AP EAMCET 2011 Chemistry Question Paper with Answer and Solution

(B) Boron reacts with concentrated $HNO_3$ to produce orthoboric acid and nitrogen dioxide,not $N_2O$.
The balanced chemical equation is: $B + 3HNO_3 \longrightarrow H_3BO_3 + 3NO_2$.

(B) The structure of $ClO_2$ is angular.
In the $ClO_2$ molecule,the central chlorine atom is $sp^3$-hybridized with one unpaired electron.
Due to the presence of the unpaired electron and the lone pair on the chlorine atom,the bond angle is observed to be $118^{\circ}$ and the $Cl-O$ bond length is $1.47 \ \mathring{A}$.

(C) The general equation of a pair of straight lines passing through the origin is $ax^2+2hxy+by^2=0$.
If one of these lines bisects the angle between the coordinate axes,its equation is $y=x$ or $y=-x$.
Substituting $y=x$ into the equation $ax^2+2hxy+by^2=0$,we get $ax^2+2hx^2+bx^2=0$,which implies $a+2h+b=0$,or $a+b=-2h$. Squaring both sides gives $(a+b)^2=4h^2$.
Similarly,if $y=-x$,we get $ax^2-2hx^2+bx^2=0$,which implies $a-2h+b=0$,or $a+b=2h$. Squaring both sides gives $(a+b)^2=4h^2$.
Thus,the condition for one of the lines to bisect the angle between the axes is $(a+b)^2=4h^2$.
For the given equation $4x^2+6xy+ky^2=0$,we have $a=4$,$2h=6$ (so $h=3$),and $b=k$.
Substituting these values into the condition $(a+b)^2=4h^2$:
$(4+k)^2=4(3)^2$
$(4+k)^2=36$
$4+k = \pm 6$
If $4+k=6$,then $k=2$.
If $4+k=-6$,then $k=-10$.
Therefore,$k \in \{-10, 2\}$.

(B) The general second-degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$ represents a pair of parallel lines if $h^2=ab$ and $bg^2=af^2$.
From $h^2=ab$,we have $h=\sqrt{ab}$.
From $bg^2=af^2$,we can write $\frac{g^2}{f^2}=\frac{a}{b}$.
For parallel lines,the condition is $\frac{a}{h}=\frac{h}{b}=\frac{g}{f}$.
Thus,$\frac{g^2}{f^2}=\frac{a^2}{h^2}=\frac{a^2}{ab}=\frac{a}{b}$.
Also,for parallel lines,the distance between them is given by $2\sqrt{\frac{g^2-ac}{a(a+b)}} = 2\sqrt{\frac{f^2-bc}{b(a+b)}}$.
This implies $\frac{g^2-ac}{a} = \frac{f^2-bc}{b}$.
Therefore,$\frac{g^2-ac}{f^2-bc} = \frac{a}{b}$.
Taking the square root on both sides,we get $\sqrt{\frac{g^2-ac}{f^2-bc}} = \sqrt{\frac{a}{b}}$.

(B) The given equation of the pair of lines is $3x^2 - 2xy - 15y^2 = 0$.
Comparing this with the general equation $ax^2 + 2hxy + by^2 = 0$,we get $a = 3$,$2h = -2$ (so $h = -1$),and $b = -15$.
Let the slopes of the lines be $m_1$ and $m_2$.
The sum of the slopes is $s = m_1 + m_2 = \frac{-2h}{b} = \frac{-2(-1)}{-15} = \frac{2}{-15} = -\frac{2}{15}$.
The product of the slopes is $p = m_1 m_2 = \frac{a}{b} = \frac{3}{-15} = -\frac{3}{15}$.
Therefore,the ratio $s:p = \left(-\frac{2}{15}\right) : \left(-\frac{3}{15}\right) = 2:3$.

(A) The given line is $y=2x+c$,which can be written as $2x-y+c=0$.
The equation of the circle is $x^2+y^2=5$,which has center $(0,0)$ and radius $r=\sqrt{5}$.
For a line $Ax+By+C=0$ to be tangent to a circle with center $(h,k)$ and radius $r$,the perpendicular distance from the center to the line must be equal to the radius:
$\frac{|A(h)+B(k)+C|}{\sqrt{A^2+B^2}} = r$
Substituting the values $A=2, B=-1, C=c, h=0, k=0, r=\sqrt{5}$:
$\frac{|2(0)-1(0)+c|}{\sqrt{2^2+(-1)^2}} = \sqrt{5}$
$\frac{|c|}{\sqrt{4+1}} = \sqrt{5}$
$\frac{|c|}{\sqrt{5}} = \sqrt{5}$
$|c| = \sqrt{5} \times \sqrt{5} = 5$
Therefore,$c = \pm 5$.
Comparing with the given options,the value of $c$ is $5$.

(C) Given,the lines $3x + 4y - 14 = 0$ and $6x + 8y + 7 = 0$ are both tangents to a circle.
We observe that both lines are parallel.
Dividing the second equation by $2$,we get $3x + 4y + \frac{7}{2} = 0$.
The distance between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.
The diameter of the circle is equal to the distance between these two parallel tangents.
Diameter $= \frac{|-14 - \frac{7}{2}|}{\sqrt{3^2 + 4^2}} = \frac{|-\frac{28}{2} - \frac{7}{2}|}{\sqrt{9 + 16}} = \frac{|-\frac{35}{2}|}{5} = \frac{35}{10} = \frac{7}{2}$.
Therefore,the radius of the circle is $r = \frac{\text{Diameter}}{2} = \frac{7/2}{2} = \frac{7}{4}$.

