AP EAMCET 2007 Chemistry Question Paper with Answer and Solution

(C) The initial horizontal velocity of the object is $u_x = u \cos \theta$. Given $\sin \theta = 3/5$,then $\cos \theta = 4/5$. So,$u_x = 100 \times (4/5) = 80 \ m/s$.
The horizontal range $R$ of the projectile is $R = \frac{u^2 \sin 2\theta}{g} = \frac{2 u^2 \sin \theta \cos \theta}{g} = \frac{2 \times 100^2 \times (3/5) \times (4/5)}{10} = 960 \ m$.
The horizontal distance to the highest point is $R/2 = 480 \ m$.
At the highest point,the momentum is conserved. The initial momentum is $P_i = (2m) u_x = (2m)(80) = 160m$.
After the explosion,the first piece $(m)$ comes to rest $(v_1 = 0)$,so $P_f = m(0) + m(v_2) = m v_2$.
Equating $P_i = P_f$,we get $160m = m v_2$,so $v_2 = 160 \ m/s$.
The second piece travels from the highest point with horizontal velocity $160 \ m/s$ for the remaining time of flight $t = \frac{u \sin \theta}{g} = \frac{100 \times (3/5)}{10} = 6 \ s$.
The additional horizontal distance covered by the second piece is $d = v_2 \times t = 160 \times 6 = 960 \ m$.
The total distance from the point of projection is $R/2 + d = 480 + 960 = 1440 \ m$.

(B) The maximum acceleration of a simple harmonic oscillator is given by $a_{max} = \omega^2 A$,where $\omega$ is the angular frequency and $A$ is the amplitude.
The maximum velocity is given by $v_{max} = \omega A$.
According to the problem,the magnitude of maximum acceleration is $\pi$ times the maximum velocity:
$a_{max} = \pi \cdot v_{max}$
$\omega^2 A = \pi \cdot \omega A$
Dividing both sides by $\omega A$ (assuming $\omega, A \neq 0$):
$\omega = \pi$
Since the angular frequency $\omega = \frac{2\pi}{T}$,where $T$ is the time period:
$\frac{2\pi}{T} = \pi$
$T = \frac{2\pi}{\pi} = 2 \ s$
Therefore,the time period of the oscillator is $2 \ s$.

(C) Metals are good conductors of electricity due to the presence of free electrons.
Electrical conductivity of semiconductors increases with an increase in temperature.
Silicon doped with boron (Group $13$) is a $p$-type semiconductor.
Silicon doped with arsenic (Group $15$) is an $n$-type semiconductor.
Therefore,option $C$ is the correct statement.

(A) Producer gas is a mixture of $CO$ and $N_2$.
Its calorific value is low primarily due to the high percentage of non-combustible nitrogen $(N_2)$ gas present in the mixture.

(C) When sodium thiosulphate solution is added to silver bromide $(AgBr)$,it dissolves to form a soluble complex called sodium argentothiosulphate.
The chemical reaction is:
$AgBr(s) + 2Na_2S_2O_3(aq) \longrightarrow Na_3[Ag(S_2O_3)_2](aq) + NaBr(aq)$
Thus,the formula of the product formed is $Na_3[Ag(S_2O_3)_2]$.

(D) peroxy acid contains an $-O-O-$ linkage in its structure.
$1$. Perphosphoric acid $(H_3PO_5)$ contains a peroxy linkage.
$2$. Pernitric acid $(HNO_4)$ contains a peroxy linkage.
$3$. Perdisulphuric acid $(H_2S_2O_8)$ contains a peroxy linkage.
$4$. Perchloric acid $(HClO_4)$ does not contain a peroxy linkage; it is an oxoacid of chlorine where chlorine is in the $+7$ oxidation state.

(D) $Krypton$ is used in miner's cap lamps because it provides a high-intensity light source suitable for underground mining conditions.

(A) polygon of $n$ sides has the number of diagonals given by the formula: $\frac{n(n-3)}{2} = 275$.
Multiplying by $2$,we get $n(n-3) = 550$.
$n^2 - 3n - 550 = 0$.
Factoring the quadratic equation: $n^2 - 25n + 22n - 550 = 0$.
$n(n - 25) + 22(n - 25) = 0$.
$(n - 25)(n + 22) = 0$.
Since the number of sides $n$ must be positive,$n = 25$ (as $n = -22$ is not possible).

(A) Since $a, b, c$ are in $AP$,we have $2b = a + c$,which implies $c - b = b - a$.
Given that $b-a, c-b, a$ are in $GP$,the square of the middle term equals the product of the extremes: $(c-b)^2 = (b-a)a$.
Substituting $c-b = b-a$ into the equation,we get $(b-a)^2 = (b-a)a$.
Assuming $b \neq a$,we divide by $(b-a)$ to get $b-a = a$,which simplifies to $b = 2a$.
Substituting $b = 2a$ into $2b = a + c$,we get $2(2a) = a + c$,so $4a = a + c$,which means $c = 3a$.
Thus,the ratio $a: b: c = a: 2a: 3a = 1: 2: 3$.

