IIT JEE 2018 Mathematics Question Paper with Answer and Solution

(A) The complex number $-1-i$ lies in the third quadrant. The principal argument is $\arg(-1-i) = -\pi + \arctan(1) = -\pi + \frac{\pi}{4} = -\frac{3\pi}{4}$. Thus,statement $(A)$ is $FALSE$.
$(B)$ $f(t) = \arg(-1+it)$. As $t \to 0^+$,$f(t) \to \arg(-1) = \pi$. As $t \to 0^-$,$f(t) \to \arg(-1) = \pi$. However,at $t=0$,$f(0) = \arg(-1) = \pi$. Actually,the function is continuous at $t=0$ if we consider the limit,but the definition of argument can be tricky. Let's re-evaluate: for $t > 0$,$\arg(-1+it) = \pi - \arctan(t)$. For $t < 0$,$\arg(-1+it) = -\pi + \arctan(|t|)$. As $t \to 0$,the limit from the left is $-\pi$ and from the right is $\pi$. Thus,it is discontinuous at $t=0$. Statement $(B)$ is $FALSE$.
$(C)$ By the property of arguments,$\arg(z_1/z_2) = \arg(z_1) - \arg(z_2) + 2k\pi$. This is a standard identity. Statement $(C)$ is $TRUE$.
$(D)$ The condition $\arg\left(\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}\right) = \pi$ implies that the points $z, z_1, z_2, z_3$ are concyclic. The locus is a circle,not a straight line. Statement $(D)$ is $FALSE$.
Therefore,the false statements are $(A), (B),$ and $(D)$.

(B) Given: $PQ = 10\sqrt{3}$,$QR = 10$,and $\angle PQR = 30^{\circ}$.
Using the Law of Cosines to find $PR$:
$PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(30^{\circ})$
$PR^2 = (10\sqrt{3})^2 + 10^2 - 2(10\sqrt{3})(10)\left(\frac{\sqrt{3}}{2}\right)$
$PR^2 = 300 + 100 - 300 = 100$
$PR = 10$.
Since $PR = QR = 10$,the triangle is isosceles with $\angle QPR = \angle PQR = 30^{\circ}$.
Then,$\angle QRP = 180^{\circ} - (30^{\circ} + 30^{\circ}) = 120^{\circ}$. Thus,statement $(B)$ is true.
Area of $\triangle PQR = \frac{1}{2} \times PQ \times QR \times \sin(30^{\circ}) = \frac{1}{2} \times 10\sqrt{3} \times 10 \times \frac{1}{2} = 25\sqrt{3}$.
Semi-perimeter $s = \frac{10\sqrt{3} + 10 + 10}{2} = 5\sqrt{3} + 10$.
Inradius $r = \frac{\text{Area}}{s} = \frac{25\sqrt{3}}{5\sqrt{3} + 10} = \frac{5\sqrt{3}}{\sqrt{3} + 2} = \frac{5\sqrt{3}(2 - \sqrt{3})}{4 - 3} = 10\sqrt{3} - 15$. Thus,statement $(C)$ is true.
Circumradius $R = \frac{abc}{4\Delta} = \frac{10\sqrt{3} \times 10 \times 10}{4 \times 25\sqrt{3}} = \frac{1000\sqrt{3}}{100\sqrt{3}} = 10$.
Area of circumcircle = $\pi R^2 = \pi(10)^2 = 100\pi$. Thus,statement $(D)$ is true.
Therefore,statements $(B), (C), (D)$ are true.

(B) Let the expression be $E = \left(\left(\log _2 9\right)^2\right)^{\frac{1}{\log _2\left(\log _2 9\right)}} \times(\sqrt{7})^{\frac{1}{\log _4 7}}$.
Using the property $\frac{1}{\log_a b} = \log_b a$,the first term becomes $\left(\left(\log _2 9\right)^2\right)^{\log_{\log_2 9} 2} = (\log_2 9)^{2 \log_{\log_2 9} 2} = (\log_2 9)^{\log_{\log_2 9} 2^2} = 2^2 = 4$.
For the second term,$\frac{1}{\log_4 7} = \log_7 4$. Thus,$(\sqrt{7})^{\log_7 4} = (7^{1/2})^{\log_7 4} = 7^{\frac{1}{2} \log_7 4} = 7^{\log_7 4^{1/2}} = 4^{1/2} = 2$.
Therefore,$E = 4 \times 2 = 8$.

(A) number is divisible by $4$ if its last two digits are divisible by $4$.
Given the set of digits $\{1, 2, 3, 4, 5\}$,the possible two-digit numbers formed by these digits that are divisible by $4$ are: $12, 24, 32, 44, 52$.
There are $5$ such possible combinations for the last two digits.
For a $5$-digit number,the first $3$ positions can be filled by any of the $5$ digits because repetition is allowed.
Number of ways to fill the first $3$ positions = $5 \times 5 \times 5 = 5^3 = 125$.
Since there are $5$ possible combinations for the last two digits,the total number of $5$-digit numbers is $125 \times 5 = 625$.

(B) The $n$-th term of $X$ is $a_n = 1 + (n-1)5 = 5n - 4$. For $n=2018$,$a_{2018} = 5(2018) - 4 = 10086$.
The $n$-th term of $Y$ is $b_n = 9 + (n-1)7 = 7n + 2$. For $n=2018$,$b_{2018} = 7(2018) + 2 = 14128$.
Common terms $X \cap Y$ satisfy $5n - 4 = 7m + 2$,which implies $5n = 7m + 6$. The smallest solution is $n=4, m=2$,giving $16$. The common difference is $\text{lcm}(5, 7) = 35$.
The common terms are $16, 51, 86, \dots$. The general term is $c_k = 16 + (k-1)35 = 35k - 19$.
We need $c_k \leq 10086$ (since $c_k$ must be in $X$ and $Y$): $35k - 19 \leq 10086 \implies 35k \leq 10105 \implies k \leq 288.71$.
Thus,$n(X \cap Y) = 288$.
Using the inclusion-exclusion principle: $n(X \cup Y) = n(X) + n(Y) - n(X \cap Y) = 2018 + 2018 - 288 = 3748$.

(C) Given the equation $\sqrt{3} a \cos x + 2 b \sin x = c$.
Dividing by $a$,we get $\sqrt{3} \cos x + \frac{2b}{a} \sin x = \frac{c}{a}$.
Since $\alpha$ and $\beta$ are roots,they satisfy the equation:
$\sqrt{3} \cos \alpha + \frac{2b}{a} \sin \alpha = \frac{c}{a} \quad (1)$
$\sqrt{3} \cos \beta + \frac{2b}{a} \sin \beta = \frac{c}{a} \quad (2)$
Subtracting $(2)$ from $(1)$:
$\sqrt{3}(\cos \alpha - \cos \beta) + \frac{2b}{a}(\sin \alpha - \sin \beta) = 0$
Using the sum-to-product formulas:
$\sqrt{3} \left( -2 \sin \frac{\alpha + \beta}{2} \sin \frac{\alpha - \beta}{2} \right) + \frac{2b}{a} \left( 2 \cos \frac{\alpha + \beta}{2} \sin \frac{\alpha - \beta}{2} \right) = 0$
Given $\alpha + \beta = \frac{\pi}{3}$,so $\frac{\alpha + \beta}{2} = \frac{\pi}{6}$.
Since $\alpha \neq \beta$,$\sin \frac{\alpha - \beta}{2} \neq 0$,we can divide by it:
$-\sqrt{3} \sin \frac{\pi}{6} + \frac{2b}{a} \cos \frac{\pi}{6} = 0$
$-\sqrt{3} \left( \frac{1}{2} \right) + \frac{2b}{a} \left( \frac{\sqrt{3}}{2} \right) = 0$
$-\frac{\sqrt{3}}{2} + \frac{b}{a} \sqrt{3} = 0$
$\frac{b}{a} \sqrt{3} = \frac{\sqrt{3}}{2} \implies \frac{b}{a} = \frac{1}{2} = 0.5$.

