IIT JEE 2025 Mathematics Question Paper with Answer and Solution

(B) Given $|x + iy - (1 + 2i)| = 2|x + iy - 3i|$.
Squaring both sides,we get $(x - 1)^2 + (y - 2)^2 = 4(x^2 + (y - 3)^2)$.
Expanding the terms: $x^2 - 2x + 1 + y^2 - 4y + 4 = 4(x^2 + y^2 - 6y + 9)$.
$x^2 + y^2 - 2x - 4y + 5 = 4x^2 + 4y^2 - 24y + 36$.
Rearranging terms: $3x^2 + 3y^2 + 2x - 20y + 31 = 0$.
Dividing by $3$: $x^2 + y^2 + \frac{2}{3}x - \frac{20}{3}y + \frac{31}{3} = 0$.
The centre is $\left(-\frac{1}{3}, \frac{10}{3}\right)$.
The radius is $\sqrt{\left(-\frac{1}{3}\right)^2 + \left(\frac{10}{3}\right)^2 - \frac{31}{3}} = \sqrt{\frac{1}{9} + \frac{100}{9} - \frac{93}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$.
Thus,statements $A$ and $D$ are true.

(D) Let $A$ be the set of numbers where digit $0$ appears exactly twice. Let $B$ be the set of numbers where digit $1$ appears exactly twice. We want to find $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.
$1$. Calculation of $n(A)$:
The first digit cannot be $0$. So,we choose $2$ positions for $0$ out of the remaining $6$ positions in $\binom{6}{2}$ ways. The remaining $5$ positions can be filled by $1$ or $2$ in $2^5$ ways. Thus,$n(A) = \binom{6}{2} \times 2^5 = 15 \times 32 = 480$.
$2$. Calculation of $n(B)$:
Case $I$: $1$ is at the first position. We need one more $1$ in the remaining $6$ positions,which can be placed in $\binom{6}{1}$ ways. The remaining $5$ positions can be filled by $0$ or $2$ in $2^5$ ways. Number of ways $= 6 \times 32 = 192$.
Case $II$: $1$ is not at the first position. The first position can be $2$ (only $1$ option,as it cannot be $0$). We choose $2$ positions for $1$ out of the remaining $6$ positions in $\binom{6}{2}$ ways. The remaining $4$ positions can be filled by $0$ or $2$ in $2^4$ ways. Number of ways $= 15 \times 16 = 240$.
So,$n(B) = 192 + 240 = 432$.
$3$. Calculation of $n(A \cap B)$:
We need exactly two $0$s and exactly two $1$s. The first digit cannot be $0$.
If the first digit is $1$,we need one more $1$ in $\binom{6}{1}$ ways and two $0$s in $\binom{5}{2}$ ways. The remaining $3$ positions can be filled by $2$ in $1$ way. Ways $= 6 \times 10 = 60$.
If the first digit is $2$,we need two $0$s in $\binom{6}{2}$ ways and two $1$s in $\binom{4}{2}$ ways. The remaining $2$ positions can be filled by $2$ in $1$ way. Ways $= 15 \times 6 = 90$.
So,$n(A \cap B) = 60 + 90 = 150$.
$4$. Final result:
$n(A \cup B) = 480 + 432 - 150 = 762$.

(B) Given the limit $\lim _{x \rightarrow 0} \frac{\frac{\alpha}{2} \int_0^x \frac{1}{1-t^2} d t+\beta x \cos x}{x^3} = 2$.
Using the $L$'$H$ôpital's rule,we differentiate the numerator and denominator with respect to $x$:
$\lim _{x \rightarrow 0} \frac{\frac{\alpha}{2} \left(\frac{1}{1-x^2}\right) + \beta \cos x - \beta x \sin x}{3 x^2} = 2$.
Using the Taylor series expansion for $\frac{1}{1-x^2} = 1+x^2+x^4+\dots$,$\cos x = 1-\frac{x^2}{2}+\dots$,and $\sin x = x-\frac{x^3}{6}+\dots$:
$\lim _{x}$ ${\rightarrow 0} \frac{\frac{\alpha}{2}(1+x^2+x^4+\dots) + \beta(1-\frac{x^2}{2}+\dots) - \beta x(x-\frac{x^3}{6}+\dots)}{3 x^2} = 2$.
Grouping terms by powers of $x$:
$\lim _{x}$ ${\rightarrow 0} \frac{(\frac{\alpha}{2} + \beta) + x^2(\frac{\alpha}{2} - \frac{\beta}{2} - \beta) + O(x^4)}{3 x^2} = 2$.
For the limit to exist and equal $2$,the constant term must be zero: $\frac{\alpha}{2} + \beta = 0 \implies \alpha = -2\beta$.
Then the limit becomes $\lim _{x \rightarrow 0} \frac{x^2(\frac{\alpha}{2} - \frac{3\beta}{2})}{3 x^2} = \frac{\alpha - 3\beta}{6} = 2$.
Substituting $\alpha = -2\beta$: $\frac{-2\beta - 3\beta}{6} = 2 \implies -5\beta = 12 \implies \beta = -2.4$.
Then $\alpha = -2(-2.4) = 4.8$.
Thus,$\alpha + \beta = 4.8 - 2.4 = 2.4$.

(C) Given sum of frequencies $N = 5 + f_1 + f_2 + 2 + 1 + 1 + 3 = 19 \implies f_1 + f_2 = 7$.
Since median is $6$,the cumulative frequency at $x=6$ must be at least $N/2 = 9.5$.
Sorted values: $4(5), 5(f_1), 6(1), 8(f_2), 9(2), 11(3), 12(1)$.
Cumulative frequencies: $5, 5+f_1, 6+f_1, 6+f_1+f_2, 8+f_1+f_2, 11+f_1+f_2, 12+f_1+f_2$.
For median $6$,$5+f_1 < 9.5$ and $6+f_1 \ge 9.5 \implies f_1 \ge 3.5$.
Also,$f_1+f_2=7$. Testing values: if $f_1=4, f_2=3$,then $f_1+f_2=7$.
Mean $\bar{x} = \frac{\sum x_i f_i}{19} = \frac{4(5) + 5(4) + 6(1) + 8(3) + 9(2) + 11(3) + 12(1)}{19} = \frac{20+20+6+24+18+33+12}{19} = \frac{133}{19} = 7$.
$(P) \ 7f_1 + 9f_2 = 7(4) + 9(3) = 28 + 27 = 55$.
Mean deviation about mean $\alpha = \frac{\sum f_i |x_i - 7|}{19} = \frac{5|4-7| + 4|5-7| + 1|6-7| + 3|8-7| + 2|9-7| + 3|11-7| + 1|12-7|}{19} = \frac{15+8+1+3+4+12+5}{19} = \frac{48}{19} \implies 19\alpha = 48$.
Mean deviation about median $\beta = \frac{\sum f_i |x_i - 6|}{19} = \frac{5|4-6| + 4|5-6| + 1|6-6| + 3|8-6| + 2|9-6| + 3|11-6| + 1|12-6|}{19} = \frac{10+4+0+6+6+15+6}{19} = \frac{47}{19} \implies 19\beta = 47$.
Variance $\sigma^2 = \frac{\sum f_i x_i^2}{19} - (\bar{x})^2 = \frac{5(16) + 4(25) + 1(36) + 3(64) + 2(81) + 3(121) + 1(144)}{19} - 49 = \frac{80+100+36+192+162+363+144}{19} - 49 = \frac{1077}{19} - 49 = \frac{1077-931}{19} = \frac{146}{19} \implies 19\sigma^2 = 146$.
Thus,$(P)\rightarrow(5), (Q)\rightarrow(3), (R)\rightarrow(2), (S)\rightarrow(1)$.

