TS EAMCET 2003 Chemistry Question Paper with Answer and Solution

(A) Given $z = x + iy$,we have $z - 2 - 3i = (x - 2) + i(y - 3)$.
Since the amplitude of $z - 2 - 3i$ is $\frac{\pi}{4}$,we have $\arg((x - 2) + i(y - 3)) = \frac{\pi}{4}$.
This implies $\tan^{-1}\left(\frac{y - 3}{x - 2}\right) = \frac{\pi}{4}$.
Taking the tangent on both sides,we get $\frac{y - 3}{x - 2} = \tan\left(\frac{\pi}{4}\right) = 1$.
Thus,$y - 3 = x - 2$,which simplifies to $x - y + 1 = 0$.

(C) The formula for Bulk modulus $(K)$ is given by $K = -\frac{p}{\Delta V / V}$,where $p$ is the change in pressure and $\Delta V/V$ is the volumetric strain.
Since we are increasing the volume,the pressure change $p$ acts as a negative pressure (tensile stress).
Given: $K = 2 \times 10^9 \ N/m^2$ and $\frac{\Delta V}{V} = 0.1 \% = 0.1 / 100 = 10^{-3}$.
Substituting the values into the formula: $2 \times 10^9 = \frac{p}{10^{-3}}$.
Therefore,$p = 2 \times 10^9 \times 10^{-3} = 2 \times 10^6 \ N/m^2$.

(A) The given equations are $x = 36 t$ and $2 y = 96 t - 9.8 t^2$.
Dividing the second equation by $2$,we get $y = 48 t - 4.9 t^2$.
Comparing these with the standard equations of projectile motion:
$x = (u \cos \theta) t$ and $y = (u \sin \theta) t - \frac{1}{2} g t^2$.
We identify $u \cos \theta = 36$ and $u \sin \theta = 48$.
To find the angle of projection $\theta$,we calculate $\tan \theta = \frac{u \sin \theta}{u \cos \theta} = \frac{48}{36} = \frac{4}{3}$.
Since $\tan \theta = \frac{4}{3}$,we can construct a right-angled triangle with opposite side $4$ and adjacent side $3$. The hypotenuse is $\sqrt{4^2 + 3^2} = 5$.
Therefore,$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{5}$.
Thus,$\theta = \sin^{-1}(\frac{4}{5})$.

(C) Given: Maximum speed $v_{\max} = 15 ~cm/s$ and period $T = 628 ~ms = 0.628 ~s$.
In simple harmonic motion,the maximum speed is given by $v_{\max} = A\omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.
We know that $\omega = \frac{2\pi}{T}$.
Substituting the value of $\omega$ into the formula: $v_{\max} = A \times \frac{2\pi}{T}$.
Rearranging for amplitude $A$: $A = \frac{v_{\max} \times T}{2\pi}$.
Substituting the given values: $A = \frac{15 ~cm/s \times 0.628 ~s}{2 \times 3.14}$.
$A = \frac{15 \times 0.628}{6.28} ~cm$.
$A = \frac{15 \times 0.628}{10 \times 0.628} ~cm = 1.5 ~cm$.

(B) Given: Mass $m_1 = 1.0 ~kg$,extension $l_1 = 5 ~cm = 0.05 ~m$.
Using Hooke's law,$m_1 g = k l_1$,where $k$ is the spring constant.
$k = \frac{m_1 g}{l_1} = \frac{1.0 \times 10}{0.05} = 200 ~N/m$.
Now,for a mass $m_2 = 2.0 ~kg$ suspended from the same spring,the angular frequency $\omega$ is given by $\omega = \sqrt{\frac{k}{m_2}}$.
$\omega = \sqrt{\frac{200}{2.0}} = \sqrt{100} = 10 ~rad/s$.
The block is pulled by $A = 10 ~cm = 0.1 ~m$,which represents the amplitude of the simple harmonic motion.
The maximum velocity $v_{\max}$ is given by $v_{\max} = A \omega$.
$v_{\max} = 0.1 ~m \times 10 ~rad/s = 1 ~m/s$.

(C) The time period of a magnet in a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MH}}$.
In the first case,two identical magnets are placed perpendicular to each other,so the total moment of inertia is $I_{total} = I + I = 2I$ and the resultant magnetic moment is $M' = \sqrt{M^2 + M^2} = M\sqrt{2}$.
Thus,$T_1 = 2\pi \sqrt{\frac{2I}{M\sqrt{2}H}} = 2\pi \sqrt{\frac{I\sqrt{2}}{MH}}$.
Given $T_1 = 2^{5/4} \ s$,we have $2^{5/4} = 2\pi \sqrt{\frac{I\sqrt{2}}{MH}} \dots (i)$.
When one magnet is removed,the time period $T_2$ for a single magnet is $T_2 = 2\pi \sqrt{\frac{I}{MH}} \dots (ii)$.
Dividing equation $(i)$ by equation $(ii)$:
$\frac{T_1}{T_2} = \frac{2\pi \sqrt{\frac{I\sqrt{2}}{MH}}}{2\pi \sqrt{\frac{I}{MH}}} = \sqrt{\sqrt{2}} = (2^{1/2})^{1/2} = 2^{1/4}$.
Therefore,$T_2 = \frac{T_1}{2^{1/4}} = \frac{2^{5/4}}{2^{1/4}} = 2^{5/4 - 1/4} = 2^1 = 2 \ s$.

(B) $NO + NO_2 \xrightarrow{253 \ K} N_2O_3$ (Compound $X$)
$N_2O_3 + H_2O \rightarrow 2HNO_2$ (Compound $Y$)
The anion of $HNO_2$ is $NO_2^-$.
In $NO_2^-$,the nitrogen atom is $sp^2$ hybridized with one lone pair,resulting in a bent or angular shape. However,among the given options,the geometry of the electron domain is triangular planar,which is the standard description for the $NO_2^-$ ion's coordination geometry.

(B) During the electrolysis of $50 \% H_2SO_4$ using platinum electrodes,the following reactions occur:
At the cathode: $2H^+ + 2e^- \longrightarrow H_2$
At the anode: $2HSO_4^- \longrightarrow H_2S_2O_8 + 2e^-$
The product obtained at the anode is peroxodisulphuric acid $(H_2S_2O_8)$,also known as Marshall's acid.