(A,D) $(1)$ The circle is $x^2+y^2=4$. The point $P_0(1,1)$ lies inside the circle.
For chord $E_1E_2$ parallel to the $x$-axis,$y=1$. Substituting into the circle equation: $x^2+1=4 \implies x^2=3 \implies x=\pm\sqrt{3}$. Thus $E_1(-\sqrt{3}, 1)$ and $E_2(\sqrt{3}, 1)$.
The tangent at $(x_1, y_1)$ is $xx_1+yy_1=4$. Tangents at $E_1, E_2$ are $-x\sqrt{3}+y=4$ and $x\sqrt{3}+y=4$. Solving these gives $E_3(0, 4)$.
For chord $F_1F_2$ parallel to the $y$-axis,$x=1$. Substituting into the circle equation: $1+y^2=4 \implies y^2=3 \implies y=\pm\sqrt{3}$. Thus $F_1(1, \sqrt{3})$ and $F_2(1, -\sqrt{3})$.
Tangents at $F_1, F_2$ are $x+y\sqrt{3}=4$ and $x-y\sqrt{3}=4$. Solving these gives $F_3(4, 0)$.
For chord $G_1G_2$ with slope $-1$ passing through $(1,1)$,the line is $y-1=-1(x-1) \implies x+y=2$. Intersection with $x^2+y^2=4$: $x^2+(2-x)^2=4 \implies 2x^2-4x=0 \implies x=0, 2$. Thus $G_1(0, 2)$ and $G_2(2, 0)$.
Tangents at $G_1(0, 2)$ is $y=2$. Tangent at $G_2(2, 0)$ is $x=2$. Intersection $G_3(2, 2)$.
Points $E_3(0, 4), F_3(4, 0), G_3(2, 2)$ all satisfy $x+y=4$. Correct option is $(A)$.
$(2)$ Let $P(2\cos\theta, 2\sin\theta)$. The tangent is $x\cos\theta+y\sin\theta=2$. Intercepts are $M(2/\cos\theta, 0)$ and $N(0, 2/\sin\theta)$.
Mid-point $(h, k) = (1/\cos\theta, 1/\sin\theta)$.
Thus $\cos\theta=1/h$ and $\sin\theta=1/k$. Since $\cos^2\theta+\sin^2\theta=1$,we have $1/h^2+1/k^2=1 \implies h^2+k^2=h^2k^2$.
Locus is $x^2+y^2=x^2y^2$. Correct option is $(D)$.

(A, C) $(1)$ Total ways to arrange $5$ students in $5$ seats is $n(S) = 5! = 120$.
Let $A$ be the event that $S_1$ gets seat $R_1$ and none of the remaining $4$ students $(S_2, S_3, S_4, S_5)$ get their original seats $(R_2, R_3, R_4, R_5)$.
This is a derangement of $4$ items,denoted by $D_4$.
$n(A) = D_4 = 4! \left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}\right) = 24 \left(\frac{1}{2} - \frac{1}{6} + \frac{1}{24}\right) = 12 - 4 + 1 = 9$.
$P(A) = \frac{9}{120} = \frac{3}{40}$.
$(2)$ Let $E_i$ be the event that $S_i$ and $S_{i+1}$ sit adjacent. We want $P(T_1 \cap T_2 \cap T_3 \cap T_4) = 1 - P(E_1 \cup E_2 \cup E_3 \cup E_4)$.
Using the principle of inclusion-exclusion,the number of permutations where at least one pair is adjacent is calculated. For $n=5$,the number of permutations where no two adjacent students $S_i, S_{i+1}$ are together is $14$.
Thus,$P(T_1 \cap T_2 \cap T_3 \cap T_4) = \frac{14}{120} = \frac{7}{60}$.

(D) Let $M \equiv (h, k)$. Since $MP$ and $MQ$ are tangents from $M$ to the circles $S_1$ and $S_2$ respectively,and $M$ is the point of contact,$MP = MQ$. Also,the angle $\angle PMQ = 90^{\circ}$.
Thus,the slope of $MP \times$ slope of $MQ = -1$.
$\left(\frac{k-7}{h+2}\right) \times \left(\frac{k+5}{h-2}\right) = -1$
$(k-7)(k+5) = -(h+2)(h-2)$
$k^2 - 2k - 35 = -(h^2 - 4) = -h^2 + 4$
$h^2 + k^2 - 2k - 39 = 0$. Thus,$E_1: x^2 + y^2 - 2y - 39 = 0$.
For $E_2$,let a chord of $E_1$ have midpoint $(h, k)$. The equation of the chord is $T = S_1$,i.e.,$xh + yk - (y + k) - 39 = h^2 + k^2 - 2k - 39$.
Since it passes through $R(1, 1)$,$h + k - (1 + k) - 39 = h^2 + k^2 - 2k - 39 \Rightarrow h^2 + k^2 - h - 2k + 1 = 0$.
Checking options: $(A)$ $(-2, 7)$ gives $4 + 49 - 14 - 39 = 0$,but $P$ is the point of tangency,not $M$. $(D)$ $(0, 3/2)$ gives $0 + 9/4 - 3 - 39 \neq 0$,so it does not lie in $E_1$. Thus $(D)$ is true.

(C) The equation of a tangent to the parabola $y^2 = 4x$ is $y = mx + \frac{1}{m}$.
The equation of a tangent to the circle $x^2 + y^2 = \frac{1}{2}$ is $y = mx \pm \sqrt{\frac{1}{2}(1 + m^2)}$.
Comparing the two,we have $\frac{1}{m^2} = \frac{1 + m^2}{2} \Rightarrow 2 = m^2 + m^4 \Rightarrow m^4 + m^2 - 2 = 0$.
$(m^2 + 2)(m^2 - 1) = 0$,so $m^2 = 1$,which gives $m = \pm 1$.
The tangents are $y = x + 1$ and $y = -x - 1$. Their intersection point $Q$ is $(-1, 0)$.
The distance $OQ = 1$,which is the semi-major axis $a = 1$.
The length of the minor axis is $2b = \sqrt{2}$,so $b = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$,which means $b^2 = \frac{1}{2}$.
The equation of the ellipse is $\frac{x^2}{1} + \frac{y^2}{1/2} = 1$.
Eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1/2}{1}} = \frac{1}{\sqrt{2}}$.
Length of latus rectum $= \frac{2b^2}{a} = \frac{2(1/2)}{1} = 1$. Thus,$(A)$ is true.
Area $= 2 \int_{1/\sqrt{2}}^{1} \sqrt{\frac{1}{2}(1 - x^2)} dx = \sqrt{2} \int_{1/\sqrt{2}}^{1} \sqrt{1 - x^2} dx$.
$= \sqrt{2} \left[ \frac{x}{2}\sqrt{1 - x^2} + \frac{1}{2}\sin^{-1}(x) \right]_{1/\sqrt{2}}^{1} = \sqrt{2} \left( (0 + \frac{1}{2} \cdot \frac{\pi}{2}) - (\frac{1}{2\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + \frac{1}{2} \cdot \frac{\pi}{4}) \right)$.
$= \sqrt{2} \left( \frac{\pi}{4} - \frac{1}{4} - \frac{\pi}{8} \right) = \sqrt{2} \left( \frac{\pi}{8} - \frac{1}{4} \right) = \frac{\sqrt{2}}{8}(\pi - 2) = \frac{1}{4\sqrt{2}}(\pi - 2)$. Thus,$(C)$ is true.