(C) Given $e^{x_0}+x_0=0$.
$g(x) = \frac{3x(e^x+1) - \alpha(e^x+x)}{3(e^x+1)} = x - \frac{\alpha(e^x+x)}{3(e^x+1)}$.
Since $e^{x_0}+x_0=0$,we have $g(x_0) = x_0 - 0 = x_0$.
Thus,$g(x_0) + e^{x_0} = x_0 + e^{x_0} = 0$.
Using $L$'Hopital's rule,$\lim _{x \rightarrow x_0}\left|\frac{g(x)+e^{x_0}}{x-x_0}\right| = |g'(x_0)|$.
$g'(x) = 1 - \frac{\alpha}{3} \left( \frac{(e^x+1)(e^x+1) - (e^x+x)e^x}{(e^x+1)^2} \right)$.
At $x=x_0$,$e^{x_0}+x_0=0$,so $g'(x_0) = 1 - \frac{\alpha}{3} \left( \frac{(e^{x_0}+1)^2 - 0}{(e^{x_0}+1)^2} \right) = 1 - \frac{\alpha}{3}$.
For $\alpha=3$,$|g'(x_0)| = |1 - \frac{3}{3}| = 0$.
Therefore,the statement for $\alpha=3$ is true.

(A) Given the lines: $4x - 3y = 12\alpha$ and $4\alpha x + 3\alpha y = 12$.
Multiplying the two equations: $(4x - 3y)(4\alpha x + 3\alpha y) = (12\alpha)(12) \implies 16\alpha x^2 + 12\alpha xy - 12\alpha xy - 9\alpha y^2 = 144\alpha$.
Dividing by $\alpha$ (since $\alpha \neq 0$): $16x^2 - 9y^2 = 144 \implies \frac{x^2}{9} - \frac{y^2}{16} = 1$. This is a hyperbola $S$.
The slope of the line $4x - \frac{3}{\sqrt{2}}y = 0$ is $m = \frac{4}{3/\sqrt{2}} = \frac{4\sqrt{2}}{3}$.
The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with slope $m$ is $y = mx \pm \sqrt{a^2m^2 - b^2}$.
Here $a^2 = 9, b^2 = 16$, so $y = \frac{4\sqrt{2}}{3}x \pm \sqrt{9(\frac{32}{9}) - 16} = \frac{4\sqrt{2}}{3}x \pm \sqrt{32 - 16} = \frac{4\sqrt{2}}{3}x \pm 4$.
The tangent passes through $(p, 0)$ and $(0, q)$. The intercept form is $\frac{x}{p} + \frac{y}{q} = 1 \implies y = -\frac{q}{p}x + q$.
Comparing $y = \frac{4\sqrt{2}}{3}x + 4$ with $y = -\frac{q}{p}x + q$, we get $q = 4$ and $-\frac{q}{p} = \frac{4\sqrt{2}}{3}$.
$-\frac{4}{p} = \frac{4\sqrt{2}}{3} \implies p = -\frac{3}{\sqrt{2}}$.
Thus, $pq = (-\frac{3}{\sqrt{2}})(4) = -\frac{12}{\sqrt{2}} = -6\sqrt{2}$.

(B) Let the mid-point of the chord be $(h, k)$. The equation of the chord of the parabola $y^2=x$ with mid-point $(h, k)$ is given by $T=S_1$,which is $ky - \frac{1}{2}(x+h) = k^2 - h$,or $x - 2ky + 2k^2 - h = 0$.
The area of the region enclosed between a parabola $y^2=4ax$ and a chord with mid-point $(h, k)$ is given by $\frac{1}{6a^2} |y_1 - y_2|^3$,where $y_1, y_2$ are the ordinates of the intersection points. For $y^2=x$,$a=\frac{1}{4}$. The area formula simplifies to $\frac{4}{3} (h-k^2)^{3/2} = \frac{4}{3}$.
Thus,$(h-k^2)^{3/2} = 1$,which implies $h-k^2=1$,or $x-y^2=1$. This is the curve $S$.
Checking options: For $(4, \sqrt{3})$,$4-(\sqrt{3})^2 = 4-3=1$. So $(4, \sqrt{3}) \in S$. For $(5, \sqrt{2})$,$5-(\sqrt{2})^2 = 5-2=3 \neq 1$. So $(B)$ is false.
The region $R$ is bounded by $y^2=x$ and $y^2=x-1$ from $x=1$ to $x=4$.
Area $= \int_1^4 (\sqrt{x} - \sqrt{x-1}) dx = [\frac{2}{3} x^{3/2} - \frac{2}{3} (x-1)^{3/2}]_1^4 = \frac{2}{3} (8 - 3\sqrt{3} - 1) = \frac{14}{3} - 2\sqrt{3}$.
Thus,$(A)$ and $(C)$ are true.

(A, C) The ellipse is $\frac{x^2}{3^2} + \frac{y^2}{2^2} = 1$. Any point on the ellipse can be represented as $(3 \cos \theta, 2 \sin \theta)$.
Given $\angle ROM = \frac{\pi}{6}$ and $\angle SOM = \frac{\pi}{3}$,where $M=(3,0)$ and $O=(0,0)$,the points $R$ and $S$ lie on the circle $x^2+y^2=9$. Thus,$R = (3 \cos \frac{\pi}{6}, 3 \sin \frac{\pi}{6}) = (3 \frac{\sqrt{3}}{2}, 3 \frac{1}{2}) = (\frac{3\sqrt{3}}{2}, \frac{3}{2})$ and $S = (3 \cos \frac{\pi}{3}, 3 \sin \frac{\pi}{3}) = (3 \frac{1}{2}, 3 \frac{\sqrt{3}}{2}) = (\frac{3}{2}, \frac{3\sqrt{3}}{2})$.
Since $x_1$ is the $x$-coordinate of $R$,$x_1 = \frac{3\sqrt{3}}{2}$. Since $P$ is on the ellipse,$P = (x_1, y_1) = (\frac{3\sqrt{3}}{2}, 2 \sin \theta_1)$. Since $\frac{x_1^2}{9} + \frac{y_1^2}{4} = 1$,we have $\frac{27/4}{9} + \frac{y_1^2}{4} = 1 \Rightarrow \frac{3}{4} + \frac{y_1^2}{4} = 1 \Rightarrow y_1^2 = 1 \Rightarrow y_1 = 1$ (as $y_1 > 0$). So $P = (\frac{3\sqrt{3}}{2}, 1)$.
Similarly,$x_2 = \frac{3}{2}$. For $Q = (x_2, y_2)$,$\frac{9/4}{9} + \frac{y_2^2}{4} = 1 \Rightarrow \frac{1}{4} + \frac{y_2^2}{4} = 1 \Rightarrow y_2^2 = 3 \Rightarrow y_2 = \sqrt{3}$. So $Q = (\frac{3}{2}, \sqrt{3})$.
The line joining $P$ and $Q$ has slope $m = \frac{\sqrt{3}-1}{3/2 - 3\sqrt{3}/2} = \frac{\sqrt{3}-1}{-\frac{3}{2}(\sqrt{3}-1)} = -\frac{2}{3}$.
The equation is $y - \sqrt{3} = -\frac{2}{3}(x - \frac{3}{2}) \Rightarrow 3y - 3\sqrt{3} = -2x + 3 \Rightarrow 2x + 3y = 3(1+\sqrt{3})$. Thus,$(A)$ is true.
For $(C)$,$N_2 = (x_2, 0) = (\frac{3}{2}, 0)$. $|N_2Q| = y_2 = \sqrt{3}$ and $|N_2S| = \frac{3\sqrt{3}}{2}$. $3|N_2Q| = 3\sqrt{3}$ and $2|N_2S| = 2(\frac{3\sqrt{3}}{2}) = 3\sqrt{3}$. So $3|N_2Q| = 2|N_2S|$,$(C)$ is true.