(C) Given,$S_n = 1^3 + 2^3 + \ldots + n^3 = \sum_{k=1}^{n} k^3$.
Also,$T_n = 1 + 2 + \ldots + n = \sum_{k=1}^{n} k$.
We know the formula for the sum of cubes of the first $n$ natural numbers is $S_n = \left[\frac{n(n+1)}{2}\right]^2$.
The sum of the first $n$ natural numbers is $T_n = \frac{n(n+1)}{2}$.
Substituting $T_n$ into the expression for $S_n$,we get $S_n = (T_n)^2$.
Therefore,$S_n = T_n^2$.

(C) We know that $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.
Therefore,$\tan A - \tan B = \tan(A-B)(1 + \tan A \tan B)$.
Substituting $A = 80^{\circ}$ and $B = 10^{\circ}$:
$\tan 80^{\circ} - \tan 10^{\circ} = \tan(80^{\circ}-10^{\circ})(1 + \tan 80^{\circ} \tan 10^{\circ}) = \tan 70^{\circ}(1 + \tan 80^{\circ} \tan 10^{\circ})$.
Now,$\frac{\tan 80^{\circ}-\tan 10^{\circ}}{\tan 70^{\circ}} = \frac{\tan 70^{\circ}(1 + \tan 80^{\circ} \tan 10^{\circ})}{\tan 70^{\circ}} = 1 + \tan 80^{\circ} \tan 10^{\circ}$.
Since $\tan 80^{\circ} = \cot 10^{\circ}$,we have $1 + \cot 10^{\circ} \tan 10^{\circ} = 1 + 1 = 2$.

(B) Let the points be $A(1, -2)$ and $B(3, 2)$.
The slope $m_1$ of the line passing through $A$ and $B$ is given by $m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - (-2)}{3 - 1} = \frac{4}{2} = 2$.
The equation of the second line is $x + 2y - 7 = 0$,which can be written as $2y = -x + 7$ or $y = -\frac{1}{2}x + \frac{7}{2}$.
The slope $m_2$ of this line is $-\frac{1}{2}$.
Now,calculate the product of the slopes: $m_1 \times m_2 = 2 \times (-\frac{1}{2}) = -1$.
Since the product of the slopes is $-1$,the two lines are perpendicular to each other.
Therefore,the angle between the lines is $\frac{\pi}{2}$.

(B) Let the line $AB$ pass through $A(2,-1)$ and $B(6,5)$. The slope of $AB$ is $m = \frac{5 - (-1)}{6 - 2} = \frac{6}{4} = \frac{3}{2}$.
The equation of line $AB$ is $y - (-1) = \frac{3}{2}(x - 2)$,which simplifies to $2y + 2 = 3x - 6$,or $3x - 2y - 8 = 0$.
The line $PD$ is perpendicular to $AB$ and passes through $P(4,1)$. The slope of $PD$ is $-\frac{1}{m} = -\frac{2}{3}$.
The equation of $PD$ is $y - 1 = -\frac{2}{3}(x - 4)$,which simplifies to $3y - 3 = -2x + 8$,or $2x + 3y - 11 = 0$.
Let $D$ divide $AB$ in the ratio $k:1$. The coordinates of $D$ are $\left(\frac{6k+2}{k+1}, \frac{5k-1}{k+1}\right)$.
Since $D$ lies on $PD$ $(2x + 3y - 11 = 0)$,we have $2\left(\frac{6k+2}{k+1}\right) + 3\left(\frac{5k-1}{k+1}\right) - 11 = 0$.
$12k + 4 + 15k - 3 - 11(k + 1) = 0$.
$27k + 1 - 11k - 11 = 0$.
$16k - 10 = 0 \Rightarrow k = \frac{10}{16} = \frac{5}{8}$.
Thus,the ratio is $5:8$.

(D) The given equation of the pair of lines is $x^2+2xy-35y^2-4x+44y-12=0$.
Factoring the quadratic expression,we get $(x+7y-6)(x-5y+2)=0$.
This represents two lines: $L_1: x+7y-6=0$ and $L_2: x-5y+2=0$.
The third line is $L_3: 5x+\lambda y-8=0$.
For the three lines to be concurrent,the determinant of their coefficients must be zero:
$\begin{vmatrix} 1 & 7 & -6 \\ 1 & -5 & 2 \\ 5 & \lambda & -8 \end{vmatrix} = 0$
Expanding the determinant:
$1(40-2\lambda) - 7(-8-10) - 6(\lambda+25) = 0$
$40 - 2\lambda + 126 - 6\lambda - 150 = 0$
$16 - 8\lambda = 0$
$8\lambda = 16$
$\lambda = 2$

(C) The given equation is $2x^2+4xy+5y^2-4x-22y+7=0$.
Comparing this with the general second-degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=2, h=2, b=5, g=-2, f=-11, c=7$.
To eliminate the first-degree terms,the origin $(0,0)$ must be shifted to the point $(h', k')$ given by the intersection of the partial derivatives $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$.
$\frac{\partial f}{\partial x} = 4x+4y-4 = 0 \implies x+y=1$
$\frac{\partial f}{\partial y} = 4x+10y-22 = 0 \implies 2x+5y=11$
Solving these equations:
From the first equation,$x = 1-y$.
Substituting into the second: $2(1-y)+5y=11 \implies 2-2y+5y=11 \implies 3y=9 \implies y=3$.
Then $x = 1-3 = -2$.
Thus,the origin is shifted to $(-2, 3)$.

