AP EAMCET 2002 Chemistry Question Paper with Answer and Solution

(B) An ionic bond is formed by the electrostatic attraction between oppositely charged ions,while a covalent bond is formed by the sharing of electrons between atoms.
In $K_2SO_4$ (potassium sulfate),the compound consists of $K^+$ ions and $SO_4^{2-}$ polyatomic ions,which are held together by ionic bonds.
Within the sulfate ion $(SO_4^{2-})$,the sulfur atom is covalently bonded to the four oxygen atoms.
Therefore,$K_2SO_4$ contains both ionic bonds (between $K^+$ and $SO_4^{2-}$) and covalent bonds (within the $SO_4^{2-}$ ion).

(C) At equilibrium,the rate of the forward reaction is equal to the rate of the backward reaction.
Therefore,the rate of the backward reaction at equilibrium is $0.02 \ mol \ L^{-1} \ min^{-1}$.

(C) Electronegativity generally increases across a period from left to right and decreases down a group.
Comparing the elements $N$ (Nitrogen),$O$ (Oxygen),and $P$ (Phosphorus):
$1$. $N$ and $O$ are in the $2^{nd}$ period,while $P$ is in the $3^{rd}$ period.
$2$. In the $2^{nd}$ period,electronegativity increases as $N < O$.
$3$. $P$ is below $N$ in the same group ($15^{th}$ group),so $N > P$.
$4$. Combining these,the electronegativity values on the Pauling scale are approximately: $O (3.44) > N (3.04) > P (2.19)$.
Therefore,the correct order is $O > N > P$.

(C) In Down's process,the electrolysis of molten sodium chloride $(NaCl)$ is performed.
At the anode,oxidation of chloride ions $(Cl^{-})$ into chlorine gas $(Cl_2)$ takes place.
The reaction is: $2 Cl^{-} \longrightarrow Cl_2 + 2 e^{-}$.

(A) Given the partial fraction decomposition: $\frac{1-x+6x^2}{x(1-x)(1+x)} = \frac{A}{x} + \frac{B}{1-x} + \frac{C}{1+x}$.
Multiplying both sides by $x(1-x)(1+x)$,we get:
$1-x+6x^2 = A(1-x^2) + Bx(1+x) + Cx(1-x)$.
To find $A$,we set $x = 0$:
$1 - 0 + 6(0)^2 = A(1 - 0^2) + B(0)(1+0) + C(0)(1-0)$.
$1 = A(1) + 0 + 0$.
Therefore,$A = 1$.

(B) Given the cubic equation $2x^3 + 0x^2 - 2x - 1 = 0$.
Let the roots be $\alpha, \beta, \gamma$.
According to Vieta's formulas for a cubic equation $ax^3 + bx^2 + cx + d = 0$,the sum of the roots taken two at a time is given by $\Sigma \alpha \beta = \frac{c}{a}$.
Here,$a = 2$,$b = 0$,$c = -2$,and $d = -1$.
Therefore,$\Sigma \alpha \beta = \frac{-2}{2} = -1$.
We need to find $(\Sigma \alpha \beta)^2$.
$(\Sigma \alpha \beta)^2 = (-1)^2 = 1$.

(C) Given the cubic equation $x^3+a x^2+b x+c=0$ with roots $\alpha, \beta, \gamma$.
By Vieta's formulas:
$\alpha+\beta+\gamma = -a$
$\alpha \beta+\beta \gamma+\gamma \alpha = b$
$\alpha \beta \gamma = -c$
We need to find $\alpha^{-1}+\beta^{-1}+\gamma^{-1} = \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$.
Combining the terms:
$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} = \frac{\alpha \beta+\beta \gamma+\gamma \alpha}{\alpha \beta \gamma}$.
Substituting the values from Vieta's formulas:
$\frac{b}{-c} = -\frac{b}{c}$.

(B) Let $f(x) = x^5 - 6x^2 - 4x + 5$.
For positive real roots,we check the sign changes in $f(x) = x^5 - 6x^2 - 4x + 5$. The signs are $(+, -, -, +)$. There are $2$ sign changes.
For negative real roots,we check the sign changes in $f(-x) = (-x)^5 - 6(-x)^2 - 4(-x) + 5 = -x^5 - 6x^2 + 4x + 5$. The signs are $(-, -, +, +)$. There is $1$ sign change.
By Descartes' Rule of Signs,the maximum number of positive real roots is $2$ and the maximum number of negative real roots is $1$.
Therefore,the maximum possible number of real roots is $2 + 1 = 3$.

(B) The given equation is $x^4-8x^3+x^2-x+3=0$.
To remove the second term (the term containing $x^{n-1}$),we need to diminish the roots by $h$,where $h = -\frac{a_1}{n a_0}$.
Here,$n=4$,$a_0=1$,and $a_1=-8$.
Thus,$h = -\frac{-8}{4 \times 1} = \frac{8}{4} = 2$.
Therefore,the roots should be diminished by $2$.

(D) Given $z = 3 + 5i$.
Then,the conjugate is $\bar{z} = 3 - 5i$.
Now,calculate $z^3$:
$z^2 = (3 + 5i)^2 = 9 + 25i^2 + 30i = 9 - 25 + 30i = -16 + 30i$.
$z^3 = z^2 \cdot z = (-16 + 30i)(3 + 5i) = -48 - 80i + 90i + 150i^2$.
Since $i^2 = -1$,$z^3 = -48 + 10i - 150 = -198 + 10i$.
Finally,substitute these values into the expression:
$z^3 + \bar{z} + 198 = (-198 + 10i) + (3 - 5i) + 198$.
$= (-198 + 198 + 3) + (10i - 5i) = 3 + 5i$.

