TS EAMCET 2005 Chemistry Question Paper with Answer and Solution

(B) Given that $b = \sum_{k=1}^{\infty} \frac{a^k}{k}$.
We know the logarithmic series expansion: $-\ln(1-a) = \sum_{k=1}^{\infty} \frac{a^k}{k}$ for $|a| < 1$.
Therefore,$b = -\ln(1-a)$.
This implies $e^{-b} = 1-a$,so $a = 1 - e^{-b}$.
Using the Taylor series expansion for $e^{-b} = \sum_{k=0}^{\infty} \frac{(-b)^k}{k!} = 1 - b + \frac{b^2}{2!} - \frac{b^3}{3!} + \dots$.
Substituting this into the expression for $a$:
$a = 1 - (1 - b + \frac{b^2}{2!} - \frac{b^3}{3!} + \dots) = b - \frac{b^2}{2!} + \frac{b^3}{3!} - \dots$.
This can be written as $a = \sum_{k=1}^{\infty} \frac{(-1)^{k-1} b^k}{k!}$.

(A) The given line is $5x - 2y = 7$. The slope of this line is $m_1 = \frac{5}{2}$.
Any line perpendicular to this line will have a slope $m_2 = -\frac{1}{m_1} = -\frac{2}{5}$.
Thus,the equation of the required line is of the form $2x + 5y = \lambda$ $(i)$.
Now,we find the point of intersection of the lines $2x + 3y = 1$ (ii) and $3x + 4y = 6$ (iii).
Multiplying (ii) by $3$ and (iii) by $2$,we get:
$6x + 9y = 3$
$6x + 8y = 12$
Subtracting the two equations,we get $y = -9$.
Substituting $y = -9$ into (ii): $2x + 3(-9) = 1$ $\Rightarrow 2x - 27 = 1$ $\Rightarrow 2x = 28$ $\Rightarrow x = 14$.
The point of intersection is $(14, -9)$.
Since the line $(i)$ passes through $(14, -9)$,we have:
$2(14) + 5(-9) = \lambda$
$28 - 45 = \lambda$
$\lambda = -17$.
Substituting $\lambda = -17$ into $(i)$,we get $2x + 5y = -17$,which is $2x + 5y + 17 = 0$.

(C) Let the coordinates of $M$ be $(x_1, y_1)$.
Since $PM$ is perpendicular to the line $x+y=3$,the slope of $PM$ is the negative reciprocal of the slope of the given line.
The slope of $x+y=3$ is $-1$,so the slope of $PM$ is $1$.
Thus,$\frac{y_1-3}{x_1-2} = 1$ $\Rightarrow y_1-3 = x_1-2$ $\Rightarrow x_1-y_1 = -1$ (Eq. $i$).
Since $M(x_1, y_1)$ lies on the line $x+y=3$,we have $x_1+y_1=3$ (Eq. $ii$).
Adding Eq. $i$ and Eq. $ii$: $(x_1-y_1) + (x_1+y_1) = -1+3$ $\Rightarrow 2x_1 = 2$ $\Rightarrow x_1 = 1$.
Substituting $x_1=1$ into Eq. $ii$: $1+y_1=3 \Rightarrow y_1=2$.
Therefore,the coordinates of $M$ are $(1, 2)$.

(C) Let the coordinates of $P$ be $(x, y)$. According to the given condition,the distance from $P(x, y)$ to $A(1, 1)$ equals the distance from $P(x, y)$ to the line $x+y+2=0$.
$\sqrt{(x-1)^2+(y-1)^2} = \frac{|x+y+2|}{\sqrt{1^2+1^2}}$
Squaring both sides,we get:
$(x-1)^2+(y-1)^2 = \frac{(x+y+2)^2}{2}$
$2(x^2-2x+1+y^2-2y+1) = x^2+y^2+4+2xy+4x+4y$
$2x^2+2y^2-4x-4y+4 = x^2+y^2+2xy+4x+4y+4$
$x^2+y^2-2xy-8x-8y = 0$
This equation is of the form $ax^2+2hxy+by^2+2gx+2fy+c=0$ where $h^2-ab = (-1)^2 - (1)(1) = 0$. Since $h^2=ab$ and the point $A$ does not lie on the line $x+y+2=0$,the locus represents a parabola.

(B) Given the polar equation $\theta = \tan^{-1} 2$.
Taking tangent on both sides,we get $\tan \theta = 2$.
We know that in Cartesian coordinates,$\tan \theta = \frac{y}{x}$.
Substituting this into the equation,we have $\frac{y}{x} = 2$.
Therefore,$y = 2x$.

(B) Given the equation of the pair of straight lines: $12x^2+25xy+12y^2+10x+11y+2=0$ ... $(i)$
First,we consider the homogeneous part of Eq. $(i)$:
$12x^2+25xy+12y^2 = 0$
$\Rightarrow (3x+4y)(4x+3y) = 0$
Let the lines be represented by $(3x+4y+c_1)(4x+3y+c_2) = 0$.
Expanding this,we get:
$12x^2+25xy+12y^2+(4c_1+3c_2)x+(3c_1+4c_2)y+c_1c_2 = 0$
Comparing this with Eq. $(i)$:
$4c_1+3c_2 = 10$ ... (ii)
$3c_1+4c_2 = 11$ ... (iii)
$c_1c_2 = 2$ ... (iv)
Solving (ii) and (iii):
Adding (ii) and (iii): $7(c_1+c_2) = 21 \Rightarrow c_1+c_2 = 3$
Subtracting (iii) from (ii): $c_1-c_2 = -1$
Adding these: $2c_1 = 2 \Rightarrow c_1 = 1$. Then $c_2 = 2$.
Check: $c_1c_2 = 1 \times 2 = 2$,which satisfies (iv).
The lines are $3x+4y+1=0$ and $4x+3y+2=0$.
The perpendicular distances from the origin $(0,0)$ to these lines are:
$p_1 = \frac{|0+0+1|}{\sqrt{3^2+4^2}} = \frac{1}{5}$
$p_2 = \frac{|0+0+2|}{\sqrt{4^2+3^2}} = \frac{2}{5}$
The product of the distances is $p_1 \cdot p_2 = \frac{1}{5} \cdot \frac{2}{5} = \frac{2}{25}$.

(C) Let the given circles be $S_1 \equiv x^2+y^2+2x+3y+2=0$ and $S_2 \equiv x^2+y^2+2x-3y-4=0$.
The equation of the common chord is given by $S_1 - S_2 = 0$.
$(x^2+y^2+2x+3y+2) - (x^2+y^2+2x-3y-4) = 0$
$6y + 6 = 0 \Rightarrow y = -1$.
Substituting $y = -1$ into the equation of $S_1$:
$x^2 + (-1)^2 + 2x + 3(-1) + 2 = 0$
$x^2 + 1 + 2x - 3 + 2 = 0$
$x^2 + 2x = 0$ $\Rightarrow x(x+2) = 0$ $\Rightarrow x = 0, -2$.
The endpoints of the diameter are $(0, -1)$ and $(-2, -1)$.
The equation of the circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.
$(x-0)(x-(-2)) + (y-(-1))(y-(-1)) = 0$
$x(x+2) + (y+1)^2 = 0$
$x^2 + 2x + y^2 + 2y + 1 = 0$
$x^2 + y^2 + 2x + 2y + 1 = 0$.

