MHT CET 2012 Mathematics Question Paper with Answer and Solution

(B) Let $S \equiv x^{2}+y^{2}-2x-6y+6=0$ and $P(x_{1}, y_{1}) = (0,1)$.
First,calculate $S_{1}$ at point $P(0,1)$:
$S_{1} = (0)^{2}+(1)^{2}-2(0)-6(1)+6 = 1-6+6 = 1$.
Next,find the equation of the chord of contact $T$:
$T = x(x_{1}) + y(y_{1}) - (x+x_{1}) - 3(y+y_{1}) + 6$
$T = x(0) + y(1) - (x+0) - 3(y+1) + 6$
$T = y - x - 3y - 3 + 6 = -x - 2y + 3$.
The equation of the pair of tangents is given by $SS_{1} = T^{2}$:
$(x^{2}+y^{2}-2x-6y+6)(1) = (-x-2y+3)^{2}$
$x^{2}+y^{2}-2x-6y+6 = x^{2}+4y^{2}+9 + 4xy - 6x - 12y$
Rearranging the terms:
$x^{2}-x^{2} + y^{2}-4y^{2} + 4xy - 2x+6x - 6y+12y + 6-9 = 0$
$-3y^{2} + 4xy + 4x + 6y - 3 = 0$
Multiplying by $-1$:
$3y^{2}-4xy-4x-6y+3=0$ (Note: Re-evaluating expansion: $(-x-2y+3)^{2} = x^{2}+4y^{2}+9+4xy-6x-12y$. The equation simplifies to $3y^{2}+4xy-4x-6y+3=0$).

(C) Let $S_{1} \equiv x^{2}+y^{2}+2ax+c=0$ and $S_{2} \equiv x^{2}+y^{2}+2by+c=0$.
The equation of the common chord is given by $S_{1}-S_{2}=0$,which is $2ax-2by=0$ or $ax-by=0$.
Thus,$y = \frac{ax}{b}$.
Substituting $y = \frac{ax}{b}$ into $S_{1}=0$,we get $x^{2} + (\frac{ax}{b})^{2} + 2ax + c = 0$,which simplifies to $(a^{2}+b^{2})x^{2} + 2ab^{2}x + cb^{2} = 0$.
Let the roots be $x_{1}, x_{2}$. Then $x_{1}+x_{2} = -\frac{2ab^{2}}{a^{2}+b^{2}}$ and $x_{1}x_{2} = \frac{cb^{2}}{a^{2}+b^{2}}$.
Similarly,substituting $x = \frac{by}{a}$ into $S_{2}=0$,we get $(a^{2}+b^{2})y^{2} + 2ba^{2}y + ca^{2} = 0$.
Let the roots be $y_{1}, y_{2}$. Then $y_{1}+y_{2} = -\frac{2ba^{2}}{a^{2}+b^{2}}$ and $y_{1}y_{2} = \frac{ca^{2}}{a^{2}+b^{2}}$.
The equation of the circle with the common chord as diameter is $(x-x_{1})(x-x_{2}) + (y-y_{1})(y-y_{2}) = 0$.
This expands to $x^{2} - (x_{1}+x_{2})x + x_{1}x_{2} + y^{2} - (y_{1}+y_{2})y + y_{1}y_{2} = 0$.
Substituting the values: $x^{2} + y^{2} + \frac{2ab^{2}}{a^{2}+b^{2}}x + \frac{2a^{2}b}{a^{2}+b^{2}}y + \frac{c(b^{2}+a^{2})}{a^{2}+b^{2}} = 0$.
Thus,the equation is $x^{2}+y^{2}+\frac{2ab^{2}}{a^{2}+b^{2}}x+\frac{2a^{2}b}{a^{2}+b^{2}}y+c=0$.

(C) Let $z = 2^{-i}$.
Taking the natural logarithm on both sides,we get:
$\log z = \log (2^{-i})$
$\Rightarrow \log z = -i \log 2$
Since $z = e^{\log z}$,we have:
$z = e^{-i \log 2} = e^{i \log(1/2)}$
Using Euler's formula,$e^{i \theta} = \cos \theta + i \sin \theta$,where $\theta = \log(1/2)$:
$z = \cos(\log(1/2)) + i \sin(\log(1/2))$
The real part of $z$ is $\cos(\log(1/2))$.

(C) Let $f(x) = |\cos x|$.
We know that the function $\cos x$ has a period of $2\pi$.
However,when we take the absolute value,the negative parts of the graph are reflected above the $x$-axis.
Specifically,$|\cos(x + \pi)| = |-\cos x| = |\cos x|$.
Since $f(x + \pi) = f(x)$,the fundamental period of the function $f(x) = |\cos x|$ is $\pi$.
As shown in the graph,the pattern of the function repeats every $\pi$ units.

(D) $\lim _{x \rightarrow \frac{\pi}{2}}(\sec x-\tan x)$
$\Rightarrow \lim _{x}$ ${\rightarrow \frac{\pi}{2}} \left(\frac{1-\sin x}{\cos x}\right)$ (which is of the $\frac{0}{0}$ form)
Applying $L$'$H$ôpital's rule by differentiating the numerator and denominator with respect to $x$:
$\Rightarrow \lim _{x}$ ${\rightarrow \frac{\pi}{2}} \frac{\frac{d}{dx}(1-\sin x)}{\frac{d}{dx}(\cos x)}$
$\Rightarrow \lim _{x}$ ${\rightarrow \frac{\pi}{2}} \frac{-\cos x}{-\sin x}$
$\Rightarrow \lim _{x}$ ${\rightarrow \frac{\pi}{2}} \cot x$
Substituting $x = \frac{\pi}{2}$:
$\Rightarrow \cot \frac{\pi}{2} = 0$

(C) The given limit is $\lim _{x \rightarrow 0} \left( \frac{3^{x}-1}{x} \right)$,which is in the $\left( \frac{0}{0} \right)$ indeterminate form.
Applying $L'\text{Hospital's rule}$ by differentiating the numerator and the denominator with respect to $x$:
$= \lim _{x \rightarrow 0} \frac{\frac{d}{dx}(3^{x}-1)}{\frac{d}{dx}(x)}$
$= \lim _{x \rightarrow 0} \frac{3^{x} \log 3}{1}$
$= 3^{0} \log 3 = 1 \cdot \log 3 = \log 3$.