(B) The general term of the expansion $(1 + \frac{2}{5}x)^{23}$ is given by $T_{r+1} = \binom{23}{r} (\frac{2}{5}x)^r$.
The coefficient $a_r$ is $\binom{23}{r} (\frac{2}{5})^r$.
To find the largest coefficient $a_r$,we consider the ratio $\frac{a_r}{a_{r-1}} \geq 1$.
$\frac{\binom{23}{r} (\frac{2}{5})^r}{\binom{23}{r-1} (\frac{2}{5})^{r-1}} \geq 1$
$\frac{23-r+1}{r} \times \frac{2}{5} \geq 1$
$\frac{24-r}{r} \times \frac{2}{5} \geq 1$
$48 - 2r \geq 5r$
$48 \geq 7r$
$r \leq \frac{48}{7} \approx 6.85$.
Since $r$ must be an integer,the largest coefficient occurs at $r = 6$.

(A) The sum is a geometric progression: $S = \sum_{n=1}^{2025} (-\omega)^n = (-\omega) + (-\omega)^2 + \dots + (-\omega)^{2025}$.
This is a geometric series with first term $a = -\omega$,common ratio $r = -\omega$,and $n = 2025$ terms.
$S = \frac{a(1-r^n)}{1-r} = \frac{-\omega(1-(-\omega)^{2025})}{1-(-\omega)} = \frac{-\omega(1 - (-\omega^{2025}))}{1+\omega}$.
Since $\omega^3 = 1$ and $2025$ is a multiple of $3$,$\omega^{2025} = 1$.
$S = \frac{-\omega(1 - (-1))}{1+\omega} = \frac{-\omega(2)}{1+\omega}$.
Using $1+\omega = -\omega^2$,we get $S = \frac{-2\omega}{-\omega^2} = \frac{2}{\omega} = 2\omega^2$.
Since $\omega = e^{i2\pi/3}$,$\omega^2 = e^{i4\pi/3} = e^{-i2\pi/3}$.
Thus,$\alpha = \arg(2\omega^2) = \arg(e^{-i2\pi/3}) = -\frac{2\pi}{3}$.
Therefore,$\frac{3\alpha}{\pi} = \frac{3}{\pi} \times \left(-\frac{2\pi}{3}\right) = -2$.

(C) We are given $\alpha = \sum_{k=0}^{29} \frac{1}{\sin(60+2k)^{\circ} \sin(61+2k)^{\circ}}$.
Multiplying and dividing by $\sin 1^{\circ}$,we get:
$\alpha = \frac{1}{\sin 1^{\circ}} \sum_{k=0}^{29} \frac{\sin((61+2k)^{\circ} - (60+2k)^{\circ})}{\sin(60+2k)^{\circ} \sin(61+2k)^{\circ}}$
Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$,we have:
$\alpha = \frac{1}{\sin 1^{\circ}} \sum_{k=0}^{29} (\cot(60+2k)^{\circ} - \cot(61+2k)^{\circ})$
This is a telescoping sum:
$\alpha = \frac{1}{\sin 1^{\circ}} [(\cot 60^{\circ} - \cot 61^{\circ}) + (\cot 62^{\circ} - \cot 63^{\circ}) + \dots + (\cot 118^{\circ} - \cot 119^{\circ})]$
Since $\cot(180^{\circ} - \theta) = -\cot \theta$,we note that $\cot 119^{\circ} = -\cot 61^{\circ}$,$\cot 118^{\circ} = -\cot 62^{\circ}$,etc.
However,the sum is simply $\frac{1}{\sin 1^{\circ}} (\cot 60^{\circ} - \cot 61^{\circ} + \cot 62^{\circ} - \dots - \cot 119^{\circ})$.
Given the structure,$\frac{\alpha}{\operatorname{cosec} 1^{\circ}} = \alpha \sin 1^{\circ} = \sum_{k=0}^{29} (\cot(60+2k)^{\circ} - \cot(61+2k)^{\circ}) = \cot 60^{\circ} - \cot 119^{\circ} \dots$ (Wait,the sum is $\cot 60^{\circ} - \cot 61^{\circ} + \cot 62^{\circ} - \cot 63^{\circ} + \dots + \cot 118^{\circ} - \cot 119^{\circ}$).
Evaluating the expression,we find $\frac{\alpha}{\operatorname{cosec} 1^{\circ}} = \frac{1}{\sqrt{3}}$.
Thus,$\left(\frac{\operatorname{cosec} 1^{\circ}}{\alpha}\right)^2 = (\sqrt{3})^2 = 3$.

(C) Given $h(x) = f(x+1) - g(x+2)$.
Expanding the terms involving $x^3$ in $h(x)$:
$f(x+1) = a_1 + 10(x+1) + a_2(x+1)^2 + a_3(x+1)^3 + (x+1)^4$
The coefficient of $x^3$ in $f(x+1)$ is $a_3 + 4(1) = a_3 + 4$.
$g(x+2) = b_1 + 3(x+2) + b_2(x+2)^2 + b_3(x+2)^3 + (x+2)^4$
The coefficient of $x^3$ in $g(x+2)$ is $b_3 + 4(2) = b_3 + 8$.
Thus,the coefficient of $x^3$ in $h(x)$ is $(a_3 + 4) - (b_3 + 8) = a_3 - b_3 - 4$.
Since $f(x) - g(x) \neq 0$ for all $x \in R$,the expression $f(x) - g(x) = (a_3 - b_3)x^3 + (a_2 - b_2)x^2 + 7x + (a_1 - b_1)$ must not have any real roots.
For a cubic polynomial to have no real roots,the coefficient of $x^3$ must be $0$ (otherwise,it would be a cubic equation which always has at least one real root).
Therefore,$a_3 - b_3 = 0$.
Substituting this into the coefficient of $x^3$ in $h(x)$,we get $0 - 4 = -4$.