(A) Given $x_n = \cos \left(\frac{\pi}{4^n}\right) + i \sin \left(\frac{\pi}{4^n}\right) = e^{i \frac{\pi}{4^n}}$.
The product $P = x_1 x_2 x_3 \ldots \infty$ is given by:
$P = e^{i \frac{\pi}{4^1}} \cdot e^{i \frac{\pi}{4^2}} \cdot e^{i \frac{\pi}{4^3}} \ldots = e^{i \pi \left(\frac{1}{4} + \frac{1}{4^2} + \frac{1}{4^3} + \ldots \right)}$.
The exponent is a geometric series with first term $a = \frac{1}{4}$ and common ratio $r = \frac{1}{4}$.
The sum of the infinite geometric series is $S = \frac{a}{1 - r} = \frac{1/4}{1 - 1/4} = \frac{1/4}{3/4} = \frac{1}{3}$.
Therefore,$P = e^{i \pi \left(\frac{1}{3}\right)} = e^{i \frac{\pi}{3}}$.
Using Euler's formula,$e^{i \frac{\pi}{3}} = \cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right) = \frac{1}{2} + i \frac{\sqrt{3}}{2} = \frac{1 + i \sqrt{3}}{2}$.

(A) We have,$\left|z+\frac{i}{2}\right|^2=\left|z-\frac{i}{2}\right|^2$.
Substituting $z=x+iy$:
$\left|x+i(y+\frac{1}{2})\right|^2 = \left|x+i(y-\frac{1}{2})\right|^2$.
Using the property $|a+ib|^2 = a^2+b^2$:
$x^2+(y+\frac{1}{2})^2 = x^2+(y-\frac{1}{2})^2$.
$x^2+y^2+y+\frac{1}{4} = x^2+y^2-y+\frac{1}{4}$.
$y = -y$ $\Rightarrow 2y = 0$ $\Rightarrow y=0$.
Since $y=0$,the locus of $z$ is the $x$-axis.

(B) Using the identity $C(n, r) + C(n, r-1) = C(n+1, r)$,we have $C(n, 5) + C(n, 6) = C(n+1, 6)$.
Given the inequality $C(n+1, 6) > C(n+1, 5)$.
Expanding the combinations: $\frac{(n+1)!}{6!(n+1-6)!} > \frac{(n+1)!}{5!(n+1-5)!}$.
Simplifying: $\frac{1}{6!(n-5)!} > \frac{1}{5!(n-4)!}$.
$\frac{1}{6 \times 5!(n-5)!} > \frac{1}{5!(n-4)(n-5)!}$.
$\frac{1}{6} > \frac{1}{n-4}$.
$n-4 > 6$.
$n > 10$.
Since $n$ is a natural number,the least value of $n$ is $11$.

(B) In $\triangle ABC$,let $a, b, c$ be the side lengths and $p_1, p_2, p_3$ be the corresponding altitudes.
We know that the area $\Delta = \frac{1}{2} a p_1 = \frac{1}{2} b p_2 = \frac{1}{2} c p_3$.
This implies $p_1 = \frac{2\Delta}{a}$,$p_2 = \frac{2\Delta}{b}$,and $p_3 = \frac{2\Delta}{c}$.
Given that $p_1, p_2, p_3$ are in $AP$,we have $\frac{2\Delta}{a}, \frac{2\Delta}{b}, \frac{2\Delta}{c}$ in $AP$.
Dividing by $2\Delta$,we get $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $AP$.
By definition,if the reciprocals of terms are in $AP$,then the terms themselves are in $HP$.
Therefore,$a, b, c$ are in $HP$.

(A) Let $T_n$ be the first term of the $n^{th}$ set. The first terms are $1, 2, 4, 7, 11, \ldots$
This is a sequence where the differences are $1, 2, 3, 4, \ldots$
The $n^{th}$ term is given by $T_n = T_1 + \sum_{k=1}^{n-1} k = 1 + \frac{(n-1)n}{2}$.
For the $50^{th}$ set,$n=50$,so $T_{50} = 1 + \frac{49 \times 50}{2} = 1 + 1225 = 1226$.
The $50^{th}$ set contains $50$ consecutive integers starting from $1226$.
The sum of these $50$ terms is $S = \frac{50}{2} [2(1226) + (50-1)(1)] = 25 [2452 + 49] = 25 \times 2501 = 62525$.

(A) Fluorosis is caused by the accumulation of excess fluoride in the body. This excess fluoride reacts with calcium $(Ca)$ present in bones and teeth to form calcium fluoride $(CaF_2)$,which leads to the disease known as fluorosis.
The reaction is: $Ca + F_2 \rightarrow CaF_2$ (Fluorosis disease).