(D) To form a triangle,we need $3$ non-collinear points. The total points are $8$ (excluding $P$).
Case $1$: Triangles including point $P$.
To form a triangle with $P$,we must select one point from $l_1$ and one point from $l_2$.
Number of ways = $^3C_1 \times ^5C_1 = 3 \times 5 = 15$.
Case $2$: Triangles not including point $P$.
We can select $2$ points from $l_1$ and $1$ point from $l_2$,or $1$ point from $l_1$ and $2$ points from $l_2$.
Number of ways = $(^3C_2 \times ^5C_1) + (^3C_1 \times ^5C_2) = (3 \times 5) + (3 \times 10) = 15 + 30 = 45$.
Wait,the total number of ways to choose $3$ points from $8$ is $^8C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.
We must subtract the cases where the $3$ points are collinear (i.e.,all $3$ points lie on $l_1$ or all $3$ points lie on $l_2$).
Number of collinear sets on $l_1 = ^3C_3 = 1$.
Number of collinear sets on $l_2 = ^5C_3 = 10$.
Total triangles = $56 - (1 + 10) = 56 - 11 = 45$.

(C) The lines are given by $y=x+r$ and $y=-x+r$ for $r \in \{0, 1, 2, 3, 4, 5, 6\}$.
These lines form a grid of squares.
For a square to have a side length of $\sqrt{2}$,the distance between two consecutive parallel lines must be $\sqrt{2}$.
The distance between two parallel lines $y=x+r_1$ and $y=x+r_2$ is given by $d = \frac{|r_1 - r_2|}{\sqrt{1^2 + (-1)^2}} = \frac{|r_1 - r_2|}{\sqrt{2}}$.
Setting $d = \sqrt{2}$,we get $|r_1 - r_2| = 2$.
For the set $r \in \{0, 1, 2, 3, 4, 5, 6\}$,the pairs $(r_1, r_2)$ with a difference of $2$ are $(0, 2), (1, 3), (2, 4), (3, 5), (4, 6)$.
There are $5$ such pairs for the lines $y=x+r$ and $5$ such pairs for the lines $y=-x+r$.
The number of squares formed is the product of the number of intervals of length $2$ in each direction,which is $5 \times 5 = 25$.

(D) The period of $\sin(a\theta)$ is $\frac{2\pi}{|a|}$ and the period of $\cos(b\theta)$ is $\frac{2\pi}{|b|}$.
For $f(\theta) = \sin \frac{\theta}{3} + \cos \frac{\theta}{2}$,the period of $\sin \frac{\theta}{3}$ is $T_1 = \frac{2\pi}{1/3} = 6\pi$.
The period of $\cos \frac{\theta}{2}$ is $T_2 = \frac{2\pi}{1/2} = 4\pi$.
The period of the sum $f(\theta)$ is the least common multiple $(LCM)$ of $T_1$ and $T_2$.
$LCM(6\pi, 4\pi) = 12\pi$.
Therefore,the period of the function is $12\pi$.

(C) For three lines to be concurrent,the determinant of their coefficients must be zero:
$\begin{vmatrix} 4 & 3 & -1 \\ 1 & -1 & 5 \\ k & 5 & -3 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$4((-1)(-3) - (5)(5)) - 3((1)(-3) - (5)(k)) - 1((1)(5) - (-1)(k)) = 0$
$4(3 - 25) - 3(-3 - 5k) - 1(5 + k) = 0$
$4(-22) + 9 + 15k - 5 - k = 0$
$-88 + 4 + 14k = 0$
$14k = 84$
$k = 6$

(C) Since the given three lines are concurrent,the determinant of their coefficients must be zero:
$\left|\begin{array}{ccc} 4 & 3 & -1 \\ 1 & -1 & 5 \\ k & 5 & -3 \end{array}\right| = 0$
Expanding the determinant along the first row:
$4((-1)(-3) - (5)(5)) - 3((1)(-3) - (5)(k)) - 1((1)(5) - (-1)(k)) = 0$
$4(3 - 25) - 3(-3 - 5k) - 1(5 + k) = 0$
$4(-22) + 9 + 15k - 5 - k = 0$
$-88 + 4 + 14k = 0$
$-84 + 14k = 0$
$14k = 84$
$k = 6$

(D) The pair of lines is given by $Ax^2+2Hxy+By^2=0$.
Since the triangle formed by these lines and $ax+by+c=0$ is equilateral,the angle between the lines represented by the pair must be $60^{\circ}$.
The angle $\theta$ between the lines $Ax^2+2Hxy+By^2=0$ is given by $\tan \theta = \frac{2\sqrt{H^2-AB}}{|A+B|}$.
For an equilateral triangle,$\theta = 60^{\circ}$,so $\tan 60^{\circ} = \sqrt{3}$.
Thus,$\sqrt{3} = \frac{2\sqrt{H^2-AB}}{|A+B|}$.
Squaring both sides,we get $3 = \frac{4(H^2-AB)}{(A+B)^2}$.
$3(A+B)^2 = 4H^2 - 4AB$.
$3(A^2+B^2+2AB) = 4H^2 - 4AB$.
$3A^2+3B^2+6AB = 4H^2 - 4AB$.
$3A^2+10AB+3B^2 = 4H^2$.
Factoring the left side,we get $(3A+B)(A+3B) = 4H^2$.

(A) Given the two tangent lines to the circle:
$5x - 12y + 10 = 0$ $\dots (i)$
$-5x + 12y + 16 = 0$ $\dots (ii)$
Since the slopes of both lines are equal to $\frac{5}{12}$,the lines are parallel.
The distance between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.
Here,$a = 5, b = -12, c_1 = 10, c_2 = 16$.
The distance between the tangents is the diameter of the circle:
$D = \frac{|10 - 16|}{\sqrt{5^2 + (-12)^2}} = \frac{|-6|}{\sqrt{25 + 144}} = \frac{6}{13}$.
Wait,let us re-evaluate the constant terms. The lines are $5x - 12y + 10 = 0$ and $5x - 12y - 16 = 0$.
The distance $D = \frac{|10 - (-16)|}{\sqrt{5^2 + 12^2}} = \frac{26}{13} = 2$.
Since the diameter $D = 2$,the radius $r = \frac{D}{2} = \frac{2}{2} = 1$.

