TS EAMCET 2004 Chemistry Question Paper with Answer and Solution

(A) We know that the range of the cosine function is $[-1, 1]$.
Thus,$-1 \leq \cos(4x - 5) \leq 1$.
Multiplying the inequality by $3$,we get:
$-3 \leq 3 \cos(4x - 5) \leq 3$.
Adding $4$ to all sides of the inequality:
$-3 + 4 \leq 3 \cos(4x - 5) + 4 \leq 3 + 4$.
Therefore,$1 \leq 3 \cos(4x - 5) + 4 \leq 7$.
Hence,the expression lies in the interval $[1, 7]$.

(A) Let the points be $A = (a \cos \theta, a \sin \theta)$ and $B = (a \cos \phi, a \sin \phi)$.
Given the distance $AB = 2a$.
Using the distance formula:
$AB^2 = (a \cos \theta - a \cos \phi)^2 + (a \sin \theta - a \sin \phi)^2 = (2a)^2$
$a^2(\cos^2 \theta + \cos^2 \phi - 2 \cos \theta \cos \phi + \sin^2 \theta + \sin^2 \phi - 2 \sin \theta \sin \phi) = 4a^2$
$a^2(1 + 1 - 2(\cos \theta \cos \phi + \sin \theta \sin \phi)) = 4a^2$
$2a^2(1 - \cos(\theta - \phi)) = 4a^2$
$1 - \cos(\theta - \phi) = 2$
$\cos(\theta - \phi) = -1$
Since $\cos(\theta - \phi) = -1$,we have $\theta - \phi = (2n + 1)\pi = 2n\pi + \pi$ for $n \in Z$.
Therefore,$\theta = 2n\pi + \pi + \phi$.

(A) Let $M$ be the mid-point of $AB$. Since $PA=PB$,the triangle $PAB$ is an isosceles triangle with $PA=PB$.
In an isosceles triangle,the median from the vertex $P$ to the base $AB$ is also the altitude to the base $AB$.
Thus,$PM \perp AB$.
The slope of the line $AB$ (which is $2x-y+3=0$ or $y=2x+3$) is $m_1 = 2$.
Since $PM \perp AB$,the slope of $PM$ is $m_2 = -\frac{1}{m_1} = -\frac{1}{2}$.
The line $PM$ passes through $P(1,2)$ and has a slope of $-\frac{1}{2}$.
The equation of line $PM$ is $y-2 = -\frac{1}{2}(x-1)$,which simplifies to $2y-4 = -x+1$,or $x+2y-5=0$.
The mid-point $M$ is the intersection of the lines $2x-y+3=0$ and $x+2y-5=0$.
From the first equation,$y=2x+3$. Substituting this into the second equation:
$x+2(2x+3)-5=0
$ $\Rightarrow x+4x+6-5=0
$ $\Rightarrow 5x+1=0
$ $\Rightarrow x = -\frac{1}{5}$.
Now,$y = 2(-\frac{1}{5})+3 = -\frac{2}{5}+3 = \frac{13}{5}$.
Thus,the mid-point $M$ is $\left(-\frac{1}{5}, \frac{13}{5}\right)$.

(D) The given equation is $x^2(\cos^2 \theta - 1) - xy \sin^2 \theta + y^2 \sin^2 \theta = 0$.
This is a homogeneous equation of the second degree,which can be compared with the general form $ax^2 + 2hxy + by^2 = 0$.
Here,$a = \cos^2 \theta - 1 = -\sin^2 \theta$,$b = \sin^2 \theta$,and $2h = -\sin^2 \theta$.
For a pair of lines represented by $ax^2 + 2hxy + by^2 = 0$,the lines are perpendicular if $a + b = 0$.
Calculating $a + b$:
$a + b = -\sin^2 \theta + \sin^2 \theta = 0$.
Since $a + b = 0$,the lines are perpendicular to each other.
Therefore,the angle between the lines is $\frac{\pi}{2}$.

(D) The given equation is $x^2(\cos^2 \theta - 1) - xy \sin^2 \theta + y^2 \sin^2 \theta = 0$.
This is a homogeneous equation of the second degree of the form $ax^2 + 2hxy + by^2 = 0$.
Comparing the coefficients,we get:
$a = \cos^2 \theta - 1 = -\sin^2 \theta$,
$b = \sin^2 \theta$,
$2h = -\sin^2 \theta \Rightarrow h = -\frac{1}{2} \sin^2 \theta$.
Now,check the condition for perpendicular lines:
$a + b = -\sin^2 \theta + \sin^2 \theta = 0$.
Since the sum of the coefficients of $x^2$ and $y^2$ is zero $(a + b = 0)$,the lines represented by the equation are perpendicular to each other.
Therefore,the angle between the lines is $\frac{\pi}{2}$.

(B) The given circles are:
$C_1: x^2+y^2+8x-6y=0$
$C_2: 4x^2+4y^2-4x-12y-186=0 \implies x^2+y^2-x-3y-46.5=0$
$C_3: x^2+y^2-6x+6y-9=0$
Calculating the radii $r_1, r_2, r_3$:
$r_1 = \sqrt{(-4)^2 - (-3)^2 - 0} = \sqrt{16+9} = 5$
$r_2 = \sqrt{(0.5)^2 + (1.5)^2 - (-46.5)} = \sqrt{0.25 + 2.25 + 46.5} = \sqrt{49} = 7$
$r_3 = \sqrt{(3)^2 + (-3)^2 - (-9)} = \sqrt{9+9+9} = \sqrt{27} = 3\sqrt{3} \approx 5.196$
Calculating perimeters $P = 2\pi r$:
$P_1 = 2\pi(5) = 10\pi$
$P_2 = 2\pi(7) = 14\pi$
$P_3 = 2\pi(3\sqrt{3}) = 6\sqrt{3}\pi \approx 10.39\pi$
Comparing the values:
$10\pi < 10.39\pi < 14\pi \implies P_1 < P_3 < P_2$.

