AP EAMCET 2006 Chemistry Question Paper with Answer and Solution

(A) Let $f(\theta) = 3-\cos \theta+\cos \left(\theta+\frac{\pi}{3}\right)$.
Using the identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$:
$f(\theta) = 3-\cos \theta + \left(\cos \theta \cdot \cos \frac{\pi}{3} - \sin \theta \cdot \sin \frac{\pi}{3}\right)$
$f(\theta) = 3-\cos \theta + \frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta$
$f(\theta) = 3 - \frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta$
$f(\theta) = 3 - \left(\cos \theta \cdot \cos \frac{\pi}{3} + \sin \theta \cdot \sin \frac{\pi}{3}\right)$
$f(\theta) = 3 - \cos \left(\theta - \frac{\pi}{3}\right)$
Since $-1 \leq \cos \left(\theta - \frac{\pi}{3}\right) \leq 1$,
$3 - (1) \leq f(\theta) \leq 3 - (-1)$
$2 \leq f(\theta) \leq 4$.
Thus,the interval is $[2,4]$.

(C) We know that $e^{\log (x)} = x$.
Therefore,$e^{\log (\cosh^{-1} 2)} = \cosh^{-1} 2$.
Using the logarithmic form of the inverse hyperbolic cosine function,$\cosh^{-1} x = \log (x + \sqrt{x^2 - 1})$.
Substituting $x = 2$,we get $\cosh^{-1} 2 = \log (2 + \sqrt{2^2 - 1})$.
$= \log (2 + \sqrt{4 - 1}) = \log (2 + \sqrt{3})$.

(B) Given $5 \cos x + 12 \cos y = 13$.
Squaring both sides,we get $(5 \cos x + 12 \cos y)^2 = 169$.
Let $S = 5 \sin x + 12 \sin y$.
Consider $S^2 + 13^2 = (5 \sin x + 12 \sin y)^2 + (5 \cos x + 12 \cos y)^2$.
$= 25(\sin^2 x + \cos^2 x) + 144(\sin^2 y + \cos^2 y) + 120(\sin x \sin y + \cos x \cos y)$.
$= 25 + 144 + 120 \cos(x - y) = 169 + 120 \cos(x - y)$.
Thus,$S^2 = 169 + 120 \cos(x - y) - 169 = 120 \cos(x - y)$.
Since the maximum value of $\cos(x - y)$ is $1$,the maximum value of $S^2$ is $120$.
Therefore,the maximum value of $S$ is $\sqrt{120}$.

(C) The equations of the lines are:
$x-y-2=0$ ...$(i)$
$x+y-4=0$ ...(ii)
$x+3y=6$ ...(iii)
Adding equations $(i)$ and (ii):
$(x-y-2) + (x+y-4) = 0$
$2x - 6 = 0 \implies x = 3$
Substituting $x=3$ in equation (ii):
$3 + y - 4 = 0 \implies y = 1$
Now,check if the point $(3,1)$ satisfies equation (iii):
$3 + 3(1) = 3 + 3 = 6$
Since the point $(3,1)$ satisfies all three equations,the lines are concurrent at $(3,1)$.

(D) The given equation is $x^2-y^2-x+3y-2=0$.
We can rearrange the terms as $x^2 - (y^2 - 3y + 2) - x = 0$.
Factoring the quadratic in $y$: $y^2 - 3y + 2 = (y-1)(y-2)$.
To factor the whole expression,we look for factors of the form $(x+ay+b)(x+cy+d)$.
Expanding $(x-y+1)(x+y-2) = x^2 + xy - 2x - xy - y^2 + 2y + x + y - 2 = x^2 - y^2 - x + 3y - 2$.
Thus,the equation represents the lines $x-y+1=0$ and $x+y-2=0$.

(C) The given equation is $x^2+6xy+8y^2=10$ $\dots$ $(i)$.
Since the axes are rotated through an angle $\theta = \frac{\pi}{4}$,we use the transformation formulas:
$x = x_1 \cos \frac{\pi}{4} - y_1 \sin \frac{\pi}{4} = \frac{x_1-y_1}{\sqrt{2}}$
$y = x_1 \sin \frac{\pi}{4} + y_1 \cos \frac{\pi}{4} = \frac{x_1+y_1}{\sqrt{2}}$
Substituting these into the original equation:
$\left(\frac{x_1-y_1}{\sqrt{2}}\right)^2 + 6\left(\frac{x_1-y_1}{\sqrt{2}}\right)\left(\frac{x_1+y_1}{\sqrt{2}}\right) + 8\left(\frac{x_1+y_1}{\sqrt{2}}\right)^2 = 10$
$\frac{x_1^2+y_1^2-2x_1y_1}{2} + \frac{6(x_1^2-y_1^2)}{2} + \frac{8(x_1^2+y_1^2+2x_1y_1)}{2} = 10$
Multiplying by $2$:
$(x_1^2+y_1^2-2x_1y_1) + (6x_1^2-6y_1^2) + (8x_1^2+8y_1^2+16x_1y_1) = 20$
Combining like terms:
$(1+6+8)x_1^2 + (1-6+8)y_1^2 + (-2+16)x_1y_1 = 20$
$15x_1^2 + 3y_1^2 + 14x_1y_1 = 20$
Thus,the transformed equation is $15x^2+14xy+3y^2=20$.