(D) In the expansion of $(1+x)^n$,the general term,i.e.,$(r+1)$-th term is given by $T_{r+1} = { }^n C_r x^r$.
The coefficient of the $p$-th term is ${ }^n C_{p-1}$. Given that this coefficient is $p$,we have $p = { }^n C_{p-1}$.
The coefficient of the $(p+1)$-th term is ${ }^n C_p$. Given that this coefficient is $q$,we have $q = { }^n C_p$.
Now,consider the ratio $\frac{q}{p} = \frac{{ }^n C_p}{{ }^n C_{p-1}}$.
Using the formula $\frac{{ }^n C_r}{{ }^n C_{r-1}} = \frac{n-r+1}{r}$,we get:
$\frac{q}{p} = \frac{n-p+1}{p}$.
Cross-multiplying,we get $qp = p(n-p+1)$.
Wait,let us re-evaluate the given condition: coefficients are $p$ and $q$.
Actually,the problem states the coefficients are $p$ and $q$. Let's use the property $\frac{{ }^n C_p}{{ }^n C_{p-1}} = \frac{n-p+1}{p}$.
Since $q = { }^n C_p$ and $p = { }^n C_{p-1}$,we have $\frac{q}{p} = \frac{n-p+1}{p}$.
This implies $q = n-p+1$,which simplifies to $p+q = n+1$.

(D) Given that,$x = \tan \theta + \sin \theta$ and $y = \tan \theta - \sin \theta$.
Adding and subtracting the two equations,we get:
$\tan \theta = \frac{x + y}{2}$ and $\sin \theta = \frac{x - y}{2}$.
Now,consider the expression $x^2 - y^2$:
$x^2 - y^2 = (x + y)(x - y) = (2 \tan \theta)(2 \sin \theta) = 4 \tan \theta \sin \theta$.
We know that $\tan \theta \sin \theta = \frac{\sin^2 \theta}{\cos \theta} = \frac{1 - \cos^2 \theta}{\cos \theta} = \sec \theta - \cos \theta$.
However,a more direct approach is:
$(x^2 - y^2)^2 = (4 \tan \theta \sin \theta)^2 = 16 \tan^2 \theta \sin^2 \theta$.
Since $\sin^2 \theta = \tan^2 \theta \cos^2 \theta$,we have $\sin^2 \theta = \tan^2 \theta (1 - \sin^2 \theta)$,which implies $\sin^2 \theta (1 + \tan^2 \theta) = \tan^2 \theta$,so $\sin^2 \theta = \frac{\tan^2 \theta}{\sec^2 \theta}$.
Alternatively,note that $xy = (\tan \theta + \sin \theta)(\tan \theta - \sin \theta) = \tan^2 \theta - \sin^2 \theta = \tan^2 \theta (1 - \cos^2 \theta) = \tan^2 \theta \sin^2 \theta$.
Thus,$(x^2 - y^2)^2 = 16 \tan^2 \theta \sin^2 \theta = 16xy$.

(A) Given that $\cos (\alpha+\beta) = \frac{4}{5}$. Since $0 < \alpha, \beta < \frac{\pi}{4}$,we have $0 < \alpha+\beta < \frac{\pi}{2}$. Thus,$\tan (\alpha+\beta) = \frac{\sqrt{1 - (4/5)^2}}{4/5} = \frac{3/5}{4/5} = \frac{3}{4}$.
Given $\sin (\alpha-\beta) = \frac{5}{13}$. Since $0 < \alpha, \beta < \frac{\pi}{4}$,we have $-\frac{\pi}{4} < \alpha-\beta < \frac{\pi}{4}$. Thus,$\tan (\alpha-\beta) = \frac{5/13}{\sqrt{1 - (5/13)^2}} = \frac{5/13}{12/13} = \frac{5}{12}$.
Now,$\tan 2\alpha = \tan [(\alpha+\beta) + (\alpha-\beta)]$.
Using the formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we get:
$\tan 2\alpha = \frac{\frac{3}{4} + \frac{5}{12}}{1 - (\frac{3}{4} \cdot \frac{5}{12})} = \frac{\frac{9+5}{12}}{1 - \frac{15}{48}} = \frac{14/12}{33/48} = \frac{14}{12} \cdot \frac{48}{33} = \frac{14 \cdot 4}{33} = \frac{56}{33}$.

(D) We use the trigonometric identity $\sin^2 A - \sin^2 B = \sin(A+B) \sin(A-B)$.
Let $A = \frac{\pi}{8} + \frac{x}{2}$ and $B = \frac{\pi}{8} - \frac{x}{2}$.
Then $A+B = \frac{\pi}{8} + \frac{x}{2} + \frac{\pi}{8} - \frac{x}{2} = \frac{\pi}{4}$.
And $A-B = \frac{\pi}{8} + \frac{x}{2} - \left( \frac{\pi}{8} - \frac{x}{2} \right) = x$.
Thus,$f(x) = \sin \left( \frac{\pi}{4} \right) \sin(x) = \frac{1}{\sqrt{2}} \sin(x)$.
The period of $\sin(x)$ is $2\pi$.
Therefore,the period of $f(x)$ is $2\pi$.

(C) Let the points be $D(4, 2)$ and $C(-2, 6)$. The line $L$ is the perpendicular bisector of the segment $CD$.
Slope of the line $CD = \frac{6-2}{-2-4} = \frac{4}{-6} = -\frac{2}{3}$.
Since $L$ is perpendicular to $CD$,the slope of $L$ is $m = -\frac{1}{(-2/3)} = \frac{3}{2}$.
The midpoint $O$ of $CD$ is $\left(\frac{4-2}{2}, \frac{2+6}{2}\right) = (1, 4)$.
The equation of line $L$ passing through $(1, 4)$ with slope $\frac{3}{2}$ is:
$y - 4 = \frac{3}{2}(x - 1)$
$2y - 8 = 3x - 3$
$3x - 2y + 5 = 0$.