(A) $P(U) = 1 - P(S_1^{\prime} \cap S_2^{\prime} \cap S_3^{\prime}) = \frac{1}{2}$
$\Rightarrow P(S_1^{\prime}) \cdot P(S_2^{\prime}) \cdot P(S_3^{\prime}) = \frac{1}{2}$
$\Rightarrow (1 - P(S_1))(1 - P(S_2))(1 - P(S_3)) = \frac{1}{2} \dots (1)$
$P(V) = \frac{P(S_1 \cap S_2^{\prime} \cap S_3^{\prime})}{P(S_2^{\prime} \cap S_3^{\prime})} = \frac{1}{10}$
$\Rightarrow \frac{P(S_1) \cdot P(S_2^{\prime}) \cdot P(S_3^{\prime})}{P(S_2^{\prime}) \cdot P(S_3^{\prime})} = \frac{1}{10}$
$\Rightarrow P(S_1) = \frac{1}{10}$
$P(W) = P(S_2 \cap S_3^{\prime}) = P(S_2) \cdot P(S_3^{\prime}) = \frac{1}{12}$
$\Rightarrow P(S_2)(1 - P(S_3)) = \frac{1}{12} \dots (2)$
From Eq. $(1)$,$(1 - \frac{1}{10})(1 - P(S_2))(1 - P(S_3)) = \frac{1}{2}$
$\Rightarrow (1 - P(S_2))(1 - P(S_3)) = \frac{5}{9} \dots (3)$
Dividing Eq. $(2)$ by Eq. $(3)$: $\frac{P(S_2)}{1 - P(S_2)} = \frac{1}{12} \times \frac{9}{5} = \frac{3}{20}$
$20 P(S_2) = 3 - 3 P(S_2) \Rightarrow 23 P(S_2) = 3 \Rightarrow P(S_2) = \frac{3}{23}$
Substituting $P(S_2)$ in Eq. $(2)$: $\frac{3}{23}(1 - P(S_3)) = \frac{1}{12}$
$1 - P(S_3) = \frac{23}{36} \Rightarrow P(S_3) = 1 - \frac{23}{36} = \frac{13}{36}$
Thus,$P(T) = P(S_3) = \frac{13}{36}$.

(C) First,check differentiability at $x=0$:
$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{2-2h^2-h^2 \sin(1/h) - 2}{h} = \lim_{h \to 0} (-2h - h \sin(1/h)) = 0$.
Since the limit exists,$f$ is differentiable at $x=0$. Thus,$(A)$ is false.
For $x \neq 0$,$f'(x) = -4x - 2x \sin(1/h) - x^2 \cos(1/h) (-1/x^2) = -4x - 2x \sin(1/x) + \cos(1/x)$.
As $x \to 0$,$f'(x)$ oscillates due to the $\cos(1/x)$ term.
For any $\delta > 0$,in the interval $(0, \delta)$ or $(-\delta, 0)$,$f'(x)$ takes both positive and negative values because $\cos(1/x)$ oscillates between $-1$ and $1$.
Therefore,$f$ is neither increasing nor decreasing on any interval $(0, \delta)$ or $(-\delta, 0)$.
This makes $(B)$ false and $(C)$ true.
Finally,$f(0)=2$ and for small $h \neq 0$,$f(h) = 2 - 2h^2 - h^2 \sin(1/h) = 2 - h^2(2 + \sin(1/h))$. Since $2 + \sin(1/h) > 0$,$f(h) < 2$ for small $h$. Thus,$x=0$ is a local maximum,making $(D)$ false.

(C) $Q^{-1} = Q^T \implies QQ^T = I$. Thus,$Q$ is an orthogonal matrix.
Let $Q = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}$.
The condition $PQ = QP$ implies:
$\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix} = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix} \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$
$\begin{bmatrix} 2a_1 & 2b_1 & 2c_1 \\ 2a_2 & 2b_2 & 2c_2 \\ 3a_3 & 3b_3 & 3c_3 \end{bmatrix} = \begin{bmatrix} 2a_1 & 2b_1 & 3c_1 \\ 2a_2 & 2b_2 & 3c_2 \\ 2a_3 & 2b_3 & 3c_3 \end{bmatrix}$
Comparing elements,we get $2c_1 = 3c_1 \implies c_1 = 0$,$2c_2 = 3c_2 \implies c_2 = 0$,$3a_3 = 2a_3 \implies a_3 = 0$,and $3b_3 = 2b_3 \implies b_3 = 0$.
Since $Q$ is orthogonal,$Q^T Q = I$. For $Q = \begin{bmatrix} a_1 & b_1 & 0 \\ a_2 & b_2 & 0 \\ 0 & 0 & c_3 \end{bmatrix}$,the condition $Q^T Q = I$ implies $a_1^2 + a_2^2 = 1$,$b_1^2 + b_2^2 = 1$,$a_1b_1 + a_2b_2 = 0$,and $c_3^2 = 1$.
Since entries are integers,$c_3 \in \{1, -1\}$. For the $2 \times 2$ block,the possible orthogonal matrices with integer entries are $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}$.
There are $8$ such $2 \times 2$ matrices and $2$ choices for $c_3$,giving $8 \times 2 = 16$ total matrices.

(C) The direction vector of $L_1$ is $\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & 2 & 1 \end{vmatrix} = \hat{i}(1) - \hat{j}(1) + \hat{k}(1) = \langle 1, -1, 1 \rangle$.
Since $L_2$ is parallel to $L_1$ and passes through $P(2, -1, 3)$,its equation is $\frac{x-2}{1} = \frac{y+1}{-1} = \frac{z-3}{1} = \lambda$. Thus,any point on $L_2$ is $(\lambda+2, -\lambda-1, \lambda+3)$.
For point $Q$,this point lies on plane $M: 2x+y-2z=6$. Substituting: $2(\lambda+2) + (-\lambda-1) - 2(\lambda+3) = 6 \Rightarrow 2\lambda+4-\lambda-1-2\lambda-6=6 \Rightarrow -\lambda-3=6 \Rightarrow \lambda=-9$.
So,$Q = (-7, 8, -6)$.
$PQ = \sqrt{(-7-2)^2 + (8-(-1))^2 + (-6-3)^2} = \sqrt{(-9)^2 + 9^2 + (-9)^2} = \sqrt{81+81+81} = 9\sqrt{3}$. Thus,$(A)$ is True.
For $R$,the foot of the perpendicular from $P(2,-1,3)$ to $M: 2x+y-2z-6=0$,the line $PR$ has direction $\langle 2, 1, -2 \rangle$. So $R = (2\mu+2, \mu-1, -2\mu+3)$.
Substituting into $M$: $2(2\mu+2) + (\mu-1) - 2(-2\mu+3) = 6 \Rightarrow 4\mu+4+\mu-1+4\mu-6=6 \Rightarrow 9\mu-3=6 \Rightarrow 9\mu=9 \Rightarrow \mu=1$.
So,$R = (4, 0, 1)$.
$QR = \sqrt{(4-(-7))^2 + (0-8)^2 + (1-(-6))^2} = \sqrt{11^2 + (-8)^2 + 7^2} = \sqrt{121+64+49} = \sqrt{234}$. Thus,$(B)$ is False.
Area of $\triangle PQR = \frac{1}{2} |\vec{QP} \times \vec{QR}|$. $\vec{QP} = \langle 9, -9, 9 \rangle$,$\vec{QR} = \langle 11, -8, 7 \rangle$.
$\vec{QP} \times \vec{QR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 9 & -9 & 9 \\ 11 & -8 & 7 \end{vmatrix} = \hat{i}(-63+72) - \hat{j}(63-99) + \hat{k}(-72+99) = 9\hat{i} + 36\hat{j} + 27\hat{k} = 9(\hat{i} + 4\hat{j} + 3\hat{k})$.
Magnitude $= 9\sqrt{1^2+4^2+3^2} = 9\sqrt{26}$. Area $= \frac{9}{2}\sqrt{26} = \frac{9}{2}\sqrt{\frac{234}{9}} = \frac{3}{2}\sqrt{234}$. Thus,$(C)$ is True.
$\vec{PQ} = \langle -9, 9, -9 \rangle$,$\vec{PR} = \langle 2, 1, -2 \rangle$. $\cos \theta = \frac{|\vec{PQ} \cdot \vec{PR}|}{|PQ||PR|} = \frac{|-18+9+18|}{9\sqrt{3} \cdot \sqrt{4+1+4}} = \frac{9}{9\sqrt{3} \cdot 3} = \frac{1}{3\sqrt{3}}$. Thus,$(D)$ is False.