(A) For the first circle $S_1: x^2+y^2+8x-4y+c=0$,the center is $C_1(-4, 2)$ and radius $r_1 = \sqrt{(-4)^2 + 2^2 - c} = \sqrt{20-c}$.
For the second circle $S_2: x^2+y^2+2x+4y-11=0$,the center is $C_2(-1, -2)$ and radius $r_2 = \sqrt{(-1)^2 + (-2)^2 - (-11)} = \sqrt{1+4+11} = 4$.
The distance between centers $C_1$ and $C_2$ is $d = \sqrt{(-4 - (-1))^2 + (2 - (-2))^2} = \sqrt{(-3)^2 + 4^2} = \sqrt{9+16} = 5$.
Since the circles touch externally,$d = r_1 + r_2$,so $5 = \sqrt{20-c} + 4$,which implies $\sqrt{20-c} = 1$,so $20-c = 1$,giving $c = 19$.
The third circle is $S_3: x^2+y^2-6x+8y+k=0$.
Two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ cut orthogonally if $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
For $S_1$ and $S_3$,we have $g_1=4, f_1=-2, c_1=19$ and $g_3=-3, f_3=4, c_3=k$.
Thus,$2(4)(-3) + 2(-2)(4) = 19 + k$.
$-24 - 16 = 19 + k$,which gives $-40 = 19 + k$,so $k = -59$.

(B) Given the circle $S_1: x^2+y^2+8x-4y+c=0$ with center $C_1=(-4, 2)$ and radius $r_1=\sqrt{(-4)^2+2^2-c}=\sqrt{20-c}$.
Given the circle $S_2: x^2+y^2+2x+4y-11=0$ with center $C_2=(-1, -2)$ and radius $r_2=\sqrt{(-1)^2+(-2)^2-(-11)}=\sqrt{1+4+11}=4$.
Since the circles touch externally,the distance between centers $C_1C_2 = r_1+r_2$.
$C_1C_2 = \sqrt{(-4 - (-1))^2 + (2 - (-2))^2} = \sqrt{(-3)^2 + 4^2} = \sqrt{9+16} = 5$.
Thus,$5 = \sqrt{20-c} + 4$ $\Rightarrow \sqrt{20-c} = 1$ $\Rightarrow 20-c = 1$ $\Rightarrow c = 19$.
Now,the circle $S_1$ cuts the circle $S_3: x^2+y^2-6x+8y+k=0$ orthogonally. The condition for orthogonality is $2g_1g_3 + 2f_1f_3 = c_1+c_3$.
For $S_1$,$g_1=4, f_1=-2, c_1=c=19$.
For $S_3$,$g_3=-3, f_3=4, c_3=k$.
$2(4)(-3) + 2(-2)(4) = 19 + k$.
$-24 - 16 = 19 + k$.
$-40 = 19 + k$.
$k = -59$.

(B) Given equations of circles are:
$S_1 \equiv x^2+y^2+2x+2y+1=0$
$S_2 \equiv x^2+y^2-2x+2y+1=0$
For circle $S_1$,the centre is $C_1 = (-1, -1)$ and radius $r_1 = \sqrt{(-1)^2 + (-1)^2 - 1} = \sqrt{1} = 1$.
For circle $S_2$,the centre is $C_2 = (1, -1)$ and radius $r_2 = \sqrt{(1)^2 + (-1)^2 - 1} = \sqrt{1} = 1$.
The distance between the centres $C_1$ and $C_2$ is $d = \sqrt{(1 - (-1))^2 + (-1 - (-1))^2} = \sqrt{2^2 + 0^2} = 2$.
Since $d = r_1 + r_2 = 1 + 1 = 2$,the circles touch each other externally.
The point of contact is the midpoint of the line segment joining the centres $C_1$ and $C_2$.
Point of contact $= \left( \frac{-1 + 1}{2}, \frac{-1 + (-1)}{2} \right) = (0, -1)$.

(B) Let the coordinates of point $A$ be $(a \cos \theta, a \sin \theta)$ since it lies on the circle $x^2 + y^2 = a^2$.
Since $AM$ is parallel to the $X$-axis and has length $a$,the coordinates of $M$ can be represented as $(x, y) = (a \cos \theta + a, a \sin \theta)$ or $(a \cos \theta - a, a \sin \theta)$.
Considering the case $(x, y) = (a \cos \theta + a, a \sin \theta)$,we have $x - a = a \cos \theta$ and $y = a \sin \theta$.
Squaring and adding these equations:
$(x - a)^2 + y^2 = a^2 \cos^2 \theta + a^2 \sin^2 \theta$
$(x - a)^2 + y^2 = a^2$
$x^2 - 2ax + a^2 + y^2 = a^2$
$x^2 + y^2 = 2ax$.
Similarly,for the other case,we get $x^2 + y^2 = -2ax$.

(D) The equation of the parabola is $y^2=lx$. The axis of this parabola is the $x$-axis,which has the equation $y=0$.
Since the line $y=mx+c$ is parallel to the axis of the parabola $(y=0)$,its slope $m$ must be $0$. Thus,the line is $y=c$.
Given that the point $\left(\frac{c^2}{8}, c\right)$ lies on the parabola $y^2=lx$,it must satisfy the equation:
$c^2 = l \left(\frac{c^2}{8}\right)$
$c^2 = \frac{lc^2}{8}$
Assuming $c \neq 0$,we can divide both sides by $c^2$:
$1 = \frac{l}{8} \Rightarrow l = 8$.
The length of the latus rectum of the parabola $y^2=lx$ is $l$.
Therefore,the length of the latus rectum is $8$.

(C) Let $P(t^2, 2t)$ be one end of a focal chord $PQ$ of the parabola $y^2=4x$. The coordinates of the other end $Q$ are $(\frac{1}{t^2}, \frac{-2}{t})$,where $tt' = -1$.
Given that the chord makes an angle $\theta$ with the positive direction of the $X$-axis,the slope of the chord is $\tan \theta$.
$\tan \theta = \frac{\frac{-2}{t} - 2t}{\frac{1}{t^2} - t^2} = \frac{-2(t + \frac{1}{t})}{\frac{1-t^4}{t^2}} = \frac{-2(t + \frac{1}{t})}{\frac{(1-t^2)(1+t^2)}{t^2}} = \frac{2t}{t^2-1}$.
Alternatively,using the slope formula for a focal chord with parameter $t$: $\tan \theta = \frac{2}{t - \frac{1}{t}}$.
Thus,$t - \frac{1}{t} = 2 \cot \theta$.
The length of the focal chord $PQ$ is given by $a(t + \frac{1}{t})^2$,where $a=1$.
$PQ = (t + \frac{1}{t})^2 = (t - \frac{1}{t})^2 + 4$.
Substituting $t - \frac{1}{t} = 2 \cot \theta$:
$PQ = (2 \cot \theta)^2 + 4 = 4 \cot^2 \theta + 4 = 4(\cot^2 \theta + 1) = 4 \operatorname{cosec}^2 \theta$.