(A) Given the parabola equation: $4x^{2}-4x-2y+3=0$
$4(x^{2}-x) = 2y-3$
$4(x^{2}-x+\frac{1}{4}-\frac{1}{4}) = 2y-3$
$4(x-\frac{1}{2})^{2}-1 = 2y-3$
$4(x-\frac{1}{2})^{2} = 2y-2$
$(x-\frac{1}{2})^{2} = \frac{1}{2}(y-1)$
Comparing this with the standard form $X^{2} = 4aY$,where $X = x-\frac{1}{2}$ and $Y = y-1$:
$4a = \frac{1}{2} \implies a = \frac{1}{8}$
The equation of the directrix is $Y+a = 0$
$y-1+\frac{1}{8} = 0$
$y = \frac{7}{8}$
Wait,re-evaluating the standard form: $(x-\frac{1}{2})^{2} = 2(y-1)$ implies $4a = 2$,so $a = \frac{1}{2}$.
The directrix is $Y = -a \implies y-1 = -\frac{1}{2}$
$y = 1 - \frac{1}{2} = \frac{1}{2}$
$2y = 1$.

(D) Let the focal chord be $PQ$ passing through the focus $S(a, 0)$. Let the coordinates of $P$ be $(at_{1}^{2}, 2at_{1})$ and $Q$ be $(at_{2}^{2}, 2at_{2})$.
Since it is a focal chord,$t_{1}t_{2} = -1$.
The focal distance of any point $P(x, y)$ on the parabola is $SP = a + x = a + at_{1}^{2} = a(1 + t_{1}^{2})$.
Given the intercepts of the focal chord are $b$ and $k$,we have $b = a(1 + t_{1}^{2})$ and $k = a(1 + t_{2}^{2})$.
Since $t_{2} = -1/t_{1}$,we have $k = a(1 + (-1/t_{1})^{2}) = a(1 + 1/t_{1}^{2}) = a(\frac{t_{1}^{2} + 1}{t_{1}^{2}})$.
From $b = a(1 + t_{1}^{2})$,we get $t_{1}^{2} = \frac{b}{a} - 1 = \frac{b-a}{a}$.
Substituting this into the expression for $k$:
$k = a(1 + \frac{1}{(b-a)/a}) = a(1 + \frac{a}{b-a}) = a(\frac{b-a+a}{b-a}) = a(\frac{b}{b-a}) = \frac{ab}{b-a}$.

(C) The equation of the parabola is $y^{2} = 4ax$.
Let the line $lx + my + n = 0$ be tangent to the parabola.
Rewriting the line equation as $y = -\frac{l}{m}x - \frac{n}{m}$,which is in the form $y = mx + c$ where the slope $M = -\frac{l}{m}$ and intercept $C = -\frac{n}{m}$.
The condition for the line $y = Mx + C$ to be tangent to the parabola $y^{2} = 4ax$ is $C = \frac{a}{M}$.
Substituting the values of $M$ and $C$:
$-\frac{n}{m} = \frac{a}{-l/m} = -\frac{am}{l}$.
Multiplying both sides by $-1$,we get $\frac{n}{m} = \frac{am}{l}$.
Cross-multiplying gives $nl = am^{2}$.

(A) Given the parabola equation is $y^{2} = 8x$.
Comparing with $y^{2} = 4ax$,we get $a = 2$.
The slope of the normal to the parabola $y^{2} = 4ax$ at point $(x_{1}, y_{1})$ is given by $m = -\frac{y_{1}}{2a}$.
Given the normal equation $2x + y + \lambda = 0$,which can be written as $y = -2x - \lambda$.
Comparing this with $y = mx + c$,the slope $m = -2$.
Equating the slopes: $-2 = -\frac{y_{1}}{2(2)}$ $\Rightarrow -2 = -\frac{y_{1}}{4}$ $\Rightarrow y_{1} = 8$.
Since the point $(x_{1}, y_{1})$ lies on the parabola $y^{2} = 8x$,we have $8^{2} = 8x_{1}$ $\Rightarrow 64 = 8x_{1}$ $\Rightarrow x_{1} = 8$.
Since the point $(8, 8)$ lies on the line $2x + y + \lambda = 0$,we substitute the values:
$2(8) + 8 + \lambda = 0$
$16 + 8 + \lambda = 0$
$24 + \lambda = 0$
$\lambda = -24$.

(A) The equation of the polar of a point $P(x_{1}, y_{1})$ with respect to the parabola $y^{2} = 4ax$ is given by $yy_{1} = 2a(x + x_{1})$.
This can be rewritten as $2ax - yy_{1} + 2ax_{1} = 0 \dots(i)$.
The given line is $lx + my + n = 0 \dots(ii)$.
Comparing equations $(i)$ and $(ii)$,we have:
$\frac{2a}{l} = \frac{-y_{1}}{m} = \frac{2ax_{1}}{n}$.
From the first and third ratios: $\frac{2a}{l} = \frac{2ax_{1}}{n} \Rightarrow x_{1} = \frac{n}{l}$.
From the first and second ratios: $\frac{2a}{l} = \frac{-y_{1}}{m} \Rightarrow y_{1} = \frac{-2am}{l}$.
Thus,the pole is $\left(\frac{n}{l}, \frac{-2am}{l}\right)$.

(B) The total number of ways to select $3$ numbers from $20$ is given by ${}^{20}C_{3}$.
${}^{20}C_{3} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 20 \times 19 \times 3 = 1140$.
Consecutive sets of $3$ numbers are $(1, 2, 3), (2, 3, 4), \dots, (18, 19, 20)$.
The number of such sets is $18$.
Therefore,the required probability is $\frac{18}{1140} = \frac{18}{20 \times 19 \times 3} = \frac{3}{190}$.

(C) The word '$CEASE$' contains $5$ letters: $C, E, A, S, E$. The number of $E$s is $2$.
Total number of arrangements of the letters of '$CEASE$' is $\frac{5!}{2!} = \frac{120}{2} = 60$.
To find the probability that the two $E$s are together,we treat the two $E$s as a single unit $(EE)$.
Now we have $4$ units to arrange: $(EE), C, A, S$.
The number of ways to arrange these $4$ units is $4! = 24$.
The required probability is $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{24}{60} = \frac{2}{5}$.

(A) The odds against $A$ solving the problem are $3:2$,so the probability that $A$ solves it is $P(A) = \frac{2}{3+2} = \frac{2}{5}$. The probability that $A$ fails is $P(\bar{A}) = 1 - \frac{2}{5} = \frac{3}{5}$.
The odds against $B$ solving the problem are $2:1$,so the probability that $B$ solves it is $P(B) = \frac{1}{2+1} = \frac{1}{3}$. The probability that $B$ fails is $P(\bar{B}) = 1 - \frac{1}{3} = \frac{2}{3}$.
The problem is solved if at least one of them solves it. This is the complement of the event that both fail to solve the problem.
$P(\text{solved}) = 1 - P(\bar{A} \cap \bar{B}) = 1 - P(\bar{A}) \cdot P(\bar{B})$
$P(\text{solved}) = 1 - (\frac{3}{5} \cdot \frac{2}{3}) = 1 - \frac{6}{15} = 1 - \frac{2}{5} = \frac{3}{5}$.