(A) Given the equations:
$x-y+1=0 \implies y=x+1$ $(i)$
$x^2+y^2+y-1=0$ (ii)
Substituting $(i)$ into (ii):
$x^2+(x+1)^2+(x+1)-1=0$
$x^2+x^2+2x+1+x=0$
$2x^2+3x+1=0$
$(2x+1)(x+1)=0$
So,$x=-\frac{1}{2}$ or $x=-1$.
For $x=-\frac{1}{2}$,$y=-\frac{1}{2}+1=\frac{1}{2}$. Point $A$ is $(-\frac{1}{2}, \frac{1}{2})$.
For $x=-1$,$y=-1+1=0$. Point $B$ is $(-1, 0)$.
The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$.
$(x+\frac{1}{2})(x+1)+(y-\frac{1}{2})(y-0)=0$
$(x+\frac{1}{2})(x+1)+y(y-\frac{1}{2})=0$
Multiply by $2$:
$(2x+1)(x+1)+2y(y-\frac{1}{2})=0$
$2x^2+2x+x+1+2y^2-y=0$
$2(x^2+y^2)+3x-y+1=0$.

(A) The line $y-3x=0$ is a tangent to the circle. The radius $r$ is the perpendicular distance from the centre $(1,1)$ to the line $3x-y=0$.
$r = \frac{|3(1) - 1(1)|}{\sqrt{3^2 + (-1)^2}} = \frac{|3-1|}{\sqrt{9+1}} = \frac{2}{\sqrt{10}}$.
Let the other tangent passing through the origin $(0,0)$ be $y=mx$,or $mx-y=0$.
The perpendicular distance from the centre $(1,1)$ to this line must also be equal to the radius $r$.
$\frac{|m(1) - 1(1)|}{\sqrt{m^2 + (-1)^2}} = \frac{2}{\sqrt{10}}$
$\frac{|m-1|}{\sqrt{m^2+1}} = \frac{2}{\sqrt{10}}$
Squaring both sides:
$\frac{(m-1)^2}{m^2+1} = \frac{4}{10} = \frac{2}{5}$
$5(m^2 - 2m + 1) = 2(m^2 + 1)$
$5m^2 - 10m + 5 = 2m^2 + 2$
$3m^2 - 10m + 3 = 0$
$3m^2 - 9m - m + 3 = 0$
$3m(m-3) - 1(m-3) = 0$
$(3m-1)(m-3) = 0$
Thus,$m=3$ or $m=\frac{1}{3}$.
The case $m=3$ corresponds to the given tangent $y=3x$. The other tangent is $y=\frac{1}{3}x$,which is $3y=x$.

(A) To determine which equation represents a circle,we convert the polar equations to Cartesian coordinates using $x = r \cos \theta$ and $y = r \sin \theta$,where $r^2 = x^2 + y^2$.
For option $A$: $r = 2 \sin \theta$
Multiply both sides by $r$: $r^2 = 2r \sin \theta$
Substituting $r^2 = x^2 + y^2$ and $y = r \sin \theta$:
$x^2 + y^2 = 2y$
$x^2 + y^2 - 2y = 0$
This is the equation of a circle with center $(0, 1)$ and radius $1$.
For option $B$: $r^2 \cos 2 \theta = 1$
Using $\cos 2 \theta = \cos^2 \theta - \sin^2 \theta$:
$r^2(\cos^2 \theta - \sin^2 \theta) = 1$
$(r \cos \theta)^2 - (r \sin \theta)^2 = 1$
$x^2 - y^2 = 1$,which is a rectangular hyperbola.
For option $C$: $r(4 \cos \theta + 5 \sin \theta) = 3$
$4r \cos \theta + 5r \sin \theta = 3$
$4x + 5y = 3$,which is a straight line.
For option $D$: $5 = r(1 + \sqrt{2} \cos \theta)$
This is the polar form of a conic section $r = \frac{ed}{1 + e \cos \theta}$,which represents a parabola,ellipse,or hyperbola depending on eccentricity $e$.

(B) The given equation of the parabola is $y^2=4ax$.
Let the equation of the line be $y=mx+c$.
If this line touches the parabola,the condition for tangency is $c = \frac{a}{m}$.
Substituting this into the line equation,we get $y = mx + \frac{a}{m}$.
Multiplying by $m$,we get $my = m^2x + a$.
Rearranging the terms,we get $m^2x - my + a = 0$.
Replacing $m$ with $\frac{1}{m}$ in the standard form $y = mx + \frac{a}{m}$,we get $y = \frac{1}{m}x + am$.
Multiplying by $m$,we get $my = x + am^2$,which simplifies to $x - my + am^2 = 0$.

(A) Let $P(x, y)$ be any point on the parabola.
By the definition of a parabola,the distance from $P$ to the focus $S(1, 0)$ is equal to the perpendicular distance from $P$ to the directrix $x+2y-1=0$.
$PS = PM$
$\sqrt{(x-1)^2 + (y-0)^2} = \frac{|x+2y-1|}{\sqrt{1^2+2^2}}$
Squaring both sides,we get:
$(x-1)^2 + y^2 = \frac{(x+2y-1)^2}{5}$
$5(x^2 - 2x + 1 + y^2) = x^2 + 4y^2 + 1 + 4xy - 2x - 4y$
$5x^2 - 10x + 5 + 5y^2 = x^2 + 4y^2 + 4xy - 2x - 4y + 1$
$4x^2 - 4xy + y^2 - 8x + 4y + 4 = 0$

(B) We have the multinomial expansion formula:
$(xy+yz+zx)^6 = \sum_{r+s+t=6} \frac{6!}{r!s!t!} (xy)^r (yz)^s (zx)^t$
$= \sum_{r+s+t=6} \frac{6!}{r!s!t!} x^{r+t} y^{r+s} z^{s+t}$
For the term $x^3 y^4 z^5$,we equate the exponents:
$r+t = 3$
$r+s = 4$
$s+t = 5$
Adding these equations: $2(r+s+t) = 3+4+5 = 12$,so $r+s+t = 6$.
Subtracting each equation from the sum:
$s = (r+s+t) - (r+t) = 6 - 3 = 3$
$t = (r+s+t) - (r+s) = 6 - 4 = 2$
$r = (r+s+t) - (s+t) = 6 - 5 = 1$
The coefficient is $\frac{6!}{1!3!2!} = \frac{720}{1 \times 6 \times 2} = \frac{720}{12} = 60$.