(D) The given equation of the pair of lines is $Ax^2 + 2Hxy + By^2 = 0$,where $A = \cos \theta - \sin \theta$,$2H = 2 \cos \theta$ (so $H = \cos \theta$),and $B = \cos \theta + \sin \theta$.
The formula for the acute angle $\alpha$ between the pair of lines is given by $\tan \alpha = \left| \frac{2 \sqrt{H^2 - AB}}{A + B} \right|$.
Substituting the values:
$H^2 - AB = \cos^2 \theta - (\cos \theta - \sin \theta)(\cos \theta + \sin \theta) = \cos^2 \theta - (\cos^2 \theta - \sin^2 \theta) = \sin^2 \theta$.
$A + B = (\cos \theta - \sin \theta) + (\cos \theta + \sin \theta) = 2 \cos \theta$.
Thus,$\tan \alpha = \left| \frac{2 \sqrt{\sin^2 \theta}}{2 \cos \theta} \right| = \left| \frac{2 \sin \theta}{2 \cos \theta} \right| = |\tan \theta|$.
Since the angle $2 \theta$ is acute,$\theta$ is also acute,so $\alpha = \theta$.

(A) The equation of the pair of bisectors of the angle between the lines $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.
Since the coordinate axes are the bisectors,their equations are $x = 0$ and $y = 0$,which implies their combined equation is $xy = 0$.
Comparing $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$ with $xy = 0$,we must have the coefficient of $x^2 - y^2$ equal to zero,which means $a - b$ must be undefined or the equation must reduce to $xy = 0$.
Actually,for the bisectors to be $x = 0$ and $y = 0$,the equation $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$ must represent $xy = 0$.
This occurs when $h = 0$ and $a + b = 0$ is not necessarily required,but rather the condition for the axes to be the bisectors is $h = 0$ and $a + b = 0$ is not the condition; rather,if $h=0$,the lines are $ax^2 + by^2 = 0$,and the bisectors are $\frac{x^2-y^2}{a-b} = 0$,i.e.,$x^2 = y^2$,which are $y = \pm x$. This is not the axes.
Wait,if the bisectors are $x=0$ and $y=0$,then the equation of the bisectors is $xy=0$. Comparing this with $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$,we see that for the $x^2$ and $y^2$ terms to vanish,we need $a-b$ to be such that the equation becomes $xy=0$. This happens when $a+b=0$ is not the condition,but $h=0$ is the condition for the lines to be $ax^2+by^2=0$. Actually,the condition for the axes to be the bisectors is $a+b=0$.

(B) The velocity of projection is given by $v = \frac{3}{4} v_e$,where $v_e = \sqrt{2gR}$ is the escape velocity.
Using the principle of conservation of energy:
$\frac{1}{2} m v^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h}$
Substituting $v = \frac{3}{4} \sqrt{\frac{2GM}{R}}$:
$\frac{1}{2} m (\frac{9}{16} \cdot \frac{2GM}{R}) - \frac{GMm}{R} = - \frac{GMm}{R+h}$
$\frac{9GMm}{16R} - \frac{GMm}{R} = - \frac{GMm}{R+h}$
$-\frac{7GMm}{16R} = - \frac{GMm}{R+h}$
$\frac{7}{16R} = \frac{1}{R+h}$
$16R = 7(R+h) \implies 16R = 7R + 7h$
$9R = 7h \implies h = \frac{9R}{7}$

(B) The areal velocity $A$ is defined as the rate at which the area is swept by the position vector of the planet.
$A = \frac{dA}{dt} = \frac{1}{2} r^2 \omega$
Multiplying both sides by the mass $M$ of the planet:
$M A = \frac{1}{2} M r^2 \omega$
Since the moment of inertia $I = M r^2$,we can substitute this into the equation:
$M A = \frac{1}{2} I \omega$
We know that the angular momentum $L$ is given by $L = I \omega$.
Therefore,$M A = \frac{1}{2} L$.
Rearranging for $L$,we get:
$L = 2 M A$.

(D) The Wurtz reaction for the preparation of ethane is:
$2CH_3I + 2Na \xrightarrow{\text{dry ether}} C_2H_6 + 2NaI$
From the stoichiometry of the reaction,$2 \text{ moles}$ of methyl iodide $(CH_3I)$ are required to produce $1 \text{ mole}$ of ethane $(C_2H_6)$.
The molar mass of $CH_3I = 12 + (3 \times 1) + 127 = 142 \text{ g/mol}$.
Therefore,the mass of $2 \text{ moles}$ of $CH_3I = 2 \times 142 \text{ g} = 284 \text{ g}$.

(A) The given reaction sequence is $A$ $\xrightarrow{HBr} C_2H_5Br$ $\xrightarrow{B} A$.
Step $1$: $A$ reacts with $HBr$ to form $C_2H_5Br$ (ethyl bromide). This indicates that $A$ is ethene $(C_2H_4)$.
Reaction: $CH_2=CH_2 HBr \rightarrow CH_3-CH_2Br$.
Step $2$: $C_2H_5Br$ reacts with reagent $B$ to regenerate $A$ $(C_2H_4)$.
This is a dehydrohalogenation reaction,which requires an alcoholic solution of $KOH$ and heat $(\Delta)$.
Reaction: $CH_3-CH_2Br \text{alc. } KOH \xrightarrow{\Delta} CH_2=CH_2 KBr H_2O$.
Therefore,$A$ is $C_2H_4$ and $B$ is alcoholic $KOH / \Delta$.