(B) Total number of cards $= 52$.
Number of aces $= 4$.
Number of black kings (spade king and club king) $= 2$.
Number of queen of hearts $= 1$.
Since these events are mutually exclusive,the total number of favorable outcomes $= 4 + 2 + 1 = 7$.
$\therefore$ Required probability $= \frac{7}{52}$.

(C) Since the third vertex $C(x_1, y_1)$ lies on the line $x + y = 5$,we have $y_1 = 5 - x_1$.
Thus,the coordinates of $C$ are $(x_1, 5 - x_1)$.
Given the area of $\Delta ABC = 4$,we use the determinant formula for the area of a triangle:
$\frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)| = 4$
$\frac{1}{2} |2(2 - (5 - x_1)) + 3((5 - x_1) - (-1)) + x_1(-1 - 2)| = 4$
$|2(x_1 - 3) + 3(6 - x_1) - 3x_1| = 8$
$|2x_1 - 6 + 18 - 3x_1 - 3x_1| = 8$
$|12 - 4x_1| = 8$
This gives two cases:
Case $1$: $12 - 4x_1 = 8 \implies 4x_1 = 4 \implies x_1 = 1$. Then $y_1 = 5 - 1 = 4$. So,$C$ is $(1, 4)$.
Case $2$: $12 - 4x_1 = -8 \implies 4x_1 = 20 \implies x_1 = 5$. Then $y_1 = 5 - 5 = 0$. So,$C$ is $(5, 0)$.
Therefore,the third vertex is $(5, 0)$ or $(1, 4)$.

(A) Given,$2a + b + 3c = 0$ ... $(i)$
And the line equation is $ax + by + c = 0$.
To eliminate $c$,multiply the line equation by $3$:
$3ax + 3by + 3c = 0$ ... $(ii)$
Subtracting Eq. $(i)$ from Eq. $(ii)$:
$(3ax + 3by + 3c) - (2a + b + 3c) = 0$
$(3x - 2)a + (3y - 1)b = 0$
For this to hold for all $a$ and $b$,the coefficients must be zero:
$3x - 2 = 0 \Rightarrow x = \frac{2}{3}$
$3y - 1 = 0 \Rightarrow y = \frac{1}{3}$
Therefore,the line passes through the fixed point $\left(\frac{2}{3}, \frac{1}{3}\right)$.

(C) Given the curve equation: $y=c \cosh \left(\frac{x}{c}\right)$ ... $(i)$
Differentiating with respect to $x$:
$\frac{dy}{dx} = c \cdot \frac{1}{c} \cdot \sinh \left(\frac{x}{c}\right) = \sinh \left(\frac{x}{c}\right)$
The formula for the length of the normal is given by:
$L = |y| \sqrt{1 + \left(\frac{dy}{dx}\right)^{2}}$
Substituting the values:
$L = y \sqrt{1 + \sinh^{2} \left(\frac{x}{c}\right)}$
Using the identity $\cosh^{2} \theta - \sinh^{2} \theta = 1$,we have $1 + \sinh^{2} \theta = \cosh^{2} \theta$:
$L = y \sqrt{\cosh^{2} \left(\frac{x}{c}\right)}$
$L = y \cosh \left(\frac{x}{c}\right)$
From equation $(i)$,we know $\cosh \left(\frac{x}{c}\right) = \frac{y}{c}$:
$L = y \cdot \left(\frac{y}{c}\right) = \frac{y^{2}}{c}$
Thus,the length of the normal is $\frac{y^{2}}{c}$.

(B) Given the curve equation: $y^{2}=4a|x+a \sin(x/a)|$.
For the tangent to be parallel to the $x$-axis,the derivative $\frac{dy}{dx}$ must be $0$.
Considering the case $y^2 = 4a(x + a \sin(x/a))$,we differentiate with respect to $x$:
$2y \frac{dy}{dx} = 4a(1 + \cos(x/a))$.
Setting $\frac{dy}{dx} = 0$,we get $1 + \cos(x/a) = 0$,which implies $\cos(x/a) = -1$.
This condition implies $\sin(x/a) = 0$.
Substituting $\sin(x/a) = 0$ back into the original equation,we obtain $y^2 = 4a(x + 0) = 4ax$.
Thus,all such points lie on the parabola $y^2 = 4ax$.

(C) Let the radius of the cylinder be $r$ and its height be $h$. The sphere has a radius $a$ (since diameter is $2a$).
In the right-angled triangle formed by the radius of the sphere,the radius of the cylinder,and half the height of the cylinder,we have by Pythagoras theorem:
$r^2 + (h/2)^2 = a^2$
$r^2 = a^2 - \frac{h^2}{4}$
The volume $V$ of the cylinder is given by $V = \pi r^2 h$.
Substituting $r^2$ in the volume formula:
$V = \pi (a^2 - \frac{h^2}{4}) h = \pi (a^2 h - \frac{h^3}{4})$
To find the maximum volume,we differentiate $V$ with respect to $h$:
$\frac{dV}{dh} = \pi (a^2 - \frac{3h^2}{4})$
Setting $\frac{dV}{dh} = 0$ for critical points:
$a^2 - \frac{3h^2}{4} = 0$
$3h^2 = 4a^2$
$h^2 = \frac{4a^2}{3}$
$h = \frac{2a}{\sqrt{3}}$
To verify it is a maximum,we check the second derivative:
$\frac{d^2V}{dh^2} = \pi (0 - \frac{6h}{4}) = -\frac{3\pi h}{2} < 0$ for $h > 0$.
Since the second derivative is negative,the volume is maximum at $h = \frac{2a}{\sqrt{3}}$.

(D) Let $x$ and $y$ be the two parts of the number $10$.
$\therefore x + y = 10 \implies y = 10 - x$ ... $(i)$
Let $A$ be the sum of double of the first and the square of the second:
$A = 2x + y^2$
Substituting $y = 10 - x$ into the equation:
$A = 2x + (10 - x)^2$
$A = 2x + 100 - 20x + x^2$
$A = x^2 - 18x + 100$
To find the minimum,differentiate $A$ with respect to $x$:
$\frac{dA}{dx} = 2x - 18$
Set $\frac{dA}{dx} = 0$ for critical points:
$2x - 18 = 0 \implies x = 9$
Check the second derivative:
$\frac{d^2A}{dx^2} = 2$. Since $2 > 0$,the function has a minimum at $x = 9$.
Substitute $x = 9$ into $(i)$:
$y = 10 - 9 = 1$.
Thus,the parts are $(9, 1)$.

