AP EAMCET 2009 Chemistry Question Paper with Answer and Solution

(A) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$.
Taking the logarithmic derivative, we get $\frac{dT}{T} = \frac{1}{2} \frac{dl}{l}$.
Since the change in length is given by $\frac{dl}{l} = \alpha \Delta \theta$, where $\Delta \theta$ is the change in temperature, we have $\frac{dT}{T} = \frac{1}{2} \alpha \Delta \theta$.
Given $\alpha = 9 \times 10^{-7} /{ }^{\circ} C$, $\Delta \theta = 30^{\circ} C - 20^{\circ} C = 10^{\circ} C$, and $T = 0.5 \, s$.
The change in time period (loss in time per oscillation) is $dT = T \times \frac{1}{2} \alpha \Delta \theta$.
Substituting the values: $dT = 0.5 \times \frac{1}{2} \times 9 \times 10^{-7} \times 10$.
$dT = 0.5 \times 4.5 \times 10^{-6} = 2.25 \times 10^{-6} \, s$.

(D) The displacement of the particle executing $SHM$ is given by $y=5 \sin \left(4 t+\frac{\pi}{3}\right)$.
The velocity of the particle is $v = \frac{dy}{dt} = 5 \cdot \cos \left(4 t+\frac{\pi}{3}\right) \cdot 4 = 20 \cos \left(4 t+\frac{\pi}{3}\right)$.
Comparing the given equation with the standard $SHM$ equation $y=a \sin (\omega t+\phi)$,we get angular frequency $\omega = 4 \text{ rad/s}$.
Since $\omega = \frac{2\pi}{T}$,the time period is $T = \frac{2\pi}{\omega} = \frac{2\pi}{4} = \frac{\pi}{2} \text{ s}$.
At $t = \frac{T}{4} = \frac{1}{4} \cdot \frac{\pi}{2} = \frac{\pi}{8} \text{ s}$,the velocity is:
$v = 20 \cos \left(4 \cdot \frac{\pi}{8} + \frac{\pi}{3}\right) = 20 \cos \left(\frac{\pi}{2} + \frac{\pi}{3}\right) = -20 \sin \left(\frac{\pi}{3}\right) = -20 \cdot \frac{\sqrt{3}}{2} = -10\sqrt{3} \text{ m/s}$.
Given mass $m = 2 \text{ g} = 2 \times 10^{-3} \text{ kg}$.
The kinetic energy is $KE = \frac{1}{2}mv^2 = \frac{1}{2} \cdot (2 \times 10^{-3}) \cdot (-10\sqrt{3})^2 = 10^{-3} \cdot (100 \cdot 3) = 300 \times 10^{-3} = 0.3 \text{ J}$.

(C) Aluminium reacts with aqueous $NaOH$ to form sodium tetrahydroxoaluminate$(III)$.
The reaction is: $2Al(s) + 2NaOH(aq) + 6H_2O(l) \longrightarrow 2Na[Al(OH)_4](aq) + 3H_2(g)$.
In aqueous solution,this exists as the complex ion $[Al(H_2O)_2(OH)_4]^{-}$.
In this complex,the aluminium atom is coordinated to $2$ water molecules and $4$ hydroxide ions,resulting in a coordination number of $2 + 4 = 6$.

(D) The hydrolysis of $SiCl_4$ produces orthosilicic acid $(H_4SiO_4)$,which is denoted as '$X$'.
$SiCl_4 + 4H_2O \rightarrow H_4SiO_4 (X) + 4HCl$
Upon heating at $1000^\circ C$,orthosilicic acid undergoes dehydration to form silica $(SiO_2)$,which is denoted as '$Y$'.
$H_4SiO_4 \xrightarrow{1000^\circ C} SiO_2 (Y) + 2H_2O$
Therefore,'$X$' is $H_4SiO_4$ and '$Y$' is $SiO_2$.

(C) The presence of $P-H$ bonds in phosphorus oxyacids is determined by their structures:
$1$. $H_3PO_3$ (Phosphorous acid) has one $P-H$ bond.
$2$. $H_3PO_2$ (Hypophosphorous acid) has two $P-H$ bonds.
$3$. $H_3PO_4$ (Orthophosphoric acid) has no $P-H$ bonds.
Therefore,the pair $H_3PO_3$ and $H_3PO_2$ contains $P-H$ bonds.

(B) The reaction of fluorine with dilute $NaOH$ is given by:
$2F_2 + 2NaOH \rightarrow 2NaF + OF_2 + H_2O$
Thus,the gaseous product $A$ is oxygen difluoride $(OF_2)$.
In $OF_2$,the central oxygen atom is $sp^3$ hybridized with two bond pairs and two lone pairs.
Due to the high electronegativity of fluorine atoms,the lone pair-lone pair repulsion is significant,which compresses the bond angle.
The bond angle in $OF_2$ is $103^{\circ}$.

(C) The total number of subsets of the set $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ is $2^9 = 512$.
To find the number of subsets containing at least one odd number,we use the complement method:
Total subsets - Subsets containing no odd numbers.
$A$ subset contains no odd numbers if it only contains even numbers from the set $S$.
The even numbers in $S$ are $\{2, 4, 6, 8\}$.
The number of subsets formed using only these even numbers is $2^4 = 16$.
These $16$ subsets include the empty set $\emptyset$.
Therefore,the number of subsets containing at least one odd number is $512 - 16 = 496$.