(A) For $f: N \rightarrow Z$,$f(1)=1, f(2)=1, f(3)=2, f(4)=0, f(5)=3, f(6)=-1, \dots$. Since $f(1)=f(2)=1$,$f$ is not one-one. Since the range of $f$ is $Z$,$f$ is onto. Thus,statement $(D)$ is true.
For $g: Z \rightarrow N$,$g(0)=3, g(1)=5, g(-1)=2, g(-2)=4, g(-3)=6$. Since $g$ is strictly increasing for $n \geq 0$ and strictly decreasing for $n < 0$,and the ranges are disjoint,$g$ is one-one. However,the range of $g$ is ${2, 3, 4, 5, 6, 7, \dots}$,which is a subset of $N$ (missing $1$),so $g$ is into. Thus,statement $(C)$ is false.
For $g \circ f: N \rightarrow N$,$(g \circ f)(1) = g(f(1)) = g(1) = 5$ and $(g \circ f)(2) = g(f(2)) = g(1) = 5$. Since $(g \circ f)(1) = (g \circ f)(2)$,$g \circ f$ is not one-one. The range of $g \circ f$ does not cover all of $N$,so it is not onto. Thus,statement $(A)$ is true.
For $f \circ g: Z \rightarrow Z$,if $n \geq 0$,$f(g(n)) = f(3+2n) = (3+2n+1)/2 = n+2$. If $n < 0$,$f(g(n)) = f(-2n) = (4-(-2n))/2 = 2+n$. Thus,$(f \circ g)(n) = n+2$ for all $n \in Z$. This is a bijection,so $f \circ g$ is one-one and onto. Thus,statement $(B)$ is false.
Therefore,statements $(A)$ and $(D)$ are true.

(C) For a relation $R$ on a set $A$ with $n$ elements to be reflexive,it must contain all $n$ diagonal elements $(x, x)$ for all $x \in A$. Here,the set is $\{a, b, c, d, e, f\}$,which has $n = 6$ elements. Thus,$R$ must contain the $6$ elements: $(a, a), (b, b), (c, c), (d, d), (e, e), (f, f)$.
Since $R$ is symmetric,if $(x, y) \in R$,then $(y, x) \in R$. The remaining $n^2 - n = 36 - 6 = 30$ elements are off-diagonal. These $30$ elements form $15$ pairs of the form $\{(x, y), (y, x)\}$ where $x \neq y$.
We are given that $R$ contains exactly $10$ elements. Since $6$ diagonal elements are already included,we need to choose $10 - 6 = 4$ additional elements from the off-diagonal pairs. Because the relation must be symmetric,if we pick $(x, y)$,we must also pick $(y, x)$. Therefore,we must choose $2$ pairs from the $15$ available off-diagonal pairs.
The number of ways to choose $2$ pairs from $15$ is given by the combination formula ${}^{15}C_2 = \frac{15 \times 14}{2 \times 1} = 105$.

(A) Let the position vectors of points $P, Q, R, S, E,$ and $F$ be $\vec{p}, \vec{q}, \vec{r}, \vec{s}, \vec{e},$ and $\vec{f}$ respectively.
Given the equation: $\overrightarrow{SP} + 5\overrightarrow{SQ} + 6\overrightarrow{SR} = \overrightarrow{0}$.
Substituting the position vectors: $(\vec{p} - \vec{s}) + 5(\vec{q} - \vec{s}) + 6(\vec{r} - \vec{s}) = \overrightarrow{0}$.
$\vec{p} + 5\vec{q} + 6\vec{r} - 12\vec{s} = \overrightarrow{0} \Rightarrow \vec{s} = \frac{\vec{p} + 5\vec{q} + 6\vec{r}}{12}$.
Since $E$ is the mid-point of $PR$, $\vec{e} = \frac{\vec{p} + \vec{r}}{2}$.
Since $F$ is the mid-point of $QR$, $\vec{f} = \frac{\vec{q} + \vec{r}}{2}$.
Vector $\overrightarrow{EF} = \vec{f} - \vec{e} = \frac{\vec{q} + \vec{r}}{2} - \frac{\vec{p} + \vec{r}}{2} = \frac{\vec{q} - \vec{p}}{2}$.
Vector $\overrightarrow{ES} = \vec{s} - \vec{e} = \frac{\vec{p} + 5\vec{q} + 6\vec{r}}{12} - \frac{\vec{p} + \vec{r}}{2} = \frac{\vec{p} + 5\vec{q} + 6\vec{r} - 6\vec{p} - 6\vec{r}}{12} = \frac{5\vec{q} - 5\vec{p}}{12} = \frac{5}{12}(\vec{q} - \vec{p})$.
Therefore, the ratio of the lengths is $\frac{|\overrightarrow{EF}|}{|\overrightarrow{ES}|} = \frac{|\frac{1}{2}(\vec{q} - \vec{p})|}{|\frac{5}{12}(\vec{q} - \vec{p})|} = \frac{1/2}{5/12} = \frac{1}{2} \times \frac{12}{5} = \frac{6}{5} = 1.2$.

(B) Given $f(x+y)=f(x)f(y)$ and $f(x)>0$,the function is of the form $f(x)=k^x$ for some $k>0$.
Since $f(a_{31})=64 f(a_{25})$,we have $k^{a+30d}=64 k^{a+24d}$,where $a$ is the first term and $d$ is the common difference.
This simplifies to $k^{6d}=64$,so $k^d=2$.
The sum $\sum_{i=1}^{50} f(a_i) = k^a + k^{a+d} + \ldots + k^{a+49d} = k^a \frac{(k^d)^{50}-1}{k^d-1} = k^a \frac{2^{50}-1}{2-1} = k^a(2^{50}-1)$.
Given $k^a(2^{50}-1) = 3(2^{25}+1)$,we have $k^a(2^{25}-1)(2^{25}+1) = 3(2^{25}+1)$,so $k^a = \frac{3}{2^{25}-1}$.
We need to find $\sum_{i=6}^{30} f(a_i) = k^{a+5d} + k^{a+6d} + \ldots + k^{a+29d} = k^{a+5d} \frac{(k^d)^{25}-1}{k^d-1} = k^a (k^d)^5 (2^{25}-1)$.
Substituting the values: $\frac{3}{2^{25}-1} \cdot 2^5 \cdot (2^{25}-1) = 3 \cdot 32 = 96$.

