TS EAMCET 2013 Chemistry Question Paper with Answer and Solution

(C) Given expression is $E = \frac{(1-2 x)^{-1}(1-3 x)^{-2}}{(1-4 x)^{-3}}$.
Using the binomial expansion $(1+ax)^n \approx 1+nax$ for small $x$:
$(1-2 x)^{-1} \approx 1 + (-1)(-2x) = 1 + 2x$.
$(1-3 x)^{-2} \approx 1 + (-2)(-3x) = 1 + 6x$.
$(1-4 x)^{-3} \approx 1 + (-3)(-4x) = 1 + 12x$.
Substituting these into the expression:
$E \approx \frac{(1+2x)(1+6x)}{1+12x} = \frac{1 + 6x + 2x + 12x^2}{1+12x}$.
Neglecting $x^2$ and higher powers:
$E \approx \frac{1+8x}{1+12x} = (1+8x)(1+12x)^{-1}$.
Using the binomial expansion again:
$E \approx (1+8x)(1-12x) = 1 - 12x + 8x - 96x^2$.
Neglecting $x^2$:
$E \approx 1 - 4x$.

(A) The equation of a chord of a conic $S=0$ with mid-point $(x_1, y_1)$ is given by $T=S_1$.
Given ellipse: $S: x^2+4y^2-2x+20y=0$.
Mid-point $(x_1, y_1) = (2, -4)$.
$T = x(x_1) + 4y(y_1) - (x+x_1) + 10(y+y_1) = x(2) + 4y(-4) - (x+2) + 10(y-4) = 2x - 16y - x - 2 + 10y - 40 = x - 6y - 42$.
$S_1 = (x_1)^2 + 4(y_1)^2 - 2(x_1) + 20(y_1) = (2)^2 + 4(-4)^2 - 2(2) + 20(-4) = 4 + 64 - 4 - 80 = -16$.
Equating $T=S_1$:
$x - 6y - 42 = -16$.
$x - 6y = 42 - 16$.
$x - 6y = 26$.

(B) The equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$. Here $a^2=25$ and $b^2=16$.
For an ellipse,$b^2=a^2(1-e^2)$,so $16=25(1-e^2)$,which gives $e^2=1-\frac{16}{25}=\frac{9}{25}$,so $e=\frac{3}{5}$.
The foci of the ellipse are $(\pm ae, 0) = (\pm 5 \times \frac{3}{5}, 0) = (\pm 3, 0)$.
The equation of the hyperbola is $\frac{x^2}{4}-\frac{y^2}{b^2}=1$. Here $a^2=4$,so $a=2$.
Since the foci of the hyperbola coincide with the foci of the ellipse,the foci of the hyperbola are $(\pm 3, 0)$.
For a hyperbola,the foci are $(\pm ae_1, 0)$,so $ae_1=3$. Since $a=2$,we have $2e_1=3$,which means $e_1=\frac{3}{2}$.
For a hyperbola,$b^2=a^2(e_1^2-1)$.
Substituting the values,$b^2=4((\frac{3}{2})^2-1) = 4(\frac{9}{4}-1) = 4(\frac{5}{4}) = 5$.
Thus,$b^2=5$.

(B) Given that $x=9$ is a chord of contact of the hyperbola $x^2-y^2=9$.
Substituting $x=9$ into the hyperbola equation:
$81-y^2=9$
$y^2=72$
$y = \pm 6\sqrt{2}$.
Thus,the points of contact are $(9, 6\sqrt{2})$ and $(9, -6\sqrt{2})$.
Differentiating $x^2-y^2=9$ with respect to $x$:
$2x - 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{x}{y}$.
At point $(9, 6\sqrt{2})$,the slope $m = \frac{9}{6\sqrt{2}} = \frac{3}{2\sqrt{2}}$.
The equation of the tangent at $(9, 6\sqrt{2})$ is:
$y - 6\sqrt{2} = \frac{3}{2\sqrt{2}}(x-9)$
$2\sqrt{2}y - 24 = 3x - 27$
$3x - 2\sqrt{2}y - 3 = 0$.
At point $(9, -6\sqrt{2})$,the slope $m = \frac{9}{-6\sqrt{2}} = -\frac{3}{2\sqrt{2}}$.
The equation of the tangent at $(9, -6\sqrt{2})$ is:
$y + 6\sqrt{2} = -\frac{3}{2\sqrt{2}}(x-9)$
$2\sqrt{2}y + 24 = -3x + 27$
$3x + 2\sqrt{2}y - 3 = 0$.
Comparing with the options,$3x + 2\sqrt{2}y - 3 = 0$ is option $B$.

(B) We have the limit $L = \lim _{x \rightarrow 0} \frac{\tan ^3 x-\sin ^3 x}{x^5}$.
Using the Taylor series expansions: $\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + O(x^7)$ and $\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} + O(x^7)$.
Factorizing the numerator: $\tan^3 x - \sin^3 x = (\tan x - \sin x)(\tan^2 x + \tan x \sin x + \sin^2 x)$.
Now,$\tan x - \sin x = \sin x (\sec x - 1) = \sin x \left(\frac{1-\cos x}{\cos x}\right) = \sin x \cdot \frac{2 \sin^2(x/2)}{\cos x}$.
As $x \rightarrow 0$,$\sin x \approx x$,$\sin(x/2) \approx x/2$,and $\cos x \approx 1$.
So,$\tan x - \sin x \approx x \cdot \frac{2(x/2)^2}{1} = \frac{x^3}{2}$.
Substituting this into the limit expression:
$L = \lim _{x \rightarrow 0} \frac{(\frac{x^3}{2})(\tan^2 x + \tan x \sin x + \sin^2 x)}{x^5}$.
$L = \lim _{x \rightarrow 0} \frac{1}{2} \cdot \frac{\tan^2 x + \tan x \sin x + \sin^2 x}{x^2}$.
Since $\lim _{x \rightarrow 0} \frac{\tan x}{x} = 1$ and $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$,we have:
$L = \frac{1}{2} (1^2 + 1 \cdot 1 + 1^2) = \frac{1}{2} (3) = \frac{3}{2}$.