(A) Given the equation of the ellipse: $x^2+4y^2+2x+16y+13=0$.
Rearranging the terms: $(x^2+2x) + 4(y^2+4y) + 13 = 0$.
Completing the squares: $(x^2+2x+1) + 4(y^2+4y+4) + 13 - 1 - 16 = 0$.
$(x+1)^2 + 4(y+2)^2 = 4$.
Dividing by $4$: $\frac{(x+1)^2}{4} + \frac{(y+2)^2}{1} = 1$.
Comparing with the standard form $\frac{X^2}{a^2} + \frac{Y^2}{b^2} = 1$,we get $a^2 = 4$ and $b^2 = 1$.
Since $a^2 > b^2$,the eccentricity $e$ is given by $b^2 = a^2(1 - e^2)$.
$1 = 4(1 - e^2) \Rightarrow 1 = 4 - 4e^2$.
$4e^2 = 3 \Rightarrow e^2 = \frac{3}{4}$.
Thus,$e = \frac{\sqrt{3}}{2}$.

(C) The given equation of the hyperbola is $x^2-3y^2=3$. Dividing by $3$,we get $\frac{x^2}{3}-\frac{y^2}{1}=1$.
Here,$a^2=3$ and $b^2=1$,so $a=\sqrt{3}$ and $b=1$.
The equations of the asymptotes are $y = \pm \frac{b}{a}x$,which gives $y = \frac{1}{\sqrt{3}}x$ and $y = -\frac{1}{\sqrt{3}}x$.
Let the slopes be $m_1 = \frac{1}{\sqrt{3}}$ and $m_2 = -\frac{1}{\sqrt{3}}$.
The angle $\theta$ between the asymptotes is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values,$\tan \theta = \left| \frac{\frac{1}{\sqrt{3}} - (-\frac{1}{\sqrt{3}})}{1 + (\frac{1}{\sqrt{3}})(-\frac{1}{\sqrt{3}})} \right| = \left| \frac{\frac{2}{\sqrt{3}}}{1 - \frac{1}{3}} \right| = \left| \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} \right| = \sqrt{3}$.
Since $\tan \theta = \sqrt{3}$,we have $\theta = \frac{\pi}{3}$.

(C) To evaluate the limit $\lim _{x \rightarrow 8} \frac{\sqrt{1+\sqrt{1+x}}-2}{x-8}$,we rationalize the numerator:
$\lim _{x \rightarrow 8} \frac{\sqrt{1+\sqrt{1+x}}-2}{x-8} \times \frac{\sqrt{1+\sqrt{1+x}}+2}{\sqrt{1+\sqrt{1+x}}+2}$
$= \lim _{x \rightarrow 8} \frac{1+\sqrt{1+x}-4}{(x-8)(\sqrt{1+\sqrt{1+x}}+2)}$
$= \lim _{x \rightarrow 8} \frac{\sqrt{1+x}-3}{(x-8)(\sqrt{1+\sqrt{1+x}}+2)}$
Now,rationalize the numerator again:
$= \lim _{x \rightarrow 8} \frac{\sqrt{1+x}-3}{(x-8)(\sqrt{1+\sqrt{1+x}}+2)} \times \frac{\sqrt{1+x}+3}{\sqrt{1+x}+3}$
$= \lim _{x \rightarrow 8} \frac{1+x-9}{(x-8)(\sqrt{1+\sqrt{1+x}}+2)(\sqrt{1+x}+3)}$
$= \lim _{x \rightarrow 8} \frac{x-8}{(x-8)(\sqrt{1+\sqrt{1+x}}+2)(\sqrt{1+x}+3)}$
$= \lim _{x \rightarrow 8} \frac{1}{(\sqrt{1+\sqrt{1+x}}+2)(\sqrt{1+x}+3)}$
Substitute $x = 8$:
$= \frac{1}{(\sqrt{1+\sqrt{9}}+2)(\sqrt{9}+3)} = \frac{1}{(\sqrt{4}+2)(3+3)} = \frac{1}{(2+2)(6)} = \frac{1}{4 \times 6} = \frac{1}{24}$

(B) Polymers that control various life processes in plants and animals are called biopolymers.
Cellulose,insulin,$DNA$,starch,and proteins are examples of biopolymers.
Nylon-$6$ is a synthetic polymer,not a biopolymer.

(C) Given,$\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$.
By the sine rule,we know that $a = k \sin A$,$b = k \sin B$,and $c = k \sin C$,where $k$ is a constant.
Substituting these in the given equation:
$\frac{\cos A}{k \sin A} = \frac{\cos B}{k \sin B} = \frac{\cos C}{k \sin C}$
$\cot A = \cot B = \cot C$
Since $A, B, C$ are angles of a triangle,$\cot A = \cot B = \cot C$ implies $A = B = C$.
Since all angles are equal to $60^{\circ}$,the triangle is equilateral.

(A) Given: $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$
Using the half-angle formulas $\cos^2 \frac{C}{2} = \frac{s(s-c)}{ab}$ and $\cos^2 \frac{A}{2} = \frac{s(s-a)}{bc}$:
$a \cdot \frac{s(s-c)}{ab} + c \cdot \frac{s(s-a)}{bc} = \frac{3b}{2}$
$\frac{s(s-c)}{b} + \frac{s(s-a)}{b} = \frac{3b}{2}$
$\frac{s}{b} (s - c + s - a) = \frac{3b}{2}$
Since $2s = a + b + c$,we have $2s - a - c = b$:
$\frac{s}{b} (b) = \frac{3b}{2}$
$s = \frac{3b}{2} \Rightarrow 2s = 3b$
Substituting $2s = a + b + c$:
$a + b + c = 3b$
$a + c = 2b$
Thus,$a, b, c$ are in arithmetic progression.