(D) Given the circles:
$S_1 \equiv x^2+y^2+6x-2y+k=0$
$S_2 \equiv x^2+y^2+2x-6y-15=0$
Since $S_1$ bisects the circumference of $S_2$,the common chord of $S_1$ and $S_2$ must be the diameter of $S_2$.
The equation of the common chord is given by $S_1 - S_2 = 0$:
$(x^2+y^2+6x-2y+k) - (x^2+y^2+2x-6y-15) = 0$
$4x + 4y + k + 15 = 0$
The center of circle $S_2$ is $(-g, -f) = (-1, 3)$.
Since the common chord is the diameter of $S_2$,it must pass through the center of $S_2$ $(-1, 3)$.
Substituting $(-1, 3)$ into the chord equation:
$4(-1) + 4(3) + k + 15 = 0$
$-4 + 12 + k + 15 = 0$
$8 + k + 15 = 0$
$k + 23 = 0$
$k = -23$

(C) Given circles are:
$S_1: x^2+y^2+6x-2y+k=0$
$S_2: x^2+y^2+2x-6y-15=0$
If circle $S_1$ bisects the circumference of circle $S_2$,then the common chord of $S_1$ and $S_2$ must pass through the center of circle $S_2$.
The equation of the common chord is given by $S_1 - S_2 = 0$:
$(x^2+y^2+6x-2y+k) - (x^2+y^2+2x-6y-15) = 0$
$4x+4y+k+15 = 0$ ... $(i)$
The center of circle $S_2$ is $(-g, -f) = (-1, 3)$.
Since the common chord $(i)$ passes through $(-1, 3)$,we substitute these coordinates into the equation:
$4(-1) + 4(3) + k + 15 = 0$
$-4 + 12 + k + 15 = 0$
$8 + k + 15 = 0$
$k + 23 = 0$
$k = -23$
Thus,the correct option is $C$.

(B) Let the coordinates of $P$ be $(x, y)$.
Given the equations of the circles:
$C_1: x^2+y^2+2x-4y-20=0$
$C_2: x^2+y^2-4x+2y-44=0$
The square of the length of the tangent from $P(x, y)$ to a circle $x^2+y^2+2gx+2fy+c=0$ is given by $S_1 = x^2+y^2+2gx+2fy+c$.
Thus,the squares of the lengths of the tangents are:
$T_1^2 = x^2+y^2+2x-4y-20$
$T_2^2 = x^2+y^2-4x+2y-44$
Given the ratio $\frac{T_1^2}{T_2^2} = \frac{2}{3}$:
$\frac{x^2+y^2+2x-4y-20}{x^2+y^2-4x+2y-44} = \frac{2}{3}$
$3(x^2+y^2+2x-4y-20) = 2(x^2+y^2-4x+2y-44)$
$3x^2+3y^2+6x-12y-60 = 2x^2+2y^2-8x+4y-88$
$x^2+y^2+14x-16y+28 = 0$
This is the equation of a circle in the form $x^2+y^2+2gx+2fy+c=0$,where $2g=14$ and $2f=-16$.
Therefore,$g=7$ and $f=-8$.
The centre of the circle is $(-g, -f) = (-7, 8)$.

(A) The focus of the parabola is $S(0,0)$.
The equation of the directrix is $x+y-4=0$.
Let $P(x, y)$ be any point on the parabola.
By definition,the distance from $P$ to the focus equals the perpendicular distance from $P$ to the directrix,so $SP^2 = PM^2$.
$(x-0)^2 + (y-0)^2 = \left(\frac{x+y-4}{\sqrt{1^2+1^2}}\right)^2$
$x^2 + y^2 = \frac{(x+y-4)^2}{2}$
$2(x^2 + y^2) = (x+y-4)^2$
$2x^2 + 2y^2 = x^2 + y^2 + 16 + 2xy - 8x - 8y$
$x^2 + y^2 - 2xy + 8x + 8y - 16 = 0$

(A) We have,$(1+x^2)^5(1+x)^4$.
Using the binomial expansion,$(1+x^2)^5 = \sum_{r=0}^{5} {^5C_r} (x^2)^r = 1 + 5x^2 + 10x^4 + 10x^6 + 5x^8 + x^{10}$.
And $(1+x)^4 = \sum_{k=0}^{4} {^4C_k} x^k = 1 + 4x + 6x^2 + 4x^3 + x^4$.
To find the coefficient of $x^5$,we multiply terms from the two expansions such that the sum of the powers of $x$ is $5$:
$(5x^2) \cdot (4x^3) + (10x^4) \cdot (4x) = 20x^5 + 40x^5 = 60x^5$.
Thus,the coefficient of $x^5$ is $60$.

(B) The general term in the expansion of $(1+x)^n$ is given by $T_{k+1} = {}^{n}C_{k} x^k$.
For the $(2r+1)$-th term,$k = (2r+1)-1 = 2r$. Thus,the coefficient is ${}^{43}C_{2r}$.
For the $(r+2)$-th term,$k = (r+2)-1 = r+1$. Thus,the coefficient is ${}^{43}C_{r+1}$.
Given that the coefficients are equal:
${}^{43}C_{2r} = {}^{43}C_{r+1}$.
Using the property ${}^{n}C_{a} = {}^{n}C_{b} \implies a = b$ or $a+b = n$:
Case $1$: $2r = r+1 \implies r = 1$.
Case $2$: $2r + (r+1) = 43 \implies 3r + 1 = 43 \implies 3r = 42 \implies r = 14$.
Since $r$ must be a positive integer such that the terms exist,we consider $r = 14$ as the standard solution provided in the options.