(A) To evaluate the product $P = \cos A \cos 2 A \cos 4 A \ldots \cos 2^{n-1} A$,multiply and divide by $2 \sin A$:
$P = \frac{2 \sin A \cos A \cos 2 A \cos 4 A \ldots \cos 2^{n-1} A}{2 \sin A}$
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we get:
$P = \frac{\sin 2A \cos 2A \cos 4 A \ldots \cos 2^{n-1} A}{2 \sin A}$
Multiply and divide by $2$ again:
$P = \frac{2 \sin 2A \cos 2A \cos 4 A \ldots \cos 2^{n-1} A}{2^2 \sin A} = \frac{\sin 4A \cos 4A \ldots \cos 2^{n-1} A}{2^2 \sin A}$
Repeating this process $n$ times,we obtain:
$P = \frac{\sin 2^n A}{2^n \sin A}$

(D) Step $1$: Find the point of intersection of the lines $x + 3y - 1 = 0$ and $x - 2y + 4 = 0$.
Subtracting the second equation from the first: $(x + 3y - 1) - (x - 2y + 4) = 0$ $\Rightarrow 5y - 5 = 0$ $\Rightarrow y = 1$.
Substituting $y = 1$ into $x + 3y - 1 = 0$,we get $x + 3(1) - 1 = 0 \Rightarrow x = -2$.
So,the point of intersection is $(-2, 1)$.
Step $2$: Find the equation of the line perpendicular to $3x + 2y = 0$.
The slope of $3x + 2y = 0$ is $m_1 = -3/2$.
The slope of the perpendicular line $m_2$ satisfies $m_1 \times m_2 = -1$,so $m_2 = 2/3$.
The equation of the line with slope $2/3$ passing through $(-2, 1)$ is $y - 1 = (2/3)(x + 2)$.
Multiplying by $3$: $3y - 3 = 2x + 4 \Rightarrow 2x - 3y + 7 = 0$.

(D) The given lines are $4x + 3y - 15 = 0$ and $4x + 3y - 5 = 0$. Since these lines are parallel,the distance $d$ between them is the diameter of the circle.
$d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} = \frac{|-15 - (-5)|}{\sqrt{4^2 + 3^2}} = \frac{|-10|}{\sqrt{16 + 9}} = \frac{10}{5} = 2$.
Since the diameter $d = 2$,the radius $r = \frac{d}{2} = 1$.
The area of the circle is $\pi r^2 = \pi(1)^2 = \pi \text{ square units}$.

(B) Let the point be $P(x, y)$. Since $P$ lies on the line $3x + 4y = 5$,we have $3x + 4y = 5$ ... $(i)$.
Since $P$ is equidistant from $A(1, 2)$ and $B(3, 4)$,we have $PA^2 = PB^2$.
$(x - 1)^2 + (y - 2)^2 = (x - 3)^2 + (y - 4)^2$
$x^2 - 2x + 1 + y^2 - 4y + 4 = x^2 - 6x + 9 + y^2 - 8y + 16$
$-2x - 4y + 5 = -6x - 8y + 25$
$4x + 4y = 20$
$x + y = 5$ ... $(ii)$.
From $(ii)$,$y = 5 - x$. Substituting this into $(i)$:
$3x + 4(5 - x) = 5$
$3x + 20 - 4x = 5$
$-x = -15 \Rightarrow x = 15$.
Then $y = 5 - 15 = -10$.
Thus,the point is $(15, -10)$.

(C) The given pair of straight lines is $x^2-3xy+2y^2=0$.
Factoring the equation: $x^2-2xy-xy+2y^2=0$ $\Rightarrow x(x-2y)-y(x-2y)=0$ $\Rightarrow (x-y)(x-2y)=0$.
This represents two lines: $L_1: x-y=0$ and $L_2: x-2y=0$.
The third line is $L_3: x+y+1=0$.
To find the vertices of the triangle,we find the intersection points of these lines:
$1$. Intersection of $L_1$ and $L_2$: Solving $x=y$ and $x=2y$ gives $(0,0)$.
$2$. Intersection of $L_1$ and $L_3$: Solving $x=y$ and $x+y+1=0$ gives $2x+1=0 \Rightarrow x=-\frac{1}{2}, y=-\frac{1}{2}$. Point is $(-\frac{1}{2}, -\frac{1}{2})$.
$3$. Intersection of $L_2$ and $L_3$: Solving $x=2y$ and $x+y+1=0$ gives $2y+y+1=0$ $\Rightarrow 3y=-1$ $\Rightarrow y=-\frac{1}{3}, x=-\frac{2}{3}$. Point is $(-\frac{2}{3}, -\frac{1}{3})$.
The area of the triangle with vertices $(0,0)$,$(-\frac{1}{2}, -\frac{1}{2})$,and $(-\frac{2}{3}, -\frac{1}{3})$ is given by:
Area $= \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$
Area $= \frac{1}{2} |0(-\frac{1}{2} - (-\frac{1}{3})) + (-\frac{1}{2})(-\frac{1}{3} - 0) + (-\frac{2}{3})(0 - (-\frac{1}{2}))|$
Area $= \frac{1}{2} |0 + \frac{1}{6} - \frac{1}{3}| = \frac{1}{2} |-\frac{1}{6}| = \frac{1}{12}$ square units.

(C) The given equations are $x^2-3xy+2y^2=0$ and $x^2-3xy+2y^2+x-2=0$.
Factorizing the first equation: $(x-2y)(x-y)=0$,which gives lines $L_1: x-2y=0$ and $L_2: x-y=0$.
Factorizing the second equation: $(x-2y+2)(x-y-1)=0$,which gives lines $L_3: x-2y+2=0$ and $L_4: x-y-1=0$.
Since $L_1 \parallel L_3$ and $L_2 \parallel L_4$,the figure formed is a parallelogram.
The slopes of the lines are $m_1 = 1/2$ and $m_2 = 1$.
Since $m_1 \times m_2 \neq -1$,the angle between the lines is not $90^{\circ}$.
Thus,the figure is a parallelogram.

(C) The given pairs of lines are $x^2-3xy+2y^2=0$ and $x^2-3xy+2y^2+x-2=0$.
Factorizing the first equation: $(x-2y)(x-y)=0$,which represents lines $x-2y=0$ and $x-y=0$.
Factorizing the second equation: $(x-2y+2)(x-y-1)=0$,which represents lines $x-2y+2=0$ and $x-y-1=0$.
Since the lines $x-2y=0$ and $x-2y+2=0$ are parallel,and the lines $x-y=0$ and $x-y-1=0$ are parallel,the figure formed is a parallelogram.
To check if it is a rectangle,we find the angle between $x-2y=0$ and $x-y=0$. The slopes are $m_1 = 1/2$ and $m_2 = 1$. Since $m_1 \times m_2 \neq -1$,the angle is not $90^{\circ}$.
Thus,it is a parallelogram.