(C) The given equation is $x^2+6xy+8y^2=10$ $\dots$ $(i)$.
Since the axes are rotated through an angle $\theta = \frac{\pi}{4}$,we use the transformation formulas:
$x = x' \cos \theta - y' \sin \theta = \frac{x'-y'}{\sqrt{2}}$
$y = x' \sin \theta + y' \cos \theta = \frac{x'+y'}{\sqrt{2}}$
Substituting these into $(i)$:
$\left(\frac{x'-y'}{\sqrt{2}}\right)^2 + 6\left(\frac{x'-y'}{\sqrt{2}}\right)\left(\frac{x'+y'}{\sqrt{2}}\right) + 8\left(\frac{x'+y'}{\sqrt{2}}\right)^2 = 10$
$\frac{x'^2+y'^2-2x'y'}{2} + 6\left(\frac{x'^2-y'^2}{2}\right) + 8\left(\frac{x'^2+y'^2+2x'y'}{2}\right) = 10$
Multiplying by $2$:
$(x'^2+y'^2-2x'y') + 6(x'^2-y'^2) + 8(x'^2+y'^2+2x'y') = 20$
$x'^2+y'^2-2x'y' + 6x'^2-6y'^2 + 8x'^2+8y'^2+16x'y' = 20$
$(1+6+8)x'^2 + (1-6+8)y'^2 + (-2+16)x'y' = 20$
$15x'^2 + 3y'^2 + 14x'y' = 20$
Replacing $x', y'$ with $x, y$,the transformed equation is $15x^2+14xy+3y^2=20$.

(C) The given pair of lines is $12x^2 - 20xy + 7y^2 = 0$. Factoring this,we get $(6x - 7y)(2x - y) = 0$.
Thus,the equations of the two lines are $6x - 7y = 0$ $(i)$ and $2x - y = 0$ (ii).
The third line is $2x - 3y + 4 = 0$ (iii).
Solving $(i)$ and (ii): $6x - 7y = 0$ and $2x - y = 0 \implies x=0, y=0$. Vertex $A = (0, 0)$.
Solving $(i)$ and (iii): $6x - 7y = 0$ and $2x - 3y = -4$. Multiplying (iii) by $3$: $6x - 9y = -12$. Subtracting from $(i)$: $2y = 12 \implies y = 6, x = 7$. Vertex $B = (7, 6)$.
Solving (ii) and (iii): $2x - y = 0$ and $2x - 3y = -4$. Subtracting: $2y = 4 \implies y = 2, x = 1$. Vertex $C = (1, 2)$.
The centroid is $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) = \left(\frac{0+7+1}{3}, \frac{0+6+2}{3}\right) = \left(\frac{8}{3}, \frac{8}{3}\right)$.

(C) The Cartesian coordinates of the centre $(h, k)$ are given by $h = r_0 \cos \theta_0$ and $k = r_0 \sin \theta_0$,where $(r_0, \theta_0) = (2, \frac{\pi}{2})$.
$h = 2 \cos(\frac{\pi}{2}) = 0$ and $k = 2 \sin(\frac{\pi}{2}) = 2$.
So,the centre is $(0, 2)$ and the radius $a = 3$.
The Cartesian equation of the circle is $(x - h)^2 + (y - k)^2 = a^2$.
$(x - 0)^2 + (y - 2)^2 = 3^2$.
$x^2 + y^2 - 4y + 4 = 9$.
$x^2 + y^2 - 4y = 5$.
Using the polar coordinate relations $x = r \cos \theta$,$y = r \sin \theta$,and $x^2 + y^2 = r^2$:
$r^2 - 4(r \sin \theta) = 5$.
$r^2 - 4r \sin \theta = 5$.

(C) The length of the tangent drawn from a point $(x_1, y_1)$ to a circle $x^2+y^2+2gx+2fy+c=0$ is given by $\sqrt{x_1^2+y_1^2+2gx_1+2fy_1+c}$.
Given the circle equation $x^2+y^2-2x+4y-11=0$ and the point $(1,3)$.
Substituting the point $(1,3)$ into the expression:
Length $= \sqrt{1^2+3^2-2(1)+4(3)-11}$
$= \sqrt{1+9-2+12-11}$
$= \sqrt{22-13}$
$= \sqrt{9}$
$= 3$

(B) The equations of the circles are $x^2+y^2-8x+2y=0$ and $x^2+y^2-2x-16y+25=0$.
For the first circle,the center $C_1 = (4, -1)$ and the radius $r_1 = \sqrt{4^2 + (-1)^2 - 0} = \sqrt{17}$.
For the second circle,the center $C_2 = (1, 8)$ and the radius $r_2 = \sqrt{1^2 + 8^2 - 25} = \sqrt{1 + 64 - 25} = \sqrt{40}$.
The distance between the centers is $C_1C_2 = \sqrt{(1-4)^2 + (8 - (-1))^2} = \sqrt{(-3)^2 + 9^2} = \sqrt{9 + 81} = \sqrt{90}$.
We compare $C_1C_2$ with the sum and difference of the radii:
$r_1 + r_2 = \sqrt{17} + \sqrt{40} \approx 4.12 + 6.32 = 10.44$.
$|r_1 - r_2| = |\sqrt{17} - \sqrt{40}| \approx |4.12 - 6.32| = 2.20$.
Since $|r_1 - r_2| < C_1C_2 < r_1 + r_2$ (because $2.20 < 9.48 < 10.44$),the two circles intersect at two distinct points.
Therefore,the number of common tangents is $2$.

(B) For a circle $x^2+y^2+2gx+2fy+c=0$ to touch the $y$-axis,the condition is $g^2=c$.
For a circle to touch the $x$-axis,the condition is $f^2=c$.
$I$. For $x^2+y^2-6x-4y-7=0$,$g=-3$ and $c=-7$. Since $g^2 = (-3)^2 = 9 \neq -7$,it does not touch the $y$-axis.
$II$. For $x^2+y^2+6x+4y-7=0$,$f=2$ and $c=-7$. Since $f^2 = (2)^2 = 4 \neq -7$,it does not touch the $x$-axis.
Therefore,neither $I$ nor $II$ is true.