(B) $PHBV$ and $Dextron$ are examples of biodegradable polymers.
Biodegradable polymers are those that disintegrate over time through enzymatic hydrolysis and oxidation.
$PHBV$ (Poly-$\beta$-hydroxybutyrate-co-$\beta$-hydroxyvalerate) is used in orthopaedic devices and controlled drug release.
$Dextron$ (a copolymer of glycolic acid and lactic acid) is used for stitching wounds after surgery.
Thus,the correct pair is $PHBV$ and $Dextron$.

(B) The random variable $X$ follows a discrete uniform distribution on the set $\{1, 2, \ldots, m\}$.
The mean $\bar{X} = E[X] = \sum_{n=1}^{m} n \cdot P(X=n) = \frac{1}{m} \sum_{n=1}^{m} n = \frac{1}{m} \cdot \frac{m(m+1)}{2} = \frac{m+1}{2}$.
The variance is given by $\operatorname{Var}(X) = E[X^2] - (E[X])^2$.
$E[X^2] = \sum_{n=1}^{m} n^2 \cdot P(X=n) = \frac{1}{m} \sum_{n=1}^{m} n^2 = \frac{1}{m} \cdot \frac{m(m+1)(2m+1)}{6} = \frac{(m+1)(2m+1)}{6}$.
Therefore,$\operatorname{Var}(X) = \frac{(m+1)(2m+1)}{6} - \left(\frac{m+1}{2}\right)^2$.
$\operatorname{Var}(X) = \frac{m+1}{2} \left[ \frac{2(2m+1)}{3} - \frac{m+1}{2} \right] = \frac{m+1}{2} \left[ \frac{8m+4-3m-3}{6} \right] = \frac{m+1}{2} \cdot \frac{5m+1}{6}$ (Wait,let's re-calculate).
$\operatorname{Var}(X) = \frac{(m+1)(2m+1)}{6} - \frac{(m+1)^2}{4} = \frac{m+1}{2} \left[ \frac{2(2m+1) - 3(m+1)}{6} \right] = \frac{m+1}{2} \left[ \frac{4m+2-3m-3}{6} \right] = \frac{m+1}{2} \cdot \frac{m-1}{6} = \frac{m^2-1}{12}$.

(C) Given the equation: $\frac{1}{a+c} + \frac{1}{b+c} = \frac{3}{a+b+c}$.
Taking $LCM$ on the left side: $\frac{(b+c) + (a+c)}{(a+c)(b+c)} = \frac{3}{a+b+c}$.
$\frac{a+b+2c}{ab+ac+bc+c^2} = \frac{3}{a+b+c}$.
Cross-multiplying: $(a+b+2c)(a+b+c) = 3(ab+ac+bc+c^2)$.
$(a+b)^2 + c(a+b) + 2c(a+b) + 2c^2 = 3ab + 3ac + 3bc + 3c^2$.
$a^2 + b^2 + 2ab + 3ac + 3bc + 2c^2 = 3ab + 3ac + 3bc + 3c^2$.
$a^2 + b^2 - ab = c^2$.
Using the Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos C$.
Comparing the two equations: $a^2 + b^2 - ab = a^2 + b^2 - 2ab \cos C$.
$-ab = -2ab \cos C$.
$\cos C = \frac{1}{2}$.
Therefore,$\angle C = 60^{\circ}$.

(B) Given that,$\frac{1}{b+c} + \frac{1}{c+a} = \frac{3}{a+b+c}$.
Taking common denominators on the left side:
$\frac{(c+a) + (b+c)}{(b+c)(c+a)} = \frac{3}{a+b+c}$
$\frac{a+b+2c}{bc + ab + c^2 + ac} = \frac{3}{a+b+c}$
Cross-multiplying:
$(a+b+2c)(a+b+c) = 3(bc + ab + c^2 + ac)$
$(a+b)^2 + c(a+b) + 2c(a+b) + 2c^2 = 3bc + 3ab + 3c^2 + 3ac$
$a^2 + b^2 + 2ab + 3ac + 3bc + 2c^2 = 3bc + 3ab + 3c^2 + 3ac$
Subtracting $3ac + 3bc$ from both sides:
$a^2 + b^2 + 2ab + 2c^2 = 3ab + 3c^2$
$a^2 + b^2 - c^2 = ab$
Using the Law of Cosines,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$:
$\cos C = \frac{ab}{2ab} = \frac{1}{2}$
Since $\cos C = \frac{1}{2}$,we have $C = 60^{\circ}$.

(A) We know that the exradii are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
Substituting these into the expression $r_1 r_2 + r_2 r_3 + r_3 r_1$:
$= \frac{\Delta^2}{(s-a)(s-b)} + \frac{\Delta^2}{(s-b)(s-c)} + \frac{\Delta^2}{(s-c)(s-a)}$
$= \frac{\Delta^2(s-c + s-a + s-b)}{(s-a)(s-b)(s-c)}$
Since $2s = a+b+c$,we have $s-a+s-b+s-c = 3s - (a+b+c) = 3s - 2s = s$.
Also,we know that $\Delta^2 = s(s-a)(s-b)(s-c)$,so $(s-a)(s-b)(s-c) = \frac{\Delta^2}{s}$.
Substituting these values:
$= \frac{\Delta^2 \cdot s}{\frac{\Delta^2}{s}} = s^2$
Since $r = \frac{\Delta}{s}$,we have $s = \frac{\Delta}{r}$,so $s^2 = \frac{\Delta^2}{r^2}$.