(B) The given equation is $2 x^2-10 x y+12 y^2+5 x+\lambda y-3=0$.
Comparing this with the general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a=2, h=-5, b=12, g=5/2, f=\lambda/2, c=-3$.
For the equation to represent a pair of straight lines,the determinant of the matrix must be zero:
$\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0$.
Substituting the values:
$\begin{vmatrix} 2 & -5 & 5/2 \\ -5 & 12 & \lambda/2 \\ 5/2 & \lambda/2 & -3 \end{vmatrix} = 0$.
Expanding the determinant:
$2(-36 - \lambda^2/4) + 5(15 - 5\lambda/4) + (5/2)(-5\lambda/2 - 30) = 0$.
$-72 - \lambda^2/2 + 75 - 25\lambda/4 - 25\lambda/4 - 75 = 0$.
$-\lambda^2/2 - 25\lambda/2 - 72 = 0$.
Multiplying by $-2$:
$\lambda^2 + 25\lambda + 144 = 0$.
Factoring the quadratic:
$(\lambda + 9)(\lambda + 16) = 0$.
Thus,$\lambda = -9$ or $\lambda = -16$.
Given the condition $|\lambda| < 16$,the only valid solution is $\lambda = -9$.

(B) The given equation is $2 x^2 - 10 x y + 12 y^2 + 5 x + \lambda y - 3 = 0$.
Comparing this with the general second-degree equation $a x^2 + 2 h x y + b y^2 + 2 g x + 2 f y + c = 0$,we get:
$a = 2, h = -5, b = 12, g = \frac{5}{2}, f = \frac{\lambda}{2}, c = -3$.
For the equation to represent a pair of straight lines,the determinant must be zero:
$\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0$.
Substituting the values:
$\begin{vmatrix} 2 & -5 & 5/2 \\ -5 & 12 & \lambda/2 \\ 5/2 & \lambda/2 & -3 \end{vmatrix} = 0$.
Expanding the determinant:
$2 \left( -36 - \frac{\lambda^2}{4} \right) + 5 \left( 15 - \frac{5 \lambda}{4} \right) + \frac{5}{2} \left( -\frac{5 \lambda}{2} - 30 \right) = 0$.
$-72 - \frac{\lambda^2}{2} + 75 - \frac{25 \lambda}{4} - \frac{25 \lambda}{4} - 75 = 0$.
$-\frac{\lambda^2}{2} - \frac{50 \lambda}{4} - 72 = 0$.
$-\frac{\lambda^2}{2} - \frac{25 \lambda}{2} - 72 = 0$.
Multiplying by $-2$:
$\lambda^2 + 25 \lambda + 144 = 0$.
$(\lambda + 9)(\lambda + 16) = 0$.
So,$\lambda = -9$ or $\lambda = -16$.
Given $|\lambda| < 16$,the only valid solution is $\lambda = -9$.

(A) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.
Since the circle passes through the origin $(0, 0)$,we have $c = 0$.
The circle makes an intercept of length $4$ on the $x$-axis,so $2\sqrt{g^2 - c} = 4$. Since $c = 0$,$2|g| = 4$,which implies $g = \pm 2$.
The circle makes an intercept of length $8$ on the $y$-axis,so $2\sqrt{f^2 - c} = 8$. Since $c = 0$,$2|f| = 8$,which implies $f = \pm 4$.
Substituting $g = \pm 2$ and $f = \pm 4$ into the equation $x^2 + y^2 + 2gx + 2fy = 0$,we get $x^2 + y^2 \pm 4x \pm 8y = 0$.

(C) The intersection point of the diameter lines is the center of the circle. Solving $2x+y=7$ and $x+3y=11$:
From the first equation,$y=7-2x$.
Substituting into the second: $x+3(7-2x)=11$ $\Rightarrow x+21-6x=11$ $\Rightarrow -5x=-10$ $\Rightarrow x=2$.
Then $y=7-2(2)=3$.
So,the center is $(h,k) = (2,3)$.
The radius $r$ is the distance between the center $(2,3)$ and the point $(5,7)$:
$r = \sqrt{(5-2)^2+(7-3)^2} = \sqrt{3^2+4^2} = \sqrt{9+16} = 5$.
The equation of the circle is $(x-h)^2+(y-k)^2=r^2$:
$(x-2)^2+(y-3)^2 = 5^2$
$x^2-4x+4+y^2-6y+9 = 25$
$x^2+y^2-4x-6y+13-25 = 0$
$x^2+y^2-4x-6y-12 = 0$.

(D) The given circles are $S_1: x^2+y^2-2x+8y+13=0$ and $S_2: x^2+y^2-4x+6y+11=0$.
For $S_1$,the center $C_1 = (1, -4)$ and radius $r_1 = \sqrt{1^2+(-4)^2-13} = \sqrt{1+16-13} = 2$.
For $S_2$,the center $C_2 = (2, -3)$ and radius $r_2 = \sqrt{2^2+(-3)^2-11} = \sqrt{4+9-11} = \sqrt{2}$.
The distance between the centers $d = C_1C_2 = \sqrt{(2-1)^2+(-3-(-4))^2} = \sqrt{1^2+1^2} = \sqrt{2}$.
The angle $\theta$ between the circles is given by $\cos \theta = \frac{d^2-r_1^2-r_2^2}{2r_1r_2}$.
Substituting the values,$\cos \theta = \frac{(\sqrt{2})^2 - 2^2 - (\sqrt{2})^2}{2 \times 2 \times \sqrt{2}} = \frac{2-4-2}{4\sqrt{2}} = \frac{-4}{4\sqrt{2}} = -\frac{1}{\sqrt{2}}$.
Thus,$\theta = 135^{\circ}$.

(C) Let the required equation of the circle be $x^2+y^2+2gx+2fy=0$ (since it passes through the origin,$c=0$).
For the circle $x^2+y^2+2g_1x+2f_1y+c_1=0$ to cut $x^2+y^2+2g_2x+2f_2y+c_2=0$ orthogonally,the condition is $2g_1g_2+2f_1f_2=c_1+c_2$.
For the first circle $x^2+y^2-6x+8=0$,we have $g_1=-3, f_1=0, c_1=8$. Applying the condition: $2g(-3)+2f(0)=0+8$ $\Rightarrow -6g=8$ $\Rightarrow g=-\frac{4}{3}$.
For the second circle $x^2+y^2-2x-2y-7=0$,we have $g_2=-1, f_2=-1, c_2=-7$. Applying the condition: $2g(-1)+2f(-1)=0-7 \Rightarrow -2g-2f=-7$.
Substituting $g=-\frac{4}{3}$ into the second equation: $-2(-\frac{4}{3})-2f=-7$ $\Rightarrow \frac{8}{3}-2f=-7$ $\Rightarrow 2f = 7+\frac{8}{3} = \frac{29}{3}$ $\Rightarrow f=\frac{29}{6}$.
The equation is $x^2+y^2+2(-\frac{4}{3})x+2(\frac{29}{6})y=0$.
$x^2+y^2-\frac{8}{3}x+\frac{29}{3}y=0$.
Multiplying by $3$,we get $3x^2+3y^2-8x+29y=0$.