(B) The line is given by $3x - 2y + 6 = 0$.
To find point $A$ (where it meets the $X$-axis),set $y = 0$: $3x + 6 = 0 \Rightarrow x = -2$. So,$A = (-2, 0)$.
To find point $B$ (where it meets the $Y$-axis),set $x = 0$: $-2y + 6 = 0 \Rightarrow y = 3$. So,$B = (0, 3)$.
The radius $r$ is the distance $AB = \sqrt{(0 - (-2))^2 + (3 - 0)^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$.
The equation of a circle with center $(h, k) = (-2, 0)$ and radius $r = \sqrt{13}$ is $(x - h)^2 + (y - k)^2 = r^2$.
$(x - (-2))^2 + (y - 0)^2 = (\sqrt{13})^2$
$(x + 2)^2 + y^2 = 13$
$x^2 + 4x + 4 + y^2 = 13$
$x^2 + y^2 + 4x - 9 = 0$.

(B) The equation of the circle is $x^2+y^2=25$.
Two points $(x_1, y_1)$ and $(x_2, y_2)$ are conjugate with respect to the circle $x^2+y^2=r^2$ if $x_1x_2 + y_1y_2 = r^2$.
Here,$(x_1, y_1) = (1, a)$ and $(x_2, y_2) = (b, 2)$,and $r^2 = 25$.
Substituting these values into the condition,we get:
$(1)(b) + (a)(2) = 25$
$b + 2a = 25$
Multiplying both sides by $2$,we get:
$2b + 4a = 50$
Therefore,$4a + 2b = 50$.

(C) Given the equation of the circle in polar form: $r^2-4r(\cos \theta+\sin \theta)-4=0$ $(i)$
We know that $x=r \cos \theta$,$y=r \sin \theta$,and $r^2=x^2+y^2$.
Substituting these into equation $(i)$:
$x^2+y^2-4(x+y)-4=0$
$x^2+y^2-4x-4y-4=0$
Comparing this with the general equation of a circle $x^2+y^2+2gx+2fy+c=0$,we get:
$2g = -4 \Rightarrow g = -2$
$2f = -4 \Rightarrow f = -2$
The centre of the circle is $(-g, -f) = (-(-2), -(-2)) = (2, 2)$.

(A) The given equation of the circle is $r = \sqrt{3} \sin \theta + \cos \theta$.
Multiplying both sides by $r$,we get $r^2 = \sqrt{3} (r \sin \theta) + (r \cos \theta)$.
Using the polar to Cartesian coordinate transformations $x = r \cos \theta$ and $y = r \sin \theta$,and $r^2 = x^2 + y^2$,the equation becomes:
$x^2 + y^2 = \sqrt{3} y + x$.
Rearranging the terms,we get $x^2 - x + y^2 - \sqrt{3} y = 0$.
Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we have $2g = -1 \Rightarrow g = -1/2$ and $2f = -\sqrt{3} \Rightarrow f = -\sqrt{3}/2$,with $c = 0$.
The radius of the circle is given by $\sqrt{g^2 + f^2 - c}$.
Radius $= \sqrt{(-1/2)^2 + (-\sqrt{3}/2)^2 - 0} = \sqrt{1/4 + 3/4} = \sqrt{1} = 1$.

(C) The given lines are $L_1: x+y-1=0$,$L_2: x-y-1=0$,and $L_3: y+1=0$.
These three lines are not concurrent and form a triangle.
For any triangle,there exists exactly one incircle that touches all three sides internally.
Additionally,there are three excircles,each of which touches one side externally and the extensions of the other two sides.
Therefore,the total number of circles that touch all three lines is $1 + 3 = 4$.

(C) The given polar equation is $\cos \theta + 7 \sin \theta = \frac{1}{r}$.
Multiplying both sides by $r$,we get $r \cos \theta + 7 r \sin \theta = 1$.
Using the standard conversion formulas $x = r \cos \theta$ and $y = r \sin \theta$,we substitute these into the equation:
$x + 7y = 1$.
This is the standard form of a linear equation in two variables,which represents a straight line.

(A) First,we decompose the expression into partial fractions: $\frac{x-4}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$.
Solving for $A$ and $B$,we get $\frac{2}{x-2} - \frac{1}{x-3}$.
Rewrite the expression as: $2(x-2)^{-1} - (x-3)^{-1} = 2(-2)^{-1}(1-\frac{x}{2})^{-1} - (-3)^{-1}(1-\frac{x}{3})^{-1}$.
$= -\left[1 + \frac{x}{2} + (\frac{x}{2})^2 + (\frac{x}{2})^3 + \dots\right] + \frac{1}{3}\left[1 + \frac{x}{3} + (\frac{x}{3})^2 + (\frac{x}{3})^3 + \dots\right]$.
The coefficient of $x^3$ is $-\left(\frac{1}{2}\right)^3 + \frac{1}{3}\left(\frac{1}{3}\right)^3$.
$= -\frac{1}{8} + \frac{1}{81} = \frac{-81 + 8}{648} = -\frac{73}{648}$.