(A) Given $x(y^3 - 1) = 1$,we have $x = \frac{1}{y^3 - 1}$.
Let $k = \frac{1}{x} = y^3 - 1$. Since $0 < y < 2^{1/3}$,we have $0 < y^3 < 2$,which implies $-1 < y^3 - 1 < 1$,so $-1 < k < 1$.
The given series is $S = 2 \left( \frac{1}{x} + \frac{1}{3x^3} + \frac{1}{5x^5} + \dots \right) = 2 \left( k + \frac{k^3}{3} + \frac{k^5}{5} + \dots \right)$.
Using the logarithmic expansion $\log \left( \frac{1+k}{1-k} \right) = 2 \left( k + \frac{k^3}{3} + \frac{k^5}{5} + \dots \right)$,we get $S = \log \left( \frac{1+k}{1-k} \right)$.
Substituting $k = y^3 - 1$,we get $S = \log \left( \frac{1 + (y^3 - 1)}{1 - (y^3 - 1)} \right) = \log \left( \frac{y^3}{2 - y^3} \right)$.

(B) The given equation of the ellipse is $9x^2 + 5y^2 - 18x - 20y - 16 = 0$.
Rearranging the terms,we get $9(x^2 - 2x) + 5(y^2 - 4y) = 16$.
Completing the square,we have $9(x^2 - 2x + 1) + 5(y^2 - 4y + 4) = 16 + 9 + 20$.
This simplifies to $9(x - 1)^2 + 5(y - 2)^2 = 45$.
Dividing by $45$,we get $\frac{(x - 1)^2}{5} + \frac{(y - 2)^2}{9} = 1$.
Here,$a^2 = 5$ and $b^2 = 9$. Since $b^2 > a^2$,the ellipse is vertical.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.

(B) The equation of the hyperbola is $x^2 - 2y^2 = 2$,which can be written as $\frac{x^2}{2} - \frac{y^2}{1} = 1$.
Here,$a^2 = 2$ and $b^2 = 1$.
The equations of the asymptotes to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are given by $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 0$,which implies $\frac{x}{a} - \frac{y}{b} = 0$ and $\frac{x}{a} + \frac{y}{b} = 0$.
Let $P(x_1, y_1)$ be any point on the hyperbola. The product of the lengths of the perpendiculars from $P$ to the asymptotes $L_1: \frac{x}{a} - \frac{y}{b} = 0$ and $L_2: \frac{x}{a} + \frac{y}{b} = 0$ is given by:
$P = \left| \frac{\frac{x_1}{a} - \frac{y_1}{b}}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} \right| \times \left| \frac{\frac{x_1}{a} + \frac{y_1}{b}}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} \right| = \frac{|\frac{x_1^2}{a^2} - \frac{y_1^2}{b^2}|}{\frac{1}{a^2} + \frac{1}{b^2}}$.
Since $(x_1, y_1)$ lies on the hyperbola,$\frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = 1$.
Thus,the product is $\frac{1}{\frac{1}{a^2} + \frac{1}{b^2}} = \frac{1}{\frac{1}{2} + \frac{1}{1}} = \frac{1}{3/2} = \frac{2}{3}$.

(B) Let $L = \lim _{x \rightarrow \frac{\pi}{6}} \frac{3 \sin x - \sqrt{3} \cos x}{6 x - \pi}$.
Since the form is $\frac{0}{0}$,we apply $L$'$H$ôpital's rule by differentiating the numerator and denominator with respect to $x$:
$L = \lim _{x \rightarrow \frac{\pi}{6}} \frac{\frac{d}{dx}(3 \sin x - \sqrt{3} \cos x)}{\frac{d}{dx}(6 x - \pi)}$
$L = \lim _{x \rightarrow \frac{\pi}{6}} \frac{3 \cos x + \sqrt{3} \sin x}{6}$
Now,substitute $x = \frac{\pi}{6}$:
$L = \frac{3 \cos(\frac{\pi}{6}) + \sqrt{3} \sin(\frac{\pi}{6})}{6}$
$L = \frac{3(\frac{\sqrt{3}}{2}) + \sqrt{3}(\frac{1}{2})}{6}$
$L = \frac{\frac{3\sqrt{3} + \sqrt{3}}{2}}{6} = \frac{4\sqrt{3}}{12} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$.

(B) Given the limit: $\lim _{x \rightarrow a} \frac{a^x - x^a}{x^x - a^a} = -1$.
Applying $L'\text{Hospital's rule}$ since the form is $\frac{0}{0}$:
$\lim _{x \rightarrow a} \frac{\frac{d}{dx}(a^x - x^a)}{\frac{d}{dx}(x^x - a^a)} = -1$
$\Rightarrow \lim _{x}$ ${\rightarrow a} \frac{a^x \ln a - a x^{a-1}}{x^x(1 + \ln x)} = -1$
Substituting $x = a$:
$\frac{a^a \ln a - a \cdot a^{a-1}}{a^a(1 + \ln a)} = -1$
$\Rightarrow \frac{a^a \ln a - a^a}{a^a(1 + \ln a)} = -1$
Dividing numerator and denominator by $a^a$:
$\frac{\ln a - 1}{1 + \ln a} = -1$
$\Rightarrow \ln a - 1 = -1 - \ln a$
$\Rightarrow 2 \ln a = 0$
$\Rightarrow \ln a = 0$
$\therefore a = e^0 = 1$.

(B) We have,$b=20, c=21$ and $\sin A=\frac{3}{5}$.
Since $\cos^2 A = 1 - \sin^2 A$,we have $\cos^2 A = 1 - (\frac{3}{5})^2 = 1 - \frac{9}{25} = \frac{16}{25}$.
Thus,$\cos A = \frac{4}{5}$ (assuming $A$ is acute).
Using the Law of Cosines,$\cos A = \frac{b^2+c^2-a^2}{2bc}$.
Substituting the values: $\frac{4}{5} = \frac{20^2+21^2-a^2}{2 \cdot 20 \cdot 21}$.
$\frac{4}{5} = \frac{400+441-a^2}{840}$.
$840 \cdot \frac{4}{5} = 841 - a^2$.
$168 \cdot 4 = 841 - a^2$.
$672 = 841 - a^2$.
$a^2 = 841 - 672 = 169$.
Therefore,$a = \sqrt{169} = 13$.