(C) The given equation is $2x^2 + 4xy + 5y^2 - 4x - 22y + 7 = 0$.
Comparing this with the general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a = 2, h = 2, b = 5, g = -2, f = -11, c = 7$.
To eliminate the first-degree terms,the origin $(0, 0)$ must be shifted to the point $(h_0, k_0)$ given by the intersection of the partial derivatives $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$.
These are given by the formulas:
$h_0 = \frac{hf - bg}{ab - h^2} = \frac{(2)(-11) - (5)(-2)}{(2)(5) - (2)^2} = \frac{-22 + 10}{10 - 4} = \frac{-12}{6} = -2$
$k_0 = \frac{gh - af}{ab - h^2} = \frac{(-2)(2) - (2)(-11)}{(2)(5) - (2)^2} = \frac{-4 + 22}{10 - 4} = \frac{18}{6} = 3$
Thus,the origin should be shifted to the point $(-2, 3)$.

(D) The given equation of the pair of lines is $x^2+2xy-35y^2-4x+44y-12=0$.
Comparing this with the general form $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=1, h=1, b=-35, g=-2, f=22, c=-12$.
The point of intersection $(x_0, y_0)$ of the pair of lines is given by the formula:
$x_0 = \frac{hf-bg}{ab-h^2} = \frac{(1)(22)-(-35)(-2)}{(1)(-35)-(1)^2} = \frac{22-70}{-35-1} = \frac{-48}{-36} = \frac{4}{3}$
$y_0 = \frac{gh-af}{ab-h^2} = \frac{(-2)(1)-(1)(22)}{(1)(-35)-(1)^2} = \frac{-2-22}{-36} = \frac{-24}{-36} = \frac{2}{3}$
Since the lines are concurrent,the point $(\frac{4}{3}, \frac{2}{3})$ must satisfy the line $5x+\lambda y-8=0$.
$5(\frac{4}{3}) + \lambda(\frac{2}{3}) - 8 = 0$
$\frac{20}{3} + \frac{2\lambda}{3} - 8 = 0$
Multiply by $3$: $20 + 2\lambda - 24 = 0$
$2\lambda - 4 = 0$
$2\lambda = 4$
$\lambda = 2$

(A) The equation of the circle is $x^2+y^2=4$ and the line is $y=3x+c$,which can be written as $\frac{y-3x}{c}=1$.
To find the equation of the pair of lines joining the origin to the intersection points,we homogenize the circle equation:
$x^2+y^2=4(1)^2$
$x^2+y^2=4\left(\frac{y-3x}{c}\right)^2$
$c^2(x^2+y^2)=4(y^2+9x^2-6xy)$
$c^2x^2+c^2y^2=4y^2+36x^2-24xy$
$(c^2-36)x^2+24xy+(c^2-4)y^2=0$
Since the lines are perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero:
$(c^2-36)+(c^2-4)=0$
$2c^2-40=0$
$2c^2=40$
$c^2=20$

(A) The equation of the circle is $x^2+y^2=4$ and the line is $y=3x+c$,which can be written as $\frac{y-3x}{c}=1$.
Homogenizing the equation of the circle using the line equation:
$x^2+y^2=4\left(\frac{y-3x}{c}\right)^2$
$c^2(x^2+y^2)=4(y^2+9x^2-6xy)$
$c^2x^2+c^2y^2=4y^2+36x^2-24xy$
$(c^2-36)x^2+24xy+(c^2-4)y^2=0$
Since the lines are perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero:
$(c^2-36)+(c^2-4)=0$
$2c^2-40=0$
$2c^2=40$
$c^2=20$

(A) Given,radius $r = 3$.
Since the circle lies in the fourth quadrant and touches both coordinate axes ($x=0$ and $y=0$),its center must be at $(3, -3)$.
The standard equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$.
Substituting $h=3$,$k=-3$,and $r=3$:
$(x-3)^2 + (y-(-3))^2 = 3^2$
$(x-3)^2 + (y+3)^2 = 9$
$x^2 - 6x + 9 + y^2 + 6y + 9 = 9$
$x^2 + y^2 - 6x + 6y + 9 = 0$.

(C) The inverse point of $P(x_1, y_1)$ with respect to a circle $S=0$ is the foot of the perpendicular from the center of the circle to the polar of $P$ with respect to the circle,or more generally,the point $P'$ such that $P, P'$ are collinear with the center and $CP \cdot CP' = r^2$.
Given circle: $x^2+y^2-4x-6y+9=0$. Center $C = (2,3)$,radius $r = \sqrt{2^2+3^2-9} = \sqrt{4+9-9} = 2$.
Polar of $P(1,2)$ w.r.t. the circle is $x(1)+y(2)-2(x+1)-3(y+2)+9=0$.
Simplifying: $x+2y-2x-2-3y-6+9=0$ $\Rightarrow -x-y+1=0$ $\Rightarrow x+y-1=0$.
The inverse point $P'(x', y')$ lies on the line $CP$ and satisfies $CP \cdot CP' = r^2$.
Line $CP$ passes through $(2,3)$ and $(1,2)$,so its equation is $y-3 = \frac{2-3}{1-2}(x-2)$ $\Rightarrow y-3 = x-2$ $\Rightarrow y=x+1$.
Intersection of $x+y-1=0$ and $y=x+1$: $x+(x+1)-1=0$ $\Rightarrow 2x=0$ $\Rightarrow x=0, y=1$.
Thus,the inverse point is $(0,1)$.