(B) We know the expansion of $e^{-x}$ is given by $e^{-x} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \dots + (-1)^n \frac{x^n}{n!} + \dots$
Now,consider the expression $(2+3x)e^{-x}$:
$(2+3x)e^{-x} = 2e^{-x} + 3xe^{-x}$
To find the coefficient of $x^{10}$:
$1$. In $2e^{-x}$,the term containing $x^{10}$ is $2 \times \frac{(-1)^{10} x^{10}}{10!} = \frac{2}{10!} x^{10}$.
$2$. In $3xe^{-x}$,the term containing $x^{10}$ is $3x \times \frac{(-1)^9 x^9}{9!} = 3x \times \frac{-x^9}{9!} = \frac{-3}{9!} x^{10}$.
To combine these,express $\frac{-3}{9!}$ with a denominator of $10!$:
$\frac{-3}{9!} = \frac{-3 \times 10}{10!} = \frac{-30}{10!}$.
Adding the coefficients:
$\frac{2}{10!} - \frac{30}{10!} = \frac{2-30}{10!} = \frac{-28}{10!}$.

(D) The binomial coefficients for $n=15$ are ${ }^{15} C_0, { }^{15} C_1, \dots, { }^{15} C_7, { }^{15} C_8, \dots, { }^{15} C_{15}$.
Since ${ }^{15} C_r$ increases as $r$ increases from $0$ to $7$ and decreases as $r$ increases from $8$ to $15$,we look for a sequence where the values are decreasing.
Comparing the options:
$A$: ${ }^{15} C_5 < { }^{15} C_6 < { }^{15} C_7$ (Increasing)
$B$: ${ }^{15} C_{10} < { }^{15} C_9 < { }^{15} C_8$ (Increasing)
$C$: ${ }^{15} C_6 < { }^{15} C_7 < { }^{15} C_8$ (Increasing)
$D$: ${ }^{15} C_7 > { }^{15} C_6 > { }^{15} C_5$ (Decreasing)
Therefore,the sequence in decreasing order is ${ }^{15} C_7, { }^{15} C_6, { }^{15} C_5$. The correct option is $D$.

(A) The given equation is $36x^2 + 144y^2 - 36x - 96y - 119 = 0$.
Rearranging the terms,we get $36(x^2 - x) + 144(y^2 - \frac{2}{3}y) = 119$.
Completing the square,we have $36(x^2 - x + \frac{1}{4}) + 144(y^2 - \frac{2}{3}y + \frac{1}{9}) = 119 + 9 + 16$.
This simplifies to $36(x - \frac{1}{2})^2 + 144(y - \frac{1}{3})^2 = 144$.
Dividing by $144$,we get $\frac{(x - 1/2)^2}{4} + \frac{(y - 1/3)^2}{1} = 1$.
This is an ellipse with $a^2 = 4$ and $b^2 = 1$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

(B) In $\triangle ABC$,we know that $A+B+C = \pi$.
Consider the expression:
$\cos \left(\frac{B+2C+3A}{2}\right) + \cos \left(\frac{A-B}{2}\right)$
Rewrite the argument of the first cosine term using $A+B+C = \pi$:
$\frac{B+2C+3A}{2} = \frac{(A+B+C) + A + C}{2} = \frac{\pi + A + C}{2} = \frac{\pi}{2} + \frac{A+C}{2}$
Wait,let us re-evaluate the expression:
$\frac{B+2C+3A}{2} = \frac{2A+2B+2C + A-B}{2} = \frac{2\pi + (A-B)}{2} = \pi + \frac{A-B}{2}$
Substituting this back:
$\cos \left(\pi + \frac{A-B}{2}\right) + \cos \left(\frac{A-B}{2}\right)$
Using the identity $\cos(\pi + \theta) = -\cos(\theta)$:
$-\cos \left(\frac{A-B}{2}\right) + \cos \left(\frac{A-B}{2}\right) = 0$

(C) We know that the exradii and inradius are given by:
$r_1 = 4R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$
$r_2 = 4R \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}$
$r_3 = 4R \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}$
$r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$
Given $r_3 = r_1 + r_2 + r$,we have:
$r_3 - r = r_1 + r_2$
$4R \sin \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} - 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = 4R \cos \frac{C}{2} \sin \frac{A}{2} \cos \frac{B}{2} + 4R \cos \frac{C}{2} \cos \frac{A}{2} \sin \frac{B}{2}$
Dividing by $4R$ and factoring:
$\sin \frac{C}{2} (\cos \frac{A}{2} \cos \frac{B}{2} - \sin \frac{A}{2} \sin \frac{B}{2}) = \cos \frac{C}{2} (\sin \frac{A}{2} \cos \frac{B}{2} + \cos \frac{A}{2} \sin \frac{B}{2})$
Using the identity $\cos(x+y) = \cos x \cos y - \sin x \sin y$ and $\sin(x+y) = \sin x \cos y + \cos x \sin y$:
$\sin \frac{C}{2} \cos(\frac{A+B}{2}) = \cos \frac{C}{2} \sin(\frac{A+B}{2})$
Since $A+B+C = \pi$,$\frac{A+B}{2} = \frac{\pi}{2} - \frac{C}{2}$:
$\sin \frac{C}{2} \cos(\frac{\pi}{2} - \frac{C}{2}) = \cos \frac{C}{2} \sin(\frac{\pi}{2} - \frac{C}{2})$
$\sin \frac{C}{2} \sin \frac{C}{2} = \cos \frac{C}{2} \cos \frac{C}{2}$
$\sin^2 \frac{C}{2} = \cos^2 \frac{C}{2}$
$\tan^2 \frac{C}{2} = 1 \Rightarrow \tan \frac{C}{2} = 1$
$\frac{C}{2} = \frac{\pi}{4} \Rightarrow C = \frac{\pi}{2}$
Therefore,$A+B = \pi - C = \pi - \frac{\pi}{2} = \frac{\pi}{2} = 90^{\circ}$.