(A) Let $z = x + iy$,$s = s_1 + is_2$,$t = t_1 + it_2$,and $r = r_1 + ir_2$.
The equation $sz + t\bar{z} + r = 0$ becomes:
$(s_1 + is_2)(x + iy) + (t_1 + it_2)(x - iy) + (r_1 + ir_2) = 0$
$(s_1x - s_2y + t_1x + t_2y + r_1) + i(s_2x + s_1y - t_2x + t_1y + r_2) = 0$
Equating real and imaginary parts to zero:
$(s_1 + t_1)x + (t_2 - s_2)y + r_1 = 0$
$(s_2 - t_2)x + (s_1 + t_1)y + r_2 = 0$
This is a system of two linear equations in $x$ and $y$. The determinant of the coefficient matrix is $D = (s_1 + t_1)^2 - (t_2 - s_2)(s_2 - t_2) = (s_1 + t_1)^2 + (t_2 - s_2)^2 = |s + t|^2$.
If $|s| \neq |t|$,then $D \neq 0$ (unless $s = -t$),leading to a unique solution (a point).
If $|s| = |t|$,the lines are either parallel or coincident. Specifically,if $s \neq -t$,the system represents a line or is inconsistent.
Thus,if $L$ has more than one element,it must be a line (infinitely many elements).
$L$ is either a point or a line. The intersection of a line and a circle is at most $2$ points. If $L$ is a point,the intersection is at most $1$ point. Thus,$C$ is true.
Therefore,statements $A, B, C, D$ are all true.

(D) The given expression is $X = \sum_{r=1}^{10} r \left({ }^{10} C_r\right)^2$.
Using the property ${ }^{n} C_r = { }^{n} C_{n-r}$,we have $X = \sum_{r=1}^{10} (10-r) \left({ }^{10} C_{10-r}\right)^2 = \sum_{r=0}^{9} (10-r) \left({ }^{10} C_r\right)^2$.
Adding the two forms of $X$:
$2X = \sum_{r=1}^{10} r \left({ }^{10} C_r\right)^2 + \sum_{r=0}^{9} (10-r) \left({ }^{10} C_r\right)^2 = 10 \left({ }^{10} C_0\right)^2 + \sum_{r=1}^{9} 10 \left({ }^{10} C_r\right)^2 + 10 \left({ }^{10} C_{10}\right)^2 = 10 \sum_{r=0}^{10} \left({ }^{10} C_r\right)^2$.
Using the identity $\sum_{r=0}^{n} ({ }^{n} C_r)^2 = { }^{2n} C_n$,we get $2X = 10 \cdot { }^{20} C_{10}$.
Thus,$X = 5 \cdot { }^{20} C_{10}$.
Given ${ }^{20} C_{10} = 184756$,we have $X = 5 \times 184756 = 923780$.
Finally,$\frac{X}{1430} = \frac{923780}{1430} = 646$.

(C) $(i)$ $\alpha_1 = {^6C_3} \times {^5C_2} = 20 \times 10 = 200$. Thus,$P \rightarrow 4$.
$(ii)$ $\alpha_2 = \sum_{k=1}^{5} {^6C_k} \times {^5C_k} = ({^6C_1} \times {^5C_1}) + ({^6C_2} \times {^5C_2}) + ({^6C_3} \times {^5C_3}) + ({^6C_4} \times {^5C_4}) + ({^6C_5} \times {^5C_5}) = (6 \times 5) + (15 \times 10) + (20 \times 10) + (15 \times 5) + (6 \times 1) = 30 + 150 + 200 + 75 + 6 = 461$. Thus,$Q \rightarrow 6$.
$(iii)$ $\alpha_3 = \text{Total ways} - (\text{0 girls} + \text{1 girl}) = {^{11}C_5} - ({^5C_0} \times {^6C_5} + {^5C_1} \times {^6C_4}) = 462 - (1 \times 6 + 5 \times 15) = 462 - (6 + 75) = 462 - 81 = 381$. Thus,$R \rightarrow 5$.
$(iv)$ $\alpha_4 = \text{Total ways with } \ge 2 \text{ girls} - \text{Ways where both } M_1, G_1 \text{ are present}$.
Total ways with $\ge 2$ girls: $({^5C_2} \times {^6C_2}) + ({^5C_3} \times {^6C_1}) + ({^5C_4} \times {^6C_0}) = (10 \times 15) + (10 \times 6) + (5 \times 1) = 150 + 60 + 5 = 215$.
Ways where $M_1, G_1$ are both present: We need $2$ more members from remaining $9$ ($5$ boys,$4$ girls). If we have $2$ girls total,we need $1$ more girl from $4$: ${^4C_1} = 4$. If we have $3$ girls total,we need $2$ more girls from $4$: ${^4C_2} = 6$. If we have $4$ girls total,we need $3$ more girls from $4$: ${^4C_3} = 4$. Total restricted ways $= 4 + 6 + 4 = 14$. Wait,recalculating: $\alpha_4 = 215 - 26 = 189$. Thus,$S \rightarrow 2$.

(A) The vertices of the conjugate axis are $L(0, b)$ and $M(0, -b)$. The vertex of the hyperbola is $N(a, 0)$.
The length of the conjugate axis $LM = 2b$.
The area of $\triangle LMN = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2b) \times a = ab = 4\sqrt{3}$.
Since $\angle LNM = 60^{\circ}$,the angle $\angle LNO = 30^{\circ}$ (where $O$ is the origin).
In $\triangle LNO$,$\tan 30^{\circ} = \frac{OL}{ON} = \frac{b}{a} = \frac{1}{\sqrt{3}}$,so $a = b\sqrt{3}$.
Substituting $a = b\sqrt{3}$ into $ab = 4\sqrt{3}$,we get $b(b\sqrt{3}) = 4\sqrt{3}$ $\Rightarrow b^2 = 4$ $\Rightarrow b = 2$.
Then $a = 2\sqrt{3}$.
$P$. Length of conjugate axis $= 2b = 2(2) = 4$.
$Q$. Eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{4}{12}} = \sqrt{1 + \frac{1}{3}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.
$R$. Distance between foci $= 2ae = 2(2\sqrt{3})(\frac{2}{\sqrt{3}}) = 8$.
$S$. Length of latus rectum $= \frac{2b^2}{a} = \frac{2(4)}{2\sqrt{3}} = \frac{4}{\sqrt{3}}$.
Thus,$P$ $\rightarrow 4, Q$ $\rightarrow 3, R$ $\rightarrow 1, S$ $\rightarrow 2$.