(B) Let the centre of the circle be $C(h, k)$. Since the circle passes through the origin $(0,0)$,its radius $r$ is given by $r^2 = h^2 + k^2$.
The equation of the circle is $(x-h)^2 + (y-k)^2 = h^2 + k^2$,which simplifies to $x^2 + y^2 - 2hx - 2ky = 0$.
The perpendicular distance $d$ from the centre $(h, k)$ to the line $x=3$ (or $x-3=0$) is $d = |h-3|$.
In the right-angled triangle formed by the centre,the midpoint of the chord,and an endpoint of the chord,we have $r^2 = d^2 + (L/2)^2$,where $L=4$ is the length of the chord.
Thus,$h^2 + k^2 = (h-3)^2 + 2^2$.
$h^2 + k^2 = h^2 - 6h + 9 + 4$.
$k^2 = -6h + 13$.
Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 = -6x + 13$,or $y^2 + 6x = 13$.

(B) The equation of the parabola is $y^2 = 4ax$,where $a = 1$. The focus of the parabola is $(a, 0) = (1, 0)$.
Any normal to the parabola $y^2 = 4ax$ at point $(at^2, 2at)$ is given by $y = -tx + 2at + at^3$.
Since the normal passes through $(1, 0)$,we substitute $x = 1$ and $y = 0$ into the equation:
$0 = -t(1) + 2(1)t + (1)t^3$
$0 = -t + 2t + t^3$
$0 = t + t^3$
$t(1 + t^2) = 0$
Since $t$ must be a real number,the only solution is $t = 0$.
For $t = 0$,the normal is $y = 0$,which is the $x$-axis.
Thus,only $1$ normal can be drawn from the point $(1, 0)$ to the parabola.

(B) We are given the expression $(1+x^2)^{12}(1+x^{12})(1+x^{24})$.
First,expand the product $(1+x^{12})(1+x^{24}) = 1 + x^{12} + x^{24} + x^{36}$.
Now,the expression becomes $(1+x^2)^{12}(1 + x^{12} + x^{24} + x^{36})$.
Using the binomial theorem,$(1+x^2)^{12} = \sum_{k=0}^{12} {}^{12}C_k (x^2)^k = \sum_{k=0}^{12} {}^{12}C_k x^{2k}$.
We need the coefficient of $x^{24}$ in the product $(\sum_{k=0}^{12} {}^{12}C_k x^{2k})(1 + x^{12} + x^{24} + x^{36})$.
This is obtained by summing the coefficients of terms that result in $x^{24}$:
$1$. From $1 \times ({}^{12}C_{12} x^{24}) = {}^{12}C_{12} x^{24} = 1 \cdot x^{24}$.
$2$. From $x^{12} \times ({}^{12}C_6 x^{12}) = {}^{12}C_6 x^{24}$.
$3$. From $x^{24} \times ({}^{12}C_0 x^0) = {}^{12}C_0 x^{24} = 1 \cdot x^{24}$.
Summing these,the coefficient of $x^{24}$ is ${}^{12}C_6 + 1 + 1 = {}^{12}C_6 + 2$.

(D) Given the expression $\frac{1}{(x-1)^2(x-2)}$.
We can rewrite this as $\frac{1}{(-1)^2(1-x)^2(-2)(1-\frac{x}{2})} = \frac{1}{1 \cdot (1-x)^2 \cdot (-2) \cdot (1-\frac{x}{2})} = -\frac{1}{2}(1-x)^{-2}(1-\frac{x}{2})^{-1}$.
Using the binomial expansion $(1-z)^{-n} = 1 + nz + \frac{n(n+1)}{2!}z^2 + \dots$,we have:
$(1-x)^{-2} = 1 + 2x + 3x^2 + \dots$
$(1-\frac{x}{2})^{-1} = 1 + \frac{x}{2} + \frac{x^2}{4} + \dots$
Multiplying these series:
$-\frac{1}{2}(1 + 2x + \dots)(1 + \frac{x}{2} + \dots) = -\frac{1}{2}(1 + \frac{x}{2} + 2x + \dots) = -\frac{1}{2}(1 + \frac{5x}{2} + \dots)$.
The constant term is the term independent of $x$,which is $-\frac{1}{2} \times 1 = -\frac{1}{2}$.

(D) Given,$\frac{1}{e^{3x}}(e^x + e^{5x}) = a_0 + a_1x + a_2x^2 + \ldots$
$\Rightarrow e^{-2x} + e^{2x} = a_0 + a_1x + a_2x^2 + \ldots$
Using the expansion $e^y + e^{-y} = 2(1 + \frac{y^2}{2!} + \frac{y^4}{4!} + \ldots)$,we get:
$2(1 + \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} + \ldots) = a_0 + a_1x + a_2x^2 + \ldots$
Comparing the coefficients of $x^n$ on both sides,we observe that all odd-indexed coefficients $a_1, a_3, a_5, \ldots$ are $0$.
Therefore,$2a_1 + 2^3a_3 + 2^5a_5 + \ldots = 2(0) + 2^3(0) + 2^5(0) + \ldots = 0$.