(D) The semi-latus rectum of a parabola is the harmonic mean of the segments of any focal chord.
Let $l$ be the length of the semi-latus rectum.
Since $l$ is the harmonic mean between $b$ and $c$,we have:
$l = \frac{2}{\frac{1}{b} + \frac{1}{c}}$
$l = \frac{2bc}{b+c}$

(D) Let the coordinates of point $A$ on the parabola $y^2=4x$ be $(t^2, 2t)$,where $a=1$.
Since $O$ is the origin $(0,0)$,the midpoint $M(h, k)$ of $OA$ is given by:
$h = \frac{0+t^2}{2} = \frac{t^2}{2} \implies t^2 = 2h$
$k = \frac{0+2t}{2} = t \implies t = k$
Substituting $t=k$ into $t^2=2h$,we get $k^2 = 2h$.
Replacing $(h, k)$ with $(x, y)$,the locus is $y^2=2x$.

(B) We have $\frac{1-2x}{e^x} = (1-2x)e^{-x}$.
Using the expansion $e^{-x} = \sum_{k=0}^{\infty} \frac{(-1)^k x^k}{k!}$,we get:
$(1-2x) \sum_{k=0}^{\infty} \frac{(-1)^k x^k}{k!} = \sum_{k=0}^{\infty} \frac{(-1)^k x^k}{k!} - 2x \sum_{k=0}^{\infty} \frac{(-1)^k x^k}{k!}$.
The coefficient of $x^n$ is obtained from the first term when $k=n$ and from the second term when $k=n-1$:
Coefficient $= \frac{(-1)^n}{n!} - 2 \cdot \frac{(-1)^{n-1}}{(n-1)!}$.
Since $(-1)^{n-1} = -(-1)^n$,we have:
Coefficient $= \frac{(-1)^n}{n!} + 2 \cdot \frac{(-1)^n}{(n-1)!} = \frac{(-1)^n}{n!} [1 + 2n] = (-1)^n \cdot \frac{1+2n}{n!}$.

(A) Given the series expansion: $y = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots$
This is the logarithmic series expansion for $\log(1+x)$.
So,$y = \log(1+x)$.
Taking the exponential of both sides: $e^y = 1+x$.
Therefore,$x = e^y - 1$.
The expansion of $e^y$ is $1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \ldots$.
Substituting this into the expression for $x$:
$x = (1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \ldots) - 1$
$x = y + \frac{y^2}{2!} + \frac{y^3}{3!} + \ldots$

(C) The given equation of the ellipse is $9x^2 + 4y^2 - 18x - 8y - 23 = 0$.
Completing the square,we get:
$9(x^2 - 2x + 1) - 9 + 4(y^2 - 2y + 1) - 4 - 23 = 0$
$9(x - 1)^2 + 4(y - 1)^2 = 36$
Dividing by $36$,we obtain the standard form:
$\frac{(x - 1)^2}{4} + \frac{(y - 1)^2}{9} = 1$
Here,$a^2 = 4$ and $b^2 = 9$,so $a^2 < b^2$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$.
The equations of the latus rectum for a vertical ellipse are $y - k = \pm be$,where $(h, k) = (1, 1)$.
$y - 1 = \pm 3 \times \frac{\sqrt{5}}{3}$
$y - 1 = \pm \sqrt{5}$
$y = 1 \pm \sqrt{5}$

(C) Let $e$ and $e^{\prime}$ be the eccentricities of a hyperbola and its conjugate hyperbola respectively.
We know that the relationship between them is given by $\frac{1}{e^2} + \frac{1}{(e^{\prime})^2} = 1$.
Given that $e = \sqrt{3}$,we have $e^2 = 3$.
Substituting this into the formula: $\frac{1}{3} + \frac{1}{(e^{\prime})^2} = 1$.
$\frac{1}{(e^{\prime})^2} = 1 - \frac{1}{3} = \frac{2}{3}$.
Therefore,$(e^{\prime})^2 = \frac{3}{2}$,which implies $e^{\prime} = \sqrt{\frac{3}{2}}$.

(C) For $l_1 = \lim_{x \rightarrow 2^{+}} (x + [x])$: Let $x = 2 + h$ where $h \rightarrow 0^{+}$. Then $[x] = 2$. So,$l_1 = \lim_{h \rightarrow 0} (2 + h + 2) = 4$.
For $l_2 = \lim_{x \rightarrow 2^{-}} (2x - [x])$: Let $x = 2 - h$ where $h \rightarrow 0^{+}$. Then $[x] = 1$. So,$l_2 = \lim_{h \rightarrow 0} (2(2 - h) - 1) = 4 - 1 = 3$.
For $l_3 = \lim_{x \rightarrow \pi / 2} \frac{\cos x}{x - \pi / 2}$: Using $L'\text{Hospital's rule}$,we differentiate the numerator and denominator with respect to $x$: $\lim_{x \rightarrow \pi / 2} \frac{-\sin x}{1} = -\sin(\pi / 2) = -1$.
Comparing the values: $l_3 = -1$,$l_2 = 3$,$l_1 = 4$. Thus,$-1 < 3 < 4$,which implies $l_3 < l_2 < l_1$.