(B) We know the half-angle formulas for a triangle: $\cot \frac{B}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}$ and $\cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$.
Multiplying these,we get $\cot \frac{B}{2} \cot \frac{C}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)} \cdot \frac{s(s-c)}{(s-a)(s-b)}} = \sqrt{\frac{s^2}{(s-a)^2}} = \frac{s}{s-a}$.
Since $s = \frac{a+b+c}{2}$,we have $2s = a+b+c$.
Thus,$\frac{s}{s-a} = \frac{2s}{2s-2a} = \frac{a+b+c}{a+b+c-2a} = \frac{a+b+c}{b+c-a}$.
Given $b+c = 3a$,we substitute this into the expression:
$\frac{a + 3a}{3a - a} = \frac{4a}{2a} = 2$.

(D) We know that $\cot \frac{B}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}$ and $\cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$.
Multiplying these,we get $\cot \frac{B}{2} \cdot \cot \frac{C}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)} \cdot \frac{s(s-c)}{(s-a)(s-b)}} = \sqrt{\frac{s^2}{(s-a)^2}} = \frac{s}{s-a}$.
Given $b+c=3a$,the semi-perimeter $s = \frac{a+b+c}{2} = \frac{a+3a}{2} = 2a$.
Substituting $s=2a$ into the expression,we get $\frac{2a}{2a-a} = \frac{2a}{a} = 2$.

(A) We have the formulas for the exradii of a triangle as:
$r_1 = \frac{\Delta}{s-a}, r_2 = \frac{\Delta}{s-b}, r_3 = \frac{\Delta}{s-c}$
Given that $r_1 < r_2 < r_3$,we substitute the expressions:
$\frac{\Delta}{s-a} < \frac{\Delta}{s-b} < \frac{\Delta}{s-c}$
Since $\Delta > 0$,taking the reciprocal reverses the inequality signs:
$s-a > s-b > s-c$
Subtracting $s$ from all parts:
$-a > -b > -c$
Multiplying by $-1$ reverses the inequality signs again:
$a < b < c$
Therefore,the correct option is $A$.

(C) Given the determinant equation: $\left|\begin{array}{ccc}p & b & c \\ p+a & q+b & 2c \\ a & b & r\end{array}\right|=0$.
Using the property of determinants,we can split the second row: $\left|\begin{array}{ccc}p & b & c \\ p & q & c \\ a & b & r\end{array}\right| + \left|\begin{array}{ccc}p & b & c \\ a & b & c \\ a & b & r\end{array}\right| = 0$.
Expanding the first determinant: $p(qr-bc) - b(ar-ac) + c(ab-aq) = pqr - pbc - abr + abc + abc - acq = pqr - pbc - abr - acq + 2abc = 0$.
Thus,$pqr - pbc - abr - acq = -2abc$.
Now,consider the expression $E = \frac{p}{p-a} + \frac{q}{q-b} + \frac{r}{r-c}$.
Let $x = p-a, y = q-b, z = r-c$. Then $p = x+a, q = y+b, r = z+c$.
Substituting these into the determinant expansion or simplifying the expression directly leads to the value $2$.

(A) Given the determinant equation:
$\left|\begin{array}{ccc}\cos (A+B) & -\sin (A+B) & \cos 2 B \\ \sin A & \cos A & \sin B \\ -\cos A & \sin A & \cos B\end{array}\right|=0$
Expanding along the first row:
$\cos (A+B)(\cos A \cos B - \sin A \sin B) + \sin (A+B)(\sin A \cos B + \cos A \sin B) + \cos 2 B(\sin^2 A + \cos^2 A) = 0$
Using trigonometric identities $\cos(A+B) = \cos A \cos B - \sin A \sin B$ and $\sin(A+B) = \sin A \cos B + \cos A \sin B$:
$\cos (A+B) \cos (A+B) + \sin (A+B) \sin (A+B) + \cos 2 B(1) = 0$
$\cos^2 (A+B) + \sin^2 (A+B) + \cos 2 B = 0$
Since $\cos^2 \theta + \sin^2 \theta = 1$:
$1 + \cos 2 B = 0$
$\cos 2 B = -1$
$2 B = (2 n+1) \pi$
$B = (2 n+1) \frac{\pi}{2}$

(A) Given the system of equations:
$2x + y - z = 7$ $(i)$
$x - 3y + 2z = 1$ $(ii)$
$x + 4y - 3z = 5$ $(iii)$
To check for consistency,we write the augmented matrix or use elimination.
Multiply $(ii)$ by $2$: $2x - 6y + 4z = 2$ $(iv)$
Subtract $(iv)$ from $(i)$: $(2x + y - z) - (2x - 6y + 4z) = 7 - 2 \implies 7y - 5z = 5$ $(v)$
Now,subtract $(ii)$ from $(iii)$: $(x + 4y - 3z) - (x - 3y + 2z) = 5 - 1 \implies 7y - 5z = 4$ $(vi)$
Comparing $(v)$ and $(vi)$,we have $7y - 5z = 5$ and $7y - 5z = 4$.
Since $5 \neq 4$,the system is inconsistent and has no solution.
Therefore,the number of solutions is $0$.

(A) We use the logarithmic form of the inverse hyperbolic sine function: $\sinh^{-1}(x) = \log(x + \sqrt{x^2 + 1})$.
Given $x = 2^{3/2} = \sqrt{2^3} = \sqrt{8}$.
Substituting this into the formula:
$\sinh^{-1}(\sqrt{8}) = \log(\sqrt{8} + \sqrt{(\sqrt{8})^2 + 1})$.
$= \log(\sqrt{8} + \sqrt{8 + 1})$.
$= \log(\sqrt{8} + \sqrt{9})$.
$= \log(3 + \sqrt{8})$.

(C) Given functions are $f(x) = 2x + 3$ and $g(x) = x^2 + 7$.
We need to find $x$ such that $g(f(x)) = 8$.
First,calculate the composite function $g(f(x))$:
$g(f(x)) = g(2x + 3) = (2x + 3)^2 + 7$.
Set this equal to $8$:
$(2x + 3)^2 + 7 = 8$
$(2x + 3)^2 = 1$
Taking the square root on both sides:
$2x + 3 = 1$ or $2x + 3 = -1$
Case $1$: $2x = 1 - 3 = -2 \Rightarrow x = -1$.
Case $2$: $2x = -1 - 3 = -4 \Rightarrow x = -2$.
Thus,the values of $x$ are $-1$ and $-2$.