(A) Using the binomial approximation $(1+u)^n \approx 1+nu$ for small $u$:
$\left(1+\frac{2 x}{3}\right)^{3 / 2} \approx 1 + \frac{3}{2} \cdot \frac{2x}{3} = 1+x$
For the second term:
$(32+5 x)^{-1 / 5} = 32^{-1 / 5} \left(1+\frac{5 x}{32}\right)^{-1 / 5} = \frac{1}{2} \left(1+\frac{5 x}{32}\right)^{-1 / 5}$
Using the approximation $(1+u)^n \approx 1+nu$:
$\frac{1}{2} \left(1 - \frac{1}{5} \cdot \frac{5x}{32}\right) = \frac{1}{2} \left(1 - \frac{x}{32}\right)$
Multiplying the two approximations:
$(1+x) \cdot \frac{1}{2} \left(1 - \frac{x}{32}\right) = \frac{1}{2} \left(1 - \frac{x}{32} + x - \frac{x^2}{32}\right)$
Neglecting $x^2$:
$\approx \frac{1}{2} \left(1 + \frac{31x}{32}\right) = \frac{32+31x}{64}$

(C) Given that the distance between the foci is $2ae = 6$,so $ae = 3$.
Given that the length of the minor axis is $2b = 8$,so $b = 4$.
We know the relation $b^2 = a^2(1 - e^2)$,which can be written as $b^2 = a^2 - a^2e^2$.
Substituting $b = 4$ and $ae = 3$:
$4^2 = a^2 - (ae)^2$
$16 = a^2 - 3^2$
$16 = a^2 - 9$
$a^2 = 25 \Rightarrow a = 5$.
Now,using $ae = 3$:
$5e = 3$
$e = \frac{3}{5}$.

(D) Given,$\frac{5}{r} = 2 + 3 \cos \theta + 4 \sin \theta$.
Divide by $2$: $\frac{5/2}{r} = 1 + \frac{3}{2} \cos \theta + 2 \sin \theta$.
We can write $\frac{3}{2} \cos \theta + 2 \sin \theta$ in the form $e \cos(\theta - \phi)$,where $e = \sqrt{(\frac{3}{2})^2 + 2^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2}$.
Thus,the equation becomes $\frac{5/2}{r} = 1 + \frac{5}{2} \cos(\theta - \phi)$.
Comparing this with the standard polar form of a conic $\frac{l}{r} = 1 + e \cos(\theta - \phi)$,we get the eccentricity $e = \frac{5}{2}$.

(B) The equation of the chord of a hyperbola $S = 0$ with midpoint $(x_1, y_1)$ is given by $T = S_1$,where $T$ is the tangent at $(x_1, y_1)$ and $S_1$ is the value of the hyperbola equation at $(x_1, y_1)$.
For the hyperbola $2x^2 - 3y^2 - 12 = 0$,the equation of the chord with midpoint $(x_1, y_1)$ is $2xx_1 - 3yy_1 - 12 = 2x_1^2 - 3y_1^2 - 12$,which simplifies to $2xx_1 - 3yy_1 = 2x_1^2 - 3y_1^2$.
Comparing this with the given chord equation $4x - 3y = 5$,we have:
$\frac{2x_1}{4} = \frac{-3y_1}{-3} = \frac{2x_1^2 - 3y_1^2}{5}$.
From $\frac{2x_1}{4} = y_1$,we get $x_1 = 2y_1$.
Substituting $x_1 = 2y_1$ into $\frac{2x_1}{4} = \frac{2x_1^2 - 3y_1^2}{5}$:
$\frac{y_1}{1} = \frac{2(2y_1)^2 - 3y_1^2}{5}$ $\Rightarrow 5y_1 = 8y_1^2 - 3y_1^2$ $\Rightarrow 5y_1 = 5y_1^2$.
This gives $y_1^2 - y_1 = 0$,so $y_1 = 0$ or $y_1 = 1$.
If $y_1 = 0$,then $x_1 = 0$. Checking the point $(0, 0)$ in the chord equation $4(0) - 3(0) = 0 \neq 5$. So,$y_1 = 0$ is rejected.
If $y_1 = 1$,then $x_1 = 2(1) = 2$. The point $(2, 1)$ satisfies the chord equation $4(2) - 3(1) = 8 - 3 = 5$.
Thus,the midpoint is $(2, 1)$.

(A) Given equations are $x^2+y^2=a^2$ and $xy=c^2$.
From the second equation,$x = \frac{c^2}{y}$.
Substituting this into the first equation:
$\left(\frac{c^2}{y}\right)^2 + y^2 = a^2$
$\frac{c^4}{y^2} + y^2 = a^2$
$c^4 + y^4 = a^2 y^2$
$y^4 - a^2 y^2 + c^4 = 0$
This is a biquadratic equation in $y$. Let the roots be $y_1, y_2, y_3, y_4$.
The sum of the roots of the equation $y^4 + 0y^3 - a^2 y^2 + 0y + c^4 = 0$ is given by the coefficient of $y^3$ divided by the coefficient of $y^4$ with a negative sign.
Sum $= -\frac{0}{1} = 0$.
Therefore,$y_1+y_2+y_3+y_4 = 0$.

(C) We know that $\lim _{x \rightarrow \infty} (1 + \frac{a}{x})^x = e^a$.
Given expression: $\lim _{x \rightarrow \infty} (\frac{x+5}{x+2})^{x+3}$
$= \lim _{x \rightarrow \infty} (1 + \frac{3}{x+2})^{x+3}$
$= \lim _{x \rightarrow \infty} [(1 + \frac{3}{x+2})^{\frac{x+2}{3}}]^{\frac{3(x+3)}{x+2}}$
$= e^{\lim _{x \rightarrow \infty} \frac{3x+9}{x+2}}$
$= e^{\lim _{x \rightarrow \infty} \frac{3 + 9/x}{1 + 2/x}}$
$= e^3$

(C) The mean $m$ is calculated as $E[X] = \sum p_i x_i$:
$m = (0 \times \frac{1}{3}) + (1 \times \frac{1}{2}) + (2 \times 0) + (3 \times \frac{1}{6})$
$m = 0 + \frac{1}{2} + 0 + \frac{1}{2} = 1$
The variance $\sigma^2$ is calculated as $E[X^2] - (E[X])^2$:
$E[X^2] = \sum p_i x_i^2 = (0^2 \times \frac{1}{3}) + (1^2 \times \frac{1}{2}) + (2^2 \times 0) + (3^2 \times \frac{1}{6})$
$E[X^2] = 0 + \frac{1}{2} + 0 + \frac{9}{6} = \frac{1}{2} + \frac{3}{2} = 2$
$\sigma^2 = E[X^2] - m^2 = 2 - (1)^2 = 2 - 1 = 1$
Thus,$m = 1$ and $\sigma^2 = 1$,so $m = \sigma^2 = 1$.