(D) The general equation of a coaxial system of circles is given by $x^2+y^2+2\lambda x+c=0$.
Limiting points are the centers of the circles of radius zero in the system.
Setting the radius $r = \sqrt{g^2+f^2-c} = 0$,we get $\sqrt{\lambda^2-c} = 0$,which implies $\lambda^2 = c$.
For the limiting points to be distinct and real,we must have $\lambda^2 > 0$,which implies $c > 0$.
Thus,the condition for the system to have distinct limiting points is $c > 0$.

(B) Given equation of the parabola is $y^2+6y-2x+5=0$.
Completing the square for $y$:
$y^2+6y+9-9-2x+5=0$
$(y+3)^2-4-2x=0$
$(y+3)^2=2(x+2)$.
Comparing this with the standard form $(y-k)^2=4a(x-h)$,we get:
Vertex $(h, k) = (-2, -3)$. Thus,statement $(I)$ is true.
Here,$4a=2$,so $a=\frac{1}{2}$.
The directrix of the parabola $(y-k)^2=4a(x-h)$ is given by $x=h-a$.
$x = -2 - \frac{1}{2} = -\frac{5}{2}$.
$2x+5=0$. Thus,statement $(II)$ is false.
Therefore,$I$ is true and $II$ is false.

(A) We have $\frac{1-2x-x^2}{e^{-x}} = (1-2x-x^2)e^x$.
Expanding $e^x$ as $\sum_{n=0}^{\infty} \frac{x^n}{n!}$,we get:
$(1-2x-x^2) \sum_{n=0}^{\infty} \frac{x^n}{n!} = \sum_{n=0}^{\infty} \frac{x^n}{n!} - 2 \sum_{n=0}^{\infty} \frac{x^{n+1}}{n!} - \sum_{n=0}^{\infty} \frac{x^{n+2}}{n!}$.
To find the coefficient of $x^k$,we look at the terms where the power of $x$ is $k$:
$1$. From $\sum \frac{x^n}{n!}$,the coefficient is $\frac{1}{k!}$.
$2$. From $-2 \sum \frac{x^{n+1}}{n!}$,we set $n+1=k \implies n=k-1$,so the coefficient is $-\frac{2}{(k-1)!}$.
$3$. From $-\sum \frac{x^{n+2}}{n!}$,we set $n+2=k \implies n=k-2$,so the coefficient is $-\frac{1}{(k-2)!}$.
Summing these,the coefficient of $x^k$ is $\frac{1}{k!} - \frac{2}{(k-1)!} - \frac{1}{(k-2)!}$.
Multiplying the terms to have a common denominator $k!$:
$\frac{1}{k!} - \frac{2k}{k!} - \frac{k(k-1)}{k!} = \frac{1-2k-(k^2-k)}{k!} = \frac{1-2k-k^2+k}{k!} = \frac{1-k-k^2}{k!}$.

(D) We have,$(1+x+x^2)^n = a_0 + a_1x + a_2x^2 + a_3x^3 + \ldots + a_{2n}x^{2n}$.
On differentiating both sides with respect to $x$,we get:
$n(1+x+x^2)^{n-1}(1+2x) = a_1 + 2a_2x + 3a_3x^2 + \ldots + 2na_{2n}x^{2n-1}$.
Now,putting $x=1$ in the above equation,we get:
$n(1+1+1)^{n-1}(1+2(1)) = a_1 + 2a_2 + 3a_3 + \ldots + 2na_{2n}$.
$n(3)^{n-1}(3) = a_1 + 2a_2 + 3a_3 + \ldots + 2na_{2n}$.
Therefore,$a_1 + 2a_2 + 3a_3 + \ldots + 2na_{2n} = n \cdot 3^n$.

(C) The given equation of the ellipse is $2x^2 + 3y^2 = 6$,which can be written as $\frac{x^2}{3} + \frac{y^2}{2} = 1$.
Two points $(x_1, y_1)$ and $(x_2, y_2)$ are conjugate with respect to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ if $\frac{x_1 x_2}{a^2} + \frac{y_1 y_2}{b^2} = 1$.
Here,$(x_1, y_1) = (1, 2)$,$(x_2, y_2) = (k, -1)$,$a^2 = 3$,and $b^2 = 2$.
Substituting these values into the condition:
$\frac{(1)(k)}{3} + \frac{(2)(-1)}{2} = 1$
$\frac{k}{3} - 1 = 1$
$\frac{k}{3} = 2$
$k = 6$.

(C) The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at the point $(a \sec \theta, b \tan \theta)$ is given by $\frac{ax}{\sec \theta} + \frac{by}{\tan \theta} = a^2 + b^2$.
Comparing this with the given line $lx + my = 1$,we can rewrite the normal equation as $\frac{a \cos \theta}{a^2 + b^2} x + \frac{b \cot \theta}{a^2 + b^2} y = 1$.
Thus,$l = \frac{a \cos \theta}{a^2 + b^2}$ and $m = \frac{b \cot \theta}{a^2 + b^2}$.
Now,$\frac{a^2}{l^2} - \frac{b^2}{m^2} = \frac{a^2 (a^2 + b^2)^2}{a^2 \cos^2 \theta} - \frac{b^2 (a^2 + b^2)^2}{b^2 \cot^2 \theta}$.
$= (a^2 + b^2)^2 (\sec^2 \theta - \tan^2 \theta)$.
Since $\sec^2 \theta - \tan^2 \theta = 1$,we get $\frac{a^2}{l^2} - \frac{b^2}{m^2} = (a^2 + b^2)^2$.