(B) Given the matrix $A = \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix}$.
Since $A$ satisfies the equation $x^2 + 4x - p = 0$,we have $A^2 + 4A - pI = 0$,where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
First,calculate $A^2$:
$A^2 = \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} (-8)(-8) + (5)(2) & (-8)(5) + (5)(4) \\ (2)(-8) + (4)(2) & (2)(5) + (4)(4) \end{bmatrix} = \begin{bmatrix} 64 + 10 & -40 + 20 \\ -16 + 8 & 10 + 16 \end{bmatrix} = \begin{bmatrix} 74 & -20 \\ -8 & 26 \end{bmatrix}$.
Next,calculate $4A$:
$4A = 4 \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} -32 & 20 \\ 8 & 16 \end{bmatrix}$.
Now,substitute these into the equation $A^2 + 4A - pI = 0$:
$\begin{bmatrix} 74 & -20 \\ -8 & 26 \end{bmatrix} + \begin{bmatrix} -32 & 20 \\ 8 & 16 \end{bmatrix} - \begin{bmatrix} p & 0 \\ 0 & p \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Performing the matrix addition and subtraction:
$\begin{bmatrix} 74 - 32 - p & -20 + 20 - 0 \\ -8 + 8 - 0 & 26 + 16 - p \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
$\begin{bmatrix} 42 - p & 0 \\ 0 & 42 - p \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Comparing the elements,we get $42 - p = 0$,which implies $p = 42$.

(A) Given $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$.
We calculate the powers of $A$:
$A^2 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$.
$A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}$.
By observation,$A^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$.
Now,check option $A$: $nA - (n-1)I = n \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} - (n-1) \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$= \begin{bmatrix} n & n \\ 0 & n \end{bmatrix} - \begin{bmatrix} n-1 & 0 \\ 0 & n-1 \end{bmatrix} = \begin{bmatrix} n-(n-1) & n-0 \\ 0-0 & n-(n-1) \end{bmatrix} = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} = A^n$.
Thus,the correct relation is $A^n = nA - (n-1)I$.

(D) Let $\Delta = \left|\begin{array}{ccc}x+2 & x+3 & x+5 \\ x+4 & x+6 & x+9 \\ x+8 & x+11 & x+15\end{array}\right|$.
Apply operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$:
$\Delta = \left|\begin{array}{ccc}x+2 & x+3 & x+5 \\ 2 & 3 & 4 \\ 6 & 8 & 10\end{array}\right|$.
Now,apply $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1$:
$\Delta = \left|\begin{array}{ccc}x+2 & 1 & 3 \\ 2 & 1 & 2 \\ 6 & 2 & 4\end{array}\right|$.
Expanding along $R_1$:
$\Delta = (x+2)(4-4) - 1(8-12) + 3(4-6)$
$\Delta = (x+2)(0) - 1(-4) + 3(-2)$
$\Delta = 0 + 4 - 6 = -2$.

(D) The given system of equations is:
$3x + 2y + z = 6$
$3x + 4y + 3z = 14$
$6x + 10y + 8z = a$
Let $A = \begin{bmatrix} 3 & 2 & 1 \\ 3 & 4 & 3 \\ 6 & 10 & 8 \end{bmatrix}$ and $B = \begin{bmatrix} 6 \\ 14 \\ a \end{bmatrix}$.
First,we calculate the determinant of $A$:
$|A| = 3(32 - 30) - 2(24 - 18) + 1(30 - 24) = 3(2) - 2(6) + 6 = 6 - 12 + 6 = 0$.
Since $|A| = 0$,the system has either no solution or infinite solutions.
For infinite solutions,we must have $(\text{adj } A) \cdot B = 0$.
The cofactor matrix of $A$ is:
$C_{11} = 2, C_{12} = -6, C_{13} = 6$
$C_{21} = -6, C_{22} = 18, C_{23} = -18$
$C_{31} = 2, C_{32} = -6, C_{33} = 6$
Thus,$\text{adj } A = \begin{bmatrix} 2 & -6 & 2 \\ -6 & 18 & -6 \\ 6 & -18 & 6 \end{bmatrix}$.
Now,$(\text{adj } A) \cdot B = \begin{bmatrix} 2 & -6 & 2 \\ -6 & 18 & -6 \\ 6 & -18 & 6 \end{bmatrix} \begin{bmatrix} 6 \\ 14 \\ a \end{bmatrix} = \begin{bmatrix} 12 - 84 + 2a \\ -36 + 252 - 6a \\ 36 - 252 + 6a \end{bmatrix} = \begin{bmatrix} 2a - 72 \\ 216 - 6a \\ 6a - 216 \end{bmatrix}$.
Setting this to zero vector,we get $2a - 72 = 0$,which implies $a = 36$.

(C) We use the formula $\cos ^{-1} A + \cos ^{-1} B = \cos ^{-1} (AB - \sqrt{1-A^2} \sqrt{1-B^2})$.
Given $\cos ^{-1}\left(\frac{5}{13}\right)+\cos ^{-1}\left(\frac{3}{5}\right)=\cos ^{-1} x$.
Here $A = \frac{5}{13}$ and $B = \frac{3}{5}$.
Then $\sqrt{1-A^2} = \sqrt{1-\left(\frac{5}{13}\right)^2} = \sqrt{1-\frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$.
And $\sqrt{1-B^2} = \sqrt{1-\left(\frac{3}{5}\right)^2} = \sqrt{1-\frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
Substituting these values into the formula:
$\cos ^{-1} x = \cos ^{-1} \left( \frac{5}{13} \cdot \frac{3}{5} - \frac{12}{13} \cdot \frac{4}{5} \right)$.
$\cos ^{-1} x = \cos ^{-1} \left( \frac{15}{65} - \frac{48}{65} \right)$.
$\cos ^{-1} x = \cos ^{-1} \left( \frac{-33}{65} \right)$.
Therefore,$x = \frac{-33}{65}$.

(D) We know that $\operatorname{coth}^{-1}(x) = \tanh^{-1}\left(\frac{1}{x}\right)$ for $|x| > 1$.
Therefore,$\operatorname{coth}^{-1}(2) = \tanh^{-1}\left(\frac{1}{2}\right)$.
Substituting this into the expression,we get:
$\tanh ^{-1}\left(\frac{1}{2}\right)+\operatorname{coth}^{-1}(2) = \tanh ^{-1}\left(\frac{1}{2}\right)+\tanh ^{-1}\left(\frac{1}{2}\right) = 2 \tanh ^{-1}\left(\frac{1}{2}\right)$.
Using the formula $\tanh ^{-1}(x) = \frac{1}{2} \log \left(\frac{1+x}{1-x}\right)$,we have:
$2 \tanh ^{-1}\left(\frac{1}{2}\right) = 2 \cdot \frac{1}{2} \log \left(\frac{1+\frac{1}{2}}{1-\frac{1}{2}}\right)$.
$= \log \left(\frac{\frac{3}{2}}{\frac{1}{2}}\right) = \log 3$.