(B) Given,$A(\alpha, \beta) = \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & e^\beta \end{bmatrix}$.
First,we find the determinant $|A(\alpha, \beta)| = e^\beta (\cos^2 \alpha + \sin^2 \alpha) = e^\beta$.
Next,we find the cofactor matrix $C$:
$C_{11} = e^\beta \cos \alpha, C_{12} = e^\beta \sin \alpha, C_{13} = 0$
$C_{21} = -e^\beta \sin \alpha, C_{22} = e^\beta \cos \alpha, C_{23} = 0$
$C_{31} = 0, C_{32} = 0, C_{33} = \cos^2 \alpha + \sin^2 \alpha = 1$
Thus,$\text{adj}(A(\alpha, \beta)) = C^T = \begin{bmatrix} e^\beta \cos \alpha & -e^\beta \sin \alpha & 0 \\ e^\beta \sin \alpha & e^\beta \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Then,$[A(\alpha, \beta)]^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{e^\beta} \begin{bmatrix} e^\beta \cos \alpha & -e^\beta \sin \alpha & 0 \\ e^\beta \sin \alpha & e^\beta \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & e^{-\beta} \end{bmatrix}$.
Comparing this with the original matrix $A(\alpha, \beta)$,we see that $\cos \alpha = \cos(-\alpha)$,$-\sin \alpha = \sin(-\alpha)$,and $e^{-\beta}$ corresponds to the third diagonal element. Thus,the result is $A(-\alpha, -\beta)$.

(C) Let $\Delta = \left|\begin{array}{lll}24 & 25 & 26 \\ 25 & 26 & 27 \\ 26 & 27 & 27\end{array}\right|$.
Applying row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_2$:
$\Delta = \left|\begin{array}{ccc} 24 & 25 & 26 \\ 1 & 1 & 1 \\ 1 & 1 & 0 \end{array}\right|$.
Expanding along the third row $(R_3)$:
$\Delta = 1(25 - 26) - 1(24 - 26) + 0(24 - 25)$
$\Delta = 1(-1) - 1(-2) + 0$
$\Delta = -1 + 2 = 1$.

(A) Given equation is $(\tan ^{-1} x)^2+(\cot ^{-1} x)^2=\frac{5 \pi^2}{8}$.
Using the identity $(\tan ^{-1} x + \cot ^{-1} x)^2 = (\tan ^{-1} x)^2 + (\cot ^{-1} x)^2 + 2 \tan ^{-1} x \cot ^{-1} x$,we have:
$(\tan ^{-1} x + \cot ^{-1} x)^2 - 2 \tan ^{-1} x \cot ^{-1} x = \frac{5 \pi^2}{8}$.
Since $\tan ^{-1} x + \cot ^{-1} x = \frac{\pi}{2}$,we substitute this into the equation:
$(\frac{\pi}{2})^2 - 2 \tan ^{-1} x (\frac{\pi}{2} - \tan ^{-1} x) = \frac{5 \pi^2}{8}$.
$\frac{\pi^2}{4} - \pi \tan ^{-1} x + 2(\tan ^{-1} x)^2 = \frac{5 \pi^2}{8}$.
$2(\tan ^{-1} x)^2 - \pi \tan ^{-1} x + \frac{\pi^2}{4} - \frac{5 \pi^2}{8} = 0$.
$2(\tan ^{-1} x)^2 - \pi \tan ^{-1} x - \frac{3 \pi^2}{8} = 0$.
Let $u = \tan ^{-1} x$. Then $2u^2 - \pi u - \frac{3 \pi^2}{8} = 0$.
Using the quadratic formula $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$u = \frac{\pi \pm \sqrt{\pi^2 - 4(2)(-\frac{3 \pi^2}{8})}}{4} = \frac{\pi \pm \sqrt{\pi^2 + 3 \pi^2}}{4} = \frac{\pi \pm 2 \pi}{4}$.
So,$u = \frac{3 \pi}{4}$ or $u = -\frac{\pi}{4}$.
Since $x = \tan(u)$,we have $x = \tan(\frac{3 \pi}{4}) = -1$ or $x = \tan(-\frac{\pi}{4}) = -1$.
Thus,$x = -1$.

(A) We know that the formula for the inverse hyperbolic sine function is $\sinh ^{-1}(y) = \log \left(y + \sqrt{1 + y^2}\right)$.
Substituting $y = \cot x$ into the formula,we get:
$\sinh ^{-1}(\cot x) = \log \left(\cot x + \sqrt{1 + \cot ^2 x}\right)$
Using the trigonometric identity $1 + \cot ^2 x = \operatorname{cosec}^2 x$,we have:
$= \log \left(\cot x + \sqrt{\operatorname{cosec}^2 x}\right)$
$= \log (\cot x + \operatorname{cosec} x)$
$= \log \left(\frac{\cos x}{\sin x} + \frac{1}{\sin x}\right)$
$= \log \left(\frac{1 + \cos x}{\sin x}\right)$
Using the half-angle identities $1 + \cos x = 2 \cos ^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$,we get:
$= \log \left(\frac{2 \cos ^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}\right)$
$= \log \left(\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}\right)$
$= \log \left(\cot \frac{x}{2}\right)$.