(D) We have the expression $\frac{1+2x}{(1-2x)^2} = (1+2x)(1-2x)^{-2}$.
Using the binomial expansion for negative indices,$(1-y)^{-n} = \sum_{k=0}^{\infty} \binom{n+k-1}{k} y^k$.
For $(1-2x)^{-2}$,we have $n=2$ and $y=2x$,so $(1-2x)^{-2} = \sum_{k=0}^{\infty} \binom{k+1}{1} (2x)^k = \sum_{k=0}^{\infty} (k+1) 2^k x^k$.
Now,multiply by $(1+2x)$:
$(1+2x) \sum_{k=0}^{\infty} (k+1) 2^k x^k = \sum_{k=0}^{\infty} (k+1) 2^k x^k + \sum_{k=0}^{\infty} 2(k+1) 2^k x^{k+1}$.
To find the coefficient of $x^r$,we take the term where $k=r$ from the first sum and the term where $k+1=r$ (i.e.,$k=r-1$) from the second sum:
Coefficient of $x^r = (r+1) 2^r + 2((r-1)+1) 2^{r-1}$.
$= (r+1) 2^r + 2(r) 2^{r-1} = (r+1) 2^r + r 2^r$.
$= (r+1+r) 2^r = (2r+1) 2^r$.

(C) Given that $(1+x)^{15} = \sum_{r=0}^{15} {}^{15}C_r x^r = a_0 + a_1 x + \ldots + a_{15} x^{15}$.
Comparing coefficients,we have $a_r = {}^{15}C_r$.
We need to evaluate $\sum_{r=1}^{15} r \frac{a_r}{a_{r-1}}$.
Using the property $\frac{{}^{n}C_r}{{}^{n}C_{r-1}} = \frac{n-r+1}{r}$,we get $\frac{a_r}{a_{r-1}} = \frac{15-r+1}{r} = \frac{16-r}{r}$.
Thus,$r \frac{a_r}{a_{r-1}} = r \cdot \frac{16-r}{r} = 16-r$.
Therefore,$\sum_{r=1}^{15} (16-r) = (16-1) + (16-2) + \ldots + (16-15) = 15 + 14 + \ldots + 1$.
This is the sum of the first $15$ natural numbers: $\frac{15(15+1)}{2} = \frac{15 \times 16}{2} = 15 \times 8 = 120$.

(B) We use the Squeeze Theorem to evaluate the limit.
We know that for all $x \neq 0$,$-1 \leq \sin \left(\frac{\pi}{x}\right) \leq 1$.
Multiplying the inequality by $x^2$ (since $x^2 > 0$ for $x \neq 0$),we get:
$-x^2 \leq x^2 \sin \left(\frac{\pi}{x}\right) \leq x^2$.
Now,taking the limit as $x \rightarrow 0$ on all sides:
$\lim _{x \rightarrow 0} (-x^2) \leq \lim _{x \rightarrow 0} x^2 \sin \left(\frac{\pi}{x}\right) \leq \lim _{x \rightarrow 0} x^2$.
Since $\lim _{x \rightarrow 0} (-x^2) = 0$ and $\lim _{x \rightarrow 0} x^2 = 0$,by the Squeeze Theorem:
$\lim _{x \rightarrow 0} x^2 \sin \left(\frac{\pi}{x}\right) = 0$.

(B) Given that $\frac{\tan 3A}{\tan A} = a$.
Using the formula $\tan 3A = \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A}$,we have:
$\frac{3\tan A - \tan^3 A}{\tan A(1 - 3\tan^2 A)} = a$
$\Rightarrow \frac{3 - \tan^2 A}{1 - 3\tan^2 A} = a$
$\Rightarrow 3 - \tan^2 A = a - 3a\tan^2 A$
$\Rightarrow \tan^2 A(3a - 1) = a - 3$
$\Rightarrow \tan^2 A = \frac{a - 3}{3a - 1}$
Now,consider $\frac{\sin 3A}{\sin A} = \frac{3\sin A - 4\sin^3 A}{\sin A} = 3 - 4\sin^2 A$.
Using $\sin^2 A = \frac{\tan^2 A}{1 + \tan^2 A}$,we get:
$\sin^2 A = \frac{\frac{a-3}{3a-1}}{1 + \frac{a-3}{3a-1}} = \frac{a-3}{3a-1+a-3} = \frac{a-3}{4a-4} = \frac{a-3}{4(a-1)}$.
Substituting this into the expression:
$\frac{\sin 3A}{\sin A} = 3 - 4\left(\frac{a-3}{4(a-1)}\right) = 3 - \frac{a-3}{a-1}$
$= \frac{3(a-1) - (a-3)}{a-1} = \frac{3a - 3 - a + 3}{a-1} = \frac{2a}{a-1}$.

(D) Given that $A+C=2B$ ... $(i)$
We need to evaluate the expression:
$\frac{\cos C-\cos A}{\sin A-\sin C}$
Using the trigonometric identities:
$\cos C - \cos A = 2 \sin \left(\frac{A+C}{2}\right) \sin \left(\frac{A-C}{2}\right)$
$\sin A - \sin C = 2 \cos \left(\frac{A+C}{2}\right) \sin \left(\frac{A-C}{2}\right)$
Substituting these into the expression:
$\frac{2 \sin \left(\frac{A+C}{2}\right) \sin \left(\frac{A-C}{2}\right)}{2 \cos \left(\frac{A+C}{2}\right) \sin \left(\frac{A-C}{2}\right)}$
Canceling the common terms $2$ and $\sin \left(\frac{A-C}{2}\right)$:
$= \frac{\sin \left(\frac{A+C}{2}\right)}{\cos \left(\frac{A+C}{2}\right)}$
Since $A+C=2B$,we have $\frac{A+C}{2} = B$:
$= \frac{\sin B}{\cos B} = \tan B$

(A) Given that $A+B=C$.
We need to evaluate the expression $E = \cos^2 A + \cos^2 B + \cos^2 C - 2 \cos A \cos B \cos C$.
Using the identity $\cos^2 \theta = \frac{1+\cos 2\theta}{2}$,we have:
$E = \frac{1+\cos 2A}{2} + \frac{1+\cos 2B}{2} + \cos^2 C - 2 \cos A \cos B \cos C$
$E = 1 + \frac{1}{2}(\cos 2A + \cos 2B) + \cos^2 C - 2 \cos A \cos B \cos C$
Using $\cos 2A + \cos 2B = 2 \cos(A+B) \cos(A-B)$:
$E = 1 + \cos(A+B) \cos(A-B) + \cos^2 C - 2 \cos A \cos B \cos C$
Since $A+B=C$,$\cos(A+B) = \cos C$:
$E = 1 + \cos C \cos(A-B) + \cos^2 C - 2 \cos A \cos B \cos C$
$E = 1 + \cos C [\cos(A-B) + \cos C] - 2 \cos A \cos B \cos C$
Since $C = A+B$,$\cos C = \cos(A+B) = \cos A \cos B - \sin A \sin B$:
$E = 1 + \cos C [\cos(A-B) + \cos(A+B)] - 2 \cos A \cos B \cos C$
Using $\cos(A-B) + \cos(A+B) = 2 \cos A \cos B$:
$E = 1 + \cos C [2 \cos A \cos B] - 2 \cos A \cos B \cos C$
$E = 1 + 2 \cos A \cos B \cos C - 2 \cos A \cos B \cos C = 1$.