(B) Using the projection formula or the cosine rule: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ and $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.
Substituting these into the expression:
$a(b \cos C - c \cos B) = a(b \cdot \frac{a^2 + b^2 - c^2}{2ab} - c \cdot \frac{a^2 + c^2 - b^2}{2ac})$
$= a(\frac{a^2 + b^2 - c^2}{2a} - \frac{a^2 + c^2 - b^2}{2a})$
$= \frac{a^2 + b^2 - c^2 - (a^2 + c^2 - b^2)}{2}$
$= \frac{a^2 + b^2 - c^2 - a^2 - c^2 + b^2}{2}$
$= \frac{2b^2 - 2c^2}{2}$
$= b^2 - c^2$.

(C) Let $2s = a+b+c$. Then $b+c-a = 2s-2a$,$c+a-b = 2s-2b$,and $a+b-c = 2s-2c$.
Substituting these into the expression:
$\frac{(2s)(2s-2a)(2s-2b)(2s-2c)}{4b^2c^2} = \frac{16s(s-a)(s-b)(s-c)}{4b^2c^2} = 4 \frac{s(s-a)}{bc} \cdot \frac{(s-b)(s-c)}{bc}$.
Using the half-angle formulas $\cos^2(\frac{A}{2}) = \frac{s(s-a)}{bc}$ and $\sin^2(\frac{A}{2}) = \frac{(s-b)(s-c)}{bc}$,we get:
$4 \cos^2(\frac{A}{2}) \sin^2(\frac{A}{2}) = (2 \sin(\frac{A}{2}) \cos(\frac{A}{2}))^2 = \sin^2 A$.

(D) Let $s = \frac{a+b+c}{2}$ be the semi-perimeter of the triangle,so $a+b+c = 2s$.
Then,$b+c-a = 2s-2a$,$c+a-b = 2s-2b$,and $a+b-c = 2s-2c$.
The expression becomes:
$\frac{(2s)(2s-2a)(2s-2b)(2s-2c)}{4b^2c^2} = \frac{16s(s-a)(s-b)(s-c)}{4b^2c^2}$.
Using Heron's formula,$\Delta^2 = s(s-a)(s-b)(s-c)$,so the expression is $\frac{16\Delta^2}{4b^2c^2} = \frac{4\Delta^2}{b^2c^2}$.
This can be written as $\left(\frac{2\Delta}{bc}\right)^2$.
Since the area of a triangle is $\Delta = \frac{1}{2}bc \sin A$,we have $\sin A = \frac{2\Delta}{bc}$.
Therefore,the expression equals $\sin^2 A$.

(B) Given that $A$ and $B$ are symmetric matrices,we have $A^{\prime} = A$ and $B^{\prime} = B$.
Consider the transpose of the matrix $(AB - BA)$:
$(AB - BA)^{\prime} = (AB)^{\prime} - (BA)^{\prime}$
Using the property $(XY)^{\prime} = Y^{\prime}X^{\prime}$,we get:
$(AB - BA)^{\prime} = B^{\prime}A^{\prime} - A^{\prime}B^{\prime}$
Since $A^{\prime} = A$ and $B^{\prime} = B$,we substitute these values:
$(AB - BA)^{\prime} = BA - AB$
$(AB - BA)^{\prime} = -(AB - BA)$
Since the transpose of the matrix $(AB - BA)$ is equal to the negative of the matrix itself,$AB - BA$ is a skew-symmetric matrix.

(D) matrix has no inverse if and only if its determinant is equal to $0$.
Let $A = \left[\begin{array}{rrr}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right]$.
We set $|A| = 0$:
$|A| = 1(x - (-1)) - (-1)(1 - x) + x(-1 - x^2) = 0$
$|A| = 1(x + 1) + 1(1 - x) + x(-1 - x^2) = 0$
$|A| = x + 1 + 1 - x - x - x^3 = 0$
$|A| = -x^3 - x + 2 = 0$
$x^3 + x - 2 = 0$
By inspection,if $x = 1$,then $1^3 + 1 - 2 = 0$,which satisfies the equation.
Alternatively,if $x = 1$,the matrix becomes $\left[\begin{array}{rrr}1 & -1 & 1 \\ 1 & 1 & 1 \\ 1 & -1 & 1\end{array}\right]$.
Since the first and third columns are identical,the determinant is $0$.
Thus,the real value of $x$ is $1$.

(A) Given the determinant equation: $\left|\begin{array}{lll}3 & 5 & x \\ 7 & x & 7 \\ x & 5 & 3\end{array}\right|=0$
Expanding along the first row:
$3(3x - 35) - 5(21 - 7x) + x(35 - x^2) = 0$
$9x - 105 - 105 + 35x + 35x - x^3 = 0$
$-x^3 + 79x - 210 = 0$
$x^3 - 79x + 210 = 0$
Since $x = -10$ is a root,$(x + 10)$ is a factor. Dividing $x^3 - 79x + 210$ by $(x + 10)$ gives $(x^2 - 10x + 21)$.
Factoring the quadratic: $(x^2 - 10x + 21) = (x - 3)(x - 7)$.
Thus,the equation is $(x + 10)(x - 3)(x - 7) = 0$.
The roots are $x = -10, 3, 7$.
Therefore,the other roots are $3$ and $7$.