(A) The conversion of $1, 2$-dibromoethane $(BrCH_2-CH_2Br)$ to ethylene $(H_2C=CH_2)$ is a dehalogenation reaction.
This reaction involves the removal of two bromine atoms from adjacent carbon atoms using zinc dust in the presence of alcohol under heating $(\Delta)$.
The chemical equation is: $BrCH_2-CH_2Br + Zn \xrightarrow{\text{alcohol}, \Delta} H_2C=CH_2 + ZnBr_2$.
Thus,the correct reaction conditions are $Zn$,alcohol,$\Delta$.

(D) When heavy water $(D_2O)$ reacts with magnesium nitride $(Mg_3N_2)$,the deuterium atoms replace the hydrogen atoms in the products.
The balanced chemical equation for the reaction is:
$Mg_3N_2 + 6D_2O \longrightarrow 3Mg(OD)_2 + 2ND_3$
Thus,the products formed are magnesium deuteroxide $(Mg(OD)_2)$ and deuteroammonia $(ND_3)$.

(A) Given that $A$ is a non-singular matrix,its determinant $|A| \neq 0$,which implies that the inverse matrix $A^{-1}$ exists.
Given the equation $AB = O$,where $O$ is the null matrix.
Multiply both sides by $A^{-1}$ from the left:
$A^{-1}(AB) = A^{-1}O$
$(A^{-1}A)B = O$
$IB = O$
$B = O$
Therefore,$B$ must be a null matrix.

(C) Given $f(x) = 3x - 4$. Let $f(x) = y$,then $y = 3x - 4$. Solving for $x$,we get $x = \frac{y + 4}{3}$. Thus,$f^{-1}(y) = \frac{y + 4}{3}$.
Given $g(x) = 2 + 3x$. Let $g(x) = z$,then $z = 2 + 3x$. Solving for $x$,we get $x = \frac{z - 2}{3}$. Thus,$g^{-1}(z) = \frac{z - 2}{3}$.
First,calculate $f^{-1}(5)$:
$f^{-1}(5) = \frac{5 + 4}{3} = \frac{9}{3} = 3$.
Now,calculate $g^{-1}(f^{-1}(5)) = g^{-1}(3)$:
$g^{-1}(3) = \frac{3 - 2}{3} = \frac{1}{3}$.
Therefore,the value is $\frac{1}{3}$.

(A) Given the function $f(x) = \frac{\cos^2 x + \sin^4 x}{\sin^2 x + \cos^4 x}$.
We know that $\sin^2 x = 1 - \cos^2 x$ and $\cos^2 x = 1 - \sin^2 x$.
Substituting these into the numerator and denominator:
$f(x) = \frac{\cos^2 x + \sin^2 x (\sin^2 x)}{\sin^2 x + \cos^2 x (\cos^2 x)}$
$f(x) = \frac{\cos^2 x + \sin^2 x (1 - \cos^2 x)}{\sin^2 x + \cos^2 x (1 - \sin^2 x)}$
$f(x) = \frac{\cos^2 x + \sin^2 x - \sin^2 x \cos^2 x}{\sin^2 x + \cos^2 x - \cos^2 x \sin^2 x}$
Since $\sin^2 x + \cos^2 x = 1$,we get:
$f(x) = \frac{1 - \sin^2 x \cos^2 x}{1 - \cos^2 x \sin^2 x} = 1$.
Thus,$f(2002) = 1$.

(A) According to the conditions of the question:
$2Fe_2S_3 + 9O_2 \rightarrow 2Fe_2O_3 + 6SO_2$ $(A)$
$SO_2 + H_2O \rightarrow H_2SO_3$
The acid formed is sulphurous acid $(H_2SO_3)$.
Since $H_2SO_3$ has two replaceable hydrogen atoms,its basicity is $2$.

(A) $1$. Option $A$: The Bronsted-Lowry theory defines acids as proton donors. $BCl_3$ is a Lewis acid because it accepts an electron pair,but it does not contain a proton to donate. Thus,the theory cannot explain its acidity. This statement is correct.
$2$. Option $B$: For $0.01 \ M \ NaOH$,$[OH^-] = 10^{-2} \ M$. Thus,$pOH = -\log(10^{-2}) = 2$. Therefore,$pH = 14 - 2 = 12$. The statement is incorrect.
$3$. Option $C$: The ionic product of water $(K_w)$ at $25^{\circ}C$ is $10^{-14} \ mol^2 \ L^{-2}$,not $10^{-10}$. The statement is incorrect.
$4$. Option $D$: The correct equation is $pH = -\log [H^+]$. The statement is incorrect.

(A) Step $1$: Calculate the total moles of $HCl$ in the mixture.
$n_1 = M_1 \times V_1 = 0.2 \ M \times 75 \ mL = 15 \ mmol$
$n_2 = M_2 \times V_2 = 1 \ M \times 25 \ mL = 25 \ mmol$
Total moles of $HCl = n_1 + n_2 = 15 + 25 = 40 \ mmol$.
Step $2$: Calculate the total volume of the final solution.
$V_{total} = V_1 + V_2 + V_{water} = 75 \ mL + 25 \ mL + 300 \ mL = 400 \ mL$.
Step $3$: Calculate the final molarity $(M_{final})$ of $HCl$.
$M_{final} = \frac{\text{Total moles}}{\text{Total volume}} = \frac{40 \ mmol}{400 \ mL} = 0.1 \ M$.
Step $4$: Calculate the $pH$ of the solution.
Since $HCl$ is a strong acid,$[H^+] = [HCl] = 0.1 \ M = 10^{-1} \ M$.
$pH = -\log_{10}[H^+] = -\log_{10}(10^{-1}) = 1$.

