MHT CET 2011 Mathematics Question Paper with Answer and Solution

(D) The equation of the circle is ${x^2} + {y^2} + 4x - 6y + 9{\sin ^2}\alpha + 13{\cos ^2}\alpha = 0$.
Comparing with ${x^2} + {y^2} + 2gx + 2fy + c = 0$,we have $g = 2, f = -3, c = 9{\sin ^2}\alpha + 13{\cos ^2}\alpha$.
The centre is $C(-g, -f) = (-2, 3)$.
The radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 9 - (9{\sin ^2}\alpha + 13{\cos ^2}\alpha)} = \sqrt{13 - 9{\sin ^2}\alpha - 13{\cos ^2}\alpha} = \sqrt{13(1 - {\cos ^2}\alpha) - 9{\sin ^2}\alpha} = \sqrt{13{\sin ^2}\alpha - 9{\sin ^2}\alpha} = \sqrt{4{\sin ^2}\alpha} = 2\sin \alpha$.
Let $P(h, k)$ be a point on the locus. The angle between the tangents is $2\alpha$,so $\angle APC = \alpha$.
In the right-angled triangle $\triangle PAC$,$\sin \alpha = \frac{AC}{PC} = \frac{r}{PC} = \frac{2\sin \alpha}{\sqrt{(h + 2)^2 + (k - 3)^2}}$.
Thus,$\sqrt{(h + 2)^2 + (k - 3)^2} = 2$.
Squaring both sides,$(h + 2)^2 + (k - 3)^2 = 4$.
Expanding,$h^2 + 4h + 4 + k^2 - 6k + 9 = 4$,which simplifies to $h^2 + k^2 + 4h - 6k + 9 = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is ${x^2} + {y^2} + 4x - 6y + 9 = 0$.

(B) The equation of the parabola is $y^{2}=12x$. Comparing this with $y^{2}=4ax$,we get $4a=12$,so $a=3$.
The equation of a normal to the parabola $y^{2}=4ax$ at point $(at^{2}, 2at)$ is $y = -tx + 2at + at^{3}$.
Given the normal is $x+y=k$,which can be written as $y = -x + k$.
Comparing $y = -tx + 2at + at^{3}$ with $y = -x + k$,we get $t=1$.
Substituting $t=1$ and $a=3$ into the expression for $k$:
$k = 2at + at^{3} = 2(3)(1) + 3(1)^{3} = 6 + 3 = 9$.
Therefore,$k=9$.

(C) The normals to a circle intersect at its center. Given $(x - 1)(y - 2) = 0$,the center of the circle is $(1, 2)$.
Since the line $3x + 4y = 6$ is a tangent to the circle,the radius $r$ is the perpendicular distance from the center $(1, 2)$ to the line $3x + 4y - 6 = 0$.
$r = \frac{|3(1) + 4(2) - 6|}{\sqrt{3^2 + 4^2}} = \frac{|3 + 8 - 6|}{\sqrt{25}} = \frac{5}{5} = 1$.
The equation of the circle with center $(h, k) = (1, 2)$ and radius $r = 1$ is $(x - h)^2 + (y - k)^2 = r^2$.
$(x - 1)^2 + (y - 2)^2 = 1^2$.
$x^2 - 2x + 1 + y^2 - 4y + 4 = 1$.
$x^2 + y^2 - 2x - 4y + 4 = 0$.

(A) The condition for two circles $x^{2}+y^{2}+2g_{1}x+2f_{1}y+c_{1}=0$ and $x^{2}+y^{2}+2g_{2}x+2f_{2}y+c_{2}=0$ to intersect orthogonally is $2g_{1}g_{2}+2f_{1}f_{2}=c_{1}+c_{2}$.
For the given circles:
Circle $1$: $g_{1}=1, f_{1}=k, c_{1}=6$.
Circle $2$: $g_{2}=0, f_{2}=k, c_{2}=k$.
Substituting these values into the condition:
$2(1)(0) + 2(k)(k) = 6 + k$
$0 + 2k^{2} = 6 + k$
$2k^{2} - k - 6 = 0$
Factoring the quadratic equation:
$(2k+3)(k-2) = 0$
Thus,$k = 2$ or $k = -3/2$.

(D) The equation of the chord of contact $QR$ for the point $P(8, 27)$ with respect to the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$ is given by $T = 0$:
$\frac{8x}{4} + \frac{27y}{9} = 1 \Rightarrow 2x + 3y = 1$.
The equation of the pair of lines passing through the origin and the points $Q$ and $R$ is obtained by homogenizing the ellipse equation using the chord of contact:
$\frac{x^{2}}{4} + \frac{y^{2}}{9} = (2x + 3y)^{2}$.
$9x^{2} + 4y^{2} = 36(4x^{2} + 12xy + 9y^{2})$.
$9x^{2} + 4y^{2} = 144x^{2} + 432xy + 324y^{2}$.
$135x^{2} + 432xy + 320y^{2} = 0$.
Comparing this with $ax^{2} + 2hxy + by^{2} = 0$,we have $a = 135$,$2h = 432$ $(h = 216)$,and $b = 320$.
The angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{2\sqrt{h^{2} - ab}}{a + b} \right|$.
$\tan \theta = \left| \frac{2\sqrt{216^{2} - 135 \cdot 320}}{135 + 320} \right| = \left| \frac{2\sqrt{46656 - 43200}}{455} \right| = \frac{2\sqrt{3456}}{455} = \frac{2 \cdot 24 \sqrt{6}}{455} = \frac{48 \sqrt{6}}{455}$.
Thus,$\theta = \tan^{-1} \left( \frac{48 \sqrt{6}}{455} \right)$,which is not among the given options.

(B) The parametric coordinates of a point on the hyperbola $xy = c^{2}$ are $(ct, c/t)$.
The equation of the normal at point $t$ is given by $x t^{3} - yt - ct^{4} + c = 0$.
If this normal meets the curve again at point $t'$,then the coordinates $(ct', c/t')$ must satisfy the normal equation.
Substituting $(ct', c/t')$ into the equation: $(ct')t^{3} - (c/t')t - ct^{4} + c = 0$.
Dividing by $c$ (assuming $c \neq 0$): $t't^{3} - t/t' - t^{4} + 1 = 0$.
Multiplying by $t'$: $t'^{2}t^{3} - t - t't^{4} + t' = 0$.
Rearranging terms: $t^{3}t'(t' - t) + (t' - t) = 0$.
$(t' - t)(t^{3}t' + 1) = 0$.
Since $t \neq t'$,we must have $t^{3}t' + 1 = 0$,which implies $t^{3}t' = -1$.