(A) Let $a$ be the first term and $R$ be the common ratio of the geometric progression $(GP)$.
Then,the $n$-th term is given by $T_n = a R^{n-1}$.
Given that $x, y, z$ are the $p$-th,$q$-th,and $r$-th terms respectively:
$x = a R^{p-1}$,$y = a R^{q-1}$,$z = a R^{r-1}$.
Taking the logarithm on both sides:
$\log x = \log a + (p-1) \log R$
$\log y = \log a + (q-1) \log R$
$\log z = \log a + (r-1) \log R$
Now,substitute these into the determinant $\Delta = \left|\begin{array}{lll} \log x & p & 1 \\ \log y & q & 1 \\ \log z & r & 1 \end{array}\right|$.
$\Delta = \left|\begin{array}{lll} \log a + (p-1) \log R & p & 1 \\ \log a + (q-1) \log R & q & 1 \\ \log a + (r-1) \log R & r & 1 \end{array}\right|$.
Using the property of determinants,we can split this into two determinants:
$\Delta = \left|\begin{array}{lll} \log a & p & 1 \\ \log a & q & 1 \\ \log a & r & 1 \end{array}\right| + \left|\begin{array}{lll} (p-1) \log R & p & 1 \\ (q-1) \log R & q & 1 \\ (r-1) \log R & r & 1 \end{array}\right|$.
In the first determinant,the first column is $\log a$ times the third column,so it is $0$.
In the second determinant,perform the operation $C_1 \rightarrow C_1 - C_3 \log R$ (or simply observe that $C_1$ is proportional to $C_2 - C_3$):
Specifically,$C_1 = (C_2 - C_3) \log R$. Since $C_1$ is a multiple of $(C_2 - C_3)$,the determinant is $0$.
Thus,$\Delta = 0 + 0 = 0$.

(D) We use the properties of inverse trigonometric functions:
$\cos ^{-1}(-x) = \pi - \cos ^{-1}(x)$,$\sin ^{-1}(-x) = -\sin ^{-1}(x)$,and $\tan ^{-1}(-x) = -\tan ^{-1}(x)$.
Given expression:
$E = \cos ^{-1}\left(-\frac{1}{2}\right) - 2 \sin ^{-1}\left(\frac{1}{2}\right) + 3 \cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right) - 4 \tan ^{-1}(-1)$
$E = \left(\pi - \cos ^{-1}\left(\frac{1}{2}\right)\right) - 2\left(\frac{\pi}{6}\right) + 3\left(\pi - \cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)\right) - 4\left(-\frac{\pi}{4}\right)$
$E = \left(\pi - \frac{\pi}{3}\right) - \frac{\pi}{3} + 3\left(\pi - \frac{\pi}{4}\right) + \pi$
$E = \frac{2\pi}{3} - \frac{\pi}{3} + 3\left(\frac{3\pi}{4}\right) + \pi$
$E = \frac{\pi}{3} + \frac{9\pi}{4} + \pi$
$E = \frac{4\pi + 27\pi + 12\pi}{12} = \frac{43\pi}{12}$

(C) The expression $\frac{2x-1}{x^3+4x^2+3x}$ is defined for all real numbers $x$ except where the denominator is zero.
Set the denominator equal to zero: $x^3+4x^2+3x = 0$.
Factor out $x$: $x(x^2+4x+3) = 0$.
Factor the quadratic: $x(x+1)(x+3) = 0$.
The roots are $x = 0, x = -1, x = -3$.
Thus,the expression is defined for all real numbers except $0, -1, -3$.
Therefore,the set is $R - \{0, -1, -3\}$.

(D) Let $f(x) = \sin ^4 x + \cos ^4 x$.
We can rewrite this using the identity $\sin ^2 x + \cos ^2 x = 1$:
$f(x) = (\sin ^2 x + \cos ^2 x)^2 - 2 \sin ^2 x \cos ^2 x$
$f(x) = 1 - \frac{1}{2} (2 \sin x \cos x)^2$
$f(x) = 1 - \frac{1}{2} \sin ^2 (2x)$
Using the identity $\sin ^2 \theta = \frac{1 - \cos 2\theta}{2}$:
$f(x) = 1 - \frac{1}{2} \left( \frac{1 - \cos 4x}{2} \right)$
$f(x) = 1 - \frac{1}{4} + \frac{\cos 4x}{4} = \frac{3}{4} + \frac{1}{4} \cos 4x$.
The period of $\cos(kx)$ is $\frac{2\pi}{|k|}$.
Here,$k = 4$,so the period is $\frac{2\pi}{4} = \frac{\pi}{2}$.

(B) The Trapezoidal rule for numerical integration is given by:
$\text{Distance} = \int_{0}^{10} v(t) \ dt \approx \frac{h}{2} [v_0 + 2(v_1 + v_2 + v_3 + v_4) + v_5]$
Here,the interval width $h = 2 \ s$.
The values are $v_0 = 0, v_1 = 12, v_2 = 16, v_3 = 20, v_4 = 35, v_5 = 60$.
Substituting these values into the formula:
$\text{Distance} = \frac{2}{2} [0 + 2(12 + 16 + 20 + 35) + 60]$
$\text{Distance} = 1 \times [0 + 2(83) + 60]$
$\text{Distance} = 166 + 60 = 226 \ m$.

(D) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x \rightarrow 0} f(x) = f(0)$.
Given $f(0) = a$.
Now,calculate the limit: $\lim_{x \rightarrow 0} \frac{2 \sin x - \sin 2x}{2x \cos x}$.
Using the identity $\sin 2x = 2 \sin x \cos x$,we get:
$\lim_{x \rightarrow 0} \frac{2 \sin x - 2 \sin x \cos x}{2x \cos x} = \lim_{x \rightarrow 0} \frac{2 \sin x (1 - \cos x)}{2x \cos x}$.
$= \lim_{x \rightarrow 0} \left( \frac{\sin x}{x} \right) \cdot \left( \frac{1 - \cos x}{\cos x} \right)$.
Since $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$ and $\lim_{x \rightarrow 0} \frac{1 - \cos x}{\cos x} = \frac{1 - 1}{1} = 0$.
Therefore,$\lim_{x \rightarrow 0} f(x) = 1 \cdot 0 = 0$.
Since $f(x)$ is continuous at $x = 0$,$a = 0$.

(B) Given,$x = \frac{1-\sqrt{y}}{1+\sqrt{y}}$.
Applying componendo and dividendo,we get:
$\frac{1+x}{1-x} = \frac{(1+\sqrt{y})+(1-\sqrt{y})}{(1+\sqrt{y})-(1-\sqrt{y})} = \frac{2}{2\sqrt{y}} = \frac{1}{\sqrt{y}}$.
Squaring both sides,we get $\sqrt{y} = \frac{1-x}{1+x}$,so $y = \left(\frac{1-x}{1+x}\right)^2$.
Now,differentiating $y$ with respect to $x$ using the chain rule:
$\frac{dy}{dx} = 2\left(\frac{1-x}{1+x}\right) \cdot \frac{d}{dx}\left(\frac{1-x}{1+x}\right)$.
Using the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$:
$\frac{d}{dx}\left(\frac{1-x}{1+x}\right) = \frac{(-1)(1+x) - (1-x)(1)}{(1+x)^2} = \frac{-1-x-1+x}{(1+x)^2} = \frac{-2}{(1+x)^2}$.
Therefore,$\frac{dy}{dx} = 2\left(\frac{1-x}{1+x}\right) \cdot \left(\frac{-2}{(1+x)^2}\right) = \frac{-4(1-x)}{(1+x)^3} = \frac{4(x-1)}{(1+x)^3}$.