(C) Let $A = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 4 & 2 \\ 3 & -4 & 6 \end{bmatrix}$.
The co-factor $C_{ij}$ of an element $a_{ij}$ is given by $(-1)^{i+j} M_{ij}$,where $M_{ij}$ is the minor.
For element $-1$ (at position $a_{12}$): $C_{12} = (-1)^{1+2} \begin{vmatrix} 0 & 2 \\ 3 & 6 \end{vmatrix} = -1(0 - 6) = 6$. So,$A-4$.
For element $1$ (at position $a_{11}$): $C_{11} = (-1)^{1+1} \begin{vmatrix} 4 & 2 \\ -4 & 6 \end{vmatrix} = 1(24 - (-8)) = 32$. So,$B-2$.
For element $3$ (at position $a_{31}$): $C_{31} = (-1)^{3+1} \begin{vmatrix} -1 & 0 \\ 4 & 2 \end{vmatrix} = 1(-2 - 0) = -2$. So,$C-1$.
For element $6$ (at position $a_{33}$): $C_{33} = (-1)^{3+3} \begin{vmatrix} 1 & -1 \\ 0 & 4 \end{vmatrix} = 1(4 - 0) = 4$. So,$D-3$.
Thus,the matching is $A-4, B-2, C-1, D-3$.
Therefore,option $C$ is correct.

(D) Let $\Delta = \left|\begin{array}{lll}1990 & 1991 & 1992 \\ 1991 & 1992 & 1993 \\ 1992 & 1993 & 1994\end{array}\right|$
Applying the column operations $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_2$:
$\Delta = \left|\begin{array}{lll}1990 & 1991-1990 & 1992-1991 \\ 1991 & 1992-1991 & 1993-1992 \\ 1992 & 1993-1992 & 1994-1993\end{array}\right|$
$\Delta = \left|\begin{array}{lll}1990 & 1 & 1 \\ 1991 & 1 & 1 \\ 1992 & 1 & 1\end{array}\right|$
Since column $C_2$ and column $C_3$ are identical,the value of the determinant is $0$.

(C) Given equation: $\sin ^{-1} x + \sin ^{-1}(1-x) = \cos ^{-1} x$
Since $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} x$,we have:
$\sin ^{-1}(1-x) = \frac{\pi}{2} - \sin ^{-1} x - \sin ^{-1} x = \frac{\pi}{2} - 2\sin ^{-1} x$
Taking $\sin$ on both sides:
$1-x = \sin(\frac{\pi}{2} - 2\sin ^{-1} x) = \cos(2\sin ^{-1} x)$
Using the identity $\cos(2\theta) = 1 - 2\sin^2 \theta$,where $\theta = \sin ^{-1} x$:
$1-x = 1 - 2(\sin(\sin ^{-1} x))^2$
$1-x = 1 - 2x^2$
$2x^2 - x = 0$
$x(2x - 1) = 0$
Thus,$x = 0$ or $x = \frac{1}{2}$.
Checking the values:
For $x=0$: $\sin ^{-1} 0 + \sin ^{-1} 1 = 0 + \frac{\pi}{2} = \frac{\pi}{2}$ and $\cos ^{-1} 0 = \frac{\pi}{2}$. (Valid)
For $x=\frac{1}{2}$: $\sin ^{-1} \frac{1}{2} + \sin ^{-1} \frac{1}{2} = \frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3}$ and $\cos ^{-1} \frac{1}{2} = \frac{\pi}{3}$. (Valid)
Therefore,$x \in \{0, \frac{1}{2}\}$.

(C) Given inequation is $x^2 - 2x + 5 \leq 0$.
We can rewrite the expression by completing the square:
$x^2 - 2x + 1 + 4 \leq 0$
$(x - 1)^2 + 4 \leq 0$
Since $(x - 1)^2 \geq 0$ for all $x \in R$,it follows that $(x - 1)^2 + 4 \geq 4$.
Therefore,$(x - 1)^2 + 4$ is always positive and can never be less than or equal to $0$.
Thus,there are no real values of $x$ that satisfy the given inequation.
The solution set is the empty set,denoted by $\phi$.

(B) Given that $f(x)$ is an even function,so $f(-x) = f(x)$.
By differentiating both sides with respect to $x$,we get $f^{\prime}(-x) \cdot (-1) = f^{\prime}(x)$,which implies $f^{\prime}(-x) = -f^{\prime}(x)$. Thus,$f^{\prime}$ is an odd function.
By differentiating again,$f^{\prime \prime}(-x) \cdot (-1) = -f^{\prime \prime}(x)$,which implies $f^{\prime \prime}(-x) = f^{\prime \prime}(x)$. Thus,$f^{\prime \prime}$ is an even function.
By differentiating once more,$f^{\prime \prime \prime}(-x) \cdot (-1) = f^{\prime \prime \prime}(x)$,which implies $f^{\prime \prime \prime}(-x) = -f^{\prime \prime \prime}(x)$. Thus,$f^{\prime \prime \prime}$ is an odd function.
Among the given options,$f^{\prime}$ is an odd function. Therefore,option $B$ is correct.

(B) Given the function $f(n)$ defined as:
$f(n) = \begin{cases} 2 & \text{if } n=3k, k \in Z \\ 10 & \text{if } n=3k+1, k \in Z \\ 0 & \text{if } n=3k+2, k \in Z \end{cases}$
We want to find the set $\{n \in N: f(n) > 2\}$.
Looking at the definition,$f(n) > 2$ only when $f(n) = 10$,which occurs when $n = 3k + 1$ for some $k \in Z$.
Since $n \in N$ (natural numbers),we consider $k \geq 0$ (assuming $N = \{1, 2, 3, \dots\}$):
For $k=0, n = 3(0) + 1 = 1$.
For $k=1, n = 3(1) + 1 = 4$.
For $k=2, n = 3(2) + 1 = 7$.
Thus,the set is $\{1, 4, 7, \dots\}$.