(A) Let $L = \lim_{x \to \infty} \left(1 + \frac{1}{a+bx}\right)^{c+dx}$. This is in the form of $1^{\infty}$.
Using the formula $\lim_{x \to \infty} (1 + f(x))^{g(x)} = e^{\lim_{x \to \infty} f(x)g(x)}$:
$L = e^{\lim_{x \to \infty} \left(\frac{1}{a+bx}\right) \cdot (c+dx)}$
$L = e^{\lim_{x \to \infty} \frac{c+dx}{a+bx}}$
Divide the numerator and denominator by $x$:
$L = e^{\lim_{x \to \infty} \frac{c/x + d}{a/x + b}}$
As $x \to \infty$,$c/x \to 0$ and $a/x \to 0$:
$L = e^{\frac{0+d}{0+b}} = e^{d/b}$

(D) In the given circuit,the top branch consists of switches $\sim p$ and $q$ connected in series. The Boolean expression for this branch is $(\sim p \wedge q)$.
The bottom branch consists of switches $p$ and $\sim q$ connected in series. The Boolean expression for this branch is $(p \wedge \sim q)$.
Since these two branches are connected in parallel,the total Boolean polynomial for the circuit is the disjunction of the two expressions:
$(\sim p \wedge q) \vee (p \wedge \sim q)$.

(B) Using the distributive law,we have: $x \wedge (x \vee y) = (x \wedge x) \vee (x \wedge y)$.
By the idempotent law,$x \wedge x = x$,so the expression becomes $x \vee (x \wedge y)$.
By the absorption law,$x \vee (x \wedge y) = x$.

(B) The inverse of a conditional statement $P \Rightarrow Q$ is defined as $\sim P \Rightarrow \sim Q$.
Given the proposition $(p \wedge \sim q) \Rightarrow r$,we identify $P = (p \wedge \sim q)$ and $Q = r$.
Therefore,the inverse is $\sim(p \wedge \sim q) \Rightarrow \sim r$.
Using De Morgan's Law,$\sim(p \wedge \sim q)$ is equivalent to $(\sim p \vee \sim(\sim q))$,which simplifies to $(\sim p \vee q)$.
Thus,the inverse is $(\sim p \vee q) \Rightarrow \sim r$.

(C) To determine the nature of the proposition $(p$ $\Rightarrow \sim p) \wedge (\sim p$ $\Rightarrow p)$,we construct a truth table:

$p$	$\sim p$	$p \Rightarrow \sim p$	$\sim p \Rightarrow p$	$(p$ $\Rightarrow \sim p) \wedge (\sim p$ $\Rightarrow p)$
$T$	$F$	$F$	$T$	$F$
$F$	$T$	$T$	$F$	$F$

Since the final column contains only $F$ (False) for all possible truth values of $p$,the proposition is a contradiction.

(A) The equation of the pair of angle bisectors of the pair of lines $ax^{2}+2hxy+by^{2}=0$ is given by $\frac{x^{2}-y^{2}}{a-b} = \frac{xy}{h}$.
For the pair $x^{2}-2pxy-y^{2}=0$,we have $a=1, h=-p, b=-1$.
The equation of the angle bisectors is $\frac{x^{2}-y^{2}}{1-(-1)} = \frac{xy}{-p}$,which simplifies to $\frac{x^{2}-y^{2}}{2} = \frac{xy}{-p}$.
This gives $x^{2}-y^{2} = -\frac{2}{p}xy$,or $x^{2} + \frac{2}{p}xy - y^{2} = 0$.
It is given that this pair is the same as $x^{2}-2qxy-y^{2}=0$.
Comparing the coefficients of $xy$,we get $-2q = \frac{2}{p}$.
Therefore,$pq = -1$.

(B) The line that bisects the angle between the positive $x$-axis and positive $y$-axis is $y=x$.
Since this line is a part of the pair $ax^{2}+2hxy+by^{2}=0$,it must satisfy the equation.
Substituting $y=x$ into the equation:
$ax^{2}+2hx(x)+b(x)^{2}=0$
$ax^{2}+2hx^{2}+bx^{2}=0$
$(a+2h+b)x^{2}=0$
For this to hold for all $x$,we must have $a+b+2h=0$,which implies $a+b=-2h$.

(C) The equation of a tangent to the parabola $y^{2}=4x$ is given by $y=mx+\frac{a}{m}$,where $a=1$. Thus,$y=mx+\frac{1}{m}$.
This line is also tangent to the circle $(x-3)^{2}+y^{2}=9$,which has center $(3,0)$ and radius $r=3$.
The perpendicular distance from the center $(3,0)$ to the line $mx-y+\frac{1}{m}=0$ must equal the radius $3$:
$\frac{|m(3)-0+\frac{1}{m}|}{\sqrt{m^{2}+(-1)^{2}}}=3$
$|3m+\frac{1}{m}|=3\sqrt{m^{2}+1}$
Squaring both sides:
$(3m+\frac{1}{m})^{2}=9(m^{2}+1)$
$9m^{2}+6+\frac{1}{m^{2}}=9m^{2}+9$
$\frac{1}{m^{2}}=3$ $\Rightarrow m^{2}=\frac{1}{3}$ $\Rightarrow m=\pm\frac{1}{\sqrt{3}}$.
Since the tangent touches the parabola above the $x$-axis,we choose $m=\frac{1}{\sqrt{3}}$.
Substituting $m$ into the tangent equation:
$y=\frac{1}{\sqrt{3}}x+\sqrt{3}$
Multiplying by $\sqrt{3}$ gives $\sqrt{3}y=x+3$.