(D) To evaluate the limit $\lim _{x \rightarrow \infty}\left[\sqrt{x^2+2 x-1}-x\right]$,we rationalize the expression by multiplying and dividing by the conjugate $\left(\sqrt{x^2+2 x-1}+x\right)$:
$\lim _{x}$ ${\rightarrow \infty}\left[\frac{\left(\sqrt{x^2+2 x-1}-x\right)\left(\sqrt{x^2+2 x-1}+x\right)}{\sqrt{x^2+2 x-1}+x}\right]$
$= \lim _{x \rightarrow \infty}\left[\frac{x^2+2 x-1-x^2}{\sqrt{x^2+2 x-1}+x}\right]$
$= \lim _{x \rightarrow \infty}\left[\frac{2x-1}{\sqrt{x^2+2 x-1}+x}\right]$
Divide the numerator and denominator by $x$:
$= \lim _{x \rightarrow \infty}\left[\frac{2-\frac{1}{x}}{\sqrt{1+\frac{2}{x}-\frac{1}{x^2}}+1}\right]$
As $x \rightarrow \infty$,$\frac{1}{x} \rightarrow 0$ and $\frac{1}{x^2} \rightarrow 0$:
$= \frac{2-0}{\sqrt{1+0-0}+1} = \frac{2}{1+1} = \frac{2}{2} = 1$

(C) Let $f(x) = \cos 4x + a \cos 2x + b$. For the limit to be finite as $x \rightarrow 0$,the numerator must approach $0$ at a rate of at least $x^4$.
Using the Taylor series expansion: $\cos \theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \dots$
$\cos 4x = 1 - \frac{(4x)^2}{2} + \frac{(4x)^4}{24} = 1 - 8x^2 + \frac{32}{3}x^4$
$a \cos 2x = a(1 - \frac{(2x)^2}{2} + \frac{(2x)^4}{24}) = a - 2ax^2 + \frac{2a}{3}x^4$
Substituting these into the numerator: $(1 + a + b) + (-8 - 2a)x^2 + (\frac{32}{3} + \frac{2a}{3})x^4$.
For the limit to be finite,the coefficients of $x^0$ and $x^2$ must be zero.
$1 + a + b = 0$ and $-8 - 2a = 0$.
From $-8 - 2a = 0$,we get $a = -4$.
Substituting $a = -4$ into $1 + a + b = 0$: $1 - 4 + b = 0 \implies b = 3$.
Thus,the values are $a = -4, b = 3$.

(C) Given the limit $L = \lim _{n \rightarrow \infty}\left(q^n+p^n\right)^{1 / n}$.
Since $0 < p < q$,we can factor out $q^n$ from the expression:
$L = \lim _{n \rightarrow \infty} \left[ q^n \left( 1 + \left( \frac{p}{q} \right)^n \right) \right]^{1/n}$
$L = \lim _{n \rightarrow \infty} q \left( 1 + \left( \frac{p}{q} \right)^n \right)^{1/n}$
Since $0 < \frac{p}{q} < 1$,as $n \rightarrow \infty$,the term $\left( \frac{p}{q} \right)^n \rightarrow 0$.
Therefore,$L = q \cdot (1 + 0)^0 = q \cdot 1 = q$.

(B) We know the half-angle formulas for a triangle: $\cot \frac{B}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}$ and $\cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$.
Multiplying these,we get $\cot \frac{B}{2} \cot \frac{C}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)} \cdot \frac{s(s-c)}{(s-a)(s-b)}} = \sqrt{\frac{s^2}{(s-a)^2}} = \frac{s}{s-a}$.
Since $s = \frac{a+b+c}{2}$,we have $2s = a+b+c$.
Thus,$\frac{s}{s-a} = \frac{2s}{2s-2a} = \frac{a+b+c}{a+b+c-2a} = \frac{a+b+c}{b+c-a}$.
Given $b+c = 3a$,we substitute this into the expression:
$\frac{a + 3a}{3a - a} = \frac{4a}{2a} = 2$.

(B) We know that $\tan \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ and $\tan \frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$.
Multiplying these,we get $\tan \frac{A}{2} \tan \frac{C}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)} \cdot \frac{(s-a)(s-b)}{s(s-c)}} = \sqrt{\frac{(s-b)^2}{s^2}} = \frac{s-b}{s}$.
Given $\tan \frac{A}{2} = \frac{5}{6}$ and $\tan \frac{C}{2} = \frac{2}{5}$,we have $\frac{s-b}{s} = \frac{5}{6} \cdot \frac{2}{5} = \frac{1}{3}$.
Thus,$3(s-b) = s$ $\Rightarrow 3s - 3b = s$ $\Rightarrow 2s = 3b$.
Since $2s = a + b + c$,we have $a + b + c = 3b$,which simplifies to $a + c = 2b$.

(D) Let the angles of the triangle be $3x, 5x,$ and $10x$.
Since the sum of angles in a triangle is $180^{\circ}$,we have $3x + 5x + 10x = 180^{\circ}$.
$18x = 180^{\circ} \Rightarrow x = 10^{\circ}$.
The angles are $30^{\circ}, 50^{\circ},$ and $100^{\circ}$.
By the Sine Rule,the ratio of sides is $\sin A : \sin B : \sin C$.
The smallest side is opposite the smallest angle $(30^{\circ})$ and the greatest side is opposite the greatest angle $(100^{\circ})$.
Ratio $= \sin 30^{\circ} : \sin 100^{\circ}$.
$= \frac{1}{2} : \sin(90^{\circ} + 10^{\circ})$.
$= \frac{1}{2} : \cos 10^{\circ}$.
$= 1 : 2 \cos 10^{\circ}$.

(D) We know that $\cot \frac{B}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}$ and $\cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$.
Multiplying these,we get $\cot \frac{B}{2} \cdot \cot \frac{C}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)} \cdot \frac{s(s-c)}{(s-a)(s-b)}} = \sqrt{\frac{s^2}{(s-a)^2}} = \frac{s}{s-a}$.
Given $b+c=3a$,the semi-perimeter $s = \frac{a+b+c}{2} = \frac{a+3a}{2} = 2a$.
Substituting $s=2a$ into the expression,we get $\frac{2a}{2a-a} = \frac{2a}{a} = 2$.