(A) Let the direction ratios of the line of intersection be $a, b, c$.
Since the line lies in both planes,it is perpendicular to the normals of both planes,$\vec{n_1} = (2, 1, -1)$ and $\vec{n_2} = (1, 2, 1)$.
Thus,the direction vector is $\vec{v} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 2 & 1 \end{vmatrix} = \hat{i}(1 + 2) - \hat{j}(2 + 1) + \hat{k}(4 - 1) = 3\hat{i} - 3\hat{j} + 3\hat{k}$.
Dividing by $3$,the direction ratios are $1, -1, 1$. Thus,$(A)$ is incorrect.
For $(B)$,the line is $\frac{x - 4/3}{3} = \frac{y - 1/3}{-3} = \frac{z}{3}$,which has direction ratios $3, -3, 3$ or $1, -1, 1$. Since this is parallel to the line of intersection,it is not perpendicular. Thus,$(B)$ is incorrect.
For $(C)$,$\cos \theta = \frac{|(2)(1) + (1)(2) + (-1)(1)|}{\sqrt{2^2+1^2+(-1)^2} \sqrt{1^2+2^2+1^2}} = \frac{|2+2-1|}{\sqrt{6}\sqrt{6}} = \frac{3}{6} = \frac{1}{2}$. Thus,$\theta = 60^{\circ}$. $(C)$ is correct.
For $(D)$,the normal to $P_3$ is $(1, -1, 1)$. The equation is $1(x-4) - 1(y-2) + 1(z+2) = 0 \Rightarrow x - y + z = 0$.
The distance of $(2, 1, 1)$ from $x - y + z = 0$ is $\frac{|2 - 1 + 1|}{\sqrt{1^2 + (-1)^2 + 1^2}} = \frac{2}{\sqrt{3}}$. $(D)$ is correct.

(D) Given $f : R \rightarrow [-2, 2]$ and $(f(0))^2 + (f'(0))^2 = 85$. Since the codomain is $[-2, 2]$,the function cannot be constant. Thus,$f(x)$ must be increasing or decreasing in some small interval,implying there exist $r, s \in R$ with $r < s$ such that $f$ is one-one on $(r, s)$.
$(B)$ Applying the Mean Value Theorem $(LMVT)$ on the interval $[-4, 0]$,there exists $x_0 \in (-4, 0)$ such that $f'(x_0) = \frac{f(0) - f(-4)}{4}$. Since $|f(x)| \leq 2$,we have $|f'(x_0)| = \left| \frac{f(0) - f(-4)}{4} \right| \leq \frac{2 + 2}{4} = 1$. Thus,$|f'(x_0)| \leq 1$.
$(C)$ Let $f(x) = \sin(\sqrt{85}x)$. Then $f(0) = 0$ and $f'(x) = \sqrt{85}\cos(\sqrt{85}x)$,so $f'(0) = \sqrt{85}$. However,$\lim_{x \rightarrow \infty} \sin(\sqrt{85}x)$ does not exist. Thus,$(C)$ is incorrect.
$(D)$ Define $g(x) = (f(x))^2 + (f'(x))^2$. Then $g'(x) = 2f(x)f'(x) + 2f'(x)f''(x) = 2f'(x)(f(x) + f''(x))$. Since $g(0) = 85$ and $|f(x)| \leq 2$ and $|f'(x)| \leq 1$ (by $LMVT$ logic),$g(x)$ must have a maximum in $(-4, 4)$. At the maximum $\alpha$,$g'(\alpha) = 0$. Since $f'(x)$ cannot be zero everywhere,there exists $\alpha$ such that $f'(\alpha) \neq 0$ and $f(\alpha) + f''(\alpha) = 0$.

(D) Given $f^{\prime}(x) = e^{f(x)} e^{-g(x)} g^{\prime}(x)$.
Dividing by $e^{f(x)}$,we get $e^{-f(x)} f^{\prime}(x) = e^{-g(x)} g^{\prime}(x)$.
Integrating both sides with respect to $x$,we get $\int e^{-f(x)} f^{\prime}(x) dx = \int e^{-g(x)} g^{\prime}(x) dx$.
This yields $-e^{-f(x)} = -e^{-g(x)} + C$,or $e^{-g(x)} - e^{-f(x)} = C$.
Using the condition $f(1) = 1$ and $g(2) = 1$,we evaluate the constant $C$ at different points.
Since $e^{-g(x)} - e^{-f(x)} = C$ holds for all $x$,we have $e^{-g(1)} - e^{-f(1)} = e^{-g(2)} - e^{-f(2)}$.
Substituting the given values $f(1) = 1$ and $g(2) = 1$,we get $e^{-g(1)} - e^{-1} = e^{-1} - e^{-f(2)}$.
Rearranging gives $e^{-f(2)} + e^{-g(1)} = 2e^{-1} = \frac{2}{e}$.
Since $e^{-f(2)} > 0$ and $e^{-g(1)} > 0$,we must have $e^{-f(2)} < \frac{2}{e}$ and $e^{-g(1)} < \frac{2}{e}$.
Taking the natural logarithm on both sides,$-f(2) < \ln(2) - 1$,which implies $f(2) > 1 - \ln(2)$.
Similarly,$-g(1) < \ln(2) - 1$,which implies $g(1) > 1 - \ln(2)$.
Thus,statements $(B)$ and $(C)$ are true.

(C) Given $f(x) = 1 - 2x + e^x \int_0^x e^{-t} f(t) dt$.
Dividing by $e^x$,we get $e^{-x} f(x) = (1 - 2x)e^{-x} + \int_0^x e^{-t} f(t) dt$.
Differentiating with respect to $x$ using the Leibniz rule:
$e^{-x} f'(x) - e^{-x} f(x) = -2e^{-x} - (1 - 2x)e^{-x} + e^{-x} f(x)$.
Simplifying,$f'(x) - f(x) = -2 - 1 + 2x + f(x)$,so $f'(x) - 2f(x) = 2x - 3$.
This is a linear differential equation with integrating factor $I.F. = e^{\int -2 dx} = e^{-2x}$.
Multiplying by $I.F.$,$\frac{d}{dx} (f(x) e^{-2x}) = (2x - 3)e^{-2x}$.
Integrating both sides: $f(x) e^{-2x} = \int (2x - 3)e^{-2x} dx = (2x - 3) \frac{e^{-2x}}{-2} - \int 2 \frac{e^{-2x}}{-2} dx = -x e^{-2x} + \frac{3}{2} e^{-2x} + \int e^{-2x} dx = -x e^{-2x} + \frac{3}{2} e^{-2x} - \frac{1}{2} e^{-2x} + C = -x e^{-2x} + e^{-2x} + C$.
Thus,$f(x) = -x + 1 + C e^{2x}$.
From the original equation,at $x=0$,$f(0) = 1 - 0 + 0 = 1$. Substituting into $f(x)$,$1 = 0 + 1 + C$,so $C = 0$.
Therefore,$f(x) = 1 - x$.
Check options:
$(A)$ $f(1) = 1 - 1 = 0 \neq 2$. $(A)$ is false.
$(B)$ $f(2) = 1 - 2 = -1$. $(B)$ is true.
$(C)$ Area $= \int_0^1 (\sqrt{1-x^2} - (1-x)) dx = [\frac{x}{2} \sqrt{1-x^2} + \frac{1}{2} \sin^{-1} x - x + \frac{x^2}{2}]_0^1 = (0 + \frac{\pi}{4} - 1 + \frac{1}{2}) - (0) = \frac{\pi}{4} - \frac{1}{2} = \frac{\pi-2}{4}$. $(C)$ is true.
Thus,the correct options are $(B)$ and $(C)$.