(A) The coordinates of the ends of the latus rectum of the parabola $y^{2}=4ax$ are $(a, 2a)$ and $(a, -2a)$.
For the parabola $y^{2}=4ax$,the slope of the tangent at $(x_{1}, y_{1})$ is $m = \frac{2a}{y_{1}}$.
At $(a, 2a)$,the slope of the tangent is $m = \frac{2a}{2a} = 1$,so the slope of the normal is $-1$.
The equation of the normal at $(a, 2a)$ is $y - 2a = -1(x - a)$,which simplifies to $x + y - 3a = 0$.
At $(a, -2a)$,the slope of the tangent is $m = \frac{2a}{-2a} = -1$,so the slope of the normal is $1$.
The equation of the normal at $(a, -2a)$ is $y - (-2a) = 1(x - a)$,which simplifies to $x - y - 3a = 0$.
The combined equation is $(x + y - 3a)(x - y - 3a) = 0$.
Expanding this,we get $((x - 3a) + y)((x - 3a) - y) = 0$,which is $(x - 3a)^{2} - y^{2} = 0$.
Thus,$x^{2} - 6ax + 9a^{2} - y^{2} = 0$,or $x^{2} - y^{2} - 6ax + 9a^{2} = 0$.

(A) Let $W$ denote a white ball and $B$ denote a black ball. The contents of the boxes are:
Box $I$: $3W, 1B$ (Total $4$ balls)
Box $II$: $2W, 2B$ (Total $4$ balls)
Box $III$: $1W, 3B$ (Total $4$ balls)
We need to draw $2$ white and $1$ black ball. This can happen in three mutually exclusive ways:
$1$. Box $I$ gives $B$,Box $II$ gives $W$,Box $III$ gives $W$
$2$. Box $I$ gives $W$,Box $II$ gives $B$,Box $III$ gives $W$
$3$. Box $I$ gives $W$,Box $II$ gives $W$,Box $III$ gives $B$
The required probability is:
$P = P(B_I)P(W_{II})P(W_{III}) + P(W_I)P(B_{II})P(W_{III}) + P(W_I)P(W_{II})P(B_{III})$
$P = (\frac{1}{4} \times \frac{2}{4} \times \frac{1}{4}) + (\frac{3}{4} \times \frac{2}{4} \times \frac{1}{4}) + (\frac{3}{4} \times \frac{2}{4} \times \frac{3}{4})$
$P = \frac{2}{64} + \frac{6}{64} + \frac{18}{64} = \frac{26}{64} = \frac{13}{32}$

(B) Let $y = \frac{1-x+x^{2}}{1+x+x^{2}}$.
We can rewrite the expression as $y = \frac{(x^{2}+x+1) - 2x}{x^{2}+x+1} = 1 - \frac{2x}{x^{2}+x+1}$.
To find the minimum value of $y$,we need to maximize the term $\frac{2x}{x^{2}+x+1}$.
Let $f(x) = \frac{x}{x^{2}+x+1}$.
Taking the derivative with respect to $x$ and setting it to $0$:
$f'(x) = \frac{(x^{2}+x+1)(1) - x(2x+1)}{(x^{2}+x+1)^{2}} = \frac{x^{2}+x+1-2x^{2}-x}{(x^{2}+x+1)^{2}} = \frac{1-x^{2}}{(x^{2}+x+1)^{2}}$.
Setting $f'(x) = 0$ gives $1-x^{2} = 0$,so $x = 1$ or $x = -1$.
For $x = 1$,$f(1) = \frac{1}{1+1+1} = \frac{1}{3}$.
For $x = -1$,$f(-1) = \frac{-1}{1-1+1} = -1$.
Since we want to minimize $y = 1 - 2f(x)$,we choose the maximum value of $f(x)$,which is $1/3$.
Thus,the minimum value of $y$ is $1 - 2(1/3) = 1 - 2/3 = 1/3$.

(B) The line $y=4$ meets the parabola $y^{2}=x$ at point $A$. Substituting $y=4$ into the equation of the parabola,we get $4^{2}=x$,so $x=16$. Thus,point $A$ is $(16, 4)$.
The required area is bounded by the parabola $x=y^{2}$,the $y$-axis $(x=0)$,and the line $y=4$ from $y=0$ to $y=4$.
Required Area $= \int_{0}^{4} x \, dy = \int_{0}^{4} y^{2} \, dy$
$= \left[ \frac{y^{3}}{3} \right]_{0}^{4}$
$= \frac{4^{3}}{3} - \frac{0^{3}}{3} = \frac{64}{3} \text{ sq. units}$.

(C) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$.
First,calculate the left-hand limit:
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (1+|\sin x|)^{\frac{a}{|\sin x|}}$.
Let $u = |\sin x|$. As $x \to 0$,$u \to 0^+$. The limit becomes $\lim_{u \to 0^+} (1+u)^{a/u} = e^a$.
Next,calculate the right-hand limit:
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} e^{\frac{\tan 2x}{\tan 3x}}$.
Using the limit $\lim_{\theta \to 0} \frac{\tan k\theta}{\theta} = k$,we get $\lim_{x \to 0^+} \frac{\tan 2x}{\tan 3x} = \lim_{x \to 0^+} \frac{(\tan 2x / 2x) \cdot 2x}{(\tan 3x / 3x) \cdot 3x} = \frac{1 \cdot 2}{1 \cdot 3} = \frac{2}{3}$.
Thus,$\lim_{x \to 0^+} f(x) = e^{2/3}$.
Since $f(0) = b$,for continuity,$e^a = e^{2/3} = b$.
Therefore,$a = 2/3$ and $b = e^{2/3}$.

(B) $LHL$ $= \lim_{x \rightarrow 1^-} f(x) = \lim_{x \rightarrow 1} (x-1) = 0$
$RHL$ $= \lim_{x \rightarrow 1^+} f(x) = \lim_{x \rightarrow 1} (x^3-1) = 0$
Also,$f(1) = 1-1 = 0$
Since $LHL$ $=$ $RHL$ $= f(1)$,$f$ is continuous at $x=1$.
Now,$Lf'(1) = \lim_{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h} = \lim_{h \rightarrow 0} \frac{(1-h)-1-0}{-h} = \lim_{h \rightarrow 0} \frac{-h}{-h} = 1$
And $Rf'(1) = \lim_{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} = \lim_{h \rightarrow 0} \frac{(1+h)^3-1-0}{h} = \lim_{h \rightarrow 0} \frac{1+h^3+3h+3h^2-1}{h} = \lim_{h \rightarrow 0} (h^2+3h+3) = 3$
Since $Lf'(1) \neq Rf'(1)$,$f(x)$ is not differentiable at $x=1$.