(A) We use the projection formula: $c \cos C + b \cos B = a$.
Substituting this into the expression,we get:
$a(\cos^2 B + \cos^2 C) + a \cos A$
$= a(\cos^2 B + \cos^2 C + \cos A)$
Since $A + B + C = \pi$,we have $A = \pi - (B + C)$,so $\cos A = -\cos(B + C)$.
The expression becomes $a(\cos^2 B + \cos^2 C - \cos(B + C))$
$= a(\cos^2 B + \cos^2 C - (\cos B \cos C - \sin B \sin C))$
Using the identity $\cos^2 B + \cos^2 C = 1 - \sin^2 B + \cos^2 C = 1 - (\sin^2 B - \cos^2 C) = 1 + \cos(B+C)\cos(B-C)$ is complex,let's simplify directly:
Using $\cos^2 B = \frac{1+\cos 2B}{2}$ and $\cos^2 C = \frac{1+\cos 2C}{2}$:
$a(\frac{1+\cos 2B + 1 + \cos 2C}{2}) + a \cos A$
$= a(1 + \frac{\cos 2B + \cos 2C}{2}) + a \cos A$
$= a(1 + \cos(B+C)\cos(B-C)) + a \cos A$
Since $\cos(B+C) = -\cos A$,we have:
$= a(1 - \cos A \cos(B-C) + \cos A)$
$= a(1 + \cos A(1 - \cos(B-C)))$
Actually,a simpler approach: $a(\cos^2 B + \cos^2 C) + a \cos A = a(\cos^2 B + \cos^2 C + \cos A)$
Using $\cos A = -\cos(B+C)$,$\cos^2 B + \cos^2 C - \cos(B+C) = \cos^2 B + \cos^2 C - (\cos B \cos C - \sin B \sin C) = \cos^2 B + \cos^2 C - \cos B \cos C + \sin B \sin C$.
This simplifies to $a$.

(D) We use the Napier's analogy: $\tan \left(\frac{B-C}{2}\right) = \frac{b-c}{b+c} \cot \frac{A}{2}$.
Substituting this into the expression:
$(b+c) \tan \frac{A}{2} \tan \left(\frac{B-C}{2}\right) = (b+c) \tan \frac{A}{2} \left[ \frac{b-c}{b+c} \cot \frac{A}{2} \right]$.
Since $\tan \frac{A}{2} \cot \frac{A}{2} = 1$,the expression simplifies to $(b-c)$.
Now,summing over the cyclic permutations:
$\Sigma(b-c) = (b-c) + (c-a) + (a-b) = 0$.

(D) The given equation is $x^2-5x+6=0$. \\ Solving for $x$: $(x-3)(x-2)=0$,which gives $x=3$ and $x=2$. \\ Let the two sides be $a=3$ and $b=2$,and the included angle be $C=\frac{\pi}{3}$. \\ Using the Law of Cosines: $\cos C = \frac{a^2+b^2-c^2}{2ab}$. \\ Substituting the values: $\cos(\frac{\pi}{3}) = \frac{3^2+2^2-c^2}{2 \times 3 \times 2}$. \\ $\frac{1}{2} = \frac{9+4-c^2}{12} \Rightarrow \frac{1}{2} = \frac{13-c^2}{12}$. \\ $6 = 13-c^2$ $\Rightarrow c^2 = 7$ $\Rightarrow c = \sqrt{7}$. \\ The perimeter of the triangle is $a+b+c = 3+2+\sqrt{7} = 5+\sqrt{7}$.

(D) Given that,$m \begin{bmatrix} -3 & 4 \end{bmatrix} + n \begin{bmatrix} 4 & -3 \end{bmatrix} = \begin{bmatrix} 10 & -11 \end{bmatrix}$
Multiplying the scalars $m$ and $n$ into the matrices,we get:
$\begin{bmatrix} -3m & 4m \end{bmatrix} + \begin{bmatrix} 4n & -3n \end{bmatrix} = \begin{bmatrix} 10 & -11 \end{bmatrix}$
Adding the matrices:
$\begin{bmatrix} -3m + 4n & 4m - 3n \end{bmatrix} = \begin{bmatrix} 10 & -11 \end{bmatrix}$
By equating the corresponding elements,we get the following system of linear equations:
$-3m + 4n = 10 \quad \dots (i)$
$4m - 3n = -11 \quad \dots (ii)$
To solve this system,multiply equation $(i)$ by $3$ and equation $(ii)$ by $4$:
$-9m + 12n = 30 \quad \dots (iii)$
$16m - 12n = -44 \quad \dots (iv)$
Adding equations $(iii)$ and $(iv)$:
$7m = -14 \Rightarrow m = -2$
Substituting $m = -2$ into equation $(i)$:
$-3(-2) + 4n = 10 \Rightarrow 6 + 4n = 10 \Rightarrow 4n = 4 \Rightarrow n = 1$
Now,calculate $3m + 7n$:
$3m + 7n = 3(-2) + 7(1) = -6 + 7 = 1$

(A) Given that,$A = \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}$.
First,we calculate $A^2$:
$A^2 = A \cdot A = \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} (-1)(-1) + (0)(0) & (-1)(0) + (0)(2) \\ (0)(-1) + (2)(0) & (0)(0) + (2)(2) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix}$.
Next,we calculate $A^3$:
$A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} (1)(-1) + (0)(0) & (1)(0) + (0)(2) \\ (0)(-1) + (4)(0) & (0)(0) + (4)(2) \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & 8 \end{bmatrix}$.
Now,calculate $A^3 - A^2$:
$A^3 - A^2 = \begin{bmatrix} -1 & 0 \\ 0 & 8 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} -1 - 1 & 0 - 0 \\ 0 - 0 & 8 - 4 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 0 & 4 \end{bmatrix}$.
We observe that $2A = 2 \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 0 & 4 \end{bmatrix}$.
Therefore,$A^3 - A^2 = 2A$.