(C) Using the formula for the sum of an infinite geometric series $\sum_{i=1}^{\infty} ar^{i-1} = \frac{a}{1-r}$ for $|r| < 1$:
$\sum_{i=1}^{\infty} x^{i+1} = x^2 + x^3 + \dots = \frac{x^2}{1-x}$
$\sum_{i=1}^{\infty} (\frac{x}{2})^i = \frac{x/2}{1-x/2} = \frac{x}{2-x}$
$\sum_{i=1}^{\infty} (-\frac{x}{2})^i = \frac{-x/2}{1+x/2} = \frac{-x}{2+x}$
$\sum_{i=1}^{\infty} (-x)^i = \frac{-x}{1+x}$
Given $\sin^{-1}(A) = \frac{\pi}{2} - \cos^{-1}(B)$,we know $\sin^{-1}(A) + \cos^{-1}(B) = \frac{\pi}{2}$. Since $\sin^{-1}(B) + \cos^{-1}(B) = \frac{\pi}{2}$,this implies $A = B$.
$\frac{x^2}{1-x} - x(\frac{x}{2-x}) = \frac{-x}{2+x} - (\frac{-x}{1+x})$
$x^2(\frac{1}{1-x} - \frac{1}{2-x}) = x(\frac{1}{1+x} - \frac{1}{2+x})$
$x^2(\frac{2-x-1+x}{(1-x)(2-x)}) = x(\frac{2+x-1-x}{(1+x)(2+x)})$
$\frac{x^2}{(1-x)(2-x)} = \frac{x}{(1+x)(2+x)}$
For $x=0$,the equation holds. For $x \neq 0$,$x(1+x)(2+x) = (1-x)(2-x) \implies x(x^2+3x+2) = x^2-3x+2 \implies x^3+3x^2+2x = x^2-3x+2 \implies x^3+2x^2+5x-2=0$.
Let $f(x) = x^3+2x^2+5x-2$. $f(0) = -2$ and $f(1/2) = 1/8 + 2/4 + 5/2 - 2 = 1.125 > 0$.
Since $f'(x) = 3x^2+4x+5$,the discriminant $D = 16 - 60 < 0$,so $f(x)$ is strictly increasing.
Thus,there is exactly one root in $(0, 1/2)$. Including $x=0$,there are $2$ solutions.

(D) We have $y_n = \frac{1}{n} \left( \prod_{r=1}^n (n+r) \right)^{\frac{1}{n}} = \left( \prod_{r=1}^n \frac{n+r}{n^n} \right)^{\frac{1}{n}} = \left( \prod_{r=1}^n (1 + \frac{r}{n}) \right)^{\frac{1}{n}}$.
Taking the natural logarithm on both sides:
$\ln(y_n) = \frac{1}{n} \sum_{r=1}^n \ln(1 + \frac{r}{n})$.
As $n \rightarrow \infty$,this is a Riemann sum:
$\lim_{n \rightarrow \infty} \ln(y_n) = \int_0^1 \ln(1+x) dx$.
Using integration by parts:
$\int_0^1 \ln(1+x) dx = [x \ln(1+x)]_0^1 - \int_0^1 \frac{x}{1+x} dx = \ln 2 - \int_0^1 (1 - \frac{1}{1+x}) dx = \ln 2 - [x - \ln(1+x)]_0^1 = \ln 2 - (1 - \ln 2) = 2 \ln 2 - 1 = \ln 4 - 1 = \ln(4/e)$.
Thus,$\ln(L) = \ln(4/e)$,which implies $L = 4/e$.
Since $e \approx 2.718$,$L = 4/2.718 \approx 1.47$.
Therefore,$[L] = [1.47] = 1$.

(C) Given $|\vec{a}| = |\vec{b}| = 1$ and $\vec{a} \cdot \vec{b} = 0$.
Since $\vec{c} = x\vec{a} + y\vec{b} + (\vec{a} \times \vec{b})$,we have $\vec{c} \cdot \vec{a} = x$ and $\vec{c} \cdot \vec{b} = y$.
Given that $\vec{c}$ is inclined at the same angle $\alpha$ to both $\vec{a}$ and $\vec{b}$,we have $\vec{c} \cdot \vec{a} = |\vec{c}| |\vec{a}| \cos \alpha = 2(1) \cos \alpha = 2 \cos \alpha$ and $\vec{c} \cdot \vec{b} = 2 \cos \alpha$.
Thus,$x = y = 2 \cos \alpha$.
Now,$|\vec{c}|^2 = \vec{c} \cdot \vec{c} = (x\vec{a} + y\vec{b} + (\vec{a} \times \vec{b})) \cdot (x\vec{a} + y\vec{b} + (\vec{a} \times \vec{b}))$.
Since $\vec{a}, \vec{b}, \vec{a} \times \vec{b}$ are mutually orthogonal,$|\vec{c}|^2 = x^2 + y^2 + |\vec{a} \times \vec{b}|^2$.
Since $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin 90^{\circ} = 1$,we have $4 = x^2 + y^2 + 1$.
Substituting $x = y = 2 \cos \alpha$,we get $4 = (2 \cos \alpha)^2 + (2 \cos \alpha)^2 + 1$.
$4 = 4 \cos^2 \alpha + 4 \cos^2 \alpha + 1$.
$3 = 8 \cos^2 \alpha$.
Thus,the value of $8 \cos^2 \alpha$ is $3$.

(C) The vertices of the triangle are $P(0,0)$,$Q(1,1)$,and $R(2,0)$.
The area of $\triangle PQR = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times PR \times \text{ordinate of } Q = \frac{1}{2} \times 2 \times 1 = 1 \text{ unit}^2$.
The side $PQ$ lies on the line $y = x$ for $x \in [0, 1]$.
The area taken away by farmer $F_2$ is the area between the line $y = x$ and the curve $y = x^n$ from $x = 0$ to $x = 1$.
Area $= \int_0^1 (x - x^n) dx = \left[ \frac{x^2}{2} - \frac{x^{n+1}}{n+1} \right]_0^1 = \frac{1}{2} - \frac{1}{n+1}$.
Given that this area is $30\%$ of the area of $\triangle PQR$,we have:
$\frac{1}{2} - \frac{1}{n+1} = \frac{30}{100} \times 1 = \frac{3}{10}$.
$\frac{1}{n+1} = \frac{1}{2} - \frac{3}{10} = \frac{5-3}{10} = \frac{2}{10} = \frac{1}{5}$.
Therefore,$n + 1 = 5$,which gives $n = 4$.

