta = 1$.