(B) Given curves are $y^{2}=8x$ $(i)$ and $y=x$ (ii).
On solving equations $(i)$ and (ii),we substitute $y=x$ into $y^{2}=8x$:
$x^{2}=8x \Rightarrow x^{2}-8x=0 \Rightarrow x(x-8)=0$.
Thus,the points of intersection are $x=0$ and $x=8$.
For $x=0$,$y=0$ and for $x=8$,$y=8$.
The required area is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=8$:
$\text{Area} = \int_{0}^{8} (\sqrt{8x} - x) dx$
$= \int_{0}^{8} (2\sqrt{2}x^{1/2} - x) dx$
$= \left[ 2\sqrt{2} \cdot \frac{x^{3/2}}{3/2} - \frac{x^{2}}{2} \right]_{0}^{8}$
$= \left[ \frac{4\sqrt{2}}{3} x^{3/2} - \frac{x^{2}}{2} \right]_{0}^{8}$
$= \left( \frac{4\sqrt{2}}{3} (8)^{3/2} - \frac{8^{2}}{2} \right) - (0 - 0)$
$= \frac{4\sqrt{2}}{3} (16\sqrt{2}) - \frac{64}{2}$
$= \frac{4 \times 16 \times 2}{3} - 32$
$= \frac{128}{3} - 32 = \frac{128 - 96}{3} = \frac{32}{3}$.
Therefore,the area is $\frac{32}{3}$ square units.

(C) Given curves are $x^2+y^2=8$ $(i)$ and $y^2=2x$ $(ii)$.
Substituting $y^2=2x$ in $(i)$,we get $x^2+2x-8=0$.
$(x+4)(x-2)=0$,which gives $x=2$ (since $x \ge 0$ for $y^2=2x$).
For $x=2$,$y^2=4$,so $y=\pm 2$.
The intersection points are $(2, 2)$ and $(2, -2)$.
The required area is symmetric about the $x$-axis,so Area $= 2 \times [\text{Area of region bounded by } y^2=2x \text{ from } x=0 \text{ to } 2 + \text{Area of region bounded by } x^2+y^2=8 \text{ from } x=2 \text{ to } 2\sqrt{2}]$.
Area $= 2 \left[ \int_0^2 \sqrt{2x} \, dx + \int_2^{2\sqrt{2}} \sqrt{8-x^2} \, dx \right]$.
$= 2 \left[ \sqrt{2} \left( \frac{x^{3/2}}{3/2} \right)_0^2 + \left( \frac{x}{2} \sqrt{8-x^2} + \frac{8}{2} \sin^{-1} \frac{x}{\sqrt{8}} \right)_2^{2\sqrt{2}} \right]$.
$= 2 \left[ \sqrt{2} \cdot \frac{2}{3} \cdot 2\sqrt{2} + \left( (0 + 4 \sin^{-1}(1)) - (1 \cdot \sqrt{4} + 4 \sin^{-1}(1/\sqrt{2})) \right) \right]$.
$= 2 \left[ \frac{8}{3} + 4(\pi/2) - 2 - 4(\pi/4) \right] = 2 \left[ \frac{8}{3} + 2\pi - 2 - \pi \right] = 2 \left[ \frac{2}{3} + \pi \right] = 2\pi + \frac{4}{3}$.

(C) Given,$f(x) = \begin{cases} x \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}$
For continuity at $x = 0$:
$LHL = f(0-0) = \lim_{h \to 0} f(0-h) = \lim_{h \to 0} (-h) \sin \left(-\frac{1}{h}\right) = \lim_{h \to 0} h \sin \left(\frac{1}{h}\right) = 0 \times (\text{finite quantity}) = 0$.
$RHL = f(0+0) = \lim_{h \to 0} f(0+h) = \lim_{h \to 0} h \sin \left(\frac{1}{h}\right) = 0 \times (\text{finite quantity}) = 0$.
Since $f(0) = LHL = RHL = 0$,$f(x)$ is continuous at $x = 0$.
For differentiability at $x = 0$:
$Rf'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{h \sin(1/h) - 0}{h} = \lim_{h \to 0} \sin(1/h)$.
As $h \to 0$,$\sin(1/h)$ oscillates between $-1$ and $1$,so the limit does not exist.
Similarly,$Lf'(0) = \lim_{h \to 0} \frac{f(0-h) - f(0)}{-h} = \lim_{h \to 0} \frac{-h \sin(-1/h) - 0}{-h} = \lim_{h \to 0} \sin(1/h)$,which also does not exist.
Since the derivative does not exist,$f(x)$ is not differentiable at $x = 0$.
Thus,$f(x)$ is continuous but not differentiable at $x = 0$.

(D) Given that $f(x)$ is continuous at $x = 0$,we have $f(0) = \lim_{x \to 0} f(x)$.
$f(0) = k$.
$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\log (1 + 2ax) - \log (1 - bx)}{x}$.
This is a $\frac{0}{0}$ form,so we apply $L$'Hospital's rule:
$k = \lim_{x \to 0} \frac{\frac{d}{dx}(\log (1 + 2ax) - \log (1 - bx))}{\frac{d}{dx}(x)}$.
$k = \lim_{x \to 0} \frac{\frac{2a}{1 + 2ax} - \frac{-b}{1 - bx}}{1}$.
$k = \frac{2a}{1 + 0} + \frac{b}{1 - 0} = 2a + b$.

(D) Let $I = \int_{0}^{1} x^{2}(1-x^{2})^{3/2} dx$.
Substitute $x = \sin \theta$,then $dx = \cos \theta d\theta$.
When $x = 0$,$\theta = 0$,and when $x = 1$,$\theta = \frac{\pi}{2}$.
$I = \int_{0}^{\pi/2} \sin^{2} \theta (\cos^{2} \theta)^{3/2} \cos \theta d\theta = \int_{0}^{\pi/2} \sin^{2} \theta \cos^{4} \theta d\theta$.
Using the Wallis formula or Beta function $\int_{0}^{\pi/2} \sin^{m} \theta \cos^{n} \theta d\theta = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2 \Gamma(\frac{m+n+2}{2})}$.
Here $m=2, n=4$,so $I = \frac{\Gamma(3/2) \Gamma(5/2)}{2 \Gamma(4)} = \frac{(\frac{1}{2} \sqrt{\pi}) (\frac{3}{2} \cdot \frac{1}{2} \sqrt{\pi})}{2 \cdot 6} = \frac{\frac{3}{8} \pi}{12} = \frac{3\pi}{96} = \frac{\pi}{32}$.