(B) Given curves are $x=y^2$ $(i)$ and $xy=a^3$ (ii).
For curve $(i)$,differentiating with respect to $x$: $1 = 2y \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{2y}$.
For curve (ii),differentiating with respect to $x$: $x \frac{dy}{dx} + y = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$.
To find the point of intersection,substitute $x=y^2$ into $xy=a^3$: $y^2(y) = a^3 \Rightarrow y^3 = a^3 \Rightarrow y = a$.
Then $x = a^2$. So the point of intersection is $(a^2, a)$.
The slopes at $(a^2, a)$ are $m_1 = \frac{1}{2a}$ and $m_2 = -\frac{a}{a^2} = -\frac{1}{a}$.
Since the curves cut orthogonally,$m_1 m_2 = -1$.
Therefore,$(\frac{1}{2a})(-\frac{1}{a}) = -1 \Rightarrow -\frac{1}{2a^2} = -1 \Rightarrow 2a^2 = 1 \Rightarrow a^2 = \frac{1}{2}$.

(A) Let $f(x) = \log(1+x) - \frac{2x}{2+x}$.
First,we find the derivative $f'(x)$:
$f'(x) = \frac{1}{1+x} - \frac{(2+x)(2) - (2x)(1)}{(2+x)^2}$
$f'(x) = \frac{1}{1+x} - \frac{4+2x-2x}{(2+x)^2} = \frac{1}{1+x} - \frac{4}{(2+x)^2}$
For the function to be increasing,we require $f'(x) > 0$:
$\frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2} > 0$
$\frac{4+x^2+4x - 4 - 4x}{(1+x)(2+x)^2} > 0$
$\frac{x^2}{(1+x)(2+x)^2} > 0$
Since $x^2 \ge 0$ and $(2+x)^2 > 0$ for all $x \neq -2$,the expression is positive when $1+x > 0$ and $x \neq 0$.
Thus,$x > -1$ and $x \neq 0$.

(A) Given the function $f(x) = x e^{-x}$.
To find the critical points,we calculate the first derivative:
$f'(x) = (1)e^{-x} + x(-e^{-x}) = e^{-x}(1 - x)$.
Setting $f'(x) = 0$ for extrema:
$e^{-x}(1 - x) = 0$.
Since $e^{-x} \neq 0$ for any real $x$,we have $1 - x = 0$,which gives $x = 1$.
Now,we check the second derivative to confirm the nature of the extremum:
$f''(x) = \frac{d}{dx}[e^{-x} - x e^{-x}] = -e^{-x} - (e^{-x} - x e^{-x}) = e^{-x}(x - 2)$.
Evaluating at $x = 1$:
$f''(1) = e^{-1}(1 - 2) = -e^{-1} < 0$.
Since the second derivative is negative at $x = 1$,the function attains a maximum value at $x = 1$.

(A) Let $I = \int \frac{3^x dx}{\sqrt{9^x-1}} = \int \frac{3^x dx}{\sqrt{(3^x)^2-1}}$.
Substitute $3^x = z$. Then,differentiating both sides with respect to $x$,we get $3^x \log 3 dx = dz$,which implies $3^x dx = \frac{dz}{\log 3}$.
Substituting these into the integral,we get $I = \frac{1}{\log 3} \int \frac{dz}{\sqrt{z^2-1}}$.
Using the standard integral formula $\int \frac{du}{\sqrt{u^2-a^2}} = \log |u + \sqrt{u^2-a^2}| + c$,we have $I = \frac{1}{\log 3} \log |z + \sqrt{z^2-1}| + c$.
Finally,substituting $z = 3^x$ back,we get $I = \frac{1}{\log 3} \log |3^x + \sqrt{9^x-1}| + c$.

(B) Let $I = \int \frac{d x}{7+5 \cos x}$.
Using the identities $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$ and $1 = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}$,we get:
$I = \int \frac{d x}{7(\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}) + 5(\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2})}$
$I = \int \frac{d x}{12 \cos^2 \frac{x}{2} + 2 \sin^2 \frac{x}{2}}$
Divide numerator and denominator by $\cos^2 \frac{x}{2}$:
$I = \int \frac{\sec^2 \frac{x}{2} dx}{12 + 2 \tan^2 \frac{x}{2}} = \frac{1}{2} \int \frac{\sec^2 \frac{x}{2} dx}{6 + \tan^2 \frac{x}{2}}$
Let $\tan \frac{x}{2} = z$,then $\frac{1}{2} \sec^2 \frac{x}{2} dx = dz$.
$I = \int \frac{dz}{6 + z^2} = \int \frac{dz}{(\sqrt{6})^2 + z^2}$
Using the formula $\int \frac{du}{a^2 + u^2} = \frac{1}{a} \tan^{-1} \frac{u}{a} + c$:
$I = \frac{1}{\sqrt{6}} \tan^{-1} \left(\frac{z}{\sqrt{6}}\right) + c = \frac{1}{\sqrt{6}} \tan^{-1} \left(\frac{1}{\sqrt{6}} \tan \frac{x}{2}\right) + c$.