(D) Let $I = \int_{0}^{\pi / 2} \frac{d x}{1+\tan x}$.
We can write $\tan x = \frac{\sin x}{\cos x}$,so $I = \int_{0}^{\pi / 2} \frac{\cos x}{\sin x + \cos x} d x$ ... $(i)$.
Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,we get:
$I = \int_{0}^{\pi / 2} \frac{\cos(\pi / 2 - x)}{\sin(\pi / 2 - x) + \cos(\pi / 2 - x)} d x = \int_{0}^{\pi / 2} \frac{\sin x}{\cos x + \sin x} d x$ ... $(ii)$.
Adding $(i)$ and $(ii)$:
$2I = \int_{0}^{\pi / 2} \frac{\cos x + \sin x}{\sin x + \cos x} d x = \int_{0}^{\pi / 2} 1 d x$.
$2I = [x]_{0}^{\pi / 2} = \pi / 2$.
Therefore,$I = \pi / 4$.

(A) Let $I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^{x}} d x$ $(i)$
Using the property $\int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x) d x$,where $a = -\pi / 2$ and $b = \pi / 2$,we have $a+b = 0$.
So,$I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos(-x)}{1+e^{-x}} d x = \int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^{-x}} d x$
Multiplying the numerator and denominator by $e^x$,we get:
$I = \int_{-\pi / 2}^{\pi / 2} \frac{e^x \cos x}{e^x + 1} d x$ $(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos x + e^x \cos x}{1+e^x} d x = \int_{-\pi / 2}^{\pi / 2} \frac{\cos x(1+e^x)}{1+e^x} d x$
$2I = \int_{-\pi / 2}^{\pi / 2} \cos x d x$
Since $\cos x$ is an even function,$2I = 2 \int_{0}^{\pi / 2} \cos x d x$
$2I = 2[\sin x]_{0}^{\pi / 2} = 2(1 - 0) = 2$
Therefore,$I = 1$.

(C) Given $n=4$,the interval $[0, 1]$ is divided into $4$ sub-intervals of width $h = \frac{1-0}{4} = 0.25$.
The values of $y = \frac{1}{1+x^2}$ at $x_i$ are:

$x$	$y$
$0$	$1.0$
$0.25$	$0.941176$
$0.5$	$0.8$
$0.75$	$0.64$
$1$	$0.5$

Using Simpson's $\frac{1}{3}$ rule:
$\int_{0}^{1} y dx \approx \frac{h}{3} [(y_0 + y_4) + 4(y_1 + y_3) + 2(y_2)]$
$\int_{0}^{1} y dx \approx \frac{0.25}{3} [(1.0 + 0.5) + 4(0.941176 + 0.64) + 2(0.8)]$
$\int_{0}^{1} y dx \approx \frac{0.25}{3} [1.5 + 4(1.581176) + 1.6]$
$\int_{0}^{1} y dx \approx \frac{0.25}{3} [1.5 + 6.324704 + 1.6] = \frac{0.25}{3} [9.424704] \approx 0.785392$
Rounding to three decimal places,we get $0.785$.

(C) For a system of homogeneous linear equations to have a non-zero solution,the determinant of the coefficient matrix must be zero.
$\left|\begin{array}{ccc} a^3 & (a+1)^3 & (a+2)^3 \\ a & (a+1) & (a+2) \\ 1 & 1 & 1 \end{array}\right| = 0$
Interchanging rows to simplify:
$-\left|\begin{array}{ccc} 1 & 1 & 1 \\ a & (a+1) & (a+2) \\ a^3 & (a+1)^3 & (a+2)^3 \end{array}\right| = 0$
Using the property of the Vandermonde-like determinant $\left|\begin{array}{ccc} 1 & 1 & 1 \\ x & y & z \\ x^3 & y^3 & z^3 \end{array}\right| = (x-y)(y-z)(z-x)(x+y+z)$,where $x=a, y=a+1, z=a+2$:
$-(a-(a+1))((a+1)-(a+2))((a+2)-a)(a+(a+1)+(a+2)) = 0$
$-(-1)(-1)(2)(3a+3) = 0$
$-2(3a+3) = 0$
$3a+3 = 0 \Rightarrow a = -1$.

(B) The given differential equation is $\left(\frac{d^{2} y}{d x^{2}}\right)^{5}+4 \cdot \frac{\left(\frac{d^{2} y}{d x^{2}}\right)^{3}}{\left(\frac{d^{3} y}{d x^{3}}\right)}+\left(\frac{d^{3} y}{d x^{3}}\right)=x^{2}-1$.
To find the order and degree,we first eliminate the fraction by multiplying the entire equation by $\frac{d^{3} y}{d x^{3}}$:
$\left(\frac{d^{2} y}{d x^{2}}\right)^{5} \cdot \left(\frac{d^{3} y}{d x^{3}}\right) + 4 \left(\frac{d^{2} y}{d x^{2}}\right)^{3} + \left(\frac{d^{3} y}{d x^{3}}\right)^{2} = (x^{2}-1) \left(\frac{d^{3} y}{d x^{3}}\right)$.
The highest order derivative present in the equation is $\frac{d^{3} y}{d x^{3}}$,so the order $m = 3$.
The degree is the highest power of the highest order derivative after the equation is made a polynomial in derivatives. Here,the highest power of $\frac{d^{3} y}{d x^{3}}$ is $2$,so the degree $n = 2$.