(D) Let $y = \frac{x^2-3x+4}{x^2+3x+4}$.
$y(x^2+3x+4) = x^2-3x+4$
$x^2(y-1) + x(3y+3) + (4y-4) = 0$.
Since $x$ is real,the discriminant $D \geq 0$.
$D = (3y+3)^2 - 4(y-1)(4y-4) \geq 0$.
$9(y+1)^2 - 16(y-1)^2 \geq 0$.
$9(y^2+2y+1) - 16(y^2-2y+1) \geq 0$.
$9y^2 + 18y + 9 - 16y^2 + 32y - 16 \geq 0$.
$-7y^2 + 50y - 7 \geq 0$.
$7y^2 - 50y + 7 \leq 0$.
$(7y-1)(y-7) \leq 0$.
Thus,$\frac{1}{7} \leq y \leq 7$.

(D) Given $f(x) = [\frac{x}{5}]$. We need to find the range of $f(x)$ for $|x| < 71$.
This is equivalent to $-71 < x < 71$.
Dividing by $5$,we get $-\frac{71}{5} < \frac{x}{5} < \frac{71}{5}$.
This simplifies to $-14.2 < \frac{x}{5} < 14.2$.
Since $f(x) = [\frac{x}{5}]$,the possible integer values for $f(x)$ are the integers in the interval $(-14.2, 14.2)$.
The smallest integer greater than $-14.2$ is $-14$,and the largest integer less than $14.2$ is $14$.
Wait,let us re-evaluate the bounds:
If $x$ is slightly greater than $-71$,say $x = -70.9$,then $f(x) = [\frac{-70.9}{5}] = [-14.18] = -15$.
If $x$ is slightly less than $71$,say $x = 70.9$,then $f(x) = [\frac{70.9}{5}] = [14.18] = 14$.
Thus,the set of values is $\{-15, -14, \ldots, 0, \ldots, 13, 14\}$.

(D) The function is given by $f(x) = 7 + \cos(5x + 3)$.
We know that the fundamental period of the function $\cos(ax + b)$ is given by $\frac{2\pi}{|a|}$.
Here,$a = 5$.
Therefore,the period of $\cos(5x + 3)$ is $\frac{2\pi}{|5|} = \frac{2\pi}{5}$.
Since adding a constant $7$ to the function does not change its period,the period of $f(x)$ remains $\frac{2\pi}{5}$.

(B) For the function $f(x)$ to be continuous at $x=0$,we must have $\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^+} f(x) = f(0)$.
First,consider the $LHL$: $\lim_{x \rightarrow 0^-} \frac{a+2 \cos x}{x^2}$.
For this limit to exist and be finite,the numerator must approach $0$ as $x \rightarrow 0$. Thus,$a + 2 \cos(0) = 0 \implies a + 2 = 0 \implies a = -2$.
Using the Taylor series expansion for $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$,we get:
$LHL = \lim_{x \rightarrow 0} \frac{-2 + 2(1 - \frac{x^2}{2} + \dots)}{x^2} = \lim_{x \rightarrow 0} \frac{-x^2}{x^2} = -1$.
Now,consider the $RHL$: $\lim_{x \rightarrow 0^+} b \tan \frac{\pi}{[x+4]}$.
As $x \rightarrow 0^+$,$[x+4] = 4$. So,$RHL = b \tan \frac{\pi}{4} = b(1) = b$.
Since the function is continuous,$LHL = RHL$,which implies $b = -1$.
Thus,the ordered pair $(a, b)$ is $(-2, -1)$.

(C) Given $y = (1+x)(1+x^2)(1+x^4) \dots (1+x^{2^n})$.
Multiply and divide by $(1-x)$:
$y = \frac{(1-x)(1+x)(1+x^2)(1+x^4) \dots (1+x^{2^n})}{1-x}$
Using the identity $(a-b)(a+b) = a^2-b^2$ repeatedly:
$y = \frac{(1-x^2)(1+x^2)(1+x^4) \dots (1+x^{2^n})}{1-x} = \frac{(1-x^4)(1+x^4) \dots (1+x^{2^n})}{1-x} = \dots = \frac{1-x^{2^{n+1}}}{1-x}$
Now,differentiate using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{vu' - uv'}{v^2}$:
$\frac{dy}{dx} = \frac{(1-x)(-2^{n+1}x^{2^{n+1}-1}) - (1-x^{2^{n+1}})(-1)}{(1-x)^2}$
At $x=0$:
$(\frac{dy}{dx})_{x=0} = \frac{(1-0)(0) - (1-0)(-1)}{(1-0)^2} = \frac{0 + 1}{1} = 1$.

(D) Given,$y=\frac{\log _e x}{x}$ and $z=\log _e x$.
Since $z=\log _e x$,we have $x=e^z$.
Substituting $x$ in $y$,we get $y=\frac{z}{e^z} = z e^{-z}$.
Now,differentiating $y$ with respect to $z$:
$\frac{d y}{d z} = \frac{d}{d z}(z e^{-z}) = e^{-z} - z e^{-z} = (1-z)e^{-z}$.
Next,differentiating $\frac{d y}{d z}$ with respect to $z$ to find the second derivative:
$\frac{d^2 y}{d z^2} = \frac{d}{d z}((1-z)e^{-z}) = -e^{-z} - (1-z)e^{-z} = (-1-1+z)e^{-z} = (z-2)e^{-z}$.
Finally,calculating the sum:
$\frac{d^2 y}{d z^2} + \frac{d y}{d z} = (z-2)e^{-z} + (1-z)e^{-z} = (z-2+1-z)e^{-z} = -1 \cdot e^{-z} = -e^{-z}$.