(D) For the expression $\log_{10} ((1.6)^{1-x^2} - (0.625)^{6(1+x)})$ to be defined in $R$,the argument of the logarithm must be strictly greater than $0$.
$(1.6)^{1-x^2} - (0.625)^{6(1+x)} > 0$
$(1.6)^{1-x^2} > (0.625)^{6(1+x)}$
Since $1.6 = \frac{8}{5}$ and $0.625 = \frac{5}{8} = (\frac{8}{5})^{-1}$,we have:
$(\frac{8}{5})^{1-x^2} > ((\frac{8}{5})^{-1})^{6(1+x)}$
$(\frac{8}{5})^{1-x^2} > (\frac{8}{5})^{-6(1+x)}$
Since the base $\frac{8}{5} > 1$,the inequality direction remains the same:
$1 - x^2 > -6(1 + x)$
$1 - x^2 > -6 - 6x$
$x^2 - 6x - 7 < 0$
$(x - 7)(x + 1) < 0$
Thus,$x \in (-1, 7)$.

(D) Given,$f(x)=\cos \left(\frac{x}{3}\right)+\sin \left(\frac{x}{2}\right)$.
We know that the period of $\cos(ax)$ and $\sin(ax)$ is $\frac{2\pi}{|a|}$.
For the term $\cos \left(\frac{x}{3}\right)$,the period $T_1 = \frac{2\pi}{1/3} = 6\pi$.
For the term $\sin \left(\frac{x}{2}\right)$,the period $T_2 = \frac{2\pi}{1/2} = 4\pi$.
The period of the sum of two periodic functions is the Least Common Multiple $(LCM)$ of their individual periods.
Therefore,the period of $f(x) = \text{LCM}(6\pi, 4\pi)$.
Since $6\pi = 2 \times 3\pi$ and $4\pi = 2 \times 2\pi$,the $\text{LCM}(6\pi, 4\pi) = 12\pi$.
Thus,the period of $f(x)$ is $12\pi$.

(A) Given,$f(x) = (p - x^n)^{1/n}$,where $p > 0$.
To find $f[f(x)]$,we substitute $f(x)$ into the function $f$:
$f[f(x)] = f((p - x^n)^{1/n})$
$= (p - ((p - x^n)^{1/n})^n)^{1/n}$
$= (p - (p - x^n))^{1/n}$
$= (p - p + x^n)^{1/n}$
$= (x^n)^{1/n}$
$= x$

(B) Given the functional equation $f(x+y)=f(x) \cdot f(y)$ for all $x, y \in R$.
We know that the continuous function satisfying this property is of the form $f(x) = a^x$.
Given $f(2) = 9$,we have $a^2 = 9$.
Since $f$ is a non-zero function,$a^2 = 3^2$,which implies $a = 3$ (as $a$ must be positive for $f(x) = a^x$ to be defined for all real $x$).
Thus,$f(x) = 3^x$.
Now,we need to find $f(6)$.
$f(6) = 3^6$.

(B) Given $f(x) = \frac{1}{1+\frac{1}{x}} = \frac{x}{x+1}$.
Then $g(x) = \frac{1}{1+\frac{1}{f(x)}} = \frac{1}{1+\frac{x+1}{x}} = \frac{1}{\frac{x+x+1}{x}} = \frac{x}{2x+1}$.
To find $g^{\prime}(x)$,we use the quotient rule: $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u^{\prime} - u v^{\prime}}{v^2}$.
$g^{\prime}(x) = \frac{(2x+1)(1) - x(2)}{(2x+1)^2} = \frac{2x+1-2x}{(2x+1)^2} = \frac{1}{(2x+1)^2}$.
Substituting $x = 2$,we get $g^{\prime}(2) = \frac{1}{(2(2)+1)^2} = \frac{1}{5^2} = \frac{1}{25}$.

(C) Given the equation: $\sqrt{\frac{y}{x}}+\sqrt{\frac{x}{y}}=2$
Multiplying by $\sqrt{xy}$,we get: $y+x=2\sqrt{xy}$
Squaring both sides: $(x+y)^2 = (2\sqrt{xy})^2$
$x^2+y^2+2xy = 4xy$
$x^2+y^2-2xy = 0$
$(x-y)^2 = 0$
This implies $x-y=0$,or $y=x$
Differentiating both sides with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(x) = 1$

(D) Let $f(x) = (x+1)(x^2+1)(x^4+1)(x^8+1)$.
Multiply and divide by $(x-1)$:
$f(x) = \frac{(x-1)(x+1)(x^2+1)(x^4+1)(x^8+1)}{(x-1)} = \frac{(x^2-1)(x^2+1)(x^4+1)(x^8+1)}{(x-1)} = \frac{(x^4-1)(x^4+1)(x^8+1)}{(x-1)} = \frac{(x^8-1)(x^8+1)}{(x-1)} = \frac{x^{16}-1}{x-1}$.
Now,differentiate $f(x)$ with respect to $x$ using the quotient rule:
$\frac{d}{dx} \left[ \frac{x^{16}-1}{x-1} \right] = \frac{(x-1)(16x^{15}) - (x^{16}-1)(1)}{(x-1)^2} = \frac{16x^{16} - 16x^{15} - x^{16} + 1}{(x-1)^2} = \frac{15x^{16} - 16x^{15} + 1}{(x-1)^2}$.
Comparing this with the given expression $\left(15 x^p-16 x^q+1\right)(x-1)^{-2}$,we get $p=16$ and $q=15$.
Therefore,$(p, q) = (16, 15)$.