(C) Let $A = \begin{bmatrix} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{bmatrix}$. The adjoint of a matrix is the transpose of the cofactor matrix,i.e.,$\operatorname{adj}(A) = [C_{ij}]^T$.
The cofactors $C_{ij}$ are calculated as follows:
$C_{11} = + \begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} = 1 - (-4) = 5$
$C_{12} = - \begin{vmatrix} -1 & -2 \\ 0 & 1 \end{vmatrix} = -(-1 - 0) = 1$
$C_{13} = + \begin{vmatrix} -1 & 1 \\ 0 & 2 \end{vmatrix} = -2 - 0 = -2$
$C_{21} = - \begin{vmatrix} 0 & 2 \\ 2 & 1 \end{vmatrix} = -(0 - 4) = 4$
$C_{22} = + \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = 1 - 0 = 1$
$C_{23} = - \begin{vmatrix} 1 & 0 \\ 0 & 2 \end{vmatrix} = -(2 - 0) = -2$
$C_{31} = + \begin{vmatrix} 0 & 2 \\ 1 & -2 \end{vmatrix} = 0 - 2 = -2$
$C_{32} = - \begin{vmatrix} 1 & 2 \\ -1 & -2 \end{vmatrix} = -(-2 - (-2)) = 0$
$C_{33} = + \begin{vmatrix} 1 & 0 \\ -1 & 1 \end{vmatrix} = 1 - 0 = 1$
The cofactor matrix is $\begin{bmatrix} 5 & 1 & -2 \\ 4 & 1 & -2 \\ -2 & 0 & 1 \end{bmatrix}$.
Taking the transpose,we get $\operatorname{adj}(A) = \begin{bmatrix} 5 & 4 & -2 \\ 1 & 1 & 0 \\ -2 & -2 & 1 \end{bmatrix}$.
Comparing this with the given matrix $\begin{bmatrix} 5 & a & -2 \\ 1 & 1 & 0 \\ -2 & -2 & b \end{bmatrix}$,we find $a = 4$ and $b = 1$.
Therefore,$\begin{bmatrix} a & b \end{bmatrix} = \begin{bmatrix} 4 & 1 \end{bmatrix}$.

(C) We use the identity $2 \tanh^{-1} x = \tanh^{-1} \left( \frac{2x}{1+x^2} \right)$.
Substituting $x = \frac{1}{2}$,we get:
$2 \tanh^{-1} \left( \frac{1}{2} \right) = \tanh^{-1} \left( \frac{2 \cdot \frac{1}{2}}{1 + (\frac{1}{2})^2} \right) = \tanh^{-1} \left( \frac{1}{1 + \frac{1}{4}} \right) = \tanh^{-1} \left( \frac{1}{\frac{5}{4}} \right) = \tanh^{-1} \left( \frac{4}{5} \right)$.
Now,we use the logarithmic form $\tanh^{-1} x = \frac{1}{2} \log \left( \frac{1+x}{1-x} \right)$ for $|x| < 1$.
Substituting $x = \frac{4}{5}$:
$\tanh^{-1} \left( \frac{4}{5} \right) = \frac{1}{2} \log \left( \frac{1 + \frac{4}{5}}{1 - \frac{4}{5}} \right) = \frac{1}{2} \log \left( \frac{\frac{9}{5}}{\frac{1}{5}} \right) = \frac{1}{2} \log 9 = \frac{1}{2} \log 3^2 = \log 3$.

(C) We are given the expression: $\sin ^{-1} \frac{4}{5} + 2 \tan ^{-1} \frac{1}{3}$.
First,use the formula $2 \tan ^{-1} x = \tan ^{-1} \frac{2x}{1-x^2}$:
$2 \tan ^{-1} \frac{1}{3} = \tan ^{-1} \frac{2(1/3)}{1-(1/3)^2} = \tan ^{-1} \frac{2/3}{1-1/9} = \tan ^{-1} \frac{2/3}{8/9} = \tan ^{-1} \left( \frac{2}{3} \times \frac{9}{8} \right) = \tan ^{-1} \frac{3}{4}$.
Now,the expression becomes $\sin ^{-1} \frac{4}{5} + \tan ^{-1} \frac{3}{4}$.
Since $\tan ^{-1} \frac{3}{4} = \theta$,then $\tan \theta = \frac{3}{4}$,which implies $\sin \theta = \frac{3}{5}$ and $\cos \theta = \frac{4}{5}$.
Thus,$\tan ^{-1} \frac{3}{4} = \cos ^{-1} \frac{4}{5}$.
Substituting this back,we get $\sin ^{-1} \frac{4}{5} + \cos ^{-1} \frac{4}{5}$.
Using the identity $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,we get $\frac{\pi}{2}$.

(B) Consider the expression $f(x) = x - |x|$.
Case $1$: If $x \geq 0$,then $|x| = x$. Thus,$f(x) = x - x = 0$.
Case $2$: If $x < 0$,then $|x| = -x$. Thus,$f(x) = x - (-x) = 2x$.
Since $x < 0$,$2x < 0$.
Therefore,the value of $x - |x|$ is either $0$ (for $x \geq 0$) or a negative number (for $x < 0$).
It can never be equal to $5$.
Hence,the set $\{x \in R : [x - |x|] = 5\}$ is the empty set,denoted by $\phi$.

(A) Given $x = \frac{1}{2} \left( \sqrt{3} + \frac{1}{\sqrt{3}} \right) = \frac{1}{2} \left( \frac{3+1}{\sqrt{3}} \right) = \frac{4}{2\sqrt{3}} = \frac{2}{\sqrt{3}}$.
Then $x^2 = \left( \frac{2}{\sqrt{3}} \right)^2 = \frac{4}{3}$.
Now,$x^2 - 1 = \frac{4}{3} - 1 = \frac{1}{3}$.
So,$\sqrt{x^2 - 1} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}$.
Substituting these values into the expression $\frac{\sqrt{x^2 - 1}}{x - \sqrt{x^2 - 1}}$:
$= \frac{\frac{1}{\sqrt{3}}}{\frac{2}{\sqrt{3}} - \frac{1}{\sqrt{3}}} = \frac{\frac{1}{\sqrt{3}}}{\frac{1}{\sqrt{3}}} = 1$.