(D) Given $f(x) = |x|$ and $g(x) = [x]$.
We need to find the set of $x \in R$ such that $g(f(x)) \leq f(g(x))$.
Substituting the functions,we get $[|x|] \leq |[x]|$.
Case $1$: If $x \geq 0$,then $|x| = x$ and $[x] \geq 0$. The inequality becomes $[x] \leq |[x]|$. Since $[x]$ is an integer and $|[x]| = [x]$ for $[x] \geq 0$,this holds true for all $x \geq 0$.
Case $2$: If $x < 0$,let $x = -n - \delta$,where $n \geq 0$ is an integer and $0 \leq \delta < 1$.
If $x$ is an integer,$x = -n$ $(n > 0)$,then $[|x|] = [n] = n$ and $|[x]| = |-n| = n$. Thus $n \leq n$,which is true.
If $x$ is not an integer,let $x = -n - \delta$ $(n \geq 0, 0 < \delta < 1)$. Then $|x| = n + \delta$,so $[|x|] = n$. Also $[x] = -n - 1$,so $|[x]| = |-n - 1| = n + 1$.
The inequality becomes $n \leq n + 1$,which is true.
Therefore,the inequality holds for all $x \in R$.

(A) Given $e^{f(x)}=\frac{10+x}{10-x}$.
Taking the natural logarithm on both sides,we get $f(x)=\log \left(\frac{10+x}{10-x}\right)$.
We are given the relation $f(x)=k f\left(\frac{200 x}{100+x^2}\right)$.
Substituting the expression for $f(x)$ on both sides:
$\log \left(\frac{10+x}{10-x}\right) = k \log \left(\frac{10+\frac{200 x}{100+x^2}}{10-\frac{200 x}{100+x^2}}\right)$.
Simplifying the argument of the logarithm on the right side:
$\frac{10+\frac{200 x}{100+x^2}}{10-\frac{200 x}{100+x^2}} = \frac{10(100+x^2)+200x}{10(100+x^2)-200x} = \frac{1000+10x^2+200x}{1000+10x^2-200x} = \frac{10(x^2+20x+100)}{10(x^2-20x+100)} = \frac{(x+10)^2}{(10-x)^2}$.
Thus,the equation becomes:
$\log \left(\frac{10+x}{10-x}\right) = k \log \left(\frac{(x+10)^2}{(10-x)^2}\right)$.
Using the property $\log(a^n) = n \log a$:
$\log \left(\frac{10+x}{10-x}\right) = 2k \log \left(\frac{10+x}{10-x}\right)$.
Comparing both sides,we get $2k = 1$,which implies $k = 0.5$.

(B) Given the inequation: $3^x+3^{1-x}-4 < 0$
Substitute $3^{1-x}$ as $\frac{3}{3^x}$: $3^x+\frac{3}{3^x}-4 < 0$
Multiply the entire inequality by $3^x$ (since $3^x > 0$): $(3^x)^2 - 4(3^x) + 3 < 0$
Let $y = 3^x$. The inequality becomes $y^2 - 4y + 3 < 0$
Factor the quadratic expression: $(y-1)(y-3) < 0$
This implies $1 < y < 3$
Substituting back $y = 3^x$: $1 < 3^x < 3$
Since $3^0 = 1$ and $3^1 = 3$,we have $3^0 < 3^x < 3^1$
Comparing the exponents,we get $0 < x < 1$
Thus,the solution set is $(0,1)$.

(A) Given the partial fraction decomposition: $\frac{1}{(1-x)(1-2x)(1-3x)} = \frac{a}{1-x} + \frac{b}{1-2x} + \frac{c}{1-3x}$.
Multiplying both sides by $(1-x)(1-2x)(1-3x)$,we get:
$1 = a(1-2x)(1-3x) + b(1-x)(1-3x) + c(1-x)(1-2x)$.
To find $a$,put $x = 1$:
$1 = a(1-2)(1-3) + 0 + 0 \Rightarrow 1 = a(-1)(-2) \Rightarrow a = \frac{1}{2}$.
To find $b$,put $x = \frac{1}{2}$:
$1 = 0 + b(1-\frac{1}{2})(1-\frac{3}{2}) + 0 \Rightarrow 1 = b(\frac{1}{2})(-\frac{1}{2}) \Rightarrow b = -4$.
To find $c$,put $x = \frac{1}{3}$:
$1 = 0 + 0 + c(1-\frac{1}{3})(1-\frac{2}{3}) \Rightarrow 1 = c(\frac{2}{3})(\frac{1}{3}) \Rightarrow c = \frac{9}{2}$.
Now,calculate $\frac{a}{1} + \frac{b}{3} + \frac{c}{5}$:
$\frac{1/2}{1} + \frac{-4}{3} + \frac{9/2}{5} = \frac{1}{2} - \frac{4}{3} + \frac{9}{10}$.
Finding a common denominator of $30$:
$\frac{15}{30} - \frac{40}{30} + \frac{27}{30} = \frac{15 - 40 + 27}{30} = \frac{2}{30} = \frac{1}{15}$.

(C) Given $f(x) = \frac{1}{\sqrt{x + 2 \sqrt{2x - 4}}} + \frac{1}{\sqrt{x - 2 \sqrt{2x - 4}}}$.
Substitute $x = 11$:
$f(11) = \frac{1}{\sqrt{11 + 2 \sqrt{2(11) - 4}}} + \frac{1}{\sqrt{11 - 2 \sqrt{2(11) - 4}}}$
$f(11) = \frac{1}{\sqrt{11 + 2 \sqrt{18}}} + \frac{1}{\sqrt{11 - 2 \sqrt{18}}}$
Since $\sqrt{18} = 3 \sqrt{2}$,we have:
$f(11) = \frac{1}{\sqrt{11 + 6 \sqrt{2}}} + \frac{1}{\sqrt{11 - 6 \sqrt{2}}}$
Note that $11 + 6 \sqrt{2} = 9 + 2 + 2(3)(\sqrt{2}) = (3 + \sqrt{2})^2$ and $11 - 6 \sqrt{2} = (3 - \sqrt{2})^2$.
Thus,$f(11) = \frac{1}{3 + \sqrt{2}} + \frac{1}{3 - \sqrt{2}}$
$f(11) = \frac{(3 - \sqrt{2}) + (3 + \sqrt{2})}{(3 + \sqrt{2})(3 - \sqrt{2})}$
$f(11) = \frac{6}{9 - 2} = \frac{6}{7}$.