(B) Given equation is $(x-h)^{2}+(y-k)^{2}=a^{2}$ ... $(i)$
Differentiating with respect to $x$:
$2(x-h) + 2(y-k) \frac{dy}{dx} = 0$
$(x-h) + (y-k) \frac{dy}{dx} = 0$ ... (ii)
Differentiating again with respect to $x$:
$1 + \left(\frac{dy}{dx}\right)^{2} + (y-k) \frac{d^{2}y}{dx^{2}} = 0$
$(y-k) = -\frac{1 + (dy/dx)^{2}}{d^{2}y/dx^{2}}$ ... (iii)
From (ii),$(x-h) = -(y-k) \frac{dy}{dx}$. Substituting (iii) into this:
$(x-h) = \frac{[1 + (dy/dx)^{2}]}{d^{2}y/dx^{2}} \cdot \frac{dy}{dx}$ ... (iv)
Substituting (iii) and (iv) into $(i)$:
$\left[ \frac{[1 + (dy/dx)^{2}] \cdot (dy/dx)}{d^{2}y/dx^{2}} \right]^{2} + \left[ -\frac{1 + (dy/dx)^{2}}{d^{2}y/dx^{2}} \right]^{2} = a^{2}$
$\frac{[1 + (dy/dx)^{2}]^{2}}{ (d^{2}y/dx^{2})^{2} } \left[ (dy/dx)^{2} + 1 \right] = a^{2}$
$\left[1 + \left(\frac{dy}{dx}\right)^{2}\right]^{3} = a^{2} \left(\frac{d^{2}y}{dx^{2}}\right)^{2}$

(A) The general equation of a circle passing through the origin with its center on the $y$-axis is given by $x^{2} + (y-a)^{2} = a^{2}$,where $a$ is an arbitrary constant.
Expanding this,we get $x^{2} + y^{2} - 2ay + a^{2} = a^{2}$,which simplifies to $x^{2} + y^{2} - 2ay = 0$.
To eliminate the arbitrary constant $a$,we differentiate with respect to $x$:
$2x + 2y \frac{dy}{dx} - 2a \frac{dy}{dx} = 0$.
Dividing by $2$,we get $x + y \frac{dy}{dx} - a \frac{dy}{dx} = 0$,which implies $a = \frac{x + y \frac{dy}{dx}}{\frac{dy}{dx}} = x \frac{dx}{dy} + y$.
Substituting $a$ back into the original equation $x^{2} + y^{2} = 2ay$:
$x^{2} + y^{2} = 2(x \frac{dx}{dy} + y)y = 2xy \frac{dx}{dy} + 2y^{2}$.
Rearranging,$x^{2} - y^{2} = 2xy \frac{dx}{dy}$.
Since $\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$,we have $x^{2} - y^{2} = \frac{2xy}{\frac{dy}{dx}}$.
Thus,$(x^{2} - y^{2}) \frac{dy}{dx} = 2xy$,or $(x^{2} - y^{2}) \frac{dy}{dx} - 2xy = 0$.

(B) The given differential equation is $\frac{dy}{dx}(x \log x) + y = 2 \log x$.
Dividing both sides by $(x \log x)$,we get:
$\frac{dy}{dx} + \frac{y}{x \log x} = \frac{2 \log x}{x \log x} = \frac{2}{x}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x \log x}$ and $Q = \frac{2}{x}$.
The integrating factor $(IF)$ is given by $e^{\int P dx}$.
$IF = e^{\int \frac{1}{x \log x} dx}$.
Let $u = \log x$,then $du = \frac{1}{x} dx$.
$IF = e^{\int \frac{1}{u} du} = e^{\log u} = u = \log x$.
Thus,the integrating factor is $\log x$.

(B) Given the displacement equation: $s = 16 - 2t + 3t^{3}$
The velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{ds}{dt} = \frac{d}{dt}(16 - 2t + 3t^{3}) = -2 + 9t^{2}$
The acceleration $a$ is the second derivative of displacement with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(-2 + 9t^{2}) = 18t$
At $t = 2 \ s$,the acceleration is: $a = 18 \times 2 = 36 \ m/s^{2}$

(A) Given,$x^{p} y^{q}=(x+y)^{p+q}$.
Taking the natural logarithm on both sides,we get:
$p \ln x + q \ln y = (p+q) \ln (x+y)$.
Differentiating both sides with respect to $x$:
$\frac{p}{x} + \frac{q}{y} \frac{dy}{dx} = \frac{p+q}{x+y} \left(1 + \frac{dy}{dx}\right)$.
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{q}{y} \frac{dy}{dx} - \frac{p+q}{x+y} \frac{dy}{dx} = \frac{p+q}{x+y} - \frac{p}{x}$.
$\frac{dy}{dx} \left( \frac{q(x+y) - y(p+q)}{y(x+y)} \right) = \frac{x(p+q) - p(x+y)}{x(x+y)}$.
$\frac{dy}{dx} \left( \frac{qx + qy - py - qy}{y(x+y)} \right) = \frac{px + qx - px - py}{x(x+y)}$.
$\frac{dy}{dx} \left( \frac{qx - py}{y(x+y)} \right) = \frac{qx - py}{x(x+y)}$.
Canceling the common term $(qx - py)$ and $(x+y)$ from both sides:
$\frac{dy}{dx} = \frac{y}{x}$.

(D) Given $x = 2 \cos t - \cos 2t$ and $y = 2 \sin t - \sin 2t$.
First,find $\frac{dx}{dt}$ and $\frac{dy}{dt}$:
$\frac{dx}{dt} = -2 \sin t + 2 \sin 2t$
$\frac{dy}{dt} = 2 \cos t - 2 \cos 2t$
Now,find $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2 \cos t - 2 \cos 2t}{-2 \sin t + 2 \sin 2t} = \frac{\cos t - \cos 2t}{\sin 2t - \sin t}$.
Using trigonometric identities $\cos A - \cos B = -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$ and $\sin A - \sin B = 2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}$:
$\frac{dy}{dx} = \frac{2 \sin \frac{3t}{2} \sin \frac{t}{2}}{2 \cos \frac{3t}{2} \sin \frac{t}{2}} = \tan \frac{3t}{2}$.
Now,differentiate with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dt} \left( \tan \frac{3t}{2} \right) \cdot \frac{dt}{dx} = \sec^2 \frac{3t}{2} \cdot \frac{3}{2} \cdot \frac{1}{-2 \sin t + 2 \sin 2t}$.
At $t = \pi/2$:
$\frac{d^2y}{dx^2} = \frac{3}{2} \sec^2 \left( \frac{3\pi}{4} \right) \cdot \frac{1}{-2 \sin(\pi/2) + 2 \sin(\pi)} = \frac{3}{2} (-\sqrt{2})^2 \cdot \frac{1}{-2(1) + 2(0)} = \frac{3}{2} (2) \cdot \frac{1}{-2} = -3/2$.