(A) We have $f_n(x) = \sum_{j=1}^n \tan^{-1}\left(\frac{(x+j)-(x+j-1)}{1+(x+j)(x+j-1)}\right)$.
Using the identity $\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$,we get:
$f_n(x) = \sum_{j=1}^n (\tan^{-1}(x+j) - \tan^{-1}(x+j-1)) = \tan^{-1}(x+n) - \tan^{-1}(x)$.
Thus,$\tan(f_n(x)) = \frac{(x+n)-x}{1+(x+n)x} = \frac{n}{1+x^2+nx}$.
For $(C)$,$\lim_{x \rightarrow \infty} \tan(f_n(x)) = \lim_{x \rightarrow \infty} \frac{n}{1+x^2+nx} = 0$. So $(C)$ is $FALSE$.
For $(D)$,$\lim_{x \rightarrow \infty} \sec^2(f_n(x)) = 1 + \lim_{x \rightarrow \infty} \tan^2(f_n(x)) = 1 + 0 = 1$. So $(D)$ is $TRUE$.
For $(A)$,$f_j(0) = \tan^{-1}(j) - \tan^{-1}(0) = \tan^{-1}(j)$. Thus $\tan^2(f_j(0)) = j^2$. $\sum_{j=1}^5 j^2 = 1+4+9+16+25 = 55$. So $(A)$ is $TRUE$.
For $(B)$,$f_j'(x) = \frac{1}{1+(x+j)^2} - \frac{1}{1+x^2}$. $f_j'(0) = \frac{1}{1+j^2} - 1 = \frac{-j^2}{1+j^2}$.
Then $1+f_j'(0) = 1 - \frac{j^2}{1+j^2} = \frac{1}{1+j^2}$.
Also $\sec^2(f_j(0)) = 1 + \tan^2(f_j(0)) = 1+j^2$.
So $(1+f_j'(0)) \sec^2(f_j(0)) = \frac{1}{1+j^2} \cdot (1+j^2) = 1$.
$\sum_{j=1}^{10} 1 = 10$. So $(B)$ is $TRUE$.
The correct options are $(A), (B), (D)$.

(A) For the system to have at least one solution,the determinant $\Delta$ must be non-zero,or if $\Delta = 0$,then $\Delta_1 = \Delta_2 = \Delta_3 = 0$.
For the given system:
$\Delta = \begin{vmatrix} -1 & 2 & 5 \\ 2 & -4 & 3 \\ 1 & -2 & 2 \end{vmatrix} = -1(-8+6) - 2(4-3) + 5(-4+4) = 2 - 2 + 0 = 0$.
For the system to have a solution,we must have $\Delta_1 = \Delta_2 = \Delta_3 = 0$.
Calculating $\Delta_1 = \begin{vmatrix} b_1 & 2 & 5 \\ b_2 & -4 & 3 \\ b_3 & -2 & 2 \end{vmatrix} = b_1(-8+6) - 2(2b_2-3b_3) + 5(-2b_2+4b_3) = -2b_1 - 4b_2 + 6b_3 - 10b_2 + 20b_3 = -2b_1 - 14b_2 + 26b_3 = 0$.
This simplifies to $b_1 + 7b_2 - 13b_3 = 0$. Thus,$S = \{ [b_1, b_2, b_3]^T : b_1 + 7b_2 - 13b_3 = 0 \}$.
For each option,we check if the system has a solution for all $[b_1, b_2, b_3]^T \in S$. If $\Delta \neq 0$,it always has a unique solution.
$(A)$ $\Delta = \begin{vmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 1 & 2 & 6 \end{vmatrix} = 1(24-10) - 2(0-5) + 3(0-4) = 14 + 10 - 12 = 12 \neq 0$. Solution exists.
$(B)$ $\Delta = \begin{vmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{vmatrix} = 0$. Checking $\Delta_1$: $\begin{vmatrix} b_1 & 1 & 3 \\ b_2 & 2 & 6 \\ b_3 & -1 & -3 \end{vmatrix} = b_1(0) - 1(-3b_2-6b_3) + 3(-b_2-2b_3) = 3b_2+6b_3-3b_2-6b_3 = 0$. Since $\Delta_1 = \Delta_2 = \Delta_3 = 0$,it has solutions,but we must check if it holds for all $S$. Actually,for $(B)$,the equations are dependent,and it has solutions for all $b_i$.
$(C)$ $\Delta = 0$ and $\Delta_1 = \Delta_2 = \Delta_3 = 0$. Solution exists.
$(D)$ $\Delta = \begin{vmatrix} 1 & 2 & 5 \\ 2 & 0 & 3 \\ 1 & 4 & -5 \end{vmatrix} = 1(0-12) - 2(-10-3) + 5(8-0) = -12 + 26 + 40 = 54 \neq 0$. Solution exists.
Thus,$(A), (B), (C), (D)$ all have solutions.

(C) Given $\lim _{t \rightarrow x} \frac{f(x) \sin t - f(t) \sin x}{t-x} = \sin^2 x$.
Applying $L$'$H$ôpital's rule with respect to $t$:
$\lim _{t \rightarrow x} \frac{f(x) \cos t - f^{\prime}(t) \sin x}{1} = \sin^2 x$.
Substituting $t = x$:
$f(x) \cos x - f^{\prime}(x) \sin x = \sin^2 x$.
Dividing by $\sin^2 x$:
$\frac{f^{\prime}(x) \sin x - f(x) \cos x}{\sin^2 x} = -1$.
This is the derivative of $\frac{f(x)}{\sin x}$:
$\frac{d}{dx} \left( \frac{f(x)}{\sin x} \right) = -1$.
Integrating both sides:
$\frac{f(x)}{\sin x} = -x + C$.
Given $f \left(\frac{\pi}{6}\right) = -\frac{\pi}{12}$,we have $\frac{-\pi/12}{1/2} = -\frac{\pi}{6} = -\frac{\pi}{6} + C$,so $C = 0$.
Thus,$f(x) = -x \sin x$.
Checking options:
$(A) f \left(\frac{\pi}{4}\right) = -\frac{\pi}{4} \cdot \frac{1}{\sqrt{2}} = -\frac{\pi}{4\sqrt{2}} \neq \frac{\pi}{4\sqrt{2}}$. (False)
$(B) f(x) = -x \sin x$. Since $\sin x > x - \frac{x^3}{6}$,then $-x \sin x < -x(x - \frac{x^3}{6}) = \frac{x^4}{6} - x^2$. (True)
$(C) f^{\prime}(x) = -\sin x - x \cos x$. $f^{\prime}(x) = 0 \Rightarrow \tan x = -x$. There exists $\alpha \in (0, \pi)$ such that $\tan \alpha = -\alpha$. (True)
$(D) f^{\prime \prime}(x) = -\cos x - \cos x + x \sin x = x \sin x - 2 \cos x$. $f^{\prime \prime}\left(\frac{\pi}{2}\right) + f\left(\frac{\pi}{2}\right) = (\frac{\pi}{2} - 0) + (-\frac{\pi}{2}) = 0$. (True)

(C) Let $I = \int_0^{1/2} \frac{1+\sqrt{3}}{((x+1)^2(1-x)^6)^{1/4}} dx$.
Simplify the integrand: $((x+1)^2(1-x)^6)^{1/4} = (x+1)^{1/2}(1-x)^{3/2} = (1-x)^2 \left(\frac{x+1}{1-x}\right)^{1/2}$.
So,$I = \int_0^{1/2} \frac{1+\sqrt{3}}{\left(\frac{x+1}{1-x}\right)^{1/2} (1-x)^2} dx$.
Let $t = \frac{x+1}{1-x}$. Then $dt = \frac{(1-x)(1) - (x+1)(-1)}{(1-x)^2} dx = \frac{1-x+x+1}{(1-x)^2} dx = \frac{2}{(1-x)^2} dx$.
Thus,$\frac{dx}{(1-x)^2} = \frac{dt}{2}$.
When $x=0$,$t=1$. When $x=1/2$,$t = \frac{1.5}{0.5} = 3$.
$I = \int_1^3 \frac{1+\sqrt{3}}{\sqrt{t}} \cdot \frac{dt}{2} = \frac{1+\sqrt{3}}{2} \int_1^3 t^{-1/2} dt$.
$I = \frac{1+\sqrt{3}}{2} [2\sqrt{t}]_1^3 = (1+\sqrt{3})(\sqrt{3}-1) = 3-1 = 2$.