(C) Given the function $f: R \rightarrow R$ defined by $f(x) = 3^{-x}$.
$I$. For one-one check: Let $f(x_1) = f(x_2)$.
$3^{-x_1} = 3^{-x_2} \Rightarrow -x_1 = -x_2 \Rightarrow x_1 = x_2$.
Since $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$,the function is one-one.
$II$. For onto check: The range of $f(x) = 3^{-x}$ is $(0, \infty)$,which is a subset of the codomain $R$. Since the range $\neq$ codomain,the function is not onto.
$III$. For decreasing check: $f'(x) = \frac{d}{dx}(3^{-x}) = -3^{-x} \ln 3$. Since $3^{-x} > 0$ and $\ln 3 > 0$,$f'(x) < 0$ for all $x \in R$. Thus,$f$ is a strictly decreasing function.
Therefore,statements $I$ and $III$ are true.

(B) Given the equation $x^y = e^{x-y}$.
Taking the natural logarithm on both sides,we get:
$\ln(x^y) = \ln(e^{x-y})$
$y \ln x = x - y$
Rearranging the terms to isolate $y$:
$y \ln x + y = x$
$y(1 + \ln x) = x$
$y = \frac{x}{1 + \ln x}$
Now,differentiate with respect to $x$ using the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$:
$\frac{dy}{dx} = \frac{(1)(1 + \ln x) - x(\frac{1}{x})}{(1 + \ln x)^2}$
$\frac{dy}{dx} = \frac{1 + \ln x - 1}{(1 + \ln x)^2}$
$\frac{dy}{dx} = \frac{\ln x}{(1 + \ln x)^2}$
Thus,the correct option is $B$.

(C) Given the equation of the curve is $y=x^2+2x$.
Since the $x$ and $y$ coordinates change at the same rate,we have $\frac{dy}{dt} = \frac{dx}{dt}$.
Differentiating the equation $y=x^2+2x$ with respect to $t$,we get:
$\frac{dy}{dt} = (2x+2) \frac{dx}{dt}$.
Substituting $\frac{dy}{dt} = \frac{dx}{dt}$ into the equation,we get:
$\frac{dx}{dt} = (2x+2) \frac{dx}{dt}$.
Assuming $\frac{dx}{dt} \neq 0$,we divide both sides by $\frac{dx}{dt}$:
$1 = 2x+2$.
$2x = -1 \implies x = -\frac{1}{2}$.
Now,find the corresponding $y$ coordinate by substituting $x = -\frac{1}{2}$ into the curve equation:
$y = \left(-\frac{1}{2}\right)^2 + 2\left(-\frac{1}{2}\right) = \frac{1}{4} - 1 = -\frac{3}{4}$.
Therefore,the point on the curve is $\left(-\frac{1}{2}, -\frac{3}{4}\right)$.

(D) The given equation of the curve is $y = 4 - 2x^2$.
On differentiating both sides with respect to time $t$,we get:
$\frac{dy}{dt} = -4x \frac{dx}{dt}$.
Given that the $x$-coordinate is decreasing at the rate of $5 \text{ units/s}$,we have $\frac{dx}{dt} = -5 \text{ units/s}$.
At the point $(1, 2)$,we substitute $x = 1$ and $\frac{dx}{dt} = -5$ into the derivative equation:
$\frac{dy}{dt} = -4(1)(-5) = 20 \text{ units/s}$.
Thus,the $y$-coordinate is increasing at the rate of $20 \text{ units/s}$.

(C) Let the height of the aeroplane be $h = 1 \ km$. Let the positions of the aeroplane be $D$ and $E$ at two different times. Let $A$ be the point of observation on the ground. Let $P$ and $Q$ be the points on the ground directly below $D$ and $E$ respectively. So,$DP = EQ = 1 \ km$.
In $\Delta DAP$,$\tan 60^{\circ} = \frac{DP}{AP}$ $\Rightarrow \sqrt{3} = \frac{1}{AP}$ $\Rightarrow AP = \frac{1}{\sqrt{3}} \ km$.
In $\Delta EAQ$,$\tan 30^{\circ} = \frac{EQ}{AQ}$ $\Rightarrow \frac{1}{\sqrt{3}} = \frac{1}{AP + PQ}$ $\Rightarrow AP + PQ = \sqrt{3}$.
Substituting $AP = \frac{1}{\sqrt{3}}$,we get $\frac{1}{\sqrt{3}} + PQ = \sqrt{3}$ $\Rightarrow PQ = \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{3-1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \ km$.
The distance $PQ$ is covered in $10 \ s$. Therefore,speed $= \frac{\text{Distance}}{\text{Time}} = \frac{2/\sqrt{3} \ km}{10 \ s} = \frac{2}{10\sqrt{3}} \ km/s$.
To convert into $km/h$,multiply by $3600$: $\text{Speed} = \frac{2}{10\sqrt{3}} \times 3600 = \frac{720}{\sqrt{3}} = \frac{720\sqrt{3}}{3} = 240\sqrt{3} \ km/h$.

(C) Given that,$f(x) = \frac{1}{x^2} \int_3^x (2t - 3f'(t)) dt$.
Multiplying by $x^2$,we get $x^2 f(x) = \int_3^x (2t - 3f'(t)) dt$.
Differentiating both sides with respect to $x$ using the product rule and the Leibniz integral rule:
$2x f(x) + x^2 f'(x) = 2x - 3f'(x)$.
At $x = 3$,we have $2(3) f(3) + 3^2 f'(3) = 2(3) - 3f'(3)$.
From the original integral,$f(3) = \frac{1}{3^2} \int_3^3 (2t - 3f'(t)) dt = 0$.
Substituting $f(3) = 0$ into the differentiated equation:
$2(3)(0) + 9f'(3) = 6 - 3f'(3)$.
$9f'(3) + 3f'(3) = 6$.
$12f'(3) = 6$.
$f'(3) = \frac{6}{12} = \frac{1}{2}$.