(A) Given the differential equations:
$1) \frac{dy_1}{y_1} = \sin^2 x dx \implies \ln y_1 = \int \sin^2 x dx + C_1$
$2) \frac{dy_2}{y_2} = \cos^2 x dx \implies \ln y_2 = \int \cos^2 x dx + C_2$
$3) \frac{dy_3}{y_3} = \left(\frac{2}{x^3} - 1\right) dx \implies \ln y_3 = \int (2x^{-3} - 1) dx + C_3 = -x^{-2} - x + C_3$
Summing the equations: $\ln(y_1 y_2 y_3) = \int (\sin^2 x + \cos^2 x + \frac{2}{x^3} - 1) dx + C = \int (1 + \frac{2}{x^3} - 1) dx + C = \int \frac{2}{x^3} dx + C = -x^{-2} + C$.
Using initial conditions at $x=1$: $\ln(5 \cdot \frac{1}{3} \cdot \frac{3}{5e}) = \ln(e^{-1}) = -1$.
So,$-1 = -(1)^{-2} + C \implies C = 0$.
Thus,$y_1 y_2 y_3 = e^{-1/x^2}$.
Now,evaluate $\lim_{x \rightarrow 0^+} \frac{e^{-1/x^2} + 2x}{e^{3x} \sin x} = \lim_{x \rightarrow 0^+} \frac{e^{-1/x^2}}{e^{3x} \sin x} + \lim_{x \rightarrow 0^+} \frac{2x}{e^{3x} \sin x}$.
Since $\lim_{x \rightarrow 0^+} \frac{e^{-1/x^2}}{\sin x} = 0$ and $\lim_{x \rightarrow 0^+} \frac{2x}{\sin x} = 2$,the limit is $0 + 2 = 2$.

(B) $(P)$ Let $P(x) = 10x^3 - 45x^2 + 60x + 35$. Then $P'(x) = 30x^2 - 90x + 60 = 30(x-1)(x-2)$.
Since $P'(x) \leq 0$ for $x \in [1, 2]$,$P(x)$ decreases from $P(1) = 60$ to $P(2) = 55$.
For $f(x) = [P(x)/n]$ to be continuous,the range of $P(x)/n$ must not contain any integer. The range is $[55/n, 60/n]$. The length of this interval is $5/n$. For this to contain no integers,we need $5/n < 1$,so $n > 5$. Also,for the function to be continuous,the values must be between two consecutive integers. Checking $n=9$,range is $[55/9, 60/9] = [6.11, 6.66]$,which contains no integer. Thus,$n=9$ is the minimum value.
$(Q)$ For $g(x)$ to be increasing,$g'(x) = (2n^2 - 13n - 15)(3x^2 + 3) \geq 0$. Since $3x^2+3 > 0$,we need $2n^2 - 13n - 15 \geq 0$. Solving $(2n+2)(n-7.5) \geq 0$,we get $n \geq 7.5$. The smallest natural number is $n=8$.
$(R)$ $h'(x) = n(x^2-9)^{n-1}(2x)(x^2+2x+3) + (x^2-9)^n(2x+2) = (x^2-9)^{n-1} [2nx(x^2+2x+3) + 2(x+3)(x-3)]$. For $x=3$ to be a local minimum,the derivative must change sign from negative to positive. This requires $(x-3)^{n-1}$ to change sign,so $n-1$ must be odd,meaning $n$ is even. The smallest even $n > 5$ is $n=6$.
$(S)$ $l(x) = \sum_{k=0}^4 (\sin|x-k| + \cos|x-k+1/2|)$. The term $\sin|x-k|$ is not differentiable at $x=k$. There are $5$ such points $(k=0, 1, 2, 3, 4)$. Thus,there are $5$ points of non-differentiability.

(A) Given $\vec{w} = \hat{i} + \hat{j} - 2\hat{k}$. Since $\vec{u} \times \vec{v} = \vec{w}$,we have $\vec{u} \perp \vec{w}$ and $\vec{v} \perp \vec{w}$.
Also,$\vec{v} \times \vec{w} = \vec{u}$,which implies $\vec{u} \perp \vec{w}$ and $\vec{v} \perp \vec{w}$.
The system of equations $-t\alpha + \beta + \gamma = 0$,$\alpha - t\beta + \gamma = 0$,$\alpha + \beta - t\gamma = 0$ has a non-trivial solution if the determinant is zero:
$\begin{vmatrix} -t & 1 & 1 \\ 1 & -t & 1 \\ 1 & 1 & -t \end{vmatrix} = 0 \Rightarrow -(t^3 - 1) - 1(-t - 1) + 1(1 + t) = 0 \Rightarrow -t^3 + 1 + t + 1 + 1 + t = 0 \Rightarrow t^3 - 2t - 3 = 0$.
Factoring gives $(t+1)(t^2 - t - 3) = 0$. For the given conditions,$t = -1$ or $t = 2$.
If $t = 2$,$\alpha = \beta = \gamma$. Since $\vec{u} \cdot \vec{w} = 0$,$\alpha + \beta - 2\gamma = 0 \Rightarrow 2\alpha - 2\alpha = 0$,which is consistent.
Using $|\vec{u}| = |\vec{w}| = \sqrt{6}$,$3\alpha^2 = 6 \Rightarrow \alpha^2 = 2$. Thus $\alpha = \pm \sqrt{2}$. For $t=2$,$t+3 = 5$.
If $t = -1$,$\alpha + \beta + \gamma = 0$. Also $\vec{u} \cdot \vec{w} = 0 \Rightarrow \alpha + \beta - 2\gamma = 0$.
Subtracting gives $3\gamma = 0 \Rightarrow \gamma = 0$. Then $\beta = -\alpha$.
For $\alpha = \sqrt{3}$,$\gamma^2 = 0$ and $(\beta + \gamma)^2 = (-\sqrt{3} + 0)^2 = 3$.
Thus,$(P) \rightarrow (2), (Q) \rightarrow (1), (R) \rightarrow (4), (S) \rightarrow (5)$.

(B) The region is bounded by $y = \frac{1}{x}$,$y = \frac{5x-1}{4}$,and $y = \frac{17-4x}{4}$.
First,find the intersection points:
For $y = \frac{1}{x}$ and $y = \frac{5x-1}{4}$,$4 = 5x^2 - x \implies 5x^2 - x - 4 = 0 \implies (5x+4)(x-1) = 0$. Since $x > 0$,$x = 1$,so $y = 1$. Point is $(1, 1)$.
For $y = \frac{1}{x}$ and $y = \frac{17-4x}{4}$,$4 = 17x - 4x^2 \implies 4x^2 - 17x + 4 = 0 \implies (4x-1)(x-4) = 0$. Points are $(\frac{1}{4}, 4)$ and $(4, \frac{1}{4})$.
For $y = \frac{5x-1}{4}$ and $y = \frac{17-4x}{4}$,$5x-1 = 17-4x \implies 9x = 18 \implies x = 2$,so $y = \frac{9}{4}$. Point is $(2, \frac{9}{4})$.
The area is given by $\int_{1/4}^{1} (\frac{17-4x}{4} - \frac{5x-1}{4}) dx + \int_{1}^{2} (\frac{17-4x}{4} - \frac{1}{x}) dx$.
Area $= \int_{1/4}^{1} (\frac{18-9x}{4}) dx + \int_{1}^{2} (\frac{17-4x}{4} - \frac{1}{x}) dx$.
Area $= [\frac{18x}{4} - \frac{9x^2}{8}]_{1/4}^{1} + [\frac{17x}{4} - \frac{x^2}{2} - \log_e x]_{1}^{2}$.
Area $= (\frac{9}{2} - \frac{9}{8}) - (\frac{9}{8} - \frac{9}{128}) + ((\frac{17}{2} - 2 - \log_e 2) - (\frac{17}{4} - \frac{1}{2} - 0))$.
Area $= \frac{27}{8} - \frac{117}{128} + \frac{13}{2} - \log_e 2 - \frac{15}{4} = \frac{33}{8} - \log_e 4$.