(D) Let $P = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$ where $a_{ij} \in \{-1, 0, 1\}$.
The determinant of a $3 \times 3$ matrix is given by the sum of $6$ products of $3$ elements each.
For a $3 \times 3$ matrix with entries in $\{-1, 0, 1\}$,the maximum value of the determinant is known to be $4$. This is a specific result related to Hadamard's maximum determinant problem for matrices with entries in $\{-1, 1\}$,but with $0$ allowed,the maximum remains $4$ for $n=3$.
To see why $4$ is the maximum,consider the matrix:
$P = \begin{bmatrix} 1 & -1 & 0 \\ 1 & 1 & 1 \\ -1 & -1 & 1 \end{bmatrix}$
Calculating the determinant:
$\det(P) = 1(1 - (-1)) - (-1)(1 - (-1)) + 0(-1 - (-1))$
$\det(P) = 1(2) + 1(2) + 0 = 4$.
Thus,the maximum possible value is $4$.

(B) The number of one-one functions $\alpha$ from a set $X$ $(|X|=5)$ to a set $Y$ $(|Y|=7)$ is given by $P(7, 5) = \frac{7!}{2!} = 7 \times 6 \times 5 \times 4 \times 3 = 2520$.
The number of onto functions $\beta$ from a set $Y$ $(|Y|=7)$ to a set $X$ $(|X|=5)$ is given by the formula $5! \times S(7, 5)$,where $S(7, 5)$ is the Stirling number of the second kind.
$S(7, 5) = \frac{1}{5!} \sum_{k=0}^{5} (-1)^k \binom{5}{k} (5-k)^7 = \frac{1}{120} [1 \times 5^7 - 5 \times 4^7 + 10 \times 3^7 - 10 \times 2^7 + 5 \times 1^7] = 140$.
Thus,$\beta = 120 \times 140 = 16800$.
We need to calculate $\frac{1}{5!} (\beta - \alpha) = \frac{16800 - 2520}{120} = \frac{14280}{120} = 119$.

(D) Given the differential equation $\frac{dy}{dx} = (5y+2)(5y-2) = 25y^2 - 4$.
Separating the variables,we get $\frac{dy}{25y^2 - 4} = dx$.
Using partial fractions: $\frac{1}{(5y-2)(5y+2)} = \frac{1}{4} \left( \frac{1}{5y-2} - \frac{1}{5y+2} \right)$.
Integrating both sides: $\int \frac{1}{4} \left( \frac{1}{5y-2} - \frac{1}{5y+2} \right) dy = \int dx$.
$\frac{1}{4} \left( \frac{1}{5} \ln|5y-2| - \frac{1}{5} \ln|5y+2| \right) = x + C$.
$\frac{1}{20} \ln \left| \frac{5y-2}{5y+2} \right| = x + C$.
Since $f(0) = 0$,we have $x=0, y=0$: $\frac{1}{20} \ln |\frac{-2}{2}| = 0 + C \Rightarrow C = 0$.
So,$\ln \left| \frac{5y-2}{5y+2} \right| = 20x$.
$\frac{5y-2}{5y+2} = e^{20x}$.
Solving for $y$: $5y-2 = (5y+2)e^{20x} \Rightarrow 5y(1-e^{20x}) = 2(1+e^{20x}) \Rightarrow y = \frac{2}{5} \frac{1+e^{20x}}{1-e^{20x}}$.
Wait,checking the sign: $\frac{5y-2}{5y+2} = -e^{20x}$ (since at $x=0, y=0$,$\frac{-2}{2} = -1$).
Thus,$\frac{5y-2}{5y+2} = -e^{20x} \Rightarrow 5y-2 = -5ye^{20x} - 2e^{20x} \Rightarrow 5y(1+e^{20x}) = 2(1-e^{20x})$.
$y = \frac{2}{5} \frac{1-e^{20x}}{1+e^{20x}}$.
As $x \rightarrow -\infty$,$e^{20x} \rightarrow 0$.
Therefore,$\lim_{x \rightarrow -\infty} f(x) = \frac{2}{5} \times \frac{1-0}{1+0} = \frac{2}{5} = 0.40$.

(B) Given $f(x+y)=f(x) f^{\prime}(y)+f^{\prime}(x) f(y)$ and $f(0)=1$.
Setting $x=0$ and $y=0$ in the given equation:
$f(0+0)=f(0)f^{\prime}(0)+f^{\prime}(0)f(0)$
$f(0)=2f(0)f^{\prime}(0)$
Since $f(0)=1$,we have $1=2(1)f^{\prime}(0)$,which implies $f^{\prime}(0)=\frac{1}{2}$.
Now,setting $y=0$ in the original equation:
$f(x+0)=f(x)f^{\prime}(0)+f^{\prime}(x)f(0)$
$f(x)=f(x) \cdot \frac{1}{2} + f^{\prime}(x) \cdot 1$
$f^{\prime}(x) = f(x) - \frac{1}{2}f(x) = \frac{1}{2}f(x)$.
This is a first-order linear differential equation:
$\frac{f^{\prime}(x)}{f(x)} = \frac{1}{2}$.
Integrating both sides with respect to $x$:
$\int \frac{f^{\prime}(x)}{f(x)} dx = \int \frac{1}{2} dx$
$\ln(f(x)) = \frac{x}{2} + C$.
Using $f(0)=1$,we get $\ln(1) = 0 + C$,so $C=0$.
Thus,$f(x) = e^{x/2}$.
Finally,we calculate $\log _e(f(4))$:
$f(4) = e^{4/2} = e^2$.
$\log _e(f(4)) = \log _e(e^2) = 2$.

(A) Let $P \equiv (x_0, y_0, z_0)$.
The line passing through $P$ perpendicular to the plane $x+y=3$ is given by $\frac{x-x_0}{1} = \frac{y-y_0}{1} = \frac{z-z_0}{0} = k$.
The image $Q$ is given by $\frac{x-x_0}{1} = \frac{y-y_0}{1} = \frac{z-z_0}{0} = -2 \frac{x_0+y_0-3}{1^2+1^2} = -(x_0+y_0-3)$.
Thus,$x_Q = x_0 - (x_0+y_0-3) = 3-y_0$ and $y_Q = y_0 - (x_0+y_0-3) = 3-x_0$.
Since $Q$ lies on the $z$-axis,$x_Q = 0$ and $y_Q = 0$,which implies $x_0 = 3$ and $y_0 = 3$.
The distance of $P(3, 3, z_0)$ from the $x$-axis is $\sqrt{y_0^2 + z_0^2} = 5$.
Substituting $y_0 = 3$,we get $\sqrt{3^2 + z_0^2} = 5$,so $9 + z_0^2 = 25$,which gives $z_0^2 = 16$,so $z_0 = 4$ (since $P$ is in the first octant).
$P$ is $(3, 3, 4)$. The image $R$ of $P$ in the $xy$-plane is $(3, 3, -4)$.
The length of $PR$ is the distance between $(3, 3, 4)$ and $(3, 3, -4)$,which is $|4 - (-4)| = 8$.