(B) We are given the function $f(x) = \begin{cases} \frac{\sin(1+[x])}{[x]}, & [x] \neq 0 \\ 0, & [x] = 0 \end{cases}$.
To find $\lim_{x \rightarrow 0^{-}} f(x)$,we consider values of $x$ slightly less than $0$.
For $x \in (-1, 0)$,the greatest integer function $[x] = -1$.
Therefore,$\lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{-}} \frac{\sin(1+[x])}{[x]}$.
Substituting $[x] = -1$,we get $\frac{\sin(1-1)}{-1} = \frac{\sin(0)}{-1} = \frac{0}{-1} = 0$.

(C) Given,$f(x) = \begin{cases} x-5 & \text{for } x \leq 1 \\ 4x^2-9 & \text{for } 1 < x < 2 \\ 3x+4 & \text{for } x \geq 2 \end{cases}$
To find the right-hand derivative $f^{\prime}(2^{+})$,we use the definition:
$f^{\prime}(2^{+}) = \lim_{h \to 0^{+}} \frac{f(2+h) - f(2)}{h}$
For $x \geq 2$,$f(x) = 3x+4$.
Thus,$f(2) = 3(2) + 4 = 10$.
$f^{\prime}(2^{+}) = \lim_{h \to 0^{+}} \frac{3(2+h) + 4 - 10}{h}$
$= \lim_{h \to 0^{+}} \frac{6 + 3h + 4 - 10}{h}$
$= \lim_{h \to 0^{+}} \frac{3h}{h}$
$= 3$

(C) Polyhydroxy butyrate-$co$-$\beta$-hydroxy valerate $(PHBV)$ is a biodegradable polymer.

(A) Let the angles of $\triangle ABC$ be $A = 45^{\circ}$,$B = 60^{\circ}$,and $C = 180^{\circ} - (45^{\circ} + 60^{\circ}) = 75^{\circ}$.
Since the side opposite to the smallest angle is the smallest side and the side opposite to the greatest angle is the greatest side,the smallest side is $a$ (opposite to $45^{\circ}$) and the greatest side is $c$ (opposite to $75^{\circ}$).
Using the Sine Rule,$\frac{a}{\sin A} = \frac{c}{\sin C}$,so the ratio $\frac{a}{c} = \frac{\sin 45^{\circ}}{\sin 75^{\circ}}$.
We know $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$ and $\sin 75^{\circ} = \sin(45^{\circ} + 30^{\circ}) = \sin 45^{\circ} \cos 30^{\circ} + \cos 45^{\circ} \sin 30^{\circ} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}}$.
Therefore,the ratio is $\frac{1/\sqrt{2}}{(\sqrt{3}+1)/(2\sqrt{2})} = \frac{2}{\sqrt{3}+1}$.
Rationalizing the denominator: $\frac{2(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{2(\sqrt{3}-1)}{3-1} = \frac{2(\sqrt{3}-1)}{2} = \sqrt{3}-1$.
Thus,the ratio is $(\sqrt{3}-1) : 1$.

(A) We know that $a+b+c = 2s$,where $s$ is the semi-perimeter of the triangle.
Using the identity $\tan \frac{A}{2} = \frac{\Delta}{s(s-a)}$ is incorrect; the correct identity is $\tan \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$.
However,a simpler approach is using $\tan \frac{A}{2} = \frac{r}{s-a}$.
Thus,$(a+b+c)\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right) = 2s \left(\frac{r}{s-a} + \frac{r}{s-b}\right)$.
$= 2sr \left(\frac{s-b+s-a}{(s-a)(s-b)}\right) = 2sr \left(\frac{c}{(s-a)(s-b)}\right)$.
Since $r = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$,we have $r^2 = \frac{(s-a)(s-b)(s-c)}{s}$.
Substituting this,the expression simplifies to $2c \cot \frac{C}{2}$.

(A) We know the standard properties of a triangle where $\Delta$ is the area,$s$ is the semi-perimeter,and $r, r_1, r_2, r_3$ are the inradius and exradii respectively.
For statement $(I)$: The product of the inradius and the three exradii is given by $r r_1 r_2 r_3 = \frac{\Delta}{s} \cdot \frac{\Delta}{s-a} \cdot \frac{\Delta}{s-b} \cdot \frac{\Delta}{s-c} = \frac{\Delta^4}{s(s-a)(s-b)(s-c)}$. Since $\Delta^2 = s(s-a)(s-b)(s-c)$,we have $r r_1 r_2 r_3 = \frac{\Delta^4}{\Delta^2} = \Delta^2$. Thus,$(I)$ is true.
For statement $(II)$: The sum of the products of the exradii taken two at a time is $r_1 r_2 + r_2 r_3 + r_3 r_1 = s^2$. This is a standard identity in triangle geometry. Thus,$(II)$ is true.
Therefore,both statements are true.