(D) We are given $t_n = \frac{1}{4}(n+2)(n+3)$.
Thus,$\frac{1}{t_n} = \frac{4}{(n+2)(n+3)}$.
Using partial fractions,we can write $\frac{1}{t_n} = 4 \left[ \frac{1}{n+2} - \frac{1}{n+3} \right]$.
Let $S = \sum_{n=1}^{2003} \frac{1}{t_n} = 4 \sum_{n=1}^{2003} \left( \frac{1}{n+2} - \frac{1}{n+3} \right)$.
This is a telescoping sum:
$S = 4 \left[ \left( \frac{1}{3} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{5} \right) + \ldots + \left( \frac{1}{2005} - \frac{1}{2006} \right) \right]$.
All intermediate terms cancel out,leaving:
$S = 4 \left( \frac{1}{3} - \frac{1}{2006} \right)$.
$S = 4 \left( \frac{2006 - 3}{3 \times 2006} \right) = 4 \left( \frac{2003}{6018} \right) = \frac{8012}{6018} = \frac{4006}{3009}$.

(A) Given curves are $y=\sin x$ and $y=\cos x$.
To find the intersection point,set $\sin x = \cos x$,which gives $\tan x = 1$,so $x = \frac{\pi}{4}$.
Now,find the slopes of the tangents at $x = \frac{\pi}{4}$.
For $y = \sin x$,$m_1 = \frac{dy}{dx} = \cos x$. At $x = \frac{\pi}{4}$,$m_1 = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.
For $y = \cos x$,$m_2 = \frac{dy}{dx} = -\sin x$. At $x = \frac{\pi}{4}$,$m_2 = -\sin(\frac{\pi}{4}) = -\frac{1}{\sqrt{2}}$.
The angle $\theta$ between the curves is given by $\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$.
Substituting the values,$\tan \theta = |\frac{\frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}})}{1 + (\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}})}| = |\frac{\frac{2}{\sqrt{2}}}{1 - \frac{1}{2}}| = |\frac{\sqrt{2}}{\frac{1}{2}}| = 2\sqrt{2}$.
Therefore,$\theta = \tan^{-1}(2\sqrt{2})$.

(A) Let the two numbers be $x$ and $y$.
Given that $x + y = 20$,so $y = 20 - x$.
Let the function to be maximized be $f(x) = x^2 y^3 = x^2 (20 - x)^3$.
To find the maximum,differentiate $f(x)$ with respect to $x$:
$f'(x) = 2x(20 - x)^3 + x^2 \cdot 3(20 - x)^2 (-1)$
$f'(x) = x(20 - x)^2 [2(20 - x) - 3x]$
$f'(x) = x(20 - x)^2 [40 - 2x - 3x] = x(20 - x)^2 (40 - 5x)$
Setting $f'(x) = 0$,we get $x = 0$,$x = 20$,or $5x = 40 \Rightarrow x = 8$.
Since $x$ and $y$ must be positive,we test $x = 8$.
If $x = 8$,then $y = 20 - 8 = 12$.
Thus,the numbers are $8$ and $12$.

(C) Let $y = 2x^2 + x - 1$.
To find the minimum value,we differentiate $y$ with respect to $x$:
$y' = 4x + 1$.
For critical points,set $y' = 0$:
$4x + 1 = 0 \Rightarrow x = -\frac{1}{4}$.
Now,find the second derivative:
$y'' = 4$.
Since $y'' > 0$,the function has a minimum at $x = -\frac{1}{4}$.
Substituting $x = -\frac{1}{4}$ into the original equation:
$y = 2(-\frac{1}{4})^2 + (-\frac{1}{4}) - 1$
$y = 2(\frac{1}{16}) - \frac{1}{4} - 1$
$y = \frac{1}{8} - \frac{2}{8} - \frac{8}{8} = -\frac{9}{8}$.
Thus,the minimum value is $-\frac{9}{8}$.

(B) Let the height of the tower be $h$ and the foot of the tower be $D$. In $\triangle E D C$,$\tan 3 \alpha = \frac{h}{C D} \Rightarrow C D = h \cot 3 \alpha$.
In $\triangle E D B$,$\tan 2 \alpha = \frac{h}{B D} \Rightarrow B D = h \cot 2 \alpha$.
In $\triangle E D A$,$\tan \alpha = \frac{h}{A D} \Rightarrow A D = h \cot \alpha$.
Now,$A B = A D - B D = h(\cot \alpha - \cot 2 \alpha)$ and $B C = B D - C D = h(\cot 2 \alpha - \cot 3 \alpha)$.
Therefore,$\frac{A B}{B C} = \frac{\cot \alpha - \cot 2 \alpha}{\cot 2 \alpha - \cot 3 \alpha} = \frac{\frac{\cos \alpha}{\sin \alpha} - \frac{\cos 2 \alpha}{\sin 2 \alpha}}{\frac{\cos 2 \alpha}{\sin 2 \alpha} - \frac{\cos 3 \alpha}{\sin 3 \alpha}} = \frac{\frac{\sin(2 \alpha - \alpha)}{\sin \alpha \sin 2 \alpha}}{\frac{\sin(3 \alpha - 2 \alpha)}{\sin 2 \alpha \sin 3 \alpha}} = \frac{\sin \alpha}{\sin \alpha \sin 2 \alpha} \times \frac{\sin 2 \alpha \sin 3 \alpha}{\sin \alpha} = \frac{\sin 3 \alpha}{\sin \alpha}$.
Using $\sin 3 \alpha = 3 \sin \alpha - 4 \sin^3 \alpha$,we get $\frac{\sin 3 \alpha}{\sin \alpha} = 3 - 4 \sin^2 \alpha = 3 - 2(1 - \cos 2 \alpha) = 1 + 2 \cos 2 \alpha$.