(A) Given the expression: $\frac{x^3}{(2 x-1)(x+2)(x-3)} = A+\frac{B}{2 x-1}+\frac{C}{x+2}+\frac{D}{x-3}$.
First,expand the denominator: $(2x-1)(x+2)(x-3) = (2x-1)(x^2-x-6) = 2x^3 - 2x^2 - 12x - x^2 + x + 6 = 2x^3 - 3x^2 - 11x + 6$.
Since the degree of the numerator $(3)$ is equal to the degree of the denominator $(3)$,we perform long division to find the constant $A$.
Dividing $x^3$ by $2x^3 - 3x^2 - 11x + 6$:
$\frac{x^3}{2x^3 - 3x^2 - 11x + 6} = \frac{1}{2} \left( \frac{2x^3}{2x^3 - 3x^2 - 11x + 6} \right) = \frac{1}{2} \left( \frac{2x^3 - 3x^2 - 11x + 6 + 3x^2 + 11x - 6}{2x^3 - 3x^2 - 11x + 6} \right) = \frac{1}{2} + \frac{\frac{3}{2}x^2 + \frac{11}{2}x - 3}{(2x-1)(x+2)(x-3)}$.
Comparing this with the given form $A+\frac{B}{2 x-1}+\frac{C}{x+2}+\frac{D}{x-3}$,we get $A = \frac{1}{2}$.

(C) Given that $f(x)$ is an even function,by definition,$f(x) = f(-x)$ for all $x \in R$.
Differentiating both sides with respect to $x$,we get $f^{\prime}(x) = -f^{\prime}(-x)$.
Differentiating again with respect to $x$,we get $f^{\prime \prime}(x) = f^{\prime \prime}(-x)$.
This shows that the second derivative of an even function is also an even function.
Since $f^{\prime \prime}(x)$ is an even function,$f^{\prime \prime}(-\pi) = f^{\prime \prime}(\pi)$.
Given $f^{\prime \prime}(\pi) = 1$,it follows that $f^{\prime \prime}(-\pi) = 1$.

(A) Given $u = \sin^{-1}\left(\frac{x}{y}\right) + \tan^{-1}\left(\frac{y}{x}\right)$.
First,we find the partial derivative with respect to $x$:
$\frac{\partial u}{\partial x} = \frac{1}{\sqrt{1 - (x/y)^2}} \cdot \frac{1}{y} + \frac{1}{1 + (y/x)^2} \cdot \left(-\frac{y}{x^2}\right)$
$= \frac{1}{\sqrt{(y^2 - x^2)/y^2}} \cdot \frac{1}{y} - \frac{x^2}{x^2 + y^2} \cdot \frac{y}{x^2}$
$= \frac{y}{\sqrt{y^2 - x^2}} \cdot \frac{1}{y} - \frac{y}{x^2 + y^2} = \frac{1}{\sqrt{y^2 - x^2}} - \frac{y}{x^2 + y^2}$
Multiplying by $x$,we get:
$x \frac{\partial u}{\partial x} = \frac{x}{\sqrt{y^2 - x^2}} - \frac{xy}{x^2 + y^2} \quad \dots(i)$
Next,we find the partial derivative with respect to $y$:
$\frac{\partial u}{\partial y} = \frac{1}{\sqrt{1 - (x/y)^2}} \cdot \left(-\frac{x}{y^2}\right) + \frac{1}{1 + (y/x)^2} \cdot \left(\frac{1}{x}\right)$
$= \frac{y}{\sqrt{y^2 - x^2}} \cdot \left(-\frac{x}{y^2}\right) + \frac{x^2}{x^2 + y^2} \cdot \frac{1}{x}$
$= -\frac{x}{y \sqrt{y^2 - x^2}} + \frac{x}{x^2 + y^2}$
Multiplying by $y$,we get:
$y \frac{\partial u}{\partial y} = -\frac{x}{\sqrt{y^2 - x^2}} + \frac{xy}{x^2 + y^2} \quad \dots(ii)$
Adding equations $(i)$ and $(ii)$:
$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \left(\frac{x}{\sqrt{y^2 - x^2}} - \frac{xy}{x^2 + y^2}\right) + \left(-\frac{x}{\sqrt{y^2 - x^2}} + \frac{xy}{x^2 + y^2}\right) = 0$.

(A) For statement $I$:
$f(x) = a x^{41} + b x^{-40}$
$f^{\prime}(x) = 41 a x^{40} - 40 b x^{-41}$
$f^{\prime \prime}(x) = 41 \times 40 a x^{39} + 40 \times 41 b x^{-42} = 1640 a x^{39} + 1640 b x^{-42}$
$\frac{f^{\prime \prime}(x)}{f(x)} = \frac{1640(a x^{39} + b x^{-42})}{a x^{41} + b x^{-40}} = \frac{1640 x^{-2}(a x^{41} + b x^{-40})}{a x^{41} + b x^{-40}} = 1640 x^{-2}$
Thus,statement $I$ is true.
For statement $II$:
Let $y = \tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$. For $|x| < 1$,we substitute $x = \tan \theta$,so $y = \tan ^{-1}(\tan 2 \theta) = 2 \theta = 2 \tan ^{-1} x$.
Then $\frac{d y}{d x} = \frac{2}{1+x^2}$.
Since the given expression is $\frac{1}{1+x^2}$,statement $II$ is false.

(B) Let $BC = x$ be the height of the tower and $CD = h$ be the height of the flagstaff. Let the point be $A$ at a distance $y$ from the foot $B$.
In $\triangle ABC$,$\tan \theta = \frac{BC}{AB} = \frac{x}{y}$.
In $\triangle ABD$,the total angle is $2\theta$,so $\tan 2\theta = \frac{BD}{AB} = \frac{x+h}{y}$.
Using the formula $\tan 2\theta = \frac{2\tan \theta}{1-\tan^2 \theta}$,we have:
$\frac{2(x/y)}{1-(x/y)^2} = \frac{x+h}{y}$
$\frac{2x/y}{(y^2-x^2)/y^2} = \frac{x+h}{y}$
$\frac{2xy}{y^2-x^2} = \frac{x+h}{y}$
$2xy^2 = (y^2-x^2)(x+h)$
$h = \frac{2xy^2}{y^2-x^2} - x = \frac{2xy^2 - x(y^2-x^2)}{y^2-x^2} = \frac{2xy^2 - xy^2 + x^3}{y^2-x^2} = \frac{xy^2 + x^3}{y^2-x^2} = \frac{x(x^2+y^2)}{y^2-x^2}$.
Note: The option $C$ is $\frac{x(x^2+y^2)}{x^2-y^2}$,which is the negative of our result. Given the standard form of such problems,the correct length is $\frac{x(x^2+y^2)}{y^2-x^2}$.

(A) Let $I = \int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x$.
Substitute $x = \tan \theta$,then $d x = \sec ^2 \theta d \theta$.
The integral becomes $I = \int \sin ^{-1}(\sin 2 \theta) \cdot \sec ^2 \theta d \theta$.
Assuming $|x| \le 1$,we have $I = 2 \int \theta \sec ^2 \theta d \theta$.
Using integration by parts: $I = 2 \left[ \theta \tan \theta - \int \tan \theta d \theta \right]$.
$I = 2 [\theta \tan \theta + \log |\cos \theta|] + C$.
Since $\tan \theta = x$,we have $\cos \theta = \frac{1}{\sqrt{1+x^2}}$,so $\log |\cos \theta| = \log (1+x^2)^{-1/2} = -\frac{1}{2} \log (1+x^2)$.
Substituting back: $I = 2 [x \tan ^{-1} x - \frac{1}{2} \log (1+x^2)] + C$.
$I = 2 x \tan ^{-1} x - \log (1+x^2) + C$.
Comparing this with $f(x) - \log (1+x^2)$,we get $f(x) = 2 x \tan ^{-1} x$.