(B) Given $A(\operatorname{adj} A) = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}$.
Taking the determinant on both sides,we get $|A(\operatorname{adj} A)| = \begin{vmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{vmatrix}$.
Using the property $|AB| = |A||B|$,we have $|A| |\operatorname{adj} A| = 4^3 = 64$.
We know that $|\operatorname{adj} A| = |A|^{n-1}$,where $n$ is the order of the matrix. Here $n = 3$,so $|\operatorname{adj} A| = |A|^{3-1} = |A|^2$.
Substituting this into the equation,we get $|A| \cdot |A|^2 = 64$,which implies $|A|^3 = 64$.
Thus,$|A| = 4$.
Finally,$|\operatorname{adj} A| = |A|^2 = 4^2 = 16$.

(A) The given system of linear equations is homogeneous,which can be written in matrix form $AX = O$ as:
$\begin{bmatrix} 1 & -1 & 1 \\ 1 & 2 & -1 \\ 2 & 1 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$
For a homogeneous system $AX = O$,non-trivial solutions exist if and only if the determinant of the coefficient matrix $A$ is zero,i.e.,$|A| = 0$.
Let us calculate the determinant $|A|$:
$|A| = \begin{vmatrix} 1 & -1 & 1 \\ 1 & 2 & -1 \\ 2 & 1 & 3 \end{vmatrix}$
$|A| = 1(2 \times 3 - (-1) \times 1) - (-1)(1 \times 3 - (-1) \times 2) + 1(1 \times 1 - 2 \times 2)$
$|A| = 1(6 + 1) + 1(3 + 2) + 1(1 - 4)$
$|A| = 7 + 5 - 3 = 9$
Since $|A| = 9 \neq 0$,the matrix $A$ is non-singular.
$A$ homogeneous system with a non-singular coefficient matrix has only the trivial solution $(x=0, y=0, z=0)$.
Therefore,the number of non-trivial solutions is $0$.

(C) matrix $A$ is singular if its determinant is zero,i.e.,$|A| = 0$.
Given $A = \left[\begin{array}{rrr}1 & 2 & x \\ 4 & -1 & 7 \\ 2 & 4 & -6\end{array}\right]$.
Calculating the determinant along the first row:
$|A| = 1((-1)(-6) - (7)(4)) - 2((4)(-6) - (7)(2)) + x((4)(4) - (-1)(2)) = 0$
$|A| = 1(6 - 28) - 2(-24 - 14) + x(16 + 2) = 0$
$|A| = 1(-22) - 2(-38) + x(18) = 0$
$-22 + 76 + 18x = 0$
$54 + 18x = 0$
$18x = -54$
$x = -3$

(D) Let $y = \operatorname{sech}^{-1}(\sin \theta)$.
Then,$\operatorname{sech} y = \sin \theta$.
Since $\operatorname{sech} y = \frac{1}{\cosh y}$,we have $\cosh y = \frac{1}{\sin \theta} = \operatorname{cosec} \theta$.
Using the logarithmic form of the inverse hyperbolic cosine function,$\cosh^{-1}(x) = \log(x + \sqrt{x^2 - 1})$.
Substituting $x = \operatorname{cosec} \theta$,we get:
$y = \log(\operatorname{cosec} \theta + \sqrt{\operatorname{cosec}^2 \theta - 1})$.
Since $\operatorname{cosec}^2 \theta - 1 = \cot^2 \theta$,we have:
$y = \log(\operatorname{cosec} \theta + \cot \theta)$.
Using the trigonometric identities $\operatorname{cosec} \theta = \frac{1}{\sin \theta} = \frac{1}{2 \sin(\theta/2) \cos(\theta/2)}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\cos^2(\theta/2) - \sin^2(\theta/2)}{2 \sin(\theta/2) \cos(\theta/2)}$,we simplify:
$\operatorname{cosec} \theta + \cot \theta = \frac{1 + \cos \theta}{\sin \theta} = \frac{2 \cos^2(\theta/2)}{2 \sin(\theta/2) \cos(\theta/2)} = \cot(\theta/2)$.
Thus,$y = \log(\cot(\theta/2))$.

(B) Given equation: $\tan \left(\sec ^{-1}\left(\frac{1}{x}\right)\right)=\sin \left(\tan ^{-1} 2\right)$.
First,simplify the left side: Let $\theta = \sec^{-1}(\frac{1}{x})$,then $\sec \theta = \frac{1}{x}$,which implies $\cos \theta = x$. Since $\tan^2 \theta = \sec^2 \theta - 1$,we have $\tan \theta = \sqrt{(\frac{1}{x})^2 - 1} = \frac{\sqrt{1-x^2}}{x}$.
Next,simplify the right side: Let $\phi = \tan^{-1}(2)$,then $\tan \phi = 2$. Using the identity $\sin(\tan^{-1} y) = \frac{y}{\sqrt{1+y^2}}$,we get $\sin \phi = \frac{2}{\sqrt{1+2^2}} = \frac{2}{\sqrt{5}}$.
Equating both sides: $\frac{\sqrt{1-x^2}}{x} = \frac{2}{\sqrt{5}}$.
Squaring both sides: $\frac{1-x^2}{x^2} = \frac{4}{5}$.
$5(1-x^2) = 4x^2 \Rightarrow 5 - 5x^2 = 4x^2 \Rightarrow 9x^2 = 5 \Rightarrow x^2 = \frac{5}{9}$.
Since $x>0$,we have $x = \frac{\sqrt{5}}{3}$.