(A) Given the relation $p V^{1/4} = C$,where $C$ is a constant.
Taking the natural logarithm on both sides: $\ln p + \frac{1}{4} \ln V = \ln C$.
Differentiating both sides with respect to $V$: $\frac{1}{p} \frac{dp}{dV} + \frac{1}{4V} = 0$.
This gives $\frac{dp}{p} = -\frac{1}{4} \frac{dV}{V}$.
We are given that the percentage decrease in volume is $\frac{dV}{V} \times 100 = -\frac{1}{2} \%$.
Substituting this into the differential relation: $\frac{dp}{p} \times 100 = -\frac{1}{4} \left( \frac{dV}{V} \times 100 \right)$.
$\frac{dp}{p} \times 100 = -\frac{1}{4} \left( -\frac{1}{2} \% \right) = \frac{1}{8} \%$.
Thus,the percentage increase in pressure is $\frac{1}{8} \%$.

(B) Given the equation: $\frac{2}{f} = \frac{1}{v} - \frac{1}{u}$ $(i)$
Taking the differential of both sides:
$-\frac{2}{f^2} df = -\frac{1}{v^2} dv + \frac{1}{u^2} du$
Given that the errors in $u$ and $v$ are $p$,we have $dv = p$ and $du = p$.
Substituting these into the differential equation:
$-\frac{2}{f^2} df = -\frac{1}{v^2} p + \frac{1}{u^2} p$
$-\frac{2}{f^2} df = p \left( \frac{1}{u^2} - \frac{1}{v^2} \right)$
Using the identity $a^2 - b^2 = (a - b)(a + b)$:
$-\frac{2}{f^2} df = p \left( \frac{1}{u} - \frac{1}{v} \right) \left( \frac{1}{u} + \frac{1}{v} \right)$
From equation $(i)$,$\frac{1}{u} - \frac{1}{v} = -\frac{2}{f}$. Substituting this:
$-\frac{2}{f^2} df = p \left( -\frac{2}{f} \right) \left( \frac{1}{u} + \frac{1}{v} \right)$
Dividing both sides by $-\frac{2}{f}$:
$\frac{df}{f} = p \left( \frac{1}{u} + \frac{1}{v} \right)$
Thus,the relative error in $f$ is $p \left( \frac{1}{u} + \frac{1}{v} \right)$.

(D) Let the height of the tower be $h$ and the distance of point $A$ from the base of the tower $(B)$ be $x$.
In $\triangle ABD$,$\tan 60^{\circ} = \frac{h}{x}$ $\Rightarrow h = x \sqrt{3}$ $\Rightarrow x = \frac{h}{\sqrt{3}}$.
The person moves $60 \ m$ perpendicular to $AB$ to reach point $C$. Thus,$AC = 60 \ m$ and $\angle CAB = 90^{\circ}$.
In $\triangle ABC$,the distance from $C$ to the base $B$ is $CB = \sqrt{AC^2 + AB^2} = \sqrt{60^2 + x^2}$.
In $\triangle CBD$,the angle of elevation is $45^{\circ}$,so $\tan 45^{\circ} = \frac{h}{CB} = 1$.
Therefore,$h = CB = \sqrt{3600 + x^2} \Rightarrow h^2 = 3600 + x^2$.
Substituting $x^2 = \frac{h^2}{3}$ into the equation: $h^2 = 3600 + \frac{h^2}{3}$.
$\frac{2h^2}{3} = 3600 \Rightarrow h^2 = 1800 \times 3 = 5400$.
$h = \sqrt{5400} = \sqrt{900 \times 6} = 30 \sqrt{6} \ m$.

(B) Let $I_1 = \int \frac{d x}{x(\log x-2)(\log x-3)}$.
Substitute $t = \log x$,then $d t = \frac{1}{x} d x$.
The integral becomes $I_1 = \int \frac{d t}{(t-2)(t-3)}$.
Using partial fractions,$\frac{1}{(t-2)(t-3)} = \frac{A}{t-2} + \frac{B}{t-3}$.
$1 = A(t-3) + B(t-2)$.
For $t=3$,$B=1$. For $t=2$,$A=-1$.
So,$I_1 = \int \left( \frac{1}{t-3} - \frac{1}{t-2} \right) d t$.
$I_1 = \log |t-3| - \log |t-2| + C = \log \left| \frac{t-3}{t-2} \right| + C$.
Substituting $t = \log x$ back,we get $I = \log \left| \frac{\log x - 3}{\log x - 2} \right|$.

(D) Let $I = \int e^x \left( \frac{2+\sin 2x}{1+\cos 2x} \right) dx$.
Using the trigonometric identities $1+\cos 2x = 2\cos^2 x$ and $\sin 2x = 2\sin x \cos x$,we get:
$I = \int e^x \left( \frac{2 + 2\sin x \cos x}{2\cos^2 x} \right) dx$
$I = \int e^x \left( \frac{2}{2\cos^2 x} + \frac{2\sin x \cos x}{2\cos^2 x} \right) dx$
$I = \int e^x (\sec^2 x + \tan x) dx$.
We know the standard integral form $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$.
Here,let $f(x) = \tan x$,then $f'(x) = \sec^2 x$.
Therefore,$I = e^x \tan x + C$.

(A) From the lens maker's formula:
$\frac{1}{f} = \left( \frac{\mu_g}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Given that the refractive index of the glass lens is $\mu_g = 1.5$ and the refractive index of the medium is $\mu_m = 1.75$.
For a concave lens,the radii of curvature are $R_1 = -R$ and $R_2 = +R$.
Substituting these values into the formula:
$\frac{1}{f} = \left( \frac{1.5}{1.75} - 1 \right) \left( \frac{1}{-R} - \frac{1}{R} \right)$
$\frac{1}{f} = \left( \frac{1.5 - 1.75}{1.75} \right) \left( -\frac{2}{R} \right)$
$\frac{1}{f} = \left( \frac{-0.25}{1.75} \right) \left( -\frac{2}{R} \right)$
$\frac{1}{f} = \left( -\frac{1}{7} \right) \left( -\frac{2}{R} \right) = \frac{2}{7R}$
$f = +3.5 R$
The positive sign of the focal length indicates that the lens behaves as a convergent lens.