(A) Given $y = \log(\tan(x/2)) + \sin^{-1}(\cos x)$.
First,differentiate $\log(\tan(x/2))$ with respect to $x$:
$\frac{d}{dx}(\log(\tan(x/2))) = \frac{1}{\tan(x/2)} \cdot \sec^2(x/2) \cdot \frac{1}{2} = \frac{\cos(x/2)}{\sin(x/2)} \cdot \frac{1}{\cos^2(x/2)} \cdot \frac{1}{2} = \frac{1}{2 \sin(x/2) \cos(x/2)} = \frac{1}{\sin x} = \operatorname{cosec} x$.
Next,differentiate $\sin^{-1}(\cos x)$ with respect to $x$:
Since $\cos x = \sin(\pi/2 - x)$,we have $\sin^{-1}(\cos x) = \sin^{-1}(\sin(\pi/2 - x)) = \pi/2 - x$.
Therefore,$\frac{d}{dx}(\sin^{-1}(\cos x)) = \frac{d}{dx}(\pi/2 - x) = -1$.
Combining these,$\frac{dy}{dx} = \operatorname{cosec} x - 1$.

(B) Given $f(x) = e^{x} g(x)$.
Applying the product rule for differentiation,we get:
$f^{\prime}(x) = e^{x} g^{\prime}(x) + e^{x} g(x)$.
Substituting $x = 0$:
$f^{\prime}(0) = e^{0} \cdot g^{\prime}(0) + e^{0} \cdot g(0)$.
Since $e^{0} = 1$,$g^{\prime}(0) = 1$,and $g(0) = 2$:
$f^{\prime}(0) = 1 \cdot 1 + 1 \cdot 2$.
$f^{\prime}(0) = 1 + 2 = 3$.

(D) We are given the functional equation $f(x+y)=f(x) \cdot f(y)$.
By the definition of the derivative,$f^{\prime}(3)=\lim _{h \rightarrow 0} \frac{f(3+h)-f(3)}{h}$.
Using the given functional equation,$f(3+h)=f(3) \cdot f(h)$.
Substituting this into the limit expression:
$f^{\prime}(3)=\lim _{h \rightarrow 0} \frac{f(3) \cdot f(h)-f(3)}{h} = f(3) \lim _{h \rightarrow 0} \frac{f(h)-1}{h}$.
Since $f(0+0)=f(0) \cdot f(0)$,we have $f(0)=f(0)^2$,which implies $f(0)=1$ (assuming $f(x)$ is not identically zero).
Thus,$f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h} = \lim _{h \rightarrow 0} \frac{f(h)-1}{h} = 11$.
Substituting $f(3)=3$ and $\lim _{h \rightarrow 0} \frac{f(h)-1}{h} = 11$ into the expression for $f^{\prime}(3)$:
$f^{\prime}(3) = 3 \times 11 = 33$.

(B) Let $I = \int (\sqrt{\tan x} + \sqrt{\cot x}) \, dx$.
We can write this as $I = \int \left( \sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}} \right) \, dx = \int \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \, dx$.
Multiply numerator and denominator by $\sqrt{2}$:
$I = \int \frac{\sqrt{2}(\sin x + \cos x)}{\sqrt{2 \sin x \cos x}} \, dx = \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{\sin 2x}} \, dx$.
Since $\sin 2x = 1 - (\sin x - \cos x)^2$,let $t = \sin x - \cos x$.
Then $dt = (\cos x + \sin x) \, dx$.
Substituting these into the integral:
$I = \sqrt{2} \int \frac{dt}{\sqrt{1 - t^2}} = \sqrt{2} \sin^{-1}(t) + C = \sqrt{2} \sin^{-1}(\sin x - \cos x) + C$.
Using the identity $\sin^{-1}(u) = \tan^{-1}\left(\frac{u}{\sqrt{1-u^2}}\right)$,we get:
$I = \sqrt{2} \tan^{-1}\left(\frac{\sin x - \cos x}{\sqrt{1 - (\sin x - \cos x)^2}}\right) + C = \sqrt{2} \tan^{-1}\left(\frac{\sin x - \cos x}{\sqrt{\sin 2x}}\right) + C$.
Dividing numerator and denominator by $\sqrt{\cos x}$,we obtain:
$I = \sqrt{2} \tan^{-1}\left(\frac{\tan x - 1}{\sqrt{2 \tan x}}\right) + C$.

(C) Let $I = \int \frac{x^{2}}{(x \sin x+\cos x)^{2}} d x$.
We know that $\frac{d}{d x}(x \sin x+\cos x) = \sin x + x \cos x - \sin x = x \cos x$.
We can rewrite the integral as:
$I = \int \left( \frac{x}{\cos x} \right) \left( \frac{x \cos x}{(x \sin x+\cos x)^{2}} \right) d x$.
Using integration by parts,let $u = \frac{x}{\cos x}$ and $dv = \frac{x \cos x}{(x \sin x+\cos x)^{2}} d x$.
Then $du = \frac{\cos x - x(-\sin x)}{\cos^{2} x} d x = \frac{\cos x + x \sin x}{\cos^{2} x} d x$ and $v = \frac{-1}{x \sin x + \cos x}$.
$I = \left( \frac{x}{\cos x} \right) \left( \frac{-1}{x \sin x + \cos x} \right) - \int \left( \frac{\cos x + x \sin x}{\cos^{2} x} \right) \left( \frac{-1}{x \sin x + \cos x} \right) d x$.
$I = \frac{-x}{\cos x(x \sin x + \cos x)} + \int \frac{1}{\cos^{2} x} d x$.
$I = \frac{-x}{\cos x(x \sin x + \cos x)} + \tan x + C$.
$I = \frac{-x + \tan x \cos x (x \sin x + \cos x)}{\cos x(x \sin x + \cos x)} + C$.
$I = \frac{-x + \sin x (x \sin x + \cos x)}{\cos x(x \sin x + \cos x)} + C$.
$I = \frac{-x + x \sin^{2} x + \sin x \cos x}{\cos x(x \sin x + \cos x)} + C$.
$I = \frac{-x(1 - \sin^{2} x) + \sin x \cos x}{\cos x(x \sin x + \cos x)} + C$.
$I = \frac{-x \cos^{2} x + \sin x \cos x}{\cos x(x \sin x + \cos x)} + C$.
$I = \frac{\cos x(\sin x - x \cos x)}{\cos x(x \sin x + \cos x)} + C$.
$I = \frac{\sin x - x \cos x}{x \sin x + \cos x} + C$.