(B) Given,$f(x) = \frac{1}{2 - \cos 3x}$.
We know that for any $x \in R$,$-1 \leq \cos 3x \leq 1$.
Multiplying by $-1$,we get $-1 \leq -\cos 3x \leq 1$.
Adding $2$ to all parts,we get $2 - 1 \leq 2 - \cos 3x \leq 2 + 1$,which simplifies to $1 \leq 2 - \cos 3x \leq 3$.
Taking the reciprocal,the inequality signs reverse: $\frac{1}{3} \leq \frac{1}{2 - \cos 3x} \leq \frac{1}{1}$.
Thus,$\frac{1}{3} \leq f(x) \leq 1$.
Therefore,the range of $f$ is $[1/3, 1]$.

(B) Given,$f(x) = x - [x]$ and $g(x) = [x]$ for $x \in R$.
We need to find the value of $f(g(x))$.
By substituting $g(x)$ into $f(x)$,we get:
$f(g(x)) = f([x])$.
Since $f(x) = x - [x]$,substituting $[x]$ for $x$ gives:
$f([x]) = [x] - [[x]]$.
Since $[x]$ is an integer,the greatest integer function of an integer is the integer itself,i.e.,$[[x]] = [x]$.
Therefore,$f([x]) = [x] - [x] = 0$.
Thus,for every $x \in R, f(g(x)) = 0$.

(D) Given,$f\left(\frac{p}{q}\right)=\sqrt{p^2-q^2}$ for $\frac{p}{q} \in Q$.
If we take $\frac{p}{q} = \frac{1}{2}$,then $p=1$ and $q=2$.
$f\left(\frac{1}{2}\right) = \sqrt{1^2 - 2^2} = \sqrt{1 - 4} = \sqrt{-3} = i\sqrt{3}$.
Since $\sqrt{-3}$ is not a real number,statement $I$ is false.
Since $\sqrt{-3}$ is a complex number,statement $II$ is true for this specific case,but the statement says it is complex for *each* $\frac{p}{q} \in Q$. If we take $\frac{p}{q} = \frac{2}{1}$,then $f(2) = \sqrt{2^2 - 1^2} = \sqrt{3}$,which is a real number (and every real number is also a complex number).
However,if the question implies $f$ must be defined for all rational numbers,the function is not well-defined as a real-valued function. Given the options,$I$ is false because it is not real for all $p < q$,and $II$ is false because it is not complex for all cases (it is real for $p \ge q$). Thus,both are false.

(D) Given the curve $y=x^2+x-1$ and the point $(x_1, y_1) = (1, 1)$.
First,find the derivative $\frac{dy}{dx} = 2x + 1$.
At the point $(1, 1)$,the slope $m = \frac{dy}{dx} = 2(1) + 1 = 3$.
Now,calculate the lengths:
$1$. Length of tangent $A = \left|\frac{y_1 \sqrt{1+m^2}}{m}\right| = \left|\frac{1 \cdot \sqrt{1+3^2}}{3}\right| = \frac{\sqrt{10}}{3} \approx \frac{3.16}{3} \approx 1.05$.
$2$. Length of subtangent $B = \left|\frac{y_1}{m}\right| = \left|\frac{1}{3}\right| = \frac{1}{3} \approx 0.33$.
$3$. Length of normal $C = \left|y_1 \sqrt{1+m^2}\right| = |1 \cdot \sqrt{1+3^2}| = \sqrt{10} \approx 3.16$.
$4$. Length of subnormal $D = |y_1 m| = |1 \cdot 3| = 3$.
Comparing the values: $B = 0.33$,$A \approx 1.05$,$D = 3$,$C \approx 3.16$.
Thus,the increasing order is $B, A, D, C$.

(A) function $f(x)$ has no extreme value if its derivative $f^{\prime}(x)$ does not change sign,which means $f^{\prime}(x)$ is either always non-negative or always non-positive.
Given $f(x) = x^3 + p x^2 + q x + r$.
The derivative is $f^{\prime}(x) = 3x^2 + 2px + q$.
For $f(x)$ to have no extreme value,$f^{\prime}(x)$ must not have distinct real roots.
This implies the discriminant $D \le 0$.
The discriminant of the quadratic $3x^2 + 2px + q$ is $D = (2p)^2 - 4(3)(q) = 4p^2 - 12q$.
Setting $D \le 0$:
$4p^2 - 12q \le 0$
$4p^2 \le 12q$
$p^2 \le 3q$.
Since the options provided represent the strict inequality condition for the derivative to be strictly monotonic,the correct condition is $p^2 < 3q$.

(A) Given function is $f(x) = x e^{-x}$.
First,we find the first derivative: $f^{\prime}(x) = (1)e^{-x} + x(-e^{-x}) = e^{-x}(1-x)$.
Setting $f^{\prime}(x) = 0$,we get $1-x = 0$,which implies $x = 1$.
Next,we find the second derivative: $f^{\prime \prime}(x) = -e^{-x}(1-x) + e^{-x}(-1) = e^{-x}(x-1-1) = e^{-x}(x-2)$.
Evaluating at $x=1$: $f^{\prime \prime}(1) = e^{-1}(1-2) = -e^{-1} = -\frac{1}{e}$.
Since $f^{\prime \prime}(1) < 0$,the function has a local maximum at $x=1$.
Thus,both Assertion $(A)$ and Reason $(R)$ are true,and $(R)$ is the correct explanation for $(A)$.