(D) Let the radius of the sector be $r$ and the sectorial angle be $\theta$ (in radians).
The arc length $l = r\theta$.
The perimeter $P = l + 2r = r\theta + 2r = r(\theta + 2)$.
Thus,$r = \frac{P}{\theta + 2}$.
The area of the sector $A = \frac{1}{2}r^2\theta$.
Substituting $r$,we get $A = \frac{1}{2} \left( \frac{P}{\theta + 2} \right)^2 \theta = \frac{P^2}{2} \cdot \frac{\theta}{(\theta + 2)^2}$.
To maximize $A$,we differentiate with respect to $\theta$ and set to $0$:
$\frac{dA}{d\theta} = \frac{P^2}{2} \left[ \frac{(\theta + 2)^2 - \theta \cdot 2(\theta + 2)}{(\theta + 2)^4} \right] = 0$.
This implies $(\theta + 2) - 2\theta = 0$,so $2 - \theta = 0$,which gives $\theta = 2 \text{ radians}$.
Therefore,the sectorial angle for maximum area is $2^c$.

(C) Given $A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$.
First,calculate $A^2 = A \cdot A$:
$A^2 = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix}$.
Next,calculate $A^3 = A^2 \cdot A$:
$A^3 = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 41 & 42 & 42 \\ 42 & 41 & 42 \\ 42 & 42 & 41 \end{bmatrix}$.
Now,compute the expression $A^3 - 4A^2 - 6A$:
$A^3 - 4A^2 - 6A = \begin{bmatrix} 41 & 42 & 42 \\ 42 & 41 & 42 \\ 42 & 42 & 41 \end{bmatrix} - 4 \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} - 6 \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$
$= \begin{bmatrix} 41-36-6 & 42-32-12 & 42-32-12 \\ 42-32-12 & 41-36-6 & 42-32-12 \\ 42-32-12 & 42-32-12 & 41-36-6 \end{bmatrix}$
$= \begin{bmatrix} -1 & -2 & -2 \\ -2 & -1 & -2 \\ -2 & -2 & -1 \end{bmatrix} = -\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} = -A$.

(C) We know that for any square matrix $A$ of order $n$,the property of the adjoint matrix is given by $A(\operatorname{adj} A) = |A|I_n$,where $I_n$ is the identity matrix of order $n$.
Taking the determinant on both sides,we get $|A(\operatorname{adj} A)| = ||A|I_n|$.
Using the property $|AB| = |A||B|$ and $|kA| = k^n|A|$ for a matrix of order $n$,we have $|A| \cdot |\operatorname{adj} A| = |A|^n \cdot |I_n|$.
Since $|I_n| = 1$,we have $|A| \cdot |\operatorname{adj} A| = |A|^n$.
Since $A$ is invertible,$|A| \neq 0$,so we can divide by $|A|$ to get $|\operatorname{adj} A| = |A|^{n-1}$.

(A) Given determinant is $\Delta = \left|\begin{array}{ccc} \log e & \log e^2 & \log e^3 \\ \log e^2 & \log e^3 & \log e^4 \\ \log e^3 & \log e^4 & \log e^5 \end{array}\right|$.
Using the property $\log a^n = n \log a$,we get:
$\Delta = \left|\begin{array}{ccc} \log e & 2 \log e & 3 \log e \\ 2 \log e & 3 \log e & 4 \log e \\ 3 \log e & 4 \log e & 5 \log e \end{array}\right|$.
Taking $\log e$ common from each column,we get:
$\Delta = (\log e)^3 \left|\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{array}\right|$.
Applying $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_2$:
$\Delta = (\log e)^3 \left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 1 & 1 \\ 3 & 1 & 1 \end{array}\right|$.
Since column $C_2$ and column $C_3$ are identical,the value of the determinant is $0$.

(B) Given $f(x) = [2x] - 2[x]$.
Case $1$: If $x$ is an integer,let $x = n$ where $n \in Z$.
Then $f(n) = [2n] - 2[n] = 2n - 2n = 0$.
Case $2$: If $x$ is not an integer,let $x = n + f$ where $n \in Z$ and $0 < f < 1$.
Then $f(x) = [2(n + f)] - 2[n + f] = [2n + 2f] - 2n = 2n + [2f] - 2n = [2f]$.
Since $0 < f < 1$,$0 < 2f < 2$. Thus,$[2f]$ can be $0$ (if $0 < f < 0.5$) or $1$ (if $0.5 \leq f < 1$).
However,for any non-integer $x$,$[2x] - 2[x]$ always evaluates to $1$.
Therefore,the range of $f$ is $\{0, 1\}$.

(C) Given the function $f(x) = x - [x] - \frac{1}{2}$.
We are looking for $x$ such that $f(x) = \frac{1}{2}$.
Substituting the expression for $f(x)$,we get $\frac{1}{2} = x - [x] - \frac{1}{2}$.
Adding $\frac{1}{2}$ to both sides,we get $x - [x] = 1$.
By definition of the fractional part function,${x} = x - [x]$,where $0 \le \{x\} < 1$.
Thus,the equation becomes ${x} = 1$.
Since the fractional part of any real number $x$ must be strictly less than $1$,there is no real number $x$ that satisfies this condition.
Therefore,the set $\{x \in R: f(x) = \frac{1}{2}\}$ is the empty set,denoted by $\phi$.