(A) Let $I = \int \frac{\sin x}{\cos x(1+\cos x)} dx$.
Substitute $\cos x = t$,then $-\sin x dx = dt$,which implies $\sin x dx = -dt$.
Substituting these into the integral,we get:
$I = \int \frac{-dt}{t(1+t)}$.
Using partial fractions,$\frac{1}{t(1+t)} = \frac{1}{t} - \frac{1}{1+t}$.
Therefore,$I = -\int \left( \frac{1}{t} - \frac{1}{1+t} \right) dt$.
$I = -(\log |t| - \log |1+t|) + c$.
$I = \log |1+t| - \log |t| + c = \log \left| \frac{1+t}{t} \right| + c$.
Substituting $t = \cos x$ back,we get $I = \log \left| \frac{1+\cos x}{\cos x} \right| + c$.
Since $I = f(x) + c$,we have $f(x) = \log \left| \frac{1+\cos x}{\cos x} \right|$.

(C) Let $I = \int \frac{x^{49} \tan ^{-1}\left(x^{50}\right)}{1+x^{100}} d x$.
Substitute $t = x^{50}$,then $dt = 50x^{49} dx$,which implies $x^{49} dx = \frac{1}{50} dt$.
The integral becomes $I = \frac{1}{50} \int \frac{\tan ^{-1} t}{1+t^2} dt$.
Now,substitute $u = \tan ^{-1} t$,then $du = \frac{1}{1+t^2} dt$.
Substituting this into the integral,we get $I = \frac{1}{50} \int u du = \frac{1}{50} \cdot \frac{u^2}{2} + c = \frac{u^2}{100} + c$.
Replacing $u$ back,we get $I = \frac{(\tan ^{-1} x^{50})^2}{100} + c$.
Comparing this with the given expression $k(\tan ^{-1} x^{50})^2 + c$,we find $k = \frac{1}{100}$.

(D) Let $I = \int_0^\pi \frac{\theta \sin \theta}{1+\cos ^2 \theta} d \theta$ ... $(i)$
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:
$I = \int_0^\pi \frac{(\pi-\theta) \sin (\pi-\theta)}{1+\cos ^2(\pi-\theta)} d \theta$
Since $\sin(\pi-\theta) = \sin\theta$ and $\cos(\pi-\theta) = -\cos\theta$,we get:
$I = \int_0^\pi \frac{(\pi-\theta) \sin \theta}{1+\cos ^2 \theta} d \theta$ ... (ii)
Adding $(i)$ and (ii):
$2I = \int_0^\pi \frac{\theta \sin \theta + (\pi-\theta) \sin \theta}{1+\cos ^2 \theta} d \theta = \int_0^\pi \frac{\pi \sin \theta}{1+\cos ^2 \theta} d \theta$
$2I = \pi \int_0^\pi \frac{\sin \theta}{1+\cos ^2 \theta} d \theta$
Let $\cos \theta = t$,then $-\sin \theta d \theta = dt$. When $\theta = 0, t = 1$ and when $\theta = \pi, t = -1$.
$2I = \pi \int_1^{-1} \frac{-dt}{1+t^2} = \pi \int_{-1}^1 \frac{dt}{1+t^2}$
$2I = \pi [\tan^{-1} t]_{-1}^1 = \pi (\tan^{-1}(1) - \tan^{-1}(-1)) = \pi (\frac{\pi}{4} - (-\frac{\pi}{4})) = \pi (\frac{\pi}{2}) = \frac{\pi^2}{2}$
$2I = \frac{\pi^2}{2} \Rightarrow I = \frac{\pi^2}{4}$

(C) Let $I = \int_0^{\pi / 2} \frac{200 \sin x+100 \cos x}{\sin x+\cos x} d x$.
We can rewrite the numerator as $100(\sin x + \cos x) + 100 \sin x$.
So,$I = \int_0^{\pi / 2} \frac{100(\sin x + \cos x) + 100 \sin x}{\sin x + \cos x} d x = 100 \int_0^{\pi / 2} 1 d x + 100 \int_0^{\pi / 2} \frac{\sin x}{\sin x + \cos x} d x$.
Let $I_1 = \int_0^{\pi / 2} \frac{\sin x}{\sin x + \cos x} d x$.
Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$,we have $I_1 = \int_0^{\pi / 2} \frac{\sin(\pi/2 - x)}{\sin(\pi/2 - x) + \cos(\pi/2 - x)} d x = \int_0^{\pi / 2} \frac{\cos x}{\cos x + \sin x} d x$.
Adding the two expressions for $I_1$: $2I_1 = \int_0^{\pi / 2} \frac{\sin x + \cos x}{\sin x + \cos x} d x = \int_0^{\pi / 2} 1 d x = \frac{\pi}{2}$.
Thus,$I_1 = \frac{\pi}{4}$.
Substituting back into the expression for $I$: $I = 100[x]_0^{\pi/2} + 100(I_1) = 100(\frac{\pi}{2}) + 100(\frac{\pi}{4}) = 50\pi + 25\pi = 75\pi$.

(B) The given curves are:
$y^2 = 4x$ ...$(i)$
$x^2 = 4y$ ...$(ii)$
From $(ii)$,we have $y = \frac{x^2}{4}$. Substituting this into $(i)$:
$(\frac{x^2}{4})^2 = 4x \implies \frac{x^4}{16} = 4x \implies x^4 = 64x \implies x(x^3 - 64) = 0$.
So,$x = 0$ or $x = 4$. The intersection points are $(0,0)$ and $(4,4)$.
The required area is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=4$:
$Area = \int_0^4 (\sqrt{4x} - \frac{x^2}{4}) dx$
$= \int_0^4 (2\sqrt{x} - \frac{x^2}{4}) dx$
$= [2 \cdot \frac{x^{3/2}}{3/2} - \frac{x^3}{12}]_0^4$
$= [\frac{4}{3} x^{3/2} - \frac{x^3}{12}]_0^4$
$= [\frac{4}{3} (4)^{3/2} - \frac{4^3}{12}] - [0]$
$= [\frac{4}{3} \cdot 8 - \frac{64}{12}]$
$= \frac{32}{3} - \frac{16}{3} = \frac{16}{3}$ sq. units.