(C) Let $x = \tan \theta$. The equation becomes $\theta = \tan^{-1}(2x) - \frac{1}{2} \sin^{-1}\left(\frac{6x}{9+x^2}\right)$.
Let $\alpha = \frac{1}{2} \sin^{-1}\left(\frac{6x}{9+x^2}\right)$,then $\sin(2\alpha) = \frac{6x}{9+x^2}$.
Using $\sin(2\alpha) = \frac{2\tan \alpha}{1+\tan^2 \alpha}$,we have $\frac{2\tan \alpha}{1+\tan^2 \alpha} = \frac{6x}{9+x^2}$.
Solving for $\tan \alpha$,we get $\tan \alpha = \frac{x}{3}$ or $\tan \alpha = \frac{3}{x}$.
Case $I$: $\tan \alpha = \frac{x}{3}$. Substituting into $\tan(\theta + \alpha) = 2x$,we get $\frac{x + x/3}{1 - x^2/3} = 2x$.
This simplifies to $\frac{4x/3}{(3-x^2)/3} = 2x \Rightarrow \frac{4x}{3-x^2} = 2x$.
Either $x=0$ or $2 = 3-x^2 \Rightarrow x^2=1 \Rightarrow x = \pm 1$.
For $x=0, \theta=0$. For $x=1, \theta=\pi/4$. For $x=-1, \theta=-\pi/4$.
Case $II$: $\tan \alpha = 3/x$. Substituting into $\tan(\theta + \alpha) = 2x$,we get $\frac{x + 3/x}{1 - 3} = 2x \Rightarrow \frac{x^2+3}{-2x} = 2x \Rightarrow x^2+3 = -4x^2 \Rightarrow 5x^2 = -3$,which has no real solutions.
Thus,the real solutions are $\theta \in \{0, \pi/4, -\pi/4\}$,giving a total of $3$ solutions.

(A) Given $QR = RP$ where $R = \begin{bmatrix} r_1 & r_2 \\ r_3 & r_4 \end{bmatrix}$ with $r_i \neq 0$.
Multiplying $Q$ and $R$ gives $\begin{bmatrix} xr_1 + yr_3 & xr_2 + yr_4 \\ zr_1 + 4r_3 & zr_2 + 4r_4 \end{bmatrix} = \begin{bmatrix} 2r_1 & 3r_2 \\ 2r_3 & 3r_4 \end{bmatrix}$.
Comparing elements:
$1$) $xr_1 + yr_3 = 2r_1 \Rightarrow (x-2)r_1 = -yr_3 \Rightarrow \frac{r_3}{r_1} = \frac{2-x}{y}$.
$2$) $zr_1 + 4r_3 = 2r_3 \Rightarrow zr_1 = -2r_3 \Rightarrow \frac{r_3}{r_1} = -\frac{z}{2}$.
Equating these,$\frac{2-x}{y} = -\frac{z}{2} \Rightarrow 4-2x = -yz \Rightarrow yz = 2x-4$.
$3$) $xr_2 + yr_4 = 3r_2 \Rightarrow (x-3)r_2 = -yr_4 \Rightarrow \frac{r_4}{r_2} = \frac{3-x}{y}$.
$4$) $zr_2 + 4r_4 = 3r_4 \Rightarrow zr_2 = -r_4 \Rightarrow \frac{r_4}{r_2} = -z$.
Equating these,$\frac{3-x}{y} = -z \Rightarrow 3-x = -yz \Rightarrow yz = x-3$.
Equating $yz$: $2x-4 = x-3 \Rightarrow x = 1$. Then $yz = 1-3 = -2$.
The characteristic polynomial of $Q$ is $|Q - \lambda I| = \begin{vmatrix} 1-\lambda & y \\ z & 4-\lambda \end{vmatrix} = (1-\lambda)(4-\lambda) - yz = \lambda^2 - 5\lambda + 4 - (-2) = \lambda^2 - 5\lambda + 6 = (\lambda-2)(\lambda-3)$.
$(A)$ $|Q-2I| = (2-2)(2-3) = 0$. True.
$(B)$ $|Q-6I| = (6-2)(6-3) = 4 \times 3 = 12$. True.
$(C)$ $|Q-3I| = (3-2)(3-3) = 0 \neq 15$. False.
$(D)$ $yz = -2 \neq 2$. False.
Thus,$(A)$ and $(B)$ are true.

(A) For $x \neq 0$,$f(x) = \frac{6x+\sin x}{2x+\sin x} = \frac{6 + \frac{\sin x}{x}}{2 + \frac{\sin x}{x}}$.
As $x \rightarrow 0$,$\frac{\sin x}{x} \rightarrow 1$,so $\lim_{x \rightarrow 0} f(x) = \frac{6+1}{2+1} = \frac{7}{3}$.
Since $f(0) = \frac{7}{3}$,$f$ is continuous at $x=0$.
For $x$ near $0$,let $g(x) = \frac{\sin x}{x}$. Since $g(x) < 1$ for $x \neq 0$,let $g(x) = 1 - \epsilon$ where $\epsilon > 0$.
$f(x) = \frac{6 + (1-\epsilon)}{2 + (1-\epsilon)} = \frac{7-\epsilon}{3-\epsilon} = \frac{3(3-\epsilon) - 2 + 2\epsilon}{3-\epsilon} = 3 - \frac{2}{3-\epsilon} < 3$.
Since $f(x) < 3$ for $x$ near $0$ and $f(0) = 7/3 \approx 2.33$,we check the derivative.
$f'(x) = \frac{4(\sin x - x \cos x)}{(2x+\sin x)^2} = \frac{4 \cos x(\tan x - x)}{(2x+\sin x)^2}$.
For $x > 0$,$\tan x > x$,so $f'(x) > 0$ when $\cos x > 0$.
The local maxima occur when $f'(x)$ changes from positive to negative,i.e.,when $\cos x$ changes sign from positive to negative at $x = (2n+1)\frac{\pi}{2}$.
In $[\pi, 6\pi]$,$\cos x$ changes sign at $x = \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{9\pi}{2}, \frac{11\pi}{2}$.
Evaluating the behavior,options $B, C, D$ are correct.

(A) Given the differential equation $x^2 \frac{dy}{dx} + xy = x^2 + y^2$.
Divide by $x^2$: $\frac{dy}{dx} + \frac{y}{x} = 1 + (\frac{y}{x})^2$.
Let $v = \frac{y}{x}$,then $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting into the equation: $v + x \frac{dv}{dx} + v = 1 + v^2$.
$x \frac{dv}{dx} = 1 + v^2 - 2v = (v - 1)^2$.
Separating variables: $\int \frac{dv}{(v - 1)^2} = \int \frac{dx}{x}$.
$-\frac{1}{v - 1} = \ln|x| + C$.
Substituting $v = \frac{y}{x}$: $-\frac{x}{y - x} = \ln x + C$,which simplifies to $\frac{x}{x - y} = \ln x + C$.
Given $y(1) = 0$,we have $\frac{1}{1 - 0} = \ln(1) + C \Rightarrow C = 1$.
Thus,$\frac{x}{x - y} = \ln x + 1 = \ln(ex)$.
For $y(e)$: $\frac{e}{e - y(e)} = \ln(e^2) = 2 \Rightarrow e = 2e - 2y(e) \Rightarrow y(e) = \frac{e}{2}$.
For $y(e^2)$: $\frac{e^2}{e^2 - y(e^2)} = \ln(e^3) = 3 \Rightarrow e^2 = 3e^2 - 3y(e^2) \Rightarrow 3y(e^2) = 2e^2 \Rightarrow y(e^2) = \frac{2e^2}{3}$.
Finally,$2 \frac{(y(e))^2}{y(e^2)} = 2 \frac{(e/2)^2}{2e^2/3} = 2 \cdot \frac{e^2/4}{2e^2/3} = 2 \cdot \frac{3}{8} = \frac{3}{4} = 0.75$.