(A) Given,$\cos ^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=k$
$\Rightarrow \frac{x^2-y^2}{x^2+y^2}=\cos k$
Let $\cos k = C$ (a constant).
Then,$x^2 - y^2 = C(x^2 + y^2)$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(x^2 - y^2) = \frac{d}{dx}(C(x^2 + y^2))$
$2x - 2y \frac{dy}{dx} = C(2x + 2y \frac{dy}{dx})$
$x - y \frac{dy}{dx} = C(x + y \frac{dy}{dx})$
$x - Cx = Cy \frac{dy}{dx} + y \frac{dy}{dx}$
$x(1 - C) = y \frac{dy}{dx}(1 + C)$
$\frac{dy}{dx} = \frac{x(1 - C)}{y(1 + C)}$
Since $C = \frac{x^2 - y^2}{x^2 + y^2}$,we substitute back:
$\frac{dy}{dx} = \frac{x(1 - \frac{x^2 - y^2}{x^2 + y^2})}{y(1 + \frac{x^2 - y^2}{x^2 + y^2})} = \frac{x(\frac{x^2 + y^2 - x^2 + y^2}{x^2 + y^2})}{y(\frac{x^2 + y^2 + x^2 - y^2}{x^2 + y^2})} = \frac{x(2y^2)}{y(2x^2)} = \frac{2xy^2}{2x^2y} = \frac{y}{x}$

(D) Given the curve $y = 5^x$.
First,find the derivative $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{d}{dx}(5^x) = 5^x \log_e 5$.
At the point $(x_1, y_1)$,the slope of the tangent is $\left(\frac{dy}{dx}\right)_{(x_1, y_1)} = 5^{x_1} \log_e 5$.
The length of the subtangent is given by the formula $\left| \frac{y_1}{\frac{dy}{dx}} \right|$.
Substituting the values:
Length of subtangent $= \frac{y_1}{5^{x_1} \log_e 5}$.
Since the point $(x_1, y_1)$ lies on the curve $y = 5^x$,we have $y_1 = 5^{x_1}$.
Therefore,the length of the subtangent $= \frac{5^{x_1}}{5^{x_1} \log_e 5} = \frac{1}{\log_e 5}$.

(B) Given the function $f(x) = x^2 - 4x + 5$ defined on the domain $[2, \infty)$.
To be a bijection,the codomain $B$ must be equal to the range of the function.
We find the range by analyzing the function's behavior.
First,find the derivative: $f'(x) = 2x - 4$.
Setting $f'(x) = 0$ gives $x = 2$.
For all $x \in [2, \infty)$,$f'(x) \geq 0$,which means the function is monotonically increasing on this interval.
The minimum value occurs at the boundary $x = 2$.
$f(2) = (2)^2 - 4(2) + 5 = 4 - 8 + 5 = 1$.
As $x \to \infty$,$f(x) \to \infty$.
Since the function is continuous and increasing,the range is $[f(2), \infty) = [1, \infty)$.
Therefore,$B = [1, \infty)$.

(C) Let $h = 2500 \ m$ be the height of the observation point above the lake. Let $H$ be the height of the cloud above the lake. The depth of the reflection of the cloud in the lake is also $H$ below the surface of the lake.
The distance of the cloud from the observation point is $H-h$.
The distance of the reflection from the observation point is $H+h$.
Let $d$ be the horizontal distance from the observation point to the point directly below the cloud.
In the triangle formed by the observation point,the cloud,and the point directly below it:
$\cot 15^{\circ} = \frac{d}{H-h} \Rightarrow d = (H-h)(2+\sqrt{3}) \quad \dots(i)$
In the triangle formed by the observation point,the reflection,and the point directly below it:
$\cot 45^{\circ} = \frac{d}{H+h}$ $\Rightarrow 1 = \frac{d}{H+h}$ $\Rightarrow d = H+h \quad \dots(ii)$
Equating $(i)$ and $(ii)$:
$(H-h)(2+\sqrt{3}) = H+h$
$H(2+\sqrt{3}) - h(2+\sqrt{3}) = H+h$
$H(2+\sqrt{3}-1) = h(2+\sqrt{3}+1)$
$H(1+\sqrt{3}) = h(3+\sqrt{3})$
$H = h \frac{\sqrt{3}(\sqrt{3}+1)}{\sqrt{3}+1} = h\sqrt{3}$
Given $h = 2500 \ m$,so $H = 2500\sqrt{3} \ m$.

(D) Let $I = \int \frac{1+\cos 4 x}{\cot x-\tan x} d x$.
Using the identity $1+\cos 4x = 2\cos^2 2x$ and $\cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} = \frac{\cos 2x}{\frac{1}{2}\sin 2x} = 2\cot 2x$.
Substituting these into the integral:
$I = \int \frac{2\cos^2 2x}{2\cot 2x} d x = \int \frac{\cos^2 2x}{\frac{\cos 2x}{\sin 2x}} d x = \int \cos 2x \sin 2x d x$.
Multiplying and dividing by $2$:
$I = \frac{1}{2} \int 2\sin 2x \cos 2x d x = \frac{1}{2} \int \sin 4x d x$.
Integrating $\sin 4x$:
$I = \frac{1}{2} \left( -\frac{\cos 4x}{4} \right) + C = -\frac{1}{8} \cos 4x + C$.

(B) Let $I = \int \left( \sqrt{\frac{a+x}{a-x}} + \sqrt{\frac{a-x}{a+x}} \right) dx$.
Simplify the integrand:
$\sqrt{\frac{a+x}{a-x}} + \sqrt{\frac{a-x}{a+x}} = \frac{a+x + a-x}{\sqrt{(a-x)(a+x)}} = \frac{2a}{\sqrt{a^2-x^2}}$.
Now,$I = \int \frac{2a}{\sqrt{a^2-x^2}} dx$.
Using the standard integral formula $\int \frac{1}{\sqrt{a^2-x^2}} dx = \sin^{-1}\left(\frac{x}{a}\right) + C$,we get:
$I = 2a \sin^{-1}\left(\frac{x}{a}\right) + C$.

(A) Given the values $f(0)=1, f(0.5)=\frac{5}{4}, f(1)=2, f(1.5)=\frac{13}{4}, f(2)=5$ with $n=4$ intervals.
The step size is $h = \frac{b-a}{n} = \frac{2-0}{4} = 0.5$.
According to Simpson's rule:
$\int_a^b f(x) dx = \frac{h}{3} [ (y_0 + y_4) + 4(y_1 + y_3) + 2(y_2) ]$.
Substituting the values:
$\int_0^2 f(x) dx = \frac{0.5}{3} [ (1 + 5) + 4(\frac{5}{4} + \frac{13}{4}) + 2(2) ]$.
$\int_0^2 f(x) dx = \frac{0.5}{3} [ 6 + 4(\frac{18}{4}) + 4 ]$.
$\int_0^2 f(x) dx = \frac{0.5}{3} [ 6 + 18 + 4 ] = \frac{0.5}{3} [ 28 ] = \frac{14}{3}$.