(B) Given $u=\sin ^{-1}\left(\frac{x^2+y^2}{x+y}\right)$.
This implies $\sin u = \frac{x^2+y^2}{x+y}$.
Let $f(x, y) = \sin u = \frac{x^2+y^2}{x+y}$.
Here,$f(x, y)$ is a homogeneous function of degree $n = 1$ because $f(tx, ty) = \frac{(tx)^2+(ty)^2}{tx+ty} = t \frac{x^2+y^2}{x+y} = t^1 f(x, y)$.
According to Euler's theorem for homogeneous functions,$x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = n f$.
Substituting $f = \sin u$ and $n = 1$:
$x \frac{\partial}{\partial x}(\sin u) + y \frac{\partial}{\partial y}(\sin u) = 1 \cdot \sin u$.
Using the chain rule,$\frac{\partial}{\partial x}(\sin u) = \cos u \frac{\partial u}{\partial x}$ and $\frac{\partial}{\partial y}(\sin u) = \cos u \frac{\partial u}{\partial y}$.
So,$x \cos u \frac{\partial u}{\partial x} + y \cos u \frac{\partial u}{\partial y} = \sin u$.
Dividing both sides by $\cos u$:
$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{\sin u}{\cos u} = \tan u$.

(B) Let $h$ be the height of the object and $x$ be the distance from the second point to the base of the hill.
In $\triangle ACD$,$\tan 30^{\circ} = \frac{h}{120 + x}$ $\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{120 + x}$ $\Rightarrow 120 + x = h\sqrt{3} \quad \dots(i)$
In $\triangle BCD$,$\tan 60^{\circ} = \frac{h}{x}$ $\Rightarrow \sqrt{3} = \frac{h}{x}$ $\Rightarrow x = \frac{h}{\sqrt{3}} \quad \dots(ii)$
Substituting $(ii)$ into $(i)$:
$120 + \frac{h}{\sqrt{3}} = h\sqrt{3}$
$120 = h\sqrt{3} - \frac{h}{\sqrt{3}} = h \left( \frac{3 - 1}{\sqrt{3}} \right) = \frac{2h}{\sqrt{3}}$
$h = \frac{120 \times \sqrt{3}}{2} = 60\sqrt{3} \ m$.

(C) For statement $A$: Let $I = \int \left(\frac{x^2-1}{x^2}\right) e^{\left(\frac{x^2+1}{x}\right)} d x$.
We can rewrite the integrand as $I = \int \left(1 - \frac{1}{x^2}\right) e^{\left(x + \frac{1}{x}\right)} d x$.
Let $t = x + \frac{1}{x}$,then $d t = \left(1 - \frac{1}{x^2}\right) d x$.
Substituting these into the integral,we get $I = \int e^t d t = e^t + c = e^{x + \frac{1}{x}} + c = e^{\frac{x^2+1}{x}} + c$.
Thus,statement $A$ is true.
For statement $R$: The integral $\int f^{\prime}(x) e^{f(x)} d x$ is a standard form. Let $u = f(x)$,then $d u = f^{\prime}(x) d x$.
The integral becomes $\int e^u d u = e^u + c = e^{f(x)} + c$.
The statement $R$ claims the result is $f(x) + c$,which is incorrect. Therefore,$R$ is false.

(D) For a biconvex lens of focal length $f$ and refractive index $\mu_g = 1.5$,the lens maker's formula in air is $\frac{1}{f} = (\mu_g - 1)(\frac{1}{R} - \frac{1}{-R}) = (0.5)(\frac{2}{R}) = \frac{1}{R}$. Given $f = 10 ~cm$,we have $R = 10 ~cm$.
When the lens is cut perpendicular to the principal axis,each plano-convex lens has a focal length $f'$ given by $\frac{1}{f'} = (\mu_g - 1)(\frac{1}{R} - \frac{1}{\infty}) = \frac{0.5}{R} = \frac{0.5}{10} = \frac{1}{20}$. Thus,$f' = 20 ~cm$.
When these two plano-convex lenses are joined such that their convex surfaces touch,the combination acts as a biconvex lens with the same focal length $F = 10 ~cm$ in air.
When immersed in water (refractive index $\mu_w = 4/3$),the new focal length $F'$ is given by $\frac{1}{F'} = (\frac{\mu_g}{\mu_w} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.
For the combination,the effective power is $\frac{1}{F'} = (\frac{1.5}{4/3} - 1)(\frac{1}{R} - \frac{1}{-R}) = (\frac{4.5}{4} - 1)(\frac{2}{R}) = (1.125 - 1)(\frac{2}{10}) = 0.125 \times 0.2 = 0.025$.
Therefore,$F' = \frac{1}{0.025} = 40 ~cm$.

(A) The dispersive power $(\omega)$ of a prism is defined as the ratio of the angular dispersion $(\delta_v - \delta_r)$ to the mean deviation $(\delta_y)$.
Mathematically,$\omega = \frac{\delta_v - \delta_r}{\delta_y} = \frac{(\mu_v - 1)A - (\mu_r - 1)A}{(\mu_y - 1)A} = \frac{\mu_v - \mu_r}{\mu_y - 1}$.
From this formula,it is evident that the dispersive power depends only on the refractive indices $(\mu_v, \mu_r, \mu_y)$ of the material of the prism for different colors.
It is independent of the angle of the prism $(A)$,the shape of the prism,and the size of the prism.
Therefore,the correct option is $A$.

(A) The balanced chemical equation for the reaction is: $2KMnO_4 + 5H_2C_2O_4 + 3H_2SO_4 \rightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 10CO_2$.
From the stoichiometry,$2 \text{ moles of } KMnO_4$ react with $5 \text{ moles of } H_2C_2O_4$.
Using the relation $n_{KMnO_4} / 2 = n_{H_2C_2O_4} / 5$:
$(0.05 \ M \times 16 \ mL) / 2 = (x \ M \times 40 \ mL) / 5$.
$0.4 / 2 = 40x / 5 \implies 0.2 = 8x \implies x = 0.025 \ M$.
Since oxalic acid $(H_2C_2O_4)$ is a diprotic acid and dissociates completely:
$[H^+] = 2 \times [H_2C_2O_4] = 2 \times 0.025 = 0.05 \ M$.
$pH = -\log[H^+] = -\log(0.05) = -\log(5 \times 10^{-2}) = 2 - \log 5 = 2 - 0.699 = 1.301 \approx 1.3$.