(D) Let the height of the object be $h$.
In $\triangle ABC$,we have $\tan \alpha = \frac{h}{x+d}$,which implies $x+d = h \cot \alpha$ $(i)$.
In $\triangle ABD$,we have $\tan \beta = \frac{h}{x}$,which implies $x = h \cot \beta$ (ii).
Substituting the value of $x$ from (ii) into $(i)$,we get:
$h \cot \beta + d = h \cot \alpha$
$d = h \cot \alpha - h \cot \beta$
$d = h(\cot \alpha - \cot \beta)$
$h = \frac{d}{\cot \alpha - \cot \beta}$

(C) We have the integral $I = \int \frac{e^x-1}{e^x+1} d x$.
We can rewrite the integrand as $\frac{e^x-1}{e^x+1} = \frac{2e^x - (e^x+1)}{e^x+1} = \frac{2e^x}{e^x+1} - 1$.
Now,integrating term by term,we get $I = \int \frac{2e^x}{e^x+1} d x - \int 1 d x$.
For the first part,let $u = e^x+1$,then $du = e^x d x$.
Thus,$\int \frac{2e^x}{e^x+1} d x = 2 \log |e^x+1|$.
Therefore,$I = 2 \log (e^x+1) - x + c$.
Comparing this with $f(x)+c$,we find $f(x) = 2 \log (e^x+1) - x$.

(A) Let $I = \int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x$.
Substitute $x = \cos 2\theta$,so $dx = -2\sin 2\theta d\theta$.
Then $\sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{1-\cos 2\theta}{1+\cos 2\theta}} = \sqrt{\frac{2\sin^2 \theta}{2\cos^2 \theta}} = \tan \theta$.
Thus,$I = \int \theta (-2\sin 2\theta) d\theta = -2 \int \theta \sin 2\theta d\theta$.
Using integration by parts: $I = -2 \left[ \theta \left( -\frac{\cos 2\theta}{2} \right) - \int 1 \cdot \left( -\frac{\cos 2\theta}{2} \right) d\theta \right] = \theta \cos 2\theta - \int \cos 2\theta d\theta = \theta \cos 2\theta - \frac{\sin 2\theta}{2} + c$.
Since $x = \cos 2\theta$,then $\theta = \frac{1}{2} \cos^{-1} x$ and $\sin 2\theta = \sqrt{1-x^2}$.
Substituting these back: $I = \frac{1}{2} x \cos^{-1} x - \frac{1}{2} \sqrt{1-x^2} + c = \frac{1}{2} \left( x \cos^{-1} x - \sqrt{1-x^2} \right) + c$.

(A) Given $f(t) = \int_{-t}^t \frac{e^{-|x|}}{2} dx$.
Since the integrand $\frac{e^{-|x|}}{2}$ is an even function,we can write:
$f(t) = 2 \int_0^t \frac{e^{-x}}{2} dx = \int_0^t e^{-x} dx$.
Evaluating the integral:
$f(t) = [-e^{-x}]_0^t = -e^{-t} - (-e^0) = 1 - e^{-t}$.
Now,taking the limit as $t \rightarrow \infty$:
$\lim_{t \rightarrow \infty} f(t) = \lim_{t \rightarrow \infty} (1 - e^{-t}) = 1 - 0 = 1$.

(C) Let $I = \int_0^{2 \pi} \sin^6 x \cos^5 x \, dx$.
Using the property $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$ if $f(2a-x) = f(x)$:
Here,$f(x) = \sin^6 x \cos^5 x$.
$f(2 \pi - x) = \sin^6(2 \pi - x) \cos^5(2 \pi - x) = (-\sin x)^6 (\cos x)^5 = \sin^6 x \cos^5 x = f(x)$.
So,$I = 2 \int_0^{\pi} \sin^6 x \cos^5 x \, dx$.
Now,using the property $\int_0^a f(x) \, dx = 0$ if $f(a-x) = -f(x)$:
Let $g(x) = \sin^6 x \cos^5 x$.
$g(\pi - x) = \sin^6(\pi - x) \cos^5(\pi - x) = (\sin x)^6 (-\cos x)^5 = -\sin^6 x \cos^5 x = -g(x)$.
Since $g(\pi - x) = -g(x)$,the integral $\int_0^{\pi} \sin^6 x \cos^5 x \, dx = 0$.
Therefore,$I = 2 \times 0 = 0$.

(A) The given curves are $y=x^2$ and $y=x^3$.
To find the intersection points,set $x^2 = x^3$,which implies $x^2(1-x) = 0$.
Thus,the curves intersect at $x=0$ and $x=1$.
In the interval $[0, 1]$,$x^2 \ge x^3$.
The area $A$ is given by the integral:
$A = \int_0^1 (x^2 - x^3) dx$
$A = \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_0^1$
$A = \left( \frac{1}{3} - \frac{1}{4} \right) - (0 - 0)$
$A = \frac{4-3}{12} = \frac{1}{12} \text{ sq unit}$.