(D) Given curves are $x=y^2$ and $x=3-2y^2$.
To find the points of intersection,set $y^2 = 3-2y^2$.
$3y^2 = 3 \Rightarrow y^2 = 1 \Rightarrow y = \pm 1$.
Thus,the points of intersection are $(1, 1)$ and $(1, -1)$.
The region is symmetric about the $x$-axis.
Required area $= 2 \int_0^1 (x_{right} - x_{left}) dy = 2 \int_0^1 ((3-2y^2) - y^2) dy$.
$= 2 \int_0^1 (3-3y^2) dy = 6 \int_0^1 (1-y^2) dy$.
$= 6 [y - \frac{y^3}{3}]_0^1 = 6 (1 - \frac{1}{3}) = 6 (\frac{2}{3}) = 4$ square units.

(C) According to Snell's Law at the interface between medium $2$ and $3$,the ray is incident at the critical angle $C$. Therefore,$\sin C = \frac{\mu_3}{\mu_2} = \frac{1.3}{1.8}$.
Since the angle of refraction at the first interface is $r$,and the ray is incident at the critical angle at the second interface,we have $r = C$.
Applying Snell's Law at the first interface between medium $1$ and $2$:
$\mu_1 \sin \theta = \mu_2 \sin r$
Substituting $\sin r = \sin C = \frac{1.3}{1.8}$:
$1.6 \times \sin \theta = 1.8 \times \left(\frac{1.3}{1.8}\right)$
$1.6 \times \sin \theta = 1.3$
$\sin \theta = \frac{1.3}{1.6} = \frac{13}{16}$
$\theta = \sin ^{-1}\left(\frac{13}{16}\right)$

(B) Phosphorous acid $(H_3PO_3)$ is a dibasic acid,meaning its n-factor is $2$.
For neutralization,the number of equivalents of acid must equal the number of equivalents of base: $N_1V_1 = N_2V_2$.
Normality $(N)$ is calculated as $Molarity \times n\text{-factor}$.
For $H_3PO_3$: $N = 0.3 \ M \times 2 = 0.6 \ N$.
For $NaOH$: $N = 0.1 \ M \times 1 = 0.1 \ N$.
Using the equation: $0.1 \ N \times V_{NaOH} = 0.6 \ N \times 100 \ mL$.
$V_{NaOH} = \frac{0.6 \times 100}{0.1} = 600 \ mL$.

(B) In the reaction of $NaOH$ with sulphur,sodium sulphide $(Na_2S)$ and sodium thiosulphate $(Na_2S_2O_3)$ are formed along with water $(H_2O)$,but no hydrogen gas is liberated.
$4S + 6NaOH \rightarrow 2Na_2S + Na_2S_2O_3 + 3H_2O$
In other options:
$1$. $2NaOH + C \rightarrow Na_2CO_3 + 2H_2$
$2$. $Si + 2NaOH + H_2O \rightarrow Na_2SiO_3 + 2H_2$
$3$. $Zn + 2NaOH \rightarrow Na_2ZnO_2 + H_2$

(C) Given: Barrier potential,$V = 0.3 ~V$.
Thickness of depletion layer,$d = 2 \times 10^{-6} ~m$.
The electric field $E$ is given by the relation $E = \frac{V}{d}$.
Substituting the values: $E = \frac{0.3}{2 \times 10^{-6}} = 0.15 \times 10^6 = 1.5 \times 10^5 ~V/m$.
In a $p-n$ junction,the depletion layer is formed by the diffusion of electrons from the $n$-side to the $p$-side and holes from the $p$-side to the $n$-side. This creates a positive charge on the $n$-side and a negative charge on the $p$-side. Therefore,the electric field is directed from the $n$-side to the $p$-side.

(A) Given differential equation is $\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}+\cos x=0$.
Separating the variables,we get $\frac{d y}{y+1} = -\frac{\cos x}{2+\sin x} d x$.
Integrating both sides,we have $\int \frac{d y}{y+1} = -\int \frac{\cos x}{2+\sin x} d x$.
Let $2+\sin x = t$,then $\cos x dx = dt$.
Thus,$\ln|y+1| = -\ln|2+\sin x| + C$.
This simplifies to $\ln|y+1| + \ln|2+\sin x| = C$,or $(y+1)(2+\sin x) = C'$.
Given $y(0)=1$,we substitute $x=0$ and $y=1$: $(1+1)(2+\sin 0) = C' \Rightarrow 2(2) = C' \Rightarrow C'=4$.
So,the equation is $(y+1)(2+\sin x) = 4$.
At $x=\frac{\pi}{2}$,we have $(y(\frac{\pi}{2})+1)(2+\sin \frac{\pi}{2}) = 4$.
$(y(\frac{\pi}{2})+1)(2+1) = 4$.
$3(y(\frac{\pi}{2})+1) = 4 \Rightarrow y(\frac{\pi}{2})+1 = \frac{4}{3}$.
$y(\frac{\pi}{2}) = \frac{4}{3} - 1 = \frac{1}{3}$.