(C) We have,$I = \int \frac{dx}{1-\cos x-\sin x}$.
Using the half-angle formulas $\cos x = \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$ and $\sin x = \frac{2\tan(x/2)}{1+\tan^2(x/2)}$,we get:
$I = \int \frac{dx}{1 - \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)} - \frac{2\tan(x/2)}{1+\tan^2(x/2)}}$
$I = \int \frac{(1+\tan^2(x/2)) dx}{1+\tan^2(x/2) - 1 + \tan^2(x/2) - 2\tan(x/2)}$
$I = \int \frac{\sec^2(x/2) dx}{2\tan^2(x/2) - 2\tan(x/2)} = \frac{1}{2} \int \frac{\sec^2(x/2) dx}{\tan(x/2)(\tan(x/2)-1)}$.
Let $z = \tan(x/2)$,then $dz = \frac{1}{2} \sec^2(x/2) dx$,so $\sec^2(x/2) dx = 2 dz$.
$I = \int \frac{2 dz}{2z(z-1)} = \int \frac{dz}{z(z-1)}$.
Using partial fractions: $\frac{1}{z(z-1)} = \frac{1}{z-1} - \frac{1}{z}$.
$I = \int (\frac{1}{z-1} - \frac{1}{z}) dz = \log|z-1| - \log|z| + C = \log|\frac{z-1}{z}| + C$.
Substituting $z = \tan(x/2)$ back:
$I = \log|\frac{\tan(x/2)-1}{\tan(x/2)}| + C = \log|1 - \cot(x/2)| + C$.

(B) Using the property $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$ if $f(x)$ is an even function:
Since $\sin^4(-x)\cos^6(-x) = \sin^4 x \cos^6 x$,the function is even.
$I = 2 \int_{0}^{\pi / 2} \sin^4 x \cos^6 x \, dx$
Using Wallis' Formula: $\int_{0}^{\pi / 2} \sin^m x \cos^n x \, dx = \frac{[(m-1)(m-3)...][(n-1)(n-3)...]}{(m+n)(m+n-2)...} \times \frac{\pi}{2}$ (if both $m, n$ are even).
Here $m=4, n=6$:
$I = 2 \times \left[ \frac{(3 \cdot 1) \times (5 \cdot 3 \cdot 1)}{(10 \cdot 8 \cdot 6 \cdot 4 \cdot 2)} \times \frac{\pi}{2} \right]$
$I = 2 \times \left[ \frac{3 \times 15}{3840} \times \frac{\pi}{2} \right] = \frac{45 \pi}{3840} = \frac{3 \pi}{256}$.

(B) We have,
$\int_2^3 \frac{dx}{x^2-x} = \int_2^3 \frac{1}{x(x-1)} dx$
Using partial fractions,$\frac{1}{x(x-1)} = \frac{A}{x} + \frac{B}{x-1}$.
Solving for $A$ and $B$,we get $A = -1$ and $B = 1$.
So,$\int_2^3 \left[ \frac{1}{x-1} - \frac{1}{x} \right] dx$
$= [\log|x-1| - \log|x|]_2^3$
$= [\log|\frac{x-1}{x}|]_2^3$
$= \log(\frac{3-1}{3}) - \log(\frac{2-1}{2})$
$= \log(\frac{2}{3}) - \log(\frac{1}{2})$
$= \log(\frac{2/3}{1/2}) = \log(\frac{4}{3})$

(B) Given the differential equation: $\frac{dy}{dx} = (\frac{x}{y})^{-1/3}$.
By simplifying the right side,we get: $\frac{dy}{dx} = (\frac{y}{x})^{1/3} = \frac{y^{1/3}}{x^{1/3}}$.
Separating the variables,we have: $y^{-1/3} dy = x^{-1/3} dx$.
Integrating both sides:
$\int y^{-1/3} dy = \int x^{-1/3} dx$.
Using the power rule $\int x^n dx = \frac{x^{n+1}}{n+1} + C$:
$\frac{y^{2/3}}{2/3} = \frac{x^{2/3}}{2/3} + C_1$.
Multiplying by $\frac{2}{3}$:
$y^{2/3} = x^{2/3} + \frac{2}{3}C_1$.
Let $c = \frac{2}{3}C_1$,then $y^{2/3} - x^{2/3} = c$.

(A) We have,$\frac{dy}{dx} - y = x^2$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -1$ and $Q = x^2$.
The integrating factor is $IF = e^{\int P dx} = e^{\int -1 dx} = e^{-x}$.
The general solution is given by $y \cdot IF = \int Q \cdot IF dx + c$.
Substituting the values,we get $y e^{-x} = \int x^2 e^{-x} dx + c$.
Using integration by parts,$\int x^2 e^{-x} dx = -x^2 e^{-x} - 2x e^{-x} - 2e^{-x}$.
Thus,$y e^{-x} = -e^{-x}(x^2 + 2x + 2) + c$.
Multiplying both sides by $e^x$,we get $y = -(x^2 + 2x + 2) + ce^x$.
Rearranging the terms,we obtain $y + x^2 + 2x + 2 = ce^x$.

(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{3}$ and $Q = 1$.
The integrating factor $(IF)$ is given by $IF = e^{\int P dx} = e^{\int \frac{1}{3} dx} = e^{x/3}$.
The general solution is $y \times (IF) = \int (Q \times IF) dx + c$.
Substituting the values,we get $y \times e^{x/3} = \int (1 \times e^{x/3}) dx + c$.
Integrating the right side,$y \times e^{x/3} = 3e^{x/3} + c$.
Dividing both sides by $e^{x/3}$,we get $y = 3 + ce^{-x/3}$.