(D) Given,error in diameter $\Delta D = \pm 0.04 \text{ cm}$.
Since $D = 2r$,the error in radius is $\Delta r = \frac{\Delta D}{2} = \frac{\pm 0.04}{2} = \pm 0.02 \text{ cm}$.
The volume of a sphere is $V = \frac{4}{3} \pi r^3$.
Differentiating with respect to $r$,we get $\frac{dV}{dr} = 4 \pi r^2$,so $dV = 4 \pi r^2 \Delta r$.
The percentage error in volume is given by $\frac{dV}{V} \times 100$.
Substituting the values: $\frac{dV}{V} \times 100 = \frac{4 \pi r^2 \Delta r}{\frac{4}{3} \pi r^3} \times 100 = \frac{3 \Delta r}{r} \times 100$.
Given $r = 10 \text{ cm}$ and $\Delta r = \pm 0.02 \text{ cm}$,we have:
Percentage error $= \frac{3 \times (\pm 0.02)}{10} \times 100 = \frac{\pm 0.06}{10} \times 100 = \pm 0.6 \%$.

(C) Let the two poles be $AD$ and $BC$ with heights $a$ and $b$ respectively,standing on a horizontal ground $AB$. $P$ is a point on $AB$.
In $\triangle APD$,$\tan 45^{\circ} = \frac{AD}{AP} = \frac{a}{AP} \Rightarrow AP = a$.
In $\triangle BPC$,$\tan 45^{\circ} = \frac{BC}{BP} = \frac{b}{BP} \Rightarrow BP = b$.
Draw a line $DE$ perpendicular to $BC$ at $E$. Then $DE = AB = AP + PB = a + b$ and $CE = BC - BE = BC - AD = b - a$.
In right-angled $\triangle DEC$,by Pythagoras theorem:
$DC^2 = DE^2 + CE^2$
$DC^2 = (a+b)^2 + (b-a)^2$
$DC^2 = (a^2 + 2ab + b^2) + (b^2 - 2ab + a^2)$
$DC^2 = 2(a^2 + b^2)$.

(B) Given,$\frac{d}{d x}\left[a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right)\right]=\frac{1}{x^4-1}$.
Integrating both sides with respect to $x$,we get:
$a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right) = \int \frac{1}{x^4-1} dx$.
We know that $\frac{1}{x^4-1} = \frac{1}{(x^2-1)(x^2+1)} = \frac{1}{2} \left[ \frac{1}{x^2-1} - \frac{1}{x^2+1} \right]$.
Thus,$\int \frac{1}{x^4-1} dx = \frac{1}{2} \int \frac{1}{x^2-1} dx - \frac{1}{2} \int \frac{1}{x^2+1} dx$.
Using standard integrals $\int \frac{1}{x^2-a^2} dx = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right|$ and $\int \frac{1}{x^2+1} dx = \tan^{-1} x$:
$\int \frac{1}{x^4-1} dx = \frac{1}{2} \left( \frac{1}{2} \log \left| \frac{x-1}{x+1} \right| \right) - \frac{1}{2} \tan^{-1} x = \frac{1}{4} \log \left| \frac{x-1}{x+1} \right| - \frac{1}{2} \tan^{-1} x$.
Comparing this with $a \tan^{-1} x + b \log \left( \frac{x-1}{x+1} \right)$,we get $a = -\frac{1}{2}$ and $b = \frac{1}{4}$.
Therefore,$a - 2b = -\frac{1}{2} - 2 \left( \frac{1}{4} \right) = -\frac{1}{2} - \frac{1}{2} = -1$.

(C) Let $I = \int \frac{dx}{(x+1) \sqrt{4x+3}}$.
Substitute $4x+3 = t^2$,which implies $4dx = 2tdt$ or $dx = \frac{1}{2} t dt$.
Also,$x = \frac{t^2-3}{4}$,so $x+1 = \frac{t^2-3}{4} + 1 = \frac{t^2+1}{4}$.
Substituting these into the integral:
$I = \int \frac{\frac{1}{2} t dt}{(\frac{t^2+1}{4}) t} = \int \frac{\frac{1}{2} dt}{\frac{t^2+1}{4}} = 2 \int \frac{dt}{t^2+1}$.
Integrating,we get $I = 2 \tan^{-1}(t) + c$.
Substituting back $t = \sqrt{4x+3}$,we get $I = 2 \tan^{-1} \sqrt{4x+3} + c$.

(C) We are given the reduction formula for $I_n = \int \sin^n x \, dx$.
Using integration by parts,let $u = \sin^{n-1} x$ and $dv = \sin x \, dx$.
Then $du = (n-1) \sin^{n-2} x \cos x \, dx$ and $v = -\cos x$.
Applying the formula $\int u \, dv = uv - \int v \, du$:
$I_n = -\sin^{n-1} x \cos x - \int (-\cos x) (n-1) \sin^{n-2} x \cos x \, dx$
$I_n = -\sin^{n-1} x \cos x + (n-1) \int \sin^{n-2} x \cos^2 x \, dx$
$I_n = -\sin^{n-1} x \cos x + (n-1) \int \sin^{n-2} x (1 - \sin^2 x) \, dx$
$I_n = -\sin^{n-1} x \cos x + (n-1) I_{n-2} - (n-1) I_n$
Rearranging the terms:
$I_n + (n-1) I_n = -\sin^{n-1} x \cos x + (n-1) I_{n-2}$
$n I_n = -\sin^{n-1} x \cos x + (n-1) I_{n-2}$
Therefore,$n I_n - (n-1) I_{n-2} = -\sin^{n-1} x \cos x$.