(A) $1$. $AlCl_3 + 4NaOH \longrightarrow NaAlO_2 + 3NaCl + 2H_2O$. This reaction forms aqueous sodium meta-aluminate and does not release any gas.
$2$. $3NaOH + P_4 + 3H_2O \longrightarrow PH_3 (\text{gas}) + 3NaH_2PO_2$. This releases phosphine gas.
$3$. $2Al + 2NaOH + 2H_2O \xrightarrow{\Delta} 2NaAlO_2 + 3H_2 (\text{gas})$. This releases hydrogen gas.
$4$. $Zn + 2NaOH \xrightarrow{\Delta} Na_2ZnO_2 + H_2 (\text{gas})$. This releases hydrogen gas.
Therefore,the reaction between $AlCl_3$ and $NaOH$ does not liberate a gaseous product.

(A) In an $n-p-n$ transistor,the base-emitter junction is forward-biased,while the collector-base junction is reverse-biased.
Capacitance $C$ is given by $C = \frac{\epsilon A}{d}$,where $d$ is the width of the depletion layer.
For a forward-biased junction (base-emitter),the depletion layer width $d_1$ is very small.
For a reverse-biased junction (collector-base),the depletion layer width $d_2$ is significantly larger.
Since $C \propto \frac{1}{d}$,a smaller depletion width leads to a larger capacitance.
Therefore,$C_1$ (base-emitter capacitance) is greater than $C_2$ (collector-base capacitance).
Thus,$C_1 > C_2$.

(B) Given the linear differential equation: $(1+x^2) \frac{dy}{dx} + 2xy = 4x^2$.
Divide by $(1+x^2)$ to get the standard form $\frac{dy}{dx} + Py = Q$:
$\frac{dy}{dx} + \left(\frac{2x}{1+x^2}\right)y = \frac{4x^2}{1+x^2}$.
Here,$P = \frac{2x}{1+x^2}$ and $Q = \frac{4x^2}{1+x^2}$.
The Integrating Factor ($I$.$F$.) is $e^{\int P dx} = e^{\int \frac{2x}{1+x^2} dx} = e^{\ln(1+x^2)} = 1+x^2$.
The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.
Substituting the values: $y(1+x^2) = \int \left(\frac{4x^2}{1+x^2}\right)(1+x^2) dx + c$.
$y(1+x^2) = \int 4x^2 dx + c$.
$y(1+x^2) = \frac{4x^3}{3} + c$.
Multiplying by $3$,we get $3y(1+x^2) = 4x^3 + c$.

(B) Given the differential equation: $\frac{dx}{dy} + \frac{x}{y} = x^2$
Divide both sides by $x^2$: $\frac{1}{x^2} \frac{dx}{dy} + \frac{1}{xy} = 1$
Let $t = \frac{1}{x}$,then $\frac{dt}{dy} = -\frac{1}{x^2} \frac{dx}{dy}$
Substituting this into the equation: $-\frac{dt}{dy} + \frac{t}{y} = 1$
Rearranging gives the linear differential equation: $\frac{dt}{dy} - \frac{t}{y} = -1$
Here,$P = -\frac{1}{y}$ and $Q = -1$
Integrating Factor ($I$.$F$.) $= e^{\int P dy} = e^{-\int \frac{1}{y} dy} = e^{-\log y} = \frac{1}{y}$
The general solution is $t \cdot (I.F.) = \int Q \cdot (I.F.) dy + c$
$t \cdot \frac{1}{y} = \int (-1) \cdot \frac{1}{y} dy + c$
$\frac{1}{xy} = -\log y + c$
Multiplying by $y$: $\frac{1}{x} = cy - y \log y$

(D) Given that $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$.
We can write this as $\overrightarrow{a}+\overrightarrow{b}=-\overrightarrow{c}$.
Squaring both sides,we get $(\overrightarrow{a}+\overrightarrow{b})^2=(-\overrightarrow{c})^2$.
Expanding the dot product,we have $|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+2|\overrightarrow{a}||\overrightarrow{b}| \cos \theta=|\overrightarrow{c}|^2$,where $\theta$ is the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$.
Substituting the given values: $3^2+4^2+2(3)(4) \cos \theta = (\sqrt{37})^2$.
$9+16+24 \cos \theta = 37$.
$25+24 \cos \theta = 37$.
$24 \cos \theta = 37-25 = 12$.
$\cos \theta = \frac{12}{24} = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,we have $\theta = \frac{\pi}{3}$.

(B) The volume of a parallelepiped with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by the scalar triple product $|\vec{a} \cdot (\vec{b} \times \vec{c})|$,which is equal to the absolute value of the determinant of the matrix formed by these vectors.
Given edges: $\vec{a} = 4 \hat{i} + 5 \hat{j} + \hat{k}$,$\vec{b} = 0 \hat{i} - 1 \hat{j} + 1 \hat{k}$,$\vec{c} = 3 \hat{i} + 9 \hat{j} + p \hat{k}$.
Volume $= |\text{det}(\vec{a}, \vec{b}, \vec{c})| = 34$.
$\Rightarrow \left|\begin{array}{rrr} 4 & 5 & 1 \\ 0 & -1 & 1 \\ 3 & 9 & p \end{array}\right| = \pm 34$.
Expanding the determinant along the first row:
$4(-p - 9) - 5(0 - 3) + 1(0 - (-3)) = \pm 34$.
$-4p - 36 + 15 + 3 = \pm 34$.
$-4p - 18 = \pm 34$.
Case $1$: $-4p - 18 = 34 \Rightarrow -4p = 52 \Rightarrow p = -13$.
Case $2$: $-4p - 18 = -34 \Rightarrow -4p = -16 \Rightarrow p = 4$.
Since $-13$ is one of the options,the correct value is $-13$.