(D) Let the vertices of the triangle be $A(2, -1, 1)$,$B(1, -3, -5)$,and $C(3, -4, -4)$.
First,we find the vectors representing the sides:
$\vec{AB} = (1-2)\hat{i} + (-3 - (-1))\hat{j} + (-5-1)\hat{k} = -\hat{i} - 2\hat{j} - 6\hat{k}$
$\vec{BC} = (3-1)\hat{i} + (-4 - (-3))\hat{j} + (-4 - (-5))\hat{k} = 2\hat{i} - \hat{j} + \hat{k}$
$\vec{CA} = (2-3)\hat{i} + (-1 - (-4))\hat{j} + (1 - (-4))\hat{k} = -\hat{i} + 3\hat{j} + 5\hat{k}$
Now,calculate the lengths of the sides:
$|\vec{AB}|^2 = (-1)^2 + (-2)^2 + (-6)^2 = 1 + 4 + 36 = 41$
$|\vec{BC}|^2 = (2)^2 + (-1)^2 + (1)^2 = 4 + 1 + 1 = 6$
$|\vec{CA}|^2 = (-1)^2 + (3)^2 + (5)^2 = 1 + 9 + 25 = 35$
Since $|\vec{AB}|^2 = |\vec{BC}|^2 + |\vec{CA}|^2$ $(41 = 6 + 35)$,the triangle satisfies the Pythagorean theorem.
Therefore,the triangle is a right-angled triangle.

(A) Given the position vectors of points $A, B, C$ as:
$A = \hat{i} + x\hat{j} + 3\hat{k}$
$B = 3\hat{i} + 4\hat{j} + 7\hat{k}$
$C = y\hat{i} - 2\hat{j} - 5\hat{k}$
Calculate the vectors $\vec{AB}$ and $\vec{BC}$:
$\vec{AB} = B - A = (3-1)\hat{i} + (4-x)\hat{j} + (7-3)\hat{k} = 2\hat{i} + (4-x)\hat{j} + 4\hat{k}$
$\vec{BC} = C - B = (y-3)\hat{i} + (-2-4)\hat{j} + (-5-7)\hat{k} = (y-3)\hat{i} - 6\hat{j} - 12\hat{k}$
Since $A, B, C$ are collinear,$\vec{AB} = t\vec{BC}$ for some scalar $t$:
$2\hat{i} + (4-x)\hat{j} + 4\hat{k} = t((y-3)\hat{i} - 6\hat{j} - 12\hat{k})$
Equating the coefficients of $\hat{i}, \hat{j}, \hat{k}$:
$1) \ 4 = -12t \Rightarrow t = -\frac{1}{3}$
$2) \ 4 - x = -6t \Rightarrow 4 - x = -6(-\frac{1}{3}) = 2 \Rightarrow x = 2$
$3) \ 2 = t(y-3) \Rightarrow 2 = -\frac{1}{3}(y-3) \Rightarrow -6 = y-3 \Rightarrow y = -3$
Thus,$(x, y) = (2, -3)$.

(B) We need to evaluate the scalar triple product: $(a+b) \cdot ((b+c) \times (a+b+c))$.
First,expand the cross product term: $(b+c) \times (a+b+c) = (b \times a) + (b \times b) + (b \times c) + (c \times a) + (c \times b) + (c \times c)$.
Since the cross product of any vector with itself is zero ($b \times b = 0$ and $c \times c = 0$),and using the property $c \times b = -(b \times c)$,we get:
$(b+c) \times (a+b+c) = (b \times a) + (b \times c) + (c \times a) - (b \times c) = (b \times a) + (c \times a)$.
Now,take the dot product with $(a+b)$:
$(a+b) \cdot ((b \times a) + (c \times a)) = a \cdot (b \times a) + a \cdot (c \times a) + b \cdot (b \times a) + b \cdot (c \times a)$.
Using the properties of scalar triple products,$a \cdot (b \times a) = 0$,$a \cdot (c \times a) = 0$,and $b \cdot (b \times a) = 0$.
This leaves us with $b \cdot (c \times a) = [b c a]$.
Since $[b c a] = [a b c]$,the final result is $[a b c]$.

(A) Let $a = x\hat{i} + y\hat{j} + z\hat{k}$.
Given $a \cdot \hat{i} = a \cdot (\hat{i} + \hat{j}) = a \cdot (\hat{i} + \hat{j} + \hat{k})$.
From $a \cdot \hat{i} = a \cdot (\hat{i} + \hat{j})$,we have $a \cdot \hat{i} = a \cdot \hat{i} + a \cdot \hat{j}$,which implies $a \cdot \hat{j} = 0$. Thus,$y = 0$.
From $a \cdot (\hat{i} + \hat{j}) = a \cdot (\hat{i} + \hat{j} + \hat{k})$,we have $a \cdot (\hat{i} + \hat{j}) = a \cdot (\hat{i} + \hat{j}) + a \cdot \hat{k}$,which implies $a \cdot \hat{k} = 0$. Thus,$z = 0$.
Since $a \cdot \hat{i} = x$,and the problem does not specify the magnitude,if we assume $a$ is a unit vector or simply look for the component form,$a = x\hat{i}$. Given the options,$a = \hat{i}$ satisfies the conditions.