(B) For the eyepiece,the image distance $v_e = -25 \ cm$ and focal length $f_e = 5 \ cm$. Using the lens formula $\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e}$:
$\frac{1}{-25} - \frac{1}{u_e} = \frac{1}{5} \Rightarrow \frac{1}{u_e} = -\frac{1}{25} - \frac{1}{5} = -\frac{6}{25} \Rightarrow u_e = -\frac{25}{6} \ cm$.
The distance of the image formed by the objective from the objective is $v_0 = L - |u_e| = 10.5 - \frac{25}{6} = \frac{63-25}{6} = \frac{38}{6} \ cm$.
For the objective,$f_0 = 1.9 \ cm$ and $v_0 = \frac{38}{6} \ cm$. Using the lens formula $\frac{1}{v_0} - \frac{1}{u_0} = \frac{1}{f_0}$:
$\frac{1}{38/6} - \frac{1}{u_0} = \frac{1}{1.9} \Rightarrow \frac{6}{38} - \frac{1}{u_0} = \frac{10}{19} \Rightarrow \frac{3}{19} - \frac{10}{19} = \frac{1}{u_0}$.
$\frac{1}{u_0} = -\frac{7}{19} \Rightarrow u_0 = -\frac{19}{7} \approx -2.71 \ cm$.
Thus,the object should be placed at a distance of $2.7 \ cm$ from the objective.

(C) When chloroform $(CHCl_3)$ is heated with aqueous sodium hydroxide $(NaOH)$,it undergoes hydrolysis to form an unstable intermediate,methanetriol $(HC(OH)_3)$.
This intermediate loses a water molecule to form formic acid $(HCOOH)$.
Formic acid then reacts with the remaining $NaOH$ to produce sodium formate $(HCOONa)$.
The overall reaction is: $CHCl_3 + 4NaOH \rightarrow HCOONa + 3NaCl + 2H_2O$.

(C) The current gain $\beta$ of a transistor is defined as the ratio of the change in collector current to the change in base current: $\beta = \frac{\Delta I_C}{\Delta I_B}$.
Given:
Change in base current $\Delta I_B = 140 \mu A - 45 \mu A = 95 \mu A = 95 \times 10^{-6} \text{ A}$.
Change in collector current $\Delta I_C = 4.0 \text{ mA} - 0.2 \text{ mA} = 3.8 \text{ mA} = 3.8 \times 10^{-3} \text{ A}$.
Substituting these values into the formula:
$\beta = \frac{3.8 \times 10^{-3}}{95 \times 10^{-6}} = \frac{3800 \times 10^{-6}}{95 \times 10^{-6}} = 40$.
Therefore,the current gain is $40$.

(C) Let the position vectors of the vertices be $\vec{a} = 2i+3j+4k$,$\vec{b} = 3i+4j+2k$,and $\vec{c} = 4i+2j+3k$.
The side vectors are:
$\vec{AB} = \vec{b} - \vec{a} = (3-2)i + (4-3)j + (2-4)k = i + j - 2k$.
$\vec{BC} = \vec{c} - \vec{b} = (4-3)i + (2-4)j + (3-2)k = i - 2j + k$.
$\vec{CA} = \vec{a} - \vec{c} = (2-4)i + (3-2)j + (4-3)k = -2i + j + k$.
The lengths of the sides are:
$|\vec{AB}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1+1+4} = \sqrt{6}$.
$|\vec{BC}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1+4+1} = \sqrt{6}$.
$|\vec{CA}| = \sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{4+1+1} = \sqrt{6}$.
Since $|\vec{AB}| = |\vec{BC}| = |\vec{CA}| = \sqrt{6}$,all three sides are equal in length.
Therefore,the triangle is an equilateral triangle.

(A) Let the direction ratios of the two lines $AB$ and $AC$ be $\vec{v_1} = \langle 1, -1, -1 \rangle$ and $\vec{v_2} = \langle 2, -1, 1 \rangle$ respectively.
Since the lines $AB$ and $AC$ lie in the plane $ABC$,the normal vector $\vec{n}$ to the plane is given by the cross product of the vectors along these lines:
$\vec{n} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & -1 \\ 2 & -1 & 1 \end{vmatrix}$
$\vec{n} = \hat{i}(-1 - 1) - \hat{j}(1 - (-2)) + \hat{k}(-1 - (-2))$
$\vec{n} = \hat{i}(-2) - \hat{j}(3) + \hat{k}(1)$
$\vec{n} = -2\hat{i} - 3\hat{j} + \hat{k}$
Thus,the direction ratios of the normal are $\langle -2, -3, 1 \rangle$. Multiplying by $-1$,we get $\langle 2, 3, -1 \rangle$.
Therefore,the correct option is $A$.

(C) The equation of a plane passing through the point $(x_0, y_0, z_0)$ with normal vector $\vec{n} = \langle a, b, c \rangle$ is given by $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.
Given the point is $(-1, 2, 3)$,the equation is $a(x+1) + b(y-2) + c(z-3) = 0$.
The normal makes equal angles $\alpha$ with the coordinate axes,so the direction cosines are $\cos \alpha, \cos \alpha, \cos \alpha$.
Since $\cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1$,we have $3 \cos^2 \alpha = 1$,which implies $\cos \alpha = \pm \frac{1}{\sqrt{3}}$.
Thus,the direction ratios $\langle a, b, c \rangle$ can be taken as $\langle 1, 1, 1 \rangle$.
Substituting these into the plane equation: $1(x+1) + 1(y-2) + 1(z-3) = 0$.
Simplifying,we get $x + 1 + y - 2 + z - 3 = 0$,which results in $x + y + z - 4 = 0$.

(B) Let the equation of the variable plane be $a(x - 1) + b(y - 2) + c(z - 3) = 0$,where $(a, b, c)$ are the direction ratios of the normal to the plane.
Let $P(x_0, y_0, z_0)$ be the foot of the perpendicular from the origin $(0, 0, 0)$ to the plane.
The vector $\vec{OP} = (x_0, y_0, z_0)$ is perpendicular to the plane,so it is parallel to the normal vector $(a, b, c)$.
Thus,we can write $(a, b, c) = k(x_0, y_0, z_0)$ for some constant $k$.
Since $P$ lies on the plane,it satisfies the equation: $x_0(x_0 - 1) + y_0(y_0 - 2) + z_0(z_0 - 3) = 0$.
This simplifies to $x_0^2 - x_0 + y_0^2 - 2y_0 + z_0^2 - 3z_0 = 0$.
Rearranging the terms,we get $(x_0^2 - x_0 + \frac{1}{4}) + (y_0^2 - 2y_0 + 1) + (z_0^2 - 3z_0 + \frac{9}{4}) = \frac{1}{4} + 1 + \frac{9}{4} = \frac{14}{4} = 3.5$.
This is the equation of a sphere with center $(\frac{1}{2}, 1, \frac{3}{2})$ and radius $\sqrt{3.5}$.