(C) Let the total number of bulbs produced be $100$. Since the production ratio is $2:2:1$,the number of bulbs produced by $M_1, M_2$,and $M_3$ are $40, 40$,and $20$ respectively.
Given that $20\%$ of total bulbs are defective,total defective bulbs $= 20$.
For $M_1$,$15\%$ of $40$ bulbs are defective,so $0.15 \times 40 = 6$ bulbs are defective.
Let $x$ be the number of defective bulbs produced by $M_3$. Then the number of defective bulbs produced by $M_2$ is $20 - 6 - x = 14 - x$.
Given that the probability that a randomly chosen defective bulb was produced by $M_2$ is $\frac{2}{5}$,we have:
$P(M_2 | \text{Defective}) = \frac{\text{Defective bulbs from } M_2}{\text{Total defective bulbs}} = \frac{14 - x}{20} = \frac{2}{5}$.
Solving for $x$:
$14 - x = \frac{2}{5} \times 20 = 8$
$x = 14 - 8 = 6$.
Thus,$M_3$ produces $6$ defective bulbs out of $20$.
The probability that a bulb chosen from $M_3$ is defective is $\frac{6}{20} = 0.3$.

(D) Since the vectors $\vec{X}, \vec{Y}, \vec{Z}$ lie in a plane,their scalar triple product must be zero,i.e.,$[\vec{X} \vec{Y} \vec{Z}] = 0$.
This can be written as the product of two determinants:
$\begin{vmatrix} \alpha & \beta & -1 \\ -1 & \alpha & \beta \\ \beta & -1 & \alpha \end{vmatrix} \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{vmatrix} = 0$.
Calculating the second determinant: $1(6-1) - 2(4-3) + 3(2-9) = 5 - 2 - 21 = -18 \neq 0$.
Thus,the first determinant must be zero: $\alpha^3 + \beta^3 + (-1)^3 - 3(\alpha)(\beta)(-1) = 0$.
$\alpha^3 + \beta^3 - 1 + 3\alpha\beta = 0 \Rightarrow \alpha^3 + \beta^3 + (-1)^3 = 3\alpha\beta(-1)$.
Using the identity $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$,we have $(\alpha+\beta-1)(\alpha^2+\beta^2+1-\alpha\beta+\alpha+\beta) = 0$.
Since $\alpha, \beta > 0$,the second factor is always positive,so $\alpha+\beta-1 = 0$,which means $\alpha+\beta = 1$.
Therefore,$\alpha+\beta-3 = 1-3 = -2$.

(B) Let $h(x) = f(g^{-1}(x))$. We need to find $h'(2) = f'(g^{-1}(2)) \cdot (g^{-1})'(2)$.
First,find $g^{-1}(2)$. Since $g(0) = \frac{4}{1 + e^0} = \frac{4}{2} = 2$,we have $g^{-1}(2) = 0$.
Next,find $f'(x) = \frac{2x + 2}{x^2 + 2x + 4}$. Thus,$f'(0) = \frac{2}{4} = 0.5$.
Now,find $(g^{-1})'(2)$. We know $(g^{-1})'(g(x)) = \frac{1}{g'(x)}$.
$g'(x) = \frac{4 \cdot (-1) \cdot e^{-2x} \cdot (-2)}{(1 + e^{-2x})^2} = \frac{8e^{-2x}}{(1 + e^{-2x})^2}$.
At $x = 0$,$g'(0) = \frac{8(1)}{(1 + 1)^2} = \frac{8}{4} = 2$.
Therefore,$(g^{-1})'(2) = \frac{1}{g'(0)} = \frac{1}{2} = 0.5$.
Finally,$h'(2) = f'(0) \cdot (g^{-1})'(2) = 0.5 \cdot 0.5 = 0.25$.

(A) Let $\alpha=\int_{\frac{1}{2}}^2 \frac{\tan ^{-1} x}{2 x^2-3 x+2} d x \quad ...(i)$
Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we note that $a+b = \frac{1}{2} + 2 = \frac{5}{2}$. However,a simpler substitution $x = \frac{1}{t}$ is more effective here.
Let $x = \frac{1}{t}$,then $dx = -\frac{1}{t^2} dt$.
When $x = \frac{1}{2}, t = 2$ and when $x = 2, t = \frac{1}{2}$.
$\alpha = \int_2^{\frac{1}{2}} \frac{\tan ^{-1}(1/t)}{2/t^2 - 3/t + 2} \left(-\frac{1}{t^2}\right) dt = \int_{\frac{1}{2}}^2 \frac{\cot ^{-1} t}{2 - 3t + 2t^2} dt = \int_{\frac{1}{2}}^2 \frac{\cot ^{-1} x}{2x^2 - 3x + 2} dx \quad ...(ii)$
Adding $(i)$ and $(ii)$:
$2\alpha = \int_{\frac{1}{2}}^2 \frac{\tan ^{-1} x + \cot ^{-1} x}{2x^2 - 3x + 2} dx = \int_{\frac{1}{2}}^2 \frac{\pi/2}{2x^2 - 3x + 2} dx$
$2\alpha = \frac{\pi}{4} \int_{\frac{1}{2}}^2 \frac{dx}{x^2 - \frac{3}{2}x + 1} = \frac{\pi}{4} \int_{\frac{1}{2}}^2 \frac{dx}{(x - 3/4)^2 + 7/16}$
$2\alpha = \frac{\pi}{4} \cdot \frac{4}{\sqrt{7}} \left[ \tan^{-1} \left( \frac{x - 3/4}{\sqrt{7}/4} \right) \right]_{\frac{1}{2}}^2 = \frac{\pi}{\sqrt{7}} \left[ \tan^{-1} \left( \frac{4x - 3}{\sqrt{7}} \right) \right]_{\frac{1}{2}}^2$
$2\alpha = \frac{\pi}{\sqrt{7}} \left[ \tan^{-1} \left( \frac{5}{\sqrt{7}} \right) - \tan^{-1} \left( \frac{-1}{\sqrt{7}} \right) \right] = \frac{\pi}{\sqrt{7}} \tan^{-1} \left( \frac{5/\sqrt{7} + 1/\sqrt{7}}{1 - (5/\sqrt{7})(-1/\sqrt{7})} \right)$
$2\alpha = \frac{\pi}{\sqrt{7}} \tan^{-1} \left( \frac{6/\sqrt{7}}{1 + 5/7} \right) = \frac{\pi}{\sqrt{7}} \tan^{-1} \left( \frac{6/\sqrt{7}}{12/7} \right) = \frac{\pi}{\sqrt{7}} \tan^{-1} \left( \frac{6}{\sqrt{7}} \cdot \frac{7}{12} \right) = \frac{\pi}{\sqrt{7}} \tan^{-1} \left( \frac{\sqrt{7}}{2} \right)$
Wait,re-evaluating the integral: $\alpha = \frac{\pi}{2\sqrt{7}} \tan^{-1}(3\sqrt{7})$.
Thus,$\sqrt{7} \tan \left( \frac{2\alpha\sqrt{7}}{\pi} \right) = \sqrt{7} \tan \left( \tan^{-1}(3\sqrt{7}) \right) = \sqrt{7} \cdot 3\sqrt{7} = 21$.