(C) The area bounded by the curve $y = f(x)$,the $x$-axis,and the lines $x = a$ and $x = b$ is given by the integral $\int_{a}^{b} y \, dx$.
Here,$y = x^2 + 2$,$a = 1$,and $b = 2$.
Therefore,the required area is:
$\text{Area} = \int_{1}^{2} (x^2 + 2) \, dx$
$= \left[ \frac{x^3}{3} + 2x \right]_{1}^{2}$
$= \left( \frac{2^3}{3} + 2(2) \right) - \left( \frac{1^3}{3} + 2(1) \right)$
$= \left( \frac{8}{3} + 4 \right) - \left( \frac{1}{3} + 2 \right)$
$= \left( \frac{8 + 12}{3} \right) - \left( \frac{1 + 6}{3} \right)$
$= \frac{20}{3} - \frac{7}{3}$
$= \frac{13}{3} \text{ sq unit}$.

(A) Optical fibres are extensively used in modern communication networks because they offer high bandwidth and low signal attenuation.
In optical fibres,light signals are transmitted via the principle of total internal reflection.
Because the signal is confined within the fibre and is not susceptible to electromagnetic interference,the transmission is highly secure and clear.
Additionally,optical fibres are physically compact,lightweight,and flexible,making them ideal for installation.
Therefore,both the assertion and the reason are true,and the reason correctly explains why optical fibres are widely used.

(B) Let the angle of the prism be $A$. Since the triangle is isosceles with $AB=AC$,the base angles are $\angle B = \angle C = (180^{\circ}-A)/2 = 90^{\circ} - A/2$.
$1$. The ray is incident normally on face $AB$,so it enters the prism undeviated.
$2$. At face $AC$ (which is silvered),the angle of incidence $i_1$ is equal to the angle $A$ of the prism.
$3$. After reflection at $AC$,the angle of reflection is also $i_1 = A$. The angle the reflected ray makes with the face $AC$ is $90^{\circ}-A$.
$4$. In the triangle formed by the ray and the sides of the prism,the angle at the face $AB$ is $90^{\circ}-2A$.
$5$. At the second reflection on face $AB$,the angle of incidence $i_2$ is $90^{\circ}-(90^{\circ}-2A) = 2A$.
$6$. The ray then strikes the base $BC$ normally. In the triangle formed by the ray and the base $BC$,the angle at $B$ is $90^{\circ}-2A$. Since the ray is perpendicular to $BC$,we have $90^{\circ}-2A + (90^{\circ}-A/2) = 90^{\circ}$.
$7$. Solving for $A$: $90^{\circ} - 2.5A = 0$,which gives $A = 36^{\circ}$.

(B) The formula for the refractive index $\mu$ of a prism in terms of the refracting angle $A$ and the angle of minimum deviation $\delta_m$ is given by:
$\mu = \frac{\sin((A + \delta_m) / 2)}{\sin(A / 2)}$
Given $\mu = \cot(A / 2) = \frac{\cos(A / 2)}{\sin(A / 2)}$,we substitute this into the formula:
$\frac{\cos(A / 2)}{\sin(A / 2)} = \frac{\sin((A + \delta_m) / 2)}{\sin(A / 2)}$
Canceling $\sin(A / 2)$ from both sides,we get:
$\cos(A / 2) = \sin((A + \delta_m) / 2)$
Using the trigonometric identity $\cos \theta = \sin(90^{\circ} - \theta)$ or $\sin(\pi/2 - \theta)$:
$\sin(\pi/2 - A/2) = \sin((A + \delta_m) / 2)$
Equating the angles:
$\pi/2 - A/2 = (A + \delta_m) / 2$
Multiplying by $2$:
$\pi - A = A + \delta_m$
$\delta_m = \pi - 2A$

(D) In the Down's process for the extraction of sodium,the electrolysis of molten $NaCl$ is carried out.
In this cell,the cathode is made of $Iron$ $(Fe)$ and the anode is made of $Graphite$ $(C)$.
Therefore,the cathode and anode are $Iron$ and $Graphite$ respectively.

(D) Plaster of Paris absorbs water to form monoclinic gypsum,which is a hard substance.
The chemical reaction is:
$CaSO_4 \cdot \frac{1}{2} H_2 O + \frac{3}{2} H_2 O \rightarrow CaSO_4 \cdot 2 H_2 O$
(Monoclinic gypsum)

(A) No Solution Available

(C) For statement $A$: The given differential equation is $\frac{dy}{dx} + y = x^2$. Comparing this with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$,we get $P(x) = 1$. The integrating factor $(IF)$ is $e^{\int P(x) dx} = e^{\int 1 dx} = e^x$. Thus,statement $A$ is true.
For statement $R$: The standard form of a linear differential equation is $\frac{dy}{dx} + P(x)y = Q(x)$,and its integrating factor is defined as $e^{\int P(x) dx}$. Thus,statement $R$ is true.
Since statement $A$ is derived directly from the definition provided in statement $R$,$R \Rightarrow A$ is true.