(C) Total number of ways to choose $2$ numbers from $8$ is ${}^8C_2 = \frac{8 \times 7}{2} = 28$.
Let the two numbers be $x$ and $y$ such that $x < y$. We want the smaller number $x$ to be less than $4$,i.e.,$x \in \{1, 2, 3\}$.
Case $I$: If $x = 1$,then $y$ can be any of the remaining $7$ numbers $\{2, 3, 4, 5, 6, 7, 8\}$. Number of ways $= 7$.
Case $II$: If $x = 2$,then $y$ can be any of the remaining $6$ numbers $\{3, 4, 5, 6, 7, 8\}$. Number of ways $= 6$.
Case $III$: If $x = 3$,then $y$ can be any of the remaining $5$ numbers $\{4, 5, 6, 7, 8\}$. Number of ways $= 5$.
Total favorable cases $= 7 + 6 + 5 = 18$.
Required probability $= \frac{18}{28} = \frac{9}{14}$.

(D) Total sample points,$n(S) = 6 \times 6 = 36$.
Favourable events for sum $\ge 10$ are:
$E = [(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)]$.
Total number of favourable events,$n(E) = 6$.
Required probability $P(E) = \frac{n(E)}{n(S)} = \frac{6}{36} = \frac{1}{6}$.

(A) Let $H$ be the event that the toss results in a head. The bag contains $2n+1$ coins in total.
There are $n$ coins with heads on both sides (biased) and $n+1$ fair coins.
The probability of picking a biased coin is $P(B) = \frac{n}{2n+1}$ and the probability of picking a fair coin is $P(F) = \frac{n+1}{2n+1}$.
If a biased coin is picked,the probability of getting a head is $1$.
If a fair coin is picked,the probability of getting a head is $\frac{1}{2}$.
Using the law of total probability:
$P(H) = P(H|B)P(B) + P(H|F)P(F)$
$\frac{31}{42} = (1) \times \frac{n}{2n+1} + \left(\frac{1}{2}\right) \times \frac{n+1}{2n+1}$
$\frac{31}{42} = \frac{2n + n + 1}{2(2n+1)} = \frac{3n+1}{4n+2}$
$31(4n+2) = 42(3n+1)$
$124n + 62 = 126n + 42$
$20 = 2n$
$n = 10$

(A) In a close packed structure ($hcp$ or $ccp$):
$(i)$ The number of octahedral voids is equal to the number of particles $(N)$ present in the close packing.
(ii) The number of tetrahedral voids is equal to $2 \times N$,where $N$ is the number of particles.
Given that there are '$X$' atoms,the number of octahedral voids is '$X$' and the number of tetrahedral voids is '$2X$'.
Therefore,the correct option is $A$.

(D) According to Raoult's law,the relative lowering of vapour pressure is given by: $\frac{p^{\circ} - p_s}{p^{\circ}} = \frac{n_2}{n_1 + n_2}$
where,$p^{\circ} = \text{vapour pressure of pure water at } 100^{\circ} C = 760 \ mmHg$.
$p_s = \text{vapour pressure of solution at } 100^{\circ} C$.
$n_2 = \text{moles of glucose} = \frac{18}{180} = 0.1 \ mol$.
$n_1 = \text{moles of water} = \frac{180}{18} = 10 \ mol$.
Substituting these values: $\frac{760 - p_s}{760} = \frac{0.1}{10 + 0.1} = \frac{0.1}{10.1}$.
$760 - p_s = 760 \times \frac{0.1}{10.1} = \frac{76}{10.1} \approx 7.524 \ mmHg$.
$p_s = 760 - 7.524 = 752.476 \ mmHg \approx 752.4 \ mmHg$.

(B) Given: Concentration $(C) = 0.10 \ M$,Degree of ionization $(\alpha) = 4.0 \ \% = 0.04$.
For the dissociation of lactic acid: $CH_3CH(OH)COOH \rightleftharpoons CH_3CH(OH)COO^{-} + H^{+}$.
The equilibrium constant $K_a$ is given by the expression: $K_a = \frac{C\alpha^2}{1-\alpha}$.
Substituting the values: $K_a = \frac{0.1 \times (0.04)^2}{1 - 0.04}$.
$K_a = \frac{0.1 \times 0.0016}{0.96} = \frac{0.00016}{0.96} = \frac{1.6 \times 10^{-4}}{0.96} \approx 1.66 \times 10^{-4}$.

(A) Two solutions are isotonic if they have the same molar concentration of particles (osmolarity).
For $0.15 \ M \ NaCl$: $NaCl$ dissociates into $2$ ions ($Na^+$ and $Cl^-$). The concentration of particles $= 0.15 \times 2 = 0.30 \ M$.
For $0.1 \ M \ Na_2SO_4$: $Na_2SO_4$ dissociates into $3$ ions ($2Na^+$ and $SO_4^{2-}$). The concentration of particles $= 0.1 \times 3 = 0.30 \ M$.
Since both solutions have the same concentration of particles,they are isotonic.

(D) The kinetic energy $(KE)$ of $n$ moles of an ideal gas at temperature $T$ is given by $KE = n \times \frac{3}{2} RT$.
For $4 \text{ g}$ of $H_2$: Moles $n_1 = \frac{4 \text{ g}}{2 \text{ g/mol}} = 2 \text{ mol}$.
$KE_{H_2} = 2 \times \frac{3}{2} RT = 3RT$.
For $8 \text{ g}$ of $O_2$: Moles $n_2 = \frac{8 \text{ g}}{32 \text{ g/mol}} = 0.25 \text{ mol} = \frac{1}{4} \text{ mol}$.
$KE_{O_2} = \frac{1}{4} \times \frac{3}{2} RT = \frac{3}{8} RT$.
Ratio $KE_{H_2} : KE_{O_2} = 3RT : \frac{3}{8} RT = 1 : \frac{1}{8} = 8 : 1$.