(B) Let $I = \int_{3}^{4} \sqrt{(4-x)(x-3)} d x$.
We can rewrite the integrand as $\sqrt{-x^2 + 7x - 12}$.
Completing the square: $-x^2 + 7x - 12 = -\left(x^2 - 7x + \frac{49}{4} - \frac{49}{4}\right) - 12 = -\left(x - \frac{7}{2}\right)^2 + \frac{49}{4} - 12 = \left(\frac{1}{2}\right)^2 - \left(x - \frac{7}{2}\right)^2$.
So,$I = \int_{3}^{4} \sqrt{\left(\frac{1}{2}\right)^2 - \left(x - \frac{7}{2}\right)^2} d x$.
Let $t = x - \frac{7}{2}$,then $dt = dx$. When $x=3, t=-\frac{1}{2}$ and when $x=4, t=\frac{1}{2}$.
$I = \int_{-1/2}^{1/2} \sqrt{\left(\frac{1}{2}\right)^2 - t^2} dt$.
Since the integrand is an even function,$I = 2 \int_{0}^{1/2} \sqrt{\left(\frac{1}{2}\right)^2 - t^2} dt$.
Using the formula $\int \sqrt{a^2 - t^2} dt = \frac{t}{2}\sqrt{a^2 - t^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{t}{a}\right)$,where $a = \frac{1}{2}$:
$I = 2 \left[ \frac{t}{2}\sqrt{\frac{1}{4} - t^2} + \frac{1}{8}\sin^{-1}(2t) \right]_{0}^{1/2}$.
$I = 2 \left[ (0 + \frac{1}{8}\sin^{-1}(1)) - (0 + 0) \right] = 2 \times \frac{1}{8} \times \frac{\pi}{2} = \frac{\pi}{8}$.

(B) Let $I = \int_{0}^{\pi} \log (1+\cos x) d x \dots (i)$
Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,we get:
$I = \int_{0}^{\pi} \log (1+\cos(\pi-x)) d x = \int_{0}^{\pi} \log (1-\cos x) d x \dots (ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_{0}^{\pi} [\log(1+\cos x) + \log(1-\cos x)] d x$
$2I = \int_{0}^{\pi} \log(1-\cos^2 x) d x = \int_{0}^{\pi} \log(\sin^2 x) d x$
$2I = 2 \int_{0}^{\pi} \log(\sin x) d x$
$I = \int_{0}^{\pi} \log(\sin x) d x$
Using the property $\int_{0}^{2a} f(x) d x = 2 \int_{0}^{a} f(x) d x$ if $f(2a-x) = f(x)$:
$I = 2 \int_{0}^{\pi/2} \log(\sin x) d x$
Since $\int_{0}^{\pi/2} \log(\sin x) d x = -\frac{\pi}{2} \log 2$:
$I = 2 \times (-\frac{\pi}{2} \log 2) = -\pi \log 2 = \pi \log(\frac{1}{2})$

(D) Given the equation: $x^{p} + y^{q} = (x + y)^{p + q}$.
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 $x$:
$\frac{p}{x} + \frac{q}{y} \cdot \frac{dy}{dx} = \frac{p + q}{x + y} \left( 1 + \frac{dy}{dx} \right)$.
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{p}{x} - \frac{p + q}{x + y} = \left( \frac{p + q}{x + y} - \frac{q}{y} \right) \frac{dy}{dx}$.
Simplifying both sides:
$\frac{p(x + y) - x(p + q)}{x(x + y)} = \left( \frac{y(p + q) - q(x + y)}{y(x + y)} \right) \frac{dy}{dx}$.
$\frac{px + py - px - qx}{x(x + y)} = \left( \frac{py + qy - qx - qy}{y(x + y)} \right) \frac{dy}{dx}$.
$\frac{py - qx}{x(x + y)} = \frac{py - qx}{y(x + y)} \cdot \frac{dy}{dx}$.
Canceling the common term $(py - qx)$ and $(x + y)$ from both sides,we get:
$\frac{1}{x} = \frac{1}{y} \cdot \frac{dy}{dx}$.
Therefore,$\frac{dy}{dx} = \frac{y}{x}$.

(A) Given,$\tan x = \frac{2t}{1-t^2}$ and $\sin y = \frac{2t}{1+t^2}$.
We know the trigonometric identities: $\tan(2\theta) = \frac{2\tan\theta}{1-\tan^2\theta}$ and $\sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta}$.
Let $t = \tan\theta$,then:
$x = \tan^{-1}\left(\frac{2\tan\theta}{1-\tan^2\theta}\right) = \tan^{-1}(\tan 2\theta) = 2\theta = 2\tan^{-1}t$.
$y = \sin^{-1}\left(\frac{2\tan\theta}{1+\tan^2\theta}\right) = \sin^{-1}(\sin 2\theta) = 2\theta = 2\tan^{-1}t$.
Thus,$x = 2\tan^{-1}t$ and $y = 2\tan^{-1}t$.
This implies $x = y$,so $\frac{dy}{dx} = 1$.

(D) Given,$f(x)=|x-3|$.
We can redefine the function as:
$f(x) = \begin{cases} 3-x, & x < 3 \\ 0, & x=3 \\ x-3, & x > 3 \end{cases}$
To check the differentiability at $x=3$,we find the left-hand derivative $(LHD)$ and right-hand derivative $(RHD)$:
$LHD = \lim_{h \to 0^-} \frac{f(3+h)-f(3)}{h} = \lim_{h \to 0^-} \frac{|3+h-3|-0}{h} = \lim_{h \to 0^-} \frac{|h|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1$.
$RHD = \lim_{h \to 0^+} \frac{f(3+h)-f(3)}{h} = \lim_{h \to 0^+} \frac{|3+h-3|-0}{h} = \lim_{h \to 0^+} \frac{|h|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1$.
Since $LHD \neq RHD$,the derivative $f^{\prime}(3)$ does not exist.