(A) Let $I = \int_{0}^{\pi} \frac{x \, dx}{1+\cos \alpha \sin x} \quad \dots (i)$
Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we get
$I = \int_{0}^{\pi} \frac{(\pi-x) \, dx}{1+\cos \alpha \sin(\pi-x)} = \int_{0}^{\pi} \frac{(\pi-x) \, dx}{1+\cos \alpha \sin x} \quad \dots (ii)$
Adding $(i)$ and $(ii)$:
$2I = \pi \int_{0}^{\pi} \frac{dx}{1+\cos \alpha \sin x}$
Using $\sin x = \frac{2 \tan(x/2)}{1+\tan^2(x/2)}$ and $dx = \frac{2 \sec^2(x/2) \, d(x/2)}{1}$,let $t = \tan(x/2)$,then $dt = \frac{1}{2} \sec^2(x/2) \, dx$:
$2I = \pi \int_{0}^{\infty} \frac{2 \, dt}{1+t^2 + 2t \cos \alpha} = 2\pi \int_{0}^{\infty} \frac{dt}{(t+\cos \alpha)^2 + \sin^2 \alpha}$
$I = \pi \left[ \frac{1}{\sin \alpha} \tan^{-1} \left( \frac{t+\cos \alpha}{\sin \alpha} \right) \right]_{0}^{\infty}$
$I = \frac{\pi}{\sin \alpha} \left( \frac{\pi}{2} - \tan^{-1} \left( \cot \alpha \right) \right) = \frac{\pi}{\sin \alpha} \left( \frac{\pi}{2} - (\frac{\pi}{2} - \alpha) \right) = \frac{\pi \alpha}{\sin \alpha}$

(C) The expression $\lim _{x \rightarrow 1} \frac{G(x)-G(1)}{x-1}$ represents the derivative of $G(x)$ at $x=1$,denoted as $G'(1)$.
Given $G(x) = -\sqrt{25-x^{2}}$.
Using the chain rule,$G'(x) = -\frac{1}{2\sqrt{25-x^{2}}} \cdot (-2x) = \frac{x}{\sqrt{25-x^{2}}}$.
Substituting $x=1$:
$G'(1) = \frac{1}{\sqrt{25-(1)^{2}}} = \frac{1}{\sqrt{24}}$.

(B) To determine the nature of the feasible region,we analyze the given constraints:
$1$) $-x_{1} + x_{2} \leq 1$
$2$) $-x_{1} + 3x_{2} \leq 9$
$3$) $x_{1}, x_{2} \geq 0$
Plotting these lines on the Cartesian plane:
- For $-x_{1} + x_{2} = 1$,the intercepts are $(0, 1)$ and $(-1, 0)$.
- For $-x_{1} + 3x_{2} = 9$,the intercepts are $(0, 3)$ and $(-9, 0)$.
The non-negativity constraints $x_{1}, x_{2} \geq 0$ restrict the region to the first quadrant.
By observing the intersection of the half-planes defined by the inequalities,we find that the region extends infinitely in the direction of increasing $x_{1}$.
Therefore,the feasible region is an unbounded feasible space.

(A) Let $x$ and $y$ be the number of units of food $A$ and food $B$ respectively.
The objective is to minimize the cost $z = 4x + 3y$.
Subject to the constraints based on the nutrients:
$1$. Vitamins: $200x + 100y \geq 4000$
$2$. Proteins: $x + 2y \geq 50$
$3$. Calories: $40x + 40y \geq 1400$
$4$. Non-negativity: $x \geq 0, y \geq 0$
Comparing these with the given options,option $A$ matches the formulated constraints and objective function.

(C) We have,$(1+\Delta)(1-\nabla) f(x)$
$= (1+\Delta) \{ f(x) - \nabla f(x) \}$
$= (1+\Delta) \{ f(x) - (f(x) - f(x-h)) \}$
$= (1+\Delta) f(x-h)$
Since $E = 1 + \Delta$,we have $E f(x-h) = f(x-h+h) = f(x)$.
Thus,$(1+\Delta)(1-\nabla) f(x) = f(x) = 1 \cdot f(x)$.
Therefore,$(1+\Delta)(1-\nabla) = 1$.

(B) To find the multiplicative inverse $A^{-1}$,we use the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.
First,calculate the determinant $|A|$:
$|A| = (\cos \theta)(\cos \theta) - (-\sin \theta)(\sin \theta) = \cos^2 \theta + \sin^2 \theta = 1$.
Next,find the adjoint of $A$ by swapping the diagonal elements and changing the signs of the off-diagonal elements:
$\text{adj}(A) = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$.
Therefore,$A^{-1} = \frac{1}{1} \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$.

(D) When two dice are thrown,the total number of outcomes is $6 \times 6 = 36$.
The event $E_{1}$ is getting a sum of $6$. The outcomes are $E_{1} = \{(1, 5), (5, 1), (2, 4), (4, 2), (3, 3)\}$.
The number of outcomes in $E_{1}$ is $n(E_{1}) = 5$.
The event $E_{2}$ is getting a $2$ on at least one of the two dice. The outcomes in $E_{1} \cap E_{2}$ are the outcomes in $E_{1}$ that contain at least one $2$.
These are $\{(2, 4), (4, 2)\}$.
So,$n(E_{1} \cap E_{2}) = 2$.
The conditional probability $P(E_{2} / E_{1})$ is given by the formula:
$P(E_{2} / E_{1}) = \frac{n(E_{1} \cap E_{2})}{n(E_{1})} = \frac{2}{5}$.