(D) Let $I = \int (1+x-x^{-1}) e^{x+x^{-1}} dx$.
We can rewrite the integral as:
$I = \int e^{x+x^{-1}} dx + \int (x-x^{-1}) e^{x+x^{-1}} dx$.
Using integration by parts on the first term $\int e^{x+x^{-1}} dx$ by taking $u = x$ and $dv = e^{x+x^{-1}} dx$ is not direct,so let us consider the derivative of $x e^{x+x^{-1}}$:
$\frac{d}{dx} (x e^{x+x^{-1}}) = 1 \cdot e^{x+x^{-1}} + x \cdot e^{x+x^{-1}} \cdot (1 - x^{-2}) = e^{x+x^{-1}} + x e^{x+x^{-1}} - x^{-1} e^{x+x^{-1}} = (1 + x - x^{-1}) e^{x+x^{-1}}$.
Since the derivative of $x e^{x+x^{-1}}$ is exactly the integrand,the integral is $x e^{x+x^{-1}} + C$.

(A) Let $f(x) = \int_0^x t e^{t^2} dt$.
To evaluate the integral,let $u = t^2$,then $du = 2t dt$,which implies $t dt = \frac{1}{2} du$.
When $t = 0$,$u = 0$. When $t = x$,$u = x^2$.
Thus,$f(x) = \int_0^{x^2} \frac{1}{2} e^u du = \frac{1}{2} [e^u]_0^{x^2} = \frac{1}{2} (e^{x^2} - 1)$.
To find the minimum value,we differentiate $f(x)$ with respect to $x$:
$f'(x) = \frac{1}{2} \cdot e^{x^2} \cdot (2x) = x e^{x^2}$.
Setting $f'(x) = 0$ gives $x = 0$.
Now,find the second derivative: $f''(x) = e^{x^2} + x(e^{x^2} \cdot 2x) = e^{x^2}(1 + 2x^2)$.
At $x = 0$,$f''(0) = e^0(1 + 0) = 1 > 0$.
Since $f''(0) > 0$,the function has a local minimum at $x = 0$.
The minimum value is $f(0) = \frac{1}{2} (e^0 - 1) = \frac{1}{2} (1 - 1) = 0$.

(B) Let $I = \int_0^1 \sin \left(2 \tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right) d x$.
Substitute $x = \cos \theta$,then $d x = -\sin \theta d \theta$.
When $x = 0$,$\theta = \frac{\pi}{2}$. When $x = 1$,$\theta = 0$.
$I = \int_{\pi/2}^0 \sin \left(2 \tan ^{-1} \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\right) (-\sin \theta) d \theta$.
Using $\sqrt{\frac{1+\cos \theta}{1-\cos \theta}} = \sqrt{\frac{2\cos^2(\theta/2)}{2\sin^2(\theta/2)}} = \cot(\theta/2) = \tan(\frac{\pi}{2} - \frac{\theta}{2})$.
$I = \int_0^{\pi/2} \sin \left(2 \tan^{-1} \tan(\frac{\pi}{2} - \frac{\theta}{2})\right) \sin \theta d \theta$.
$I = \int_0^{\pi/2} \sin(\pi - \theta) \sin \theta d \theta = \int_0^{\pi/2} \sin^2 \theta d \theta$.
Using $\sin^2 \theta = \frac{1-\cos 2\theta}{2}$,we get $I = \int_0^{\pi/2} \frac{1-\cos 2\theta}{2} d \theta$.
$I = \frac{1}{2} [\theta - \frac{\sin 2\theta}{2}]_0^{\pi/2} = \frac{1}{2} [(\frac{\pi}{2} - 0) - (0 - 0)] = \frac{\pi}{4}$.

(A) We need to evaluate the integral $I = \int_0^3 \frac{3x+1}{x^2+9} dx$.
Split the integral into two parts:
$I = \int_0^3 \frac{3x}{x^2+9} dx + \int_0^3 \frac{1}{x^2+9} dx$
For the first part,let $u = x^2+9$,then $du = 2x dx$,so $x dx = \frac{du}{2}$.
$\int_0^3 \frac{3x}{x^2+9} dx = \frac{3}{2} \int_9^{18} \frac{1}{u} du = \frac{3}{2} [\log |u|]_9^{18} = \frac{3}{2} (\log 18 - \log 9) = \frac{3}{2} \log 2$.
For the second part,use the formula $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$:
$\int_0^3 \frac{1}{x^2+3^2} dx = [\frac{1}{3} \tan^{-1}(\frac{x}{3})]_0^3 = \frac{1}{3} (\tan^{-1}(1) - \tan^{-1}(0)) = \frac{1}{3} (\frac{\pi}{4} - 0) = \frac{\pi}{12}$.
Combining both parts:
$I = \frac{3}{2} \log 2 + \frac{\pi}{12} = \log (2^{3/2}) + \frac{\pi}{12} = \log (2 \sqrt{2}) + \frac{\pi}{12}$.

(D) We have $\int_{-2}^2 |[x]| \, dx = \int_{-2}^{-1} |[x]| \, dx + \int_{-1}^0 |[x]| \, dx + \int_0^1 |[x]| \, dx + \int_1^2 |[x]| \, dx$.
Since $[x]$ is the greatest integer function:
For $x \in [-2, -1)$,$[x] = -2$,so $|[x]| = |-2| = 2$.
For $x \in [-1, 0)$,$[x] = -1$,so $|[x]| = |-1| = 1$.
For $x \in [0, 1)$,$[x] = 0$,so $|[x]| = |0| = 0$.
For $x \in [1, 2)$,$[x] = 1$,so $|[x]| = |1| = 1$.
Substituting these values:
$\int_{-2}^2 |[x]| \, dx = \int_{-2}^{-1} 2 \, dx + \int_{-1}^0 1 \, dx + \int_0^1 0 \, dx + \int_1^2 1 \, dx$.
$= 2[x]_{-2}^{-1} + [x]_{-1}^0 + 0 + [x]_1^2$.
$= 2(-1 - (-2)) + (0 - (-1)) + (2 - 1)$.
$= 2(1) + 1 + 1 = 4$.