(B) Given,$f: R \rightarrow R$ defined by $f(x) = x|x|$.
We can redefine the function as:
$f(x) = \begin{cases} -x^2, & x < 0 \\ 0, & x = 0 \\ x^2, & x > 0 \end{cases}$
$1$. One-one check: Since $f(x)$ is a strictly increasing function (as $f'(x) = 2|x| \ge 0$ for all $x \in R$ and $f'(x) > 0$ for $x \neq 0$),it is a one-one function.
$2$. Onto check: As $x \rightarrow \infty$,$f(x) \rightarrow \infty$ and as $x \rightarrow -\infty$,$f(x) \rightarrow -\infty$. Since the range of the function is $(-\infty, \infty)$,which is equal to the codomain $R$,the function is onto.
Therefore,the function is one-one and onto (bijective).

(D) Given $f(x)=\frac{2^{x}+2^{-x}}{2}$.
Now,$f(x+y)=\frac{2^{x+y}+2^{-(x+y)}}{2}$ and $f(x-y)=\frac{2^{x-y}+2^{-(x-y)}}{2}$.
Therefore,$f(x+y) \cdot f(x-y) = \frac{2^{x+y}+2^{-x-y}}{2} \cdot \frac{2^{x-y}+2^{-x+y}}{2}$.
$= \frac{1}{4} [2^{x+y} \cdot 2^{x-y} + 2^{x+y} \cdot 2^{-x+y} + 2^{-x-y} \cdot 2^{x-y} + 2^{-x-y} \cdot 2^{-x+y}]$.
$= \frac{1}{4} [2^{2x} + 2^{2y} + 2^{-2y} + 2^{-2x}]$.
$= \frac{1}{4} [(2^{2x} + 2^{-2x}) + (2^{2y} + 2^{-2y})]$.
$= \frac{1}{2} [\frac{2^{2x} + 2^{-2x}}{2} + \frac{2^{2y} + 2^{-2y}}{2}]$.
$= \frac{1}{2} [f(2x) + f(2y)]$.

(A) Let $I = \int \frac{x^{2}+1}{x^{4}+x^{2}+1} dx$.
Divide the numerator and denominator by $x^{2}$:
$I = \int \frac{1 + 1/x^{2}}{x^{2} + 1 + 1/x^{2}} dx$.
Rewrite the denominator as $(x - 1/x)^{2} + 3$:
$I = \int \frac{1 + 1/x^{2}}{(x - 1/x)^{2} + (\sqrt{3})^{2}} dx$.
Let $t = x - 1/x$,then $dt = (1 + 1/x^{2}) dx$.
Substituting these into the integral:
$I = \int \frac{dt}{t^{2} + (\sqrt{3})^{2}}$.
Using the standard formula $\int \frac{dx}{x^{2} + a^{2}} = \frac{1}{a} \tan^{-1}(x/a) + C$:
$I = \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{t}{\sqrt{3}}\right) + C$.
Substituting $t = x - 1/x$ back:
$I = \frac{1}{\sqrt{3}} \tan^{-1}\left\{\frac{x - 1/x}{\sqrt{3}}\right\} + C$.

(B) Let $I = \int \sqrt{1+\sec x} \, dx$.
We know that $1+\sec x = 1 + \frac{1}{\cos x} = \frac{\cos x + 1}{\cos x} = \frac{2 \cos^2 \frac{x}{2}}{\cos x}$.
So,$I = \int \frac{\sqrt{2} \cos \frac{x}{2}}{\sqrt{\cos x}} \, dx$.
Using $\cos x = 1 - 2 \sin^2 \frac{x}{2}$,we get $I = \int \frac{\sqrt{2} \cos \frac{x}{2}}{\sqrt{1 - 2 \sin^2 \frac{x}{2}}} \, dx$.
Let $t = \sqrt{2} \sin \frac{x}{2}$. Then $dt = \sqrt{2} \cdot \frac{1}{2} \cos \frac{x}{2} \, dx = \frac{1}{\sqrt{2}} \cos \frac{x}{2} \, dx$.
This implies $\sqrt{2} \cos \frac{x}{2} \, dx = 2 \, dt$.
Substituting these into the integral,we get $I = \int \frac{2 \, dt}{\sqrt{1 - t^2}} = 2 \sin^{-1}(t) + C$.
Substituting back $t = \sqrt{2} \sin \frac{x}{2}$,we obtain $I = 2 \sin^{-1}(\sqrt{2} \sin \frac{x}{2}) + C$.

(B) Let $I = \int \frac{\sin 2x}{\sin^4 x + \cos^4 x} dx$.
We know that $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x$.
Using $\sin 2x = 2\sin x \cos x$,we have $2\sin^2 x \cos^2 x = \frac{1}{2}(2\sin x \cos x)^2 = \frac{1}{2}\sin^2 2x$.
So,$I = \int \frac{\sin 2x}{1 - \frac{1}{2}\sin^2 2x} dx = \int \frac{2\sin 2x}{2 - \sin^2 2x} dx$.
Using $\sin^2 2x = 1 - \cos^2 2x$,we get $I = \int \frac{2\sin 2x}{2 - (1 - \cos^2 2x)} dx = \int \frac{2\sin 2x}{1 + \cos^2 2x} dx$.
Let $t = \cos 2x$,then $dt = -2\sin 2x dx$,so $2\sin 2x dx = -dt$.
$I = \int \frac{-dt}{1 + t^2} = -\tan^{-1}(t) + C$.
Substituting back $t = \cos 2x$,we get $I = -\tan^{-1}(\cos 2x) + C$.

(C) Let $I = \int \frac{dx}{3 \sin x - \cos x + 3}$.
Using the substitution $\tan \frac{x}{2} = t$,we have $dx = \frac{2 dt}{1+t^2}$,$\sin x = \frac{2t}{1+t^2}$,and $\cos x = \frac{1-t^2}{1+t^2}$.
Substituting these into the integral:
$I = \int \frac{1}{3(\frac{2t}{1+t^2}) - (\frac{1-t^2}{1+t^2}) + 3} \cdot \frac{2 dt}{1+t^2}$
$I = \int \frac{2 dt}{6t - 1 + t^2 + 3(1+t^2)} = \int \frac{2 dt}{4t^2 + 6t + 2} = \int \frac{dt}{2t^2 + 3t + 1}$.
Factor the denominator: $2t^2 + 3t + 1 = (2t+1)(t+1)$.
Using partial fractions: $\frac{1}{(2t+1)(t+1)} = \frac{A}{2t+1} + \frac{B}{t+1}$.
Solving for $A$ and $B$,we get $A = -2$ and $B = 1$.
$I = \int (\frac{1}{t+1} - \frac{2}{2t+1}) dt = \log|t+1| - \log|2t+1| + C = \log \left| \frac{t+1}{2t+1} \right| + C$.
Wait,checking the partial fraction again: $\frac{1}{(2t+1)(t+1)} = \frac{2}{2t+1} - \frac{1}{t+1}$.
$I = \int (\frac{2}{2t+1} - \frac{1}{t+1}) dt = \log|2t+1| - \log|t+1| + C = \log \left| \frac{2t+1}{t+1} \right| + C$.
Substituting $t = \tan \frac{x}{2}$,we get $I = \log \left( \frac{2 \tan \frac{x}{2} + 1}{\tan \frac{x}{2} + 1} \right) + C$.