(D) Let the coordinates of point $A$ be $(x, y, z)$.
Since $OA$ is equally inclined to the axes $OX, OY$ and $OZ$,the direction cosines $l, m, n$ are equal.
Thus,$l = m = n$.
Since $l^2 + m^2 + n^2 = 1$,we have $3l^2 = 1$,which implies $l = m = n = \pm \frac{1}{\sqrt{3}}$.
The coordinates of $A$ are given by $(r l, r m, r n)$,where $r$ is the distance from the origin.
Given $r = \sqrt{3}$,the coordinates are $(\sqrt{3} \times \pm \frac{1}{\sqrt{3}}, \sqrt{3} \times \pm \frac{1}{\sqrt{3}}, \sqrt{3} \times \pm \frac{1}{\sqrt{3}})$.
This simplifies to $(1, 1, 1)$ or $(-1, -1, -1)$.
Comparing with the given options,the correct coordinate is $(1, 1, 1)$.

(B) Given equations for direction cosines $(l, m, n)$ are:
$l+m+n=0$ $\dots(i)$
$l^2+m^2-n^2=0$ $\dots(ii)$
We also know that for direction cosines,$l^2+m^2+n^2=1$ $\dots(iii)$
From $(i)$,$l+m = -n$. Squaring both sides: $l^2+m^2+2lm = n^2$.
Substituting $l^2+m^2 = n^2$ from $(ii)$ into this,we get $n^2+2lm = n^2$,which implies $2lm = 0$,so $l=0$ or $m=0$.
Case $1$: If $l=0$,then $m+n=0 \implies m=-n$. Substituting into $(iii)$,$0^2+(-n)^2+n^2=1 \implies 2n^2=1 \implies n = \pm 1/\sqrt{2}$. Thus,$(l_1, m_1, n_1) = (0, -1/\sqrt{2}, 1/\sqrt{2})$ and $(0, 1/\sqrt{2}, -1/\sqrt{2})$.
Case $2$: If $m=0$,then $l+n=0 \implies l=-n$. Substituting into $(iii)$,$(-n)^2+0^2+n^2=1 \implies 2n^2=1 \implies n = \pm 1/\sqrt{2}$. Thus,$(l_2, m_2, n_2) = (-1/\sqrt{2}, 0, 1/\sqrt{2})$ and $(1/\sqrt{2}, 0, -1/\sqrt{2})$.
The angle $\theta$ between two lines with direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ is given by $\cos \theta = |l_1l_2 + m_1m_2 + n_1n_2|$.
Taking $L_1 = (0, -1/\sqrt{2}, 1/\sqrt{2})$ and $L_2 = (-1/\sqrt{2}, 0, 1/\sqrt{2})$:
$\cos \theta = |(0)(-1/\sqrt{2}) + (-1/\sqrt{2})(0) + (1/\sqrt{2})(1/\sqrt{2})| = |0 + 0 + 1/2| = 1/2$.
Therefore,$\theta = \cos^{-1}(1/2) = \pi/3$.

(C) Total number of balls $= 5 + 6 = 11$.
Total number of ways to draw $7$ balls from $11$ is ${ }^{11} C_7$.
Number of ways to draw $3$ white balls from $5$ is ${ }^5 C_3$.
Number of ways to draw $4$ green balls from $6$ is ${ }^6 C_4$.
Therefore,the number of favorable outcomes is ${ }^5 C_3 \times { }^6 C_4$.
Required probability $= \frac{{ }^5 C_3 \times { }^6 C_4}{{ }^{11} C_7}$.

(B) The set of numbers is $S = \{1, 2, 3, \ldots, 1000\}$,so $n(S) = 1000$.
We are looking for numbers $n$ such that $n \equiv 1 \pmod{7}$.
These numbers are of the form $n = 7k + 1$ for $k \ge 0$.
For $1 \le 7k + 1 \le 1000$,we have $0 \le 7k \le 999$,which implies $0 \le k \le \frac{999}{7} \approx 142.71$.
Since $k$ must be an integer,$k \in \{0, 1, 2, \ldots, 142\}$.
The number of such values is $142 - 0 + 1 = 143$.
The probability is $P = \frac{143}{1000}$.

(C) The average number of errors per page is $\lambda_{page} = \frac{250}{500} = 0.5$.
For a sample of $n = 2$ pages,the average number of errors is $\lambda = n \times \lambda_{page} = 2 \times 0.5 = 1$.
According to the Poisson distribution,the probability of finding $X$ errors is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.
For no errors,we set $k = 0$:
$P(X=0) = \frac{e^{-1} \times 1^0}{0!} = e^{-1} \times 1 = e^{-1}$.

(B) The total number of outcomes when tossing two dice is $n(S) = 6 \times 6 = 36$.
For event $E$ (sum is $8$): The outcomes are $\{(2,6), (3,5), (4,4), (5,3), (6,2)\}$. Thus,$n(E) = 5$ and $P(E) = \frac{5}{36}$. So,statement $I$ is false.
For event $F$ (even numbers on both dice): The possible even numbers on a die are $\{2, 4, 6\}$. The outcomes are $\{(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)\}$. Thus,$n(F) = 9$ and $P(F) = \frac{9}{36} = \frac{1}{4}$. So,statement $II$ is false.
Therefore,neither $I$ nor $II$ is true.