(B) The propagation of light through an optical fibre is indeed based on the principle of total internal reflection $(TIR)$.
For $TIR$ to occur,light must travel from a medium of higher refractive index to a medium of lower refractive index,and the angle of incidence must be greater than the critical angle.
In an optical fibre,the core has a higher refractive index $(n_1)$ than the cladding $(n_2)$. The interface between the core and the cladding is where $TIR$ occurs.
The reason statement mentions that the refractive index of the core is greater than that of air. While this is true,it is not the reason why $TIR$ occurs at the core-clad interface. $TIR$ occurs because the core's refractive index is greater than the cladding's refractive index.
Therefore,both $(A)$ and $(R)$ are true,but $(R)$ is not the correct explanation of $(A)$.

(A) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For an equi-convex lens,$R_1 = R$ and $R_2 = -R$.
Substituting these values,we get $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = (\mu - 1) \left( \frac{2}{R} \right)$.
Thus,$f = \frac{R}{2(\mu - 1)}$.
According to the problem,the focal length $f$ is greater than the radius of curvature $R$,so $f > R$.
Substituting the expression for $f$,we get $\frac{R}{2(\mu - 1)} > R$.
This simplifies to $\frac{1}{2(\mu - 1)} > 1$,which implies $2(\mu - 1) < 1$.
Therefore,$\mu - 1 < 0.5$,which means $\mu < 1.5$.
Since the refractive index $\mu$ must be greater than $1$ for a converging lens in air,the range is $1 < \mu < 1.5$. However,based on the provided options,the correct range is greater than zero but less than $1.5$.

(B) The reaction of sodium with oxygen at $300^{\circ} C$ forms sodium peroxide $(X)$: $2Na + O_2 \xrightarrow{300^{\circ} C} Na_2O_2$ $(X)$.
Sodium peroxide $(X)$ reacts with carbon dioxide to form sodium carbonate and oxygen $(Y)$: $2Na_2O_2 + 2CO_2 \longrightarrow 2Na_2CO_3 + O_2$ $(Y)$.
Thus,$Y$ is $O_2$.

(B) In the reaction of $Mg$ with dilute $HNO_3$,magnesium nitrate and hydrogen gas are produced: $Mg + 2HNO_3 \longrightarrow Mg(NO_3)_2 + H_2 \uparrow$.
In the other reactions:
$2Mg + CO_2 \longrightarrow 2MgO + C$
$2Mg + 2NO \longrightarrow 2MgO + N_2$
$3Mg + B_2O_3 \longrightarrow 3MgO + 2B$
Therefore,$MgO$ is not formed in the reaction with dilute $HNO_3$.

(C) In a Common-Emitter $(C-E)$ configuration, the transistor acts as an amplifier.
$1$. Current gain $(\beta)$ is defined as the ratio of collector current $(I_C)$ to base current $(I_B)$, which is typically much greater than $1$.
$2$. Voltage gain $(A_V)$ is defined as the product of current gain $(\beta)$ and the ratio of output resistance to input resistance $(R_{out}/R_{in})$.
$3$. Since both current gain and voltage gain are greater than $1$ in a $C-E$ amplifier, the device provides both current and voltage amplification.
$4$. Consequently, it also provides significant power amplification.

(C) Given that $dx + dy = (x + y)(dx - dy)$.
Dividing by $dx$,we get:
$1 + \frac{dy}{dx} = (x + y)(1 - \frac{dy}{dx})$
$1 + \frac{dy}{dx} = x + y - (x + y)\frac{dy}{dx}$
$\frac{dy}{dx}(1 + x + y) = x + y - 1$
$\frac{dy}{dx} = \frac{x + y - 1}{x + y + 1} \quad \dots(i)$
Let $x + y = t$. Then $1 + \frac{dy}{dx} = \frac{dt}{dx}$,so $\frac{dy}{dx} = \frac{dt}{dx} - 1$.
Substituting into $(i)$:
$\frac{dt}{dx} - 1 = \frac{t - 1}{t + 1}$
$\frac{dt}{dx} = \frac{t - 1}{t + 1} + 1 = \frac{t - 1 + t + 1}{t + 1} = \frac{2t}{t + 1}$
Separating variables:
$\frac{t + 1}{2t} dt = dx$
$\frac{1}{2}(1 + \frac{1}{t}) dt = dx$
Integrating both sides:
$\frac{1}{2}(t + \log|t|) = x + C_1$
$t + \log|t| = 2x + 2C_1$
Substitute $t = x + y$:
$x + y + \log(x + y) = 2x + C$
$\log(x + y) = x - y + C$

(B) Given the differential equation: $\frac{dy}{dx} = \frac{y + x \tan(\frac{y}{x})}{x} \quad (i)$
This is a homogeneous differential equation.
Let $y = vx$,then differentiating with respect to $x$,we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into Eq. $(i)$:
$v + x \frac{dv}{dx} = \frac{vx + x \tan(v)}{x}$
$v + x \frac{dv}{dx} = v + \tan(v)$
$x \frac{dv}{dx} = \tan(v)$
Separating the variables,we get:
$\cot(v) dv = \frac{1}{x} dx$
Integrating both sides:
$\int \cot(v) dv = \int \frac{1}{x} dx$
$\log|\sin(v)| = \log|x| + \log|c|$
$\log|\sin(v)| = \log|cx|$
Taking the exponential of both sides:
$\sin(v) = cx$
Substituting $v = \frac{y}{x}$ back:
$\sin(\frac{y}{x}) = cx$

(B) Given that $a = 2\hat{i} + 3\hat{j} + 6\hat{k}$.
First,calculate the magnitude of vector $a$:
$|a| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
Since $a$ and $b$ are collinear,$b$ can be written as $b = k a$ for some scalar $k$.
Given $|b| = 21$,we have $|k a| = 21$,which implies $|k| |a| = 21$.
Substituting $|a| = 7$,we get $|k| \times 7 = 21$,so $|k| = 3$,which means $k = \pm 3$.
Therefore,$b = \pm 3(2\hat{i} + 3\hat{j} + 6\hat{k})$.

(C) $I$: Two vectors $\vec{a}$ and $\vec{b}$ are linearly independent if and only if they are non-zero and non-collinear. Thus,statement $I$ is true.
$II$: Any three coplanar vectors $\vec{a}, \vec{b}, \vec{c}$ are linearly dependent because there exist scalars $x, y, z$ (not all zero) such that $x\vec{a} + y\vec{b} + z\vec{c} = \vec{0}$. Thus,statement $II$ is true.
$\therefore$ Both $I$ and $II$ are true.

(B) Statement $A$ is true because if three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar,then one can be written as $\vec{c} = x\vec{a} + y\vec{b}$ for some scalars $x$ and $y$.
Statement $R$ is also true because any set of three coplanar vectors in $3D$ space is linearly dependent,as their scalar triple product is zero.
However,$R$ is a general property of coplanar vectors,while $A$ is a specific condition for coplanarity. Thus,$R$ is not the correct explanation of $A$.