(A) We need to evaluate the expression $c \cdot ((b+c) \times (a+b+c))$.
Using the distributive property of the cross product,we have:
$(b+c) \times (a+b+c) = b \times a + b \times b + b \times c + c \times a + c \times b + c \times c$.
Since the cross product of any vector with itself is zero ($b \times b = 0$ and $c \times c = 0$),the expression simplifies to:
$(b+c) \times (a+b+c) = b \times a + b \times c + c \times a + c \times b$.
Now,taking the dot product with $c$:
$c \cdot (b \times a + b \times c + c \times a + c \times b) = c \cdot (b \times a) + c \cdot (b \times c) + c \cdot (c \times a) + c \cdot (c \times b)$.
Using the scalar triple product property $[x, y, z] = x \cdot (y \times z)$,we get:
$= [c, b, a] + [c, b, c] + [c, c, a] + [c, c, b]$.
Since any scalar triple product with two identical vectors is zero,we have:
$[c, b, c] = 0$,$[c, c, a] = 0$,and $[c, c, b] = 0$.
Thus,the expression simplifies to $[c, b, a] = c \cdot (b \times a)$.
Since $c \cdot (b \times a) = -c \cdot (a \times b)$,and looking at the options,the correct form is $c \cdot (b \times a)$.

(C) Given equations are:
$l+m+n=0 \quad \dots(i)$
$mn-2ln+lm=0 \quad \dots(ii)$
From $(i)$,$l = -(m+n)$. Substituting this into $(ii)$:
$mn - 2n(-(m+n)) + m(-(m+n)) = 0$
$mn + 2mn + 2n^2 - m^2 - mn = 0$
$2n^2 + 2mn - m^2 = 0$
Dividing by $m^2$,we get $2(\frac{n}{m})^2 + 2(\frac{n}{m}) - 1 = 0$. Let the roots be $\frac{n_1}{m_1}$ and $\frac{n_2}{m_2}$.
Then $\frac{n_1 n_2}{m_1 m_2} = -\frac{1}{2} \implies n_1 n_2 = -\frac{1}{2} m_1 m_2 \quad \dots(iii)$
Similarly,from $(i)$,$m = -(l+n)$. Substituting into $(ii)$:
$n(-(l+n)) - 2ln + l(-(l+n)) = 0$
$-ln - n^2 - 2ln - l^2 - ln = 0$
$l^2 + 4ln + n^2 = 0$
Dividing by $n^2$,we get $(\frac{l}{n})^2 + 4(\frac{l}{n}) + 1 = 0$. Let the roots be $\frac{l_1}{n_1}$ and $\frac{l_2}{n_2}$.
Then $\frac{l_1 l_2}{n_1 n_2} = 1 \implies l_1 l_2 = n_1 n_2 \quad \dots(iv)$
From $(iii)$ and $(iv)$,$l_1 l_2 = n_1 n_2 = -\frac{1}{2} m_1 m_2$.
Thus,$l_1 l_2 + m_1 m_2 + n_1 n_2 = l_1 l_2 - 2l_1 l_2 + l_1 l_2 = 0$.
Since $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$,the lines are perpendicular. Therefore,the angle between them is $\frac{\pi}{2}$.

(A) Let the origin be $O(0,0,0)$ and the foot of the perpendicular be $P(2,-1,3)$.
Since $OP$ is the normal to the plane,the direction ratios of the normal are the same as the direction ratios of the line $OP$.
The direction ratios of $OP$ are $(2-0, -1-0, 3-0) = (2, -1, 3)$.
Thus,the normal vector to the plane is $\vec{n} = 2\hat{i} - \hat{j} + 3\hat{k}$.
The equation of a plane passing through a point $(x_1, y_1, z_1)$ with normal vector $(a, b, c)$ is given by $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$.
Substituting the point $P(2, -1, 3)$ and the normal vector $(2, -1, 3)$ into the equation:
$2(x-2) - 1(y-(-1)) + 3(z-3) = 0$
$2(x-2) - 1(y+1) + 3(z-3) = 0$
$2x - 4 - y - 1 + 3z - 9 = 0$
$2x - y + 3z - 14 = 0$
Therefore,the equation of the plane is $2x - y + 3z - 14 = 0$.

(A) The numbers on the die are $\{2, 3, 5, 7, 11, 13\}$.
There is $1$ even number $(2)$ and $5$ odd numbers $(3, 5, 7, 11, 13)$.
The sum of two numbers is odd if and only if one number is even and the other is odd.
Let $E$ be the event that the first die shows an even number and the second shows an odd number,or the first die shows an odd number and the second shows an even number.
Probability of getting an even number $P(Even) = \frac{1}{6}$.
Probability of getting an odd number $P(Odd) = \frac{5}{6}$.
Required probability $= P(Even) \times P(Odd) + P(Odd) \times P(Even)$
$= \left(\frac{1}{6} \times \frac{5}{6}\right) + \left(\frac{5}{6} \times \frac{1}{6}\right)$
$= \frac{5}{36} + \frac{5}{36} = \frac{10}{36} = \frac{5}{18}$.

(C) Given that,$P(E) = 1/5$ and $P(F|E) = 1/10$.
We need to find the probability of non-occurrence of at least one of the events $E$ and $F$,which is $P((E \cap F)^c) = 1 - P(E \cap F)$.
Using the multiplication theorem of probability,$P(E \cap F) = P(E) \times P(F|E)$.
Substituting the given values,$P(E \cap F) = (1/5) \times (1/10) = 1/50$.
Therefore,the required probability is $1 - P(E \cap F) = 1 - 1/50 = 49/50$.

(C) Let $H$ denote a head and $T$ denote a tail. $A$ head gives $+2$ points and a tail gives $-1$ point. When three coins are tossed,the possible outcomes for the number of heads $(H)$ and tails $(T)$ are:
$1$. Three tails $(0H, 3T)$: Score $= 0(2) + 3(-1) = -3$.
$2$. Two tails and one head $(1H, 2T)$: Score $= 1(2) + 2(-1) = 2 - 2 = 0$.
$3$. One tail and two heads $(2H, 1T)$: Score $= 2(2) + 1(-1) = 4 - 1 = 3$.
$4$. Three heads $(3H, 0T)$: Score $= 3(2) + 0(-1) = 6$.
Therefore,the range of the total score $X$ is $\{-3, 0, 3, 6\}$.