(C) The quantum or wave mechanical model of an atom is based upon the dual nature of the electron,i.e.,the electron exhibits both particle and wave-like properties.
This model incorporates the de Broglie hypothesis regarding the wave nature of matter and the Heisenberg uncertainty principle.

(B) The number of radial nodes in an orbital is given by the formula: $n - l - 1$.
For the $3s$ orbital:
$n = 3$,$l = 0$.
Number of radial nodes $= 3 - 0 - 1 = 2$.
For the $2p$ orbital:
$n = 2$,$l = 1$.
Number of radial nodes $= 2 - 1 - 1 = 0$.
Thus,the number of radial nodes for $3s$ and $2p$ orbitals are $2$ and $0$ respectively.

(D) According to Stefan-Boltzmann law,the rate of energy radiation is $E = e A \sigma T^4$. Since the surface areas $A$ and the rates of energy radiation $E$ are equal for both bodies,we have $e_A T_A^4 = e_B T_B^4$.
Given $e_A = 0.01$,$e_B = 0.81$,and $T_A = 5802 \ K$,we get:
$T_B^4 = \frac{e_A}{e_B} T_A^4 = \frac{0.01}{0.81} (5802)^4 = \frac{1}{81} (5802)^4$.
Taking the fourth root,$T_B = \frac{5802}{3} = 1934 \ K$.
According to Wien's displacement law,$\lambda_A T_A = \lambda_B T_B = b$ (constant).
Thus,$\lambda_A = \lambda_B \frac{T_B}{T_A} = \lambda_B \frac{1934}{5802} = \frac{\lambda_B}{3}$.
Given the difference $\lambda_B - \lambda_A = 1 \mu m$,we substitute $\lambda_A$:
$\lambda_B - \frac{\lambda_B}{3} = 1 \mu m \Rightarrow \frac{2}{3} \lambda_B = 1 \mu m$.
Therefore,$\lambda_B = \frac{3}{2} \mu m$.

(C) The relationship between any temperature scale $X$ and the Kelvin scale $K$ is given by the formula: $\frac{X - X_{freezing}}{X_{boiling} - X_{freezing}} = \frac{K - 273}{373 - 273}$.
Given for scale $Y$: $Y_{freezing} = -160^{\circ} Y$ and $Y_{boiling} = -50^{\circ} Y$.
Substituting these values into the formula for $K = 340 \ K$:
$\frac{Y - (-160)}{-50 - (-160)} = \frac{340 - 273}{373 - 273}$
$\frac{Y + 160}{110} = \frac{67}{100}$
$Y + 160 = \frac{67 \times 110}{100}$
$Y + 160 = 73.7$
$Y = 73.7 - 160 = -86.3^{\circ} Y$.

(C) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature.
Given $\eta_1 = 40 \% = 0.4$ and $T_2 = 300 ~K$:
$0.4 = 1 - \frac{300}{T_1} \Rightarrow \frac{300}{T_1} = 0.6 \Rightarrow T_1 = \frac{300}{0.6} = 500 ~K$.
Now,to increase the efficiency to $\eta_2 = 60 \% = 0.6$ with the same sink temperature $T_2 = 300 ~K$:
$0.6 = 1 - \frac{300}{T_1^{\prime}} \Rightarrow \frac{300}{T_1^{\prime}} = 0.4 \Rightarrow T_1^{\prime} = \frac{300}{0.4} = 750 ~K$.
The increase in source temperature is $\Delta T = T_1^{\prime} - T_1 = 750 ~K - 500 ~K = 250 ~K$.

(A) The work done by the gas in a cyclic process on a $P-V$ diagram is equal to the area enclosed by the cycle.
For the given process $1-2-3-4-1$, the area is the area of the trapezoid formed by the points $(P_2, V_2), (P_2, V_3), (P_1, V_4), (P_1, V_1)$.
Since the lines $1-2$ and $3-4$ pass through the origin, they represent processes where $P \propto V$, meaning $P/V = \text{constant}$.
Using the ideal gas equation $PV = nRT$, we have $P = (nR/V)T$. Since $P/V$ is constant, $T/V^2$ is constant.
However, it is simpler to calculate the work done as the sum of work in individual processes:
$W_{1-2} = 0$ (since it is a line through the origin, $P/V = k$, so $W = \int P dV = \int kV dV = \frac{k}{2}(V_2^2 - V_1^2) = \frac{1}{2}(P_2 V_2 - P_1 V_1) = \frac{nR}{2}(T_2 - T_1)$).
$W_{2-3} = P_2(V_3 - V_2) = nR(T_3 - T_2)$.
$W_{3-4} = 0$ (similar to $1-2$, $W = \frac{nR}{2}(T_4 - T_3)$).
$W_{4-1} = P_1(V_1 - V_4) = nR(T_1 - T_4)$.
Total Work $W = W_{1-2} + W_{2-3} + W_{3-4} + W_{4-1} = \frac{nR}{2}(T_2 - T_1) + nR(T_3 - T_2) + \frac{nR}{2}(T_4 - T_3) + nR(T_1 - T_4)$.
$W = nR [ \frac{1}{2}T_2 - \frac{1}{2}T_1 + T_3 - T_2 + \frac{1}{2}T_4 - \frac{1}{2}T_3 + T_1 - T_4 ]$.
$W = nR [ \frac{1}{2}T_1 - \frac{1}{2}T_2 + \frac{1}{2}T_3 - \frac{1}{2}T_4 ] = \frac{nR}{2} (T_1 - T_2 + T_3 - T_4)$.
Given $n = 3, T_1 = 400 \ K, T_2 = 700 \ K, T_3 = 2500 \ K, T_4 = 1100 \ K$.
$W = \frac{3R}{2} (400 - 700 + 2500 - 1100) = \frac{3R}{2} (1100) = 1650 R$.