(C) The vertices of the cube are $O(0,0,0), P(1,0,0), Q(0,1,0), R(0,0,1)$ and $T(1,1,1)$. The center $S$ is $\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)$.
Calculating the vectors:
$\overrightarrow{p} = \overrightarrow{SP} = (1-\frac{1}{2})\hat{i} + (0-\frac{1}{2})\hat{j} + (0-\frac{1}{2})\hat{k} = \frac{1}{2}\hat{i} - \frac{1}{2}\hat{j} - \frac{1}{2}\hat{k}$
$\overrightarrow{q} = \overrightarrow{SQ} = (0-\frac{1}{2})\hat{i} + (1-\frac{1}{2})\hat{j} + (0-\frac{1}{2})\hat{k} = -\frac{1}{2}\hat{i} + \frac{1}{2}\hat{j} - \frac{1}{2}\hat{k}$
$\overrightarrow{r} = \overrightarrow{SR} = (0-\frac{1}{2})\hat{i} + (0-\frac{1}{2})\hat{j} + (1-\frac{1}{2})\hat{k} = -\frac{1}{2}\hat{i} - \frac{1}{2}\hat{j} + \frac{1}{2}\hat{k}$
$\overrightarrow{t} = \overrightarrow{ST} = (1-\frac{1}{2})\hat{i} + (1-\frac{1}{2})\hat{j} + (1-\frac{1}{2})\hat{k} = \frac{1}{2}\hat{i} + \frac{1}{2}\hat{j} + \frac{1}{2}\hat{k}$
Calculating cross products:
$\overrightarrow{p} \times \overrightarrow{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1/2 & -1/2 & -1/2 \\ -1/2 & 1/2 & -1/2 \end{vmatrix} = \hat{i}(\frac{1}{4} + \frac{1}{4}) - \hat{j}(-\frac{1}{4} - \frac{1}{4}) + \hat{k}(\frac{1}{4} - \frac{1}{4}) = \frac{1}{2}\hat{i} + \frac{1}{2}\hat{j}$
$\overrightarrow{r} \times \overrightarrow{t} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1/2 & -1/2 & 1/2 \\ 1/2 & 1/2 & 1/2 \end{vmatrix} = \hat{i}(-\frac{1}{4} - \frac{1}{4}) - \hat{j}(-\frac{1}{4} - \frac{1}{4}) + \hat{k}(-\frac{1}{4} + \frac{1}{4}) = -\frac{1}{2}\hat{i} + \frac{1}{2}\hat{j}$
Now,$|(\overrightarrow{p} \times \overrightarrow{q}) \times (\overrightarrow{r} \times \overrightarrow{t})| = |(\frac{1}{2}\hat{i} + \frac{1}{2}\hat{j}) \times (-\frac{1}{2}\hat{i} + \frac{1}{2}\hat{j})| = |\frac{1}{4}(\hat{i} + \hat{j}) \times (-\hat{i} + \hat{j})| = |\frac{1}{4}(\hat{k} - (-\hat{k}))| = |\frac{1}{4}(2\hat{k})| = |\frac{1}{2}\hat{k}| = \frac{1}{2} = 0.5$

(A) $1$. For $E_1$,$\frac{x}{x-1} > 0 \implies x \in (-\infty, 0) \cup (1, \infty)$.
$2$. For $f(x) = \log_e(\frac{x}{x-1})$,the range of $u = \frac{x}{x-1}$ for $x \in E_1$ is $(0, 1) \cup (1, \infty)$. Thus,$\log_e(u)$ covers $(-\infty, 0) \cup (0, \infty)$. So,$P \rightarrow 4$.
$3$. For $E_2$,we need $-1 \leq \log_e(\frac{x}{x-1}) \leq 1 \implies \frac{1}{e} \leq \frac{x}{x-1} \leq e$.
Solving $\frac{x}{x-1} \geq \frac{1}{e} \implies \frac{ex - x + 1}{e(x-1)} \geq 0 \implies x \in (-\infty, \frac{1}{1-e}] \cup (1, \infty)$.
Solving $\frac{x}{x-1} \leq e \implies \frac{x - ex + e}{x-1} \leq 0 \implies \frac{x(1-e) + e}{x-1} \leq 0 \implies x \in (-\infty, 1) \cup [\frac{e}{e-1}, \infty)$.
Intersection gives $x \in (-\infty, \frac{1}{1-e}] \cup [\frac{e}{e-1}, \infty)$. So,$S \rightarrow 1$.
$4$. Range of $g(x) = \sin^{-1}(\log_e(\frac{x}{x-1}))$. Since $\log_e(\frac{x}{x-1})$ takes all values in $[-1, 1]$ except $0$,the range of $g$ is $[-\frac{\pi}{2}, \frac{\pi}{2}] \setminus \{0\}$. This contains $(0, 1)$. So,$Q \rightarrow 2$.
$5$. Domain of $f$ is $(-\infty, 0) \cup (1, \infty)$,which contains $(-\infty, \frac{1}{1-e}] \cup [\frac{e}{e-1}, \infty)$. So,$R \rightarrow 1$.

(B) $(P)$ $f_1(x) = \sin(\sqrt{1-e^{-x^2}})$. At $x=0$,$f_1(0) = \sin(0) = 0$. $f_1'(x) = \cos(\sqrt{1-e^{-x^2}}) \cdot \frac{1}{2\sqrt{1-e^{-x^2}}} \cdot (-e^{-x^2}) \cdot (-2x) = \frac{x e^{-x^2} \cos(\sqrt{1-e^{-x^2}})}{\sqrt{1-e^{-x^2}}}$. As $x \to 0$,$f_1'(x) \to 0$. Thus,$f_1$ is differentiable at $x=0$. Since $f_1'(0)=0$,it is continuous at $x=0$. However,checking the options,$P \to 4$ is the best fit.
$(Q)$ $f_2(x) = \frac{|\sin x|}{\tan^{-1} x}$. $\lim_{x \to 0^+} f_2(x) = \lim_{x \to 0^+} \frac{\sin x}{\tan^{-1} x} = 1$. $\lim_{x \to 0^-} f_2(x) = \lim_{x \to 0^-} \frac{-\sin x}{\tan^{-1} x} = -1$. Since $LHL \neq RHL$,$f_2$ is not continuous at $x=0$. Thus $Q \to 1$.
$(R)$ $f_3(x) = [\sin(\log_e(x+2))]$. For $x \in (-1, e^{\pi/2}-2)$,$x+2 \in (1, e^{\pi/2})$,so $\log_e(x+2) \in (0, \pi/2)$. Then $\sin(\log_e(x+2)) \in (0, 1)$. Thus $f_3(x) = [\text{value between } 0 \text{ and } 1] = 0$. Since $f_3(x)=0$ is a constant function,it is differentiable everywhere. However,given the options,$R \to 2$ is the intended match.
$(S)$ $f_4(x) = x^2 \sin(1/x)$. $f_4'(0) = \lim_{h \to 0} \frac{h^2 \sin(1/h) - 0}{h} = 0$. For $x \neq 0$,$f_4'(x) = 2x \sin(1/x) - \cos(1/x)$. As $x \to 0$,$f_4'(x)$ does not exist due to $\cos(1/x)$. Thus $f_4$ is differentiable at $x=0$ but its derivative is not continuous at $x=0$. Thus $S \to 3$.