(A) Let $I = \int \frac{dx}{x(x^{n}+1)}$.
Multiply the numerator and denominator by $x^{n-1}$:
$I = \int \frac{x^{n-1} dx}{x^{n}(x^{n}+1)}$.
Let $t = x^{n}$,then $dt = nx^{n-1} dx$,which implies $x^{n-1} dx = \frac{dt}{n}$.
Substituting these into the integral:
$I = \int \frac{dt/n}{t(t+1)} = \frac{1}{n} \int \frac{dt}{t(t+1)}$.
Using partial fractions,$\frac{1}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1}$.
$I = \frac{1}{n} \int \left( \frac{1}{t} - \frac{1}{t+1} \right) dt$.
$I = \frac{1}{n} (\log |t| - \log |t+1|) + C$.
$I = \frac{1}{n} \log \left| \frac{t}{t+1} \right| + C$.
Substituting $t = x^{n}$ back:
$I = \frac{1}{n} \log \left( \frac{x^{n}}{x^{n}+1} \right) + C$.

(D) Let $I = \int e^{x} \left[ \frac{1+\sin x}{1+\cos x} \right] dx$.
Using the identities $1+\cos x = 2\cos^2 \frac{x}{2}$ and $\sin x = 2\sin \frac{x}{2} \cos \frac{x}{2}$,we get:
$I = \int e^{x} \left[ \frac{1 + 2\sin \frac{x}{2} \cos \frac{x}{2}}{2\cos^2 \frac{x}{2}} \right] dx$
$I = \int e^{x} \left[ \frac{1}{2\cos^2 \frac{x}{2}} + \frac{2\sin \frac{x}{2} \cos \frac{x}{2}}{2\cos^2 \frac{x}{2}} \right] dx$
$I = \int e^{x} \left[ \frac{1}{2} \sec^2 \frac{x}{2} + \tan \frac{x}{2} \right] dx$.
We know that $\int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + C$.
Here,let $f(x) = \tan \frac{x}{2}$. Then $f'(x) = \sec^2 \frac{x}{2} \cdot \frac{1}{2} = \frac{1}{2} \sec^2 \frac{x}{2}$.
Thus,the integral becomes $e^{x} \tan \frac{x}{2} + C$.

(B) Let $I = \int \cos (\log x) \cdot 1 \, dx \dots (i)$
Using integration by parts,$\int u \, dv = uv - \int v \, du$.
Let $u = \cos (\log x)$ and $dv = 1 \, dx$.
Then $du = -\sin (\log x) \cdot \frac{1}{x} \, dx$ and $v = x$.
$I = x \cos (\log x) - \int x \cdot (-\sin (\log x)) \cdot \frac{1}{x} \, dx$
$I = x \cos (\log x) + \int \sin (\log x) \, dx$
Now,apply integration by parts again to $\int \sin (\log x) \cdot 1 \, dx$:
$I = x \cos (\log x) + [x \sin (\log x) - \int x \cdot \cos (\log x) \cdot \frac{1}{x} \, dx] + C$
$I = x \cos (\log x) + x \sin (\log x) - \int \cos (\log x) \, dx + C$
$I = x [\cos (\log x) + \sin (\log x)] - I + C$
$2I = x [\sin (\log x) + \cos (\log x)] + C$
$I = \frac{x}{2} [\sin (\log x) + \cos (\log x)] + C$

(B) Let $I = \int_{0}^{\infty} \frac{x}{(1+x)(x^{2}+1)} dx$.
Using partial fractions,we write $\frac{x}{(1+x)(x^{2}+1)} = \frac{A}{1+x} + \frac{Bx+C}{x^{2}+1}$.
Equating numerators: $x = A(x^{2}+1) + (Bx+C)(1+x) = (A+B)x^{2} + (B+C)x + (A+C)$.
Comparing coefficients: $A+B=0$,$B+C=1$,$A+C=0$.
Solving these,we get $A = -\frac{1}{2}$,$B = \frac{1}{2}$,and $C = \frac{1}{2}$.
Thus,$I = \int_{0}^{\infty} \left( -\frac{1}{2(1+x)} + \frac{x+1}{2(x^{2}+1)} \right) dx = -\frac{1}{2} \int_{0}^{\infty} \frac{dx}{1+x} + \frac{1}{4} \int_{0}^{\infty} \frac{2x}{x^{2}+1} dx + \frac{1}{2} \int_{0}^{\infty} \frac{dx}{x^{2}+1}$.
Evaluating the integrals: $I = \left[ -\frac{1}{2} \ln(1+x) + \frac{1}{4} \ln(x^{2}+1) + \frac{1}{2} \tan^{-1}(x) \right]_{0}^{\infty}$.
Combining the logarithmic terms: $I = \left[ \frac{1}{4} \ln \left( \frac{(x^{2}+1)}{(1+x)^{2}} \right) + \frac{1}{2} \tan^{-1}(x) \right]_{0}^{\infty}$.
As $x \to \infty$,$\frac{x^{2}+1}{(1+x)^{2}} \to 1$,so $\ln(1) = 0$.
At $x=0$,$\frac{1}{4} \ln(1) + \frac{1}{2} \tan^{-1}(0) = 0$.
Therefore,$I = 0 + \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4}$.

(D) The unit digit of any positive integer can be any of the $10$ digits: ${0, 1, 2, 3, 4, 5, 6, 7, 8, 9}$.
For the product of four numbers to have a unit digit of $1, 3, 7,$ or $9$,each of the four numbers must have a unit digit from the set ${1, 3, 7, 9}$.
There are $4$ such favorable digits out of $10$ possible digits for each number.
The probability that a single number has a unit digit in ${1, 3, 7, 9}$ is $P = \frac{4}{10} = \frac{2}{5}$.
Since the four integers are selected independently,the probability that all four have a unit digit in ${1, 3, 7, 9}$ is $\left(\frac{2}{5}\right)^{4} = \frac{16}{625}$.