(B) Let $D$ be the midpoint of side $BC$. The median through $A$ is the vector $\vec{AD}$.
Since $D$ is the midpoint of $BC$,the position vector of $D$ relative to $A$ is given by the average of the vectors $\vec{AB}$ and $\vec{AC}$:
$\vec{AD} = \frac{1}{2}(\vec{AB} + \vec{AC})$
$\vec{AD} = \frac{1}{2}((-3\hat{i} + 4\hat{k}) + (5\hat{i} - 2\hat{j} + 4\hat{k}))$
$\vec{AD} = \frac{1}{2}(2\hat{i} - 2\hat{j} + 8\hat{k}) = \hat{i} - \hat{j} + 4\hat{k}$
The length of the median $\vec{AD}$ is the magnitude of the vector $\vec{AD}$:
$|\vec{AD}| = \sqrt{(1)^2 + (-1)^2 + (4)^2}$
$|\vec{AD}| = \sqrt{1 + 1 + 16} = \sqrt{18}$

(D) Given: $|a|=1, |b|=2$ and the angle $\theta = 120^{\circ}$.
We need to evaluate the expression ${(a+3b) \times (3a-b)}^2$.
First,expand the cross product inside the brackets:
$(a+3b) \times (3a-b) = a \times (3a) - a \times b + (3b) \times (3a) - (3b) \times b$
Since $a \times a = 0$ and $b \times b = 0$,we have:
$= 0 - (a \times b) + 9(b \times a) - 0$
Using the property $b \times a = -(a \times b)$,we get:
$= -(a \times b) - 9(a \times b) = -10(a \times b)$
Now,square the magnitude:
${-10(a \times b)}^2 = 100 |a \times b|^2$
Using the formula $|a \times b| = |a||b| \sin \theta$:
$|a \times b| = 1 \times 2 \times \sin(120^{\circ}) = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$
Therefore,$100 |a \times b|^2 = 100 \times (\sqrt{3})^2 = 100 \times 3 = 300$.

(B) Let the unit vector along the line be $\hat{v}$. Since the line makes equal angles $\alpha$ with the coordinate axes,its direction cosines are $\cos \alpha, \cos \alpha, \cos \alpha$.
We know that $\cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1$,which implies $3\cos^2 \alpha = 1$,so $\cos \alpha = \pm \frac{1}{\sqrt{3}}$.
Thus,the unit vector along the line is $\hat{v} = \pm \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})$.
The projection of vector $\vec{a} = 4\hat{i} - 3\hat{j} + 2\hat{k}$ on the line is given by the magnitude of the dot product $\vec{a} \cdot \hat{v}$.
$|\vec{a} \cdot \hat{v}| = |(4\hat{i} - 3\hat{j} + 2\hat{k}) \cdot \pm \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})|$
$= |\pm \frac{1}{\sqrt{3}}(4 - 3 + 2)| = |\pm \frac{3}{\sqrt{3}}| = \sqrt{3}$.

(C) Let the direction angles of the line be $\alpha_1, \alpha_2, \alpha_3$. We are given $\alpha_1 = \alpha$,$\alpha_2 = \frac{\pi}{2} - \alpha$,and $\alpha_3 = \beta$.
The property of direction cosines states that $\cos^2 \alpha_1 + \cos^2 \alpha_2 + \cos^2 \alpha_3 = 1$.
Substituting the given values:
$\cos^2 \alpha + \cos^2 \left(\frac{\pi}{2} - \alpha\right) + \cos^2 \beta = 1$
Since $\cos \left(\frac{\pi}{2} - \alpha\right) = \sin \alpha$,the equation becomes:
$\cos^2 \alpha + \sin^2 \alpha + \cos^2 \beta = 1$
Using the identity $\cos^2 \alpha + \sin^2 \alpha = 1$,we get:
$1 + \cos^2 \beta = 1$
$\cos^2 \beta = 0$
$\cos \beta = 0$
Therefore,$\beta = \frac{\pi}{2}$.

(A) Given polar equation of the line is $\sin \theta - \cos \theta = \frac{1}{r}$.
Multiplying by $r$,we get $r \sin \theta - r \cos \theta = 1$.
In Cartesian coordinates,$x = r \cos \theta$ and $y = r \sin \theta$,so the equation becomes $y - x = 1$ or $x - y + 1 = 0$.
The slope of this line is $m_1 = 1$.
The line perpendicular to this line will have a slope $m_2 = -\frac{1}{m_1} = -1$.
The equation of a line with slope $-1$ is $x + y = k$.
This line passes through the point $\left(2, \frac{\pi}{6}\right)$.
Converting this point to Cartesian coordinates:
$x = 2 \cos \left(\frac{\pi}{6}\right) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$
$y = 2 \sin \left(\frac{\pi}{6}\right) = 2 \cdot \frac{1}{2} = 1$
Substituting $(\sqrt{3}, 1)$ into $x + y = k$,we get $k = \sqrt{3} + 1$.
Thus,the Cartesian equation is $x + y = \sqrt{3} + 1$.
Converting back to polar form:
$r \cos \theta + r \sin \theta = \sqrt{3} + 1$
$\sin \theta + \cos \theta = \frac{\sqrt{3} + 1}{r}$.

(A) Total number of students = $15 + 5 = 20$.
Number of ways to select $3$ students out of $20$ is $^{20}C_3 = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140$.
Number of ways to select $2$ boys out of $15$ is $^{15}C_2 = \frac{15 \times 14}{2 \times 1} = 105$.
Number of ways to select $1$ girl out of $5$ is $^5C_1 = 5$.
Required probability = $\frac{^{15}C_2 \times ^5C_1}{^{20}C_3} = \frac{105 \times 5}{1140} = \frac{525}{1140} = \frac{35}{76}$.

(B) Total number of ways to arrange $7$ white balls and $3$ black balls in a row is $\frac{10!}{7!3!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.
To ensure no two black balls are adjacent,we place the $7$ white balls first,creating $8$ possible gaps (including ends) where the $3$ black balls can be placed.
The number of ways to choose $3$ gaps out of $8$ is $\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.
The required probability is $\frac{\binom{8}{3}}{\binom{10}{3}} = \frac{56}{120} = \frac{7}{15}$.

(B) Potassium crystallizes in a $BCC$ system.
Number of moles of potassium $= \frac{39 \ g}{39 \ g/mol} = 1 \ mol$.
Total number of atoms $= 1 \ mol \times N = N \ \text{atoms}$.
In a $BCC$ unit cell,the number of atoms per unit cell is $2$.
Therefore,the number of unit cells $= \frac{\text{Total number of atoms}}{2} = \frac{N}{2}$.