(A) The probability mass function of a Binomial distribution is given by $P(X=k) = {}^{n}C_{k} p^{k} (1-p)^{n-k}$.
Given $n=6$,we have $P(X=4) = {}^{6}C_{4} p^{4} (1-p)^{2}$ and $P(X=2) = {}^{6}C_{2} p^{2} (1-p)^{4}$.
According to the problem,$9 P(X=4) = P(X=2)$.
Substituting the values,we get $9 \cdot {}^{6}C_{4} p^{4} (1-p)^{2} = {}^{6}C_{2} p^{2} (1-p)^{4}$.
Since ${}^{6}C_{4} = {}^{6}C_{2} = 15$,the equation simplifies to $9 p^{4} (1-p)^{2} = p^{2} (1-p)^{4}$.
Dividing both sides by $p^{2} (1-p)^{2}$ (assuming $p \neq 0, 1$),we get $9 p^{2} = (1-p)^{2}$.
Taking the square root of both sides,$3p = 1-p$ (since $p$ must be positive).
$4p = 1$,which gives $p = 1/4$.

(A) The direction ratios of the line joining the points $(3, 1, 4)$ and $(7, 2, 12)$ are given by $\langle 7-3, 2-1, 12-4 \rangle = \langle 4, 1, 8 \rangle$. Let these be $\langle a_1, a_2, a_3 \rangle = \langle 4, 1, 8 \rangle$.
The direction ratios of the given line are $\langle b_1, b_2, b_3 \rangle = \langle 2, 2, 1 \rangle$.
Let $\theta$ be the angle between the two lines. The formula for the cosine of the angle is $\cos \theta = \frac{|a_1 b_1 + a_2 b_2 + a_3 b_3|}{\sqrt{a_1^2 + a_2^2 + a_3^2} \sqrt{b_1^2 + b_2^2 + b_3^2}}$.
Substituting the values: $\cos \theta = \frac{|(4)(2) + (1)(2) + (8)(1)|}{\sqrt{4^2 + 1^2 + 8^2} \sqrt{2^2 + 2^2 + 1^2}}$.
$\cos \theta = \frac{|8 + 2 + 8|}{\sqrt{16 + 1 + 64} \sqrt{4 + 4 + 1}} = \frac{18}{\sqrt{81} \sqrt{9}}$.
$\cos \theta = \frac{18}{9 \times 3} = \frac{18}{27} = \frac{2}{3}$.
Therefore,$\theta = \cos ^{-1}\left(\frac{2}{3}\right)$.

(B) We know that $\alpha, \beta, \gamma$ are the direction angles of a line,so the direction cosines are $\cos \alpha, \cos \beta, \cos \gamma$.
The sum of the squares of the direction cosines is given by $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma = 1$.
We want to evaluate $\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma$.
Using the identity $\sin ^{2} \theta = 1 - \cos ^{2} \theta$,we have:
$\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma = (1 - \cos ^{2} \alpha) + (1 - \cos ^{2} \beta) + (1 - \cos ^{2} \gamma)$
$= 3 - (\cos ^{2} \alpha + \cos ^{2} \beta + \cos ^{2} \gamma)$
$= 3 - 1 = 2$.

(B) Let the plane meet the coordinate axes at $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$.
The centroid of $\Delta ABC$ is given by the formula $\left(\frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3}\right) = \left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$.
Given that the centroid is $(1, 2, 3)$,we have:
$\frac{a}{3} = 1 \Rightarrow a = 3$
$\frac{b}{3} = 2 \Rightarrow b = 6$
$\frac{c}{3} = 3 \Rightarrow c = 9$
The intercept form of the equation of a plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
Substituting the values of $a, b$,and $c$,we get:
$\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1$.

(C) Let the vector $v$ make an angle $\alpha$ with each of the three axes. Then the direction cosines of $v$ are $\langle \cos \alpha, \cos \alpha, \cos \alpha \rangle$.
We know that for any vector,the sum of the squares of its direction cosines is $1$:
$\cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1$
$3 \cos^2 \alpha = 1$
$\cos^2 \alpha = \frac{1}{3}$
$\cos \alpha = \pm \frac{1}{\sqrt{3}}$
Therefore,the direction cosines are $\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \rangle$ or $\langle -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \rangle$.

(B) Since the vectors $a$,$b$,and $c$ are coplanar,there must exist scalars $x$,$y$,and $z$ (not all zero) such that $x a + y b + z c = 0$ $(i)$.
Multiplying both sides of equation $(i)$ by $a$ and $b$ respectively,we get:
$x(a \cdot a) + y(a \cdot b) + z(a \cdot c) = 0$ $(ii)$
$x(b \cdot a) + y(b \cdot b) + z(b \cdot c) = 0$ $(iii)$
Considering equations $(i)$,$(ii)$,and $(iii)$ as a system of linear equations in variables $x$,$y$,and $z$,since a non-trivial solution exists,the determinant of the coefficient matrix must be zero.
Therefore,$\left|\begin{array}{ccc} a & b & c \\ a \cdot a & a \cdot b & a \cdot c \\ b \cdot a & b \cdot b & b \cdot c \end{array}\right| = 0$.

(A) Given $u = a - b$ and $v = a + b$.
We need to find $|u \times v|$.
$|u \times v| = |(a - b) \times (a + b)|$
$= |a \times a + a \times b - b \times a - b \times b|$
Since $a \times a = 0$ and $b \times b = 0$,and $b \times a = -(a \times b)$,we have:
$|u \times v| = |0 + a \times b + a \times b - 0| = |2(a \times b)| = 2|a \times b|$.
We know that $|a \times b|^{2} + (a \cdot b)^{2} = |a|^{2} |b|^{2}$.
Given $|a| = |b| = 2$,so $|a|^{2} = 4$ and $|b|^{2} = 4$.
$|a \times b|^{2} + (a \cdot b)^{2} = 4 \times 4 = 16$.
$|a \times b| = \sqrt{16 - (a \cdot b)^{2}}$.
Therefore,$|u \times v| = 2 \sqrt{16 - (a \cdot b)^{2}}$.