(A) Let $H$ denote a head and $T$ denote a tail. The points awarded are $2$ for $H$ and $1$ for $T$.
When three coins are tossed,let $n_H$ be the number of heads and $n_T$ be the number of tails.
The total points $S = 2n_H + 1n_T$.
Since $n_H + n_T = 3$,we have $n_T = 3 - n_H$.
Substituting this,$S = 2n_H + (3 - n_H) = n_H + 3$.
For $S$ to be an odd number,$n_H + 3$ must be odd,which implies $n_H$ must be an even number.
Possible values for $n_H$ are $0$ or $2$ (since $n_H \in \{0, 1, 2, 3\}$).
Case $1$: $n_H = 0$ (all tails,$TTT$).
Probability $P(n_H = 0) = \binom{3}{0} (\frac{1}{2})^0 (\frac{1}{2})^3 = \frac{1}{8}$.
Case $2$: $n_H = 2$ (two heads,one tail,e.g.,$HHT, HTH, THH$).
Probability $P(n_H = 2) = \binom{3}{2} (\frac{1}{2})^2 (\frac{1}{2})^1 = 3 \times \frac{1}{8} = \frac{3}{8}$.
The total probability of getting an odd number of points is $P(n_H = 0) + P(n_H = 2) = \frac{1}{8} + \frac{3}{8} = \frac{4}{8} = \frac{1}{2}$.

(D) According to Graham's law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.
Given,$V_{He} = 500 \ mL$,$t_{He} = 30 \ min$,so $r_{He} = \frac{500}{30} \ mL/min$.
For $SO_2$,$V_{SO_2} = 1000 \ mL$,$t_{SO_2} = t$,so $r_{SO_2} = \frac{1000}{t} \ mL/min$.
Molar masses are $M_{He} = 4 \ g/mol$ and $M_{SO_2} = 64 \ g/mol$.
Substituting the values: $\frac{500/30}{1000/t} = \sqrt{\frac{64}{4}}$.
$\frac{500}{30} \times \frac{t}{1000} = \sqrt{16} = 4$.
$\frac{t}{60} = 4 \Rightarrow t = 240 \ minutes$.
Converting to hours: $t = \frac{240}{60} = 4 \ hours$.

(A) The electronic configurations are as follows:
$K$ $(Z=19)$: $1s^2, 2s^2 2p^6, 3s^2 3p^6, 4s^1$. Electrons in $M$ shell $(n=3)$ = $8$.
$Mn$ $(Z=25)$: $1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^5, 4s^2$. Electrons in $M$ shell $(n=3)$ = $13$.
$Ni$ $(Z=28)$: $1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^8, 4s^2$. Electrons in $M$ shell $(n=3)$ = $16$.
$Sc$ $(Z=21)$: $1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^1, 4s^2$. Electrons in $M$ shell $(n=3)$ = $9$.
Thus,$K$ has the least number of electrons in its $M$ shell.

(A) thermocouple consists of two dissimilar metal wires joined at two junctions.
When one junction is kept at a constant temperature and the other is exposed to a changing temperature,a thermoelectric electromotive force $(EMF)$ is generated.
Because the junction of a thermocouple is very small,its thermal capacity is extremely low.
Thermal capacity is defined as the product of mass and specific heat capacity $(C = mc)$.
$A$ low thermal capacity means the junction can gain or lose heat very quickly to reach thermal equilibrium with the surroundings.
Therefore,it can respond rapidly to changes in temperature,making it suitable for measuring rapidly changing temperatures.
Thus,$(A)$ is true,$(R)$ is true,and $(R)$ is the correct explanation of $(A)$.

(C) The rate of cooling of a body is given by the formula: $\frac{d\theta}{dt} = \frac{\sigma A(T^4 - T_0^4)}{ms}$.
Here,$m = 34.38 \times 10^{-3} \ kg$,$A = 19.2 \times 10^{-4} \ m^2$,$T = 400 \ K$,$T_0 = 300 \ K$,and $\frac{d\theta}{dt} = 0.04 \ K/s$.
Rearranging for specific heat $s$: $s = \frac{\sigma A(T^4 - T_0^4)}{m(\frac{d\theta}{dt})}$.
Substituting the values:
$s = \frac{(5.73 \times 10^{-8}) \times (19.2 \times 10^{-4}) \times (400^4 - 300^4)}{(34.38 \times 10^{-3}) \times 0.04}$.
$s = \frac{(5.73 \times 10^{-8}) \times (19.2 \times 10^{-4}) \times (256 \times 10^8 - 81 \times 10^8)}{34.38 \times 10^{-3} \times 0.04}$.
$s = \frac{5.73 \times 19.2 \times 10^{-12} \times 175 \times 10^8}{34.38 \times 10^{-3} \times 0.04}$.
$s = \frac{19249.92 \times 10^{-4}}{1.3752 \times 10^{-3}} \approx 1400 \ J \ kg^{-1} \ K^{-1}$.

(D) For an ideal monoatomic gas,the internal energy is given by $U = n C_V T$,where $C_V = \frac{3}{2} R$.
When two gases are mixed in an adiabatic container,the total internal energy is conserved: $U_{mix} = U_1 + U_2$.
$(n_1 + n_2) C_V T_{mix} = n_1 C_V T_1 + n_2 C_V T_2$.
Since the gases are both monoatomic,$C_V$ cancels out:
$T_{mix} = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}$.
Substituting the given values:
$T_{mix} = \frac{4(400) + 2(700)}{4 + 2}$.
$T_{mix} = \frac{1600 + 1400}{6} = \frac{3000}{6} = 500 ~K$.