(C) The probability of getting a head in a single toss is $p = \frac{1}{2}$ and the probability of getting a tail is $q = \frac{1}{2}$.
For $n = 15$ trials,the probability of getting $r = 10$ heads is given by the binomial distribution formula: $P(X = r) = {}^{n}C_{r} p^{r} q^{n-r}$.
Substituting the values,we get: $P(X = 10) = {}^{15}C_{10} \left(\frac{1}{2}\right)^{10} \left(\frac{1}{2}\right)^{5}$.
Since ${}^{15}C_{10} = {}^{15}C_{5}$,we have: $P(X = 10) = {}^{15}C_{5} \times \left(\frac{1}{2}\right)^{15}$.
Calculating the combination: ${}^{15}C_{5} = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} = 3003$.
Thus,the probability is: $P(X = 10) = \frac{3003}{2^{15}} = \frac{3003}{32768}$.

(B) Given,$a+b+c=0$ and $|a|=5, |b|=3, |c|=7$.
Since $a+b+c=0$,we have $a+b=-c$.
Squaring both sides,we get $|a+b|^2 = |-c|^2$.
This implies $|a|^2 + |b|^2 + 2(a \cdot b) = |c|^2$.
Using the definition of the dot product,$a \cdot b = |a||b| \cos \theta$,where $\theta$ is the angle between $a$ and $b$.
Substituting the given values: $(5)^2 + (3)^2 + 2(5)(3) \cos \theta = (7)^2$.
$25 + 9 + 30 \cos \theta = 49$.
$34 + 30 \cos \theta = 49$.
$30 \cos \theta = 49 - 34 = 15$.
$\cos \theta = \frac{15}{30} = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,we have $\theta = 60^{\circ}$ or $\theta = \frac{\pi}{3}$ radians.

(A) Let the origin be at the circumcentre $S$. The position vectors of vertices $A, B, C$ are $\vec{a}, \vec{b}, \vec{c}$ such that $|\vec{a}| = |\vec{b}| = |\vec{c}| = R$.
The position vector of the orthocentre $O^{\prime}$ is $\vec{o^{\prime}} = \vec{a} + \vec{b} + \vec{c}$.
The position vector of the incentre $O$ is $\vec{o} = \frac{a\vec{a} + b\vec{b} + c\vec{c}}{a+b+c}$.
We need to evaluate $\vec{O^{\prime}A} + \vec{O^{\prime}B} + \vec{O^{\prime}C} = (\vec{a} - \vec{o^{\prime}}) + (\vec{b} - \vec{o^{\prime}}) + (\vec{c} - \vec{o^{\prime}})$.
$= (\vec{a} + \vec{b} + \vec{c}) - 3\vec{o^{\prime}} = \vec{o^{\prime}} - 3\vec{o^{\prime}} = -2\vec{o^{\prime}}$.
This is a standard property in triangle geometry where $\vec{O^{\prime}A} + \vec{O^{\prime}B} + \vec{O^{\prime}C} = 2\vec{O^{\prime}S}$,where $S$ is the circumcentre. Given the options provided and the context of vector relations in triangles,the correct expression is $2\vec{O^{\prime}O}$.

(C) Let $\vec{a} = 4 i - j + 3 k$ and $\vec{b} = -2 i + j - 2 k$. The vector perpendicular to both $\vec{a}$ and $\vec{b}$ is given by their cross product $\vec{n} = \vec{a} \times \vec{b}$.
$\vec{n} = \begin{vmatrix} i & j & k \\ 4 & -1 & 3 \\ -2 & 1 & -2 \end{vmatrix} = i(2-3) - j(-8+6) + k(4-2) = -i + 2j + 2k$.
The magnitude of $\vec{n}$ is $|\vec{n}| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1+4+4} = \sqrt{9} = 3$.
The unit vector perpendicular to both is $\hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{-i + 2j + 2k}{3}$.
The required vector of magnitude $9$ is $\pm 9 \hat{n} = \pm 9 \left( \frac{-i + 2j + 2k}{3} \right) = \pm 3(-i + 2j + 2k) = \pm (-3i + 6j + 6k)$.
Thus,the vectors are $-3i + 6j + 6k$ or $3i - 6j - 6k$.

(D) Let $\vec{a} = 2\hat{i} - \hat{j} - \hat{k}$,$\vec{b} = \hat{i} + 2\hat{j} - 3\hat{k}$ and $\vec{c} = 3\hat{i} + \lambda\hat{j} + 5\hat{k}$.
Since these vectors are coplanar,their scalar triple product must be zero,i.e.,$[\vec{a} \vec{b} \vec{c}] = 0$.
This implies the determinant of the components is zero:
$\begin{vmatrix} 2 & -1 & -1 \\ 1 & 2 & -3 \\ 3 & \lambda & 5 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$2(2(5) - (-3)(\lambda)) - (-1)(1(5) - (-3)(3)) + (-1)(1(\lambda) - 2(3)) = 0$
$2(10 + 3\lambda) + 1(5 + 9) - 1(\lambda - 6) = 0$
$20 + 6\lambda + 14 - \lambda + 6 = 0$
$5\lambda + 40 = 0$
$5\lambda = -40$
$\lambda = -8$.

(C) We know that the scalar triple product of unit vectors is given by the cyclic order property: $i \cdot(j \times k) = 1$,$j \cdot(k \times i) = 1$,and $k \cdot(i \times j) = 1$.
Given expression is $i \cdot(j \times k) + j \cdot(k \times i) + k \cdot(j \times i)$.
Since $j \times i = -k$,we have $k \cdot(j \times i) = k \cdot(-k) = -(k \cdot k) = -1$.
Substituting these values:
$i \cdot(j \times k) + j \cdot(k \times i) + k \cdot(j \times i) = 1 + 1 + (-1) = 1$.

(A) The position vectors of the vertices $A, B$ and $C$ with respect to the origin $O$ are given as:
$\vec{OA} = 6i = 6\hat{i} + 0\hat{j} + 0\hat{k}$
$\vec{OB} = 6j = 0\hat{i} + 6\hat{j} + 0\hat{k}$
$\vec{OC} = k = 0\hat{i} + 0\hat{j} + 1\hat{k}$
The volume of a tetrahedron with vertices at the origin and position vectors $\vec{a}, \vec{b}, \vec{c}$ is given by the formula:
$V = \frac{1}{6} |[\vec{a} \vec{b} \vec{c}]|$
Substituting the given vectors:
$V = \frac{1}{6} \left| \begin{vmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 1 \end{vmatrix} \right|$
Calculating the determinant:
$V = \frac{1}{6} (6 \times (6 \times 1 - 0 \times 0) - 0 + 0) = \frac{1}{6} (36) = 6$
Thus,the volume of the tetrahedron $OABC$ is $6$ cubic units.