KCET 2018 Mathematics Question Paper with Answer and Solution

(B) Given the statement $P(n):$ " $2^{2n}-1$ is divisible by $k$ for all $n \in N$ ".
We can write $2^{2n}-1$ as $(2^2)^n - 1 = 4^n - 1$.
Using the binomial expansion,$4^n - 1 = (3+1)^n - 1$.
Expanding $(3+1)^n$ using the binomial theorem: $(3+1)^n = 1 + n(3) + \frac{n(n-1)}{2}(3^2) + \dots + 3^n$.
So,$4^n - 1 = (1 + 3n + 9\frac{n(n-1)}{2} + \dots + 3^n) - 1 = 3n + 9\frac{n(n-1)}{2} + \dots + 3^n$.
This expression is clearly divisible by $3$ for all $n \in N$.
Thus,the value of $k$ is $3$.

(C) Given the inequality $|x+5| \geq 10$.
By the property of absolute value inequalities,$|u| \geq a$ implies $u \leq -a$ or $u \geq a$.
Therefore,$x+5 \leq -10$ or $x+5 \geq 10$.
Solving the first part: $x \leq -10 - 5 \Rightarrow x \leq -15$.
Solving the second part: $x \geq 10 - 5 \Rightarrow x \geq 5$.
Combining these,we get $x \in (-\infty, -15] \cup [5, \infty)$.

(B) Given that,$\left(\frac{1-i}{1+i}\right)^{96}=a+ib$.
First,simplify the expression inside the bracket:
$\frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1-2i+i^2}{1-i^2} = \frac{1-2i-1}{1+1} = \frac{-2i}{2} = -i$.
Now,substitute this back into the equation:
$(-i)^{96} = a+ib$.
Since $96$ is an even number,$(-i)^{96} = i^{96}$.
$i^{96} = (i^4)^{24} = (1)^{24} = 1$.
So,$1 = a+ib$,which can be written as $1+0i = a+ib$.
Comparing the real and imaginary parts,we get $a=1$ and $b=0$.
Therefore,$(a, b) = (1, 0)$.

(B) Let the total number of persons in the room be $n$.
The number of handshakes is given by the combination formula ${}^{n}C_{2}$,as a handshake occurs between $2$ people.
Given,${}^{n}C_{2} = 45$.
Using the formula ${}^{n}C_{2} = \frac{n(n-1)}{2}$,we have:
$\frac{n(n-1)}{2} = 45$
$n(n-1) = 90$
$n^2 - n - 90 = 0$
$(n - 10)(n + 9) = 0$
Since the number of persons cannot be negative,$n = 10$.

(D) To ensure no two boys are together,we first arrange the $5$ girls in a row. The number of ways to arrange $5$ girls is $5! = 120$.
There are $6$ possible gaps (including the ends) created by the $5$ girls: $\_ G_1 \_ G_2 \_ G_3 \_ G_4 \_ G_5 \_$.
We need to place $3$ boys in these $6$ gaps. The number of ways to choose $3$ gaps out of $6$ is $^{6}C_{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
The $3$ boys can be arranged in these $3$ chosen gaps in $3! = 6$ ways.
Therefore,the total number of ways is $5! \times ^{6}C_{3} \times 3! = 120 \times 20 \times 6 = 14400$.

(D) Given that $a, b, c$ are in $A.P.$ and $x, y, z$ are in $G.P.$
Since $a, b, c$ are in $A.P.$,we have $b-a = c-b = d$,where $d$ is the common difference.
This implies $b-c = -d$,$c-a = 2d$,and $a-b = -d$.
Since $x, y, z$ are in $G.P.$,we have $y^2 = xz$.
Now,consider the expression $E = x^{b-c} \cdot y^{c-a} \cdot z^{a-b}$.
Substituting the values,we get $E = x^{-d} \cdot y^{2d} \cdot z^{-d}$.
$E = (xz)^{-d} \cdot (y^2)^d$.
Since $xz = y^2$,we have $E = (y^2)^{-d} \cdot (y^2)^d = (y^2)^{-d+d} = (y^2)^0 = 1$.

(A) The general term $T_{r+1}$ in the expansion of $(x^2 - x^{-2})^{16}$ is given by:
$T_{r+1} = ^{16}C_{r} (x^2)^{16-r} (-x^{-2})^r$
$T_{r+1} = ^{16}C_{r} (-1)^r x^{32-2r} x^{-2r}$
$T_{r+1} = ^{16}C_{r} (-1)^r x^{32-4r}$
For the constant term,the power of $x$ must be $0$:
$32 - 4r = 0 \implies 4r = 32 \implies r = 8$
Substituting $r = 8$ into the expression:
$T_{8+1} = ^{16}C_{8} (-1)^8 x^{32-4(8)} = ^{16}C_{8} (1) (1) = ^{16}C_{8}$

(A) The equation of a line parallel to $ax + by + c = 0$ is of the form $ax + by + k = 0$.
Given the line $3x - 4y + 2 = 0$,the parallel line is $3x - 4y + k = 0$.
Since this line passes through the point $(-2, 3)$,we substitute $x = -2$ and $y = 3$ into the equation:
$3(-2) - 4(3) + k = 0$
$-6 - 12 + k = 0$
$-18 + k = 0$
$k = 18$
Substituting $k = 18$ back into the equation,we get $3x - 4y + 18 = 0$.

(D) Given,the distance between the foci is $2ae = 16$.
Since the eccentricity $e = \sqrt{2}$,we have $2a(\sqrt{2}) = 16$,which implies $a = \frac{8}{\sqrt{2}} = 4\sqrt{2}$.
For a hyperbola,$b^2 = a^2(e^2 - 1)$.
Substituting the values,$b^2 = (4\sqrt{2})^2 ((\sqrt{2})^2 - 1) = 32(2 - 1) = 32$.
The standard equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Substituting $a^2 = 32$ and $b^2 = 32$,we get $\frac{x^2}{32} - \frac{y^2}{32} = 1$,which simplifies to $x^2 - y^2 = 32$.

(D) We are given the limit $\lim_{x \rightarrow 0} \frac{|x|}{x}$.
To evaluate this,we check the left-hand limit and the right-hand limit.
For the right-hand limit $(x \rightarrow 0^{+})$,$|x| = x$,so $\lim_{x \rightarrow 0^{+}} \frac{x}{x} = 1$.
For the left-hand limit $(x \rightarrow 0^{-})$,$|x| = -x$,so $\lim_{x \rightarrow 0^{-}} \frac{-x}{x} = -1$.
Since the left-hand limit $\neq$ the right-hand limit,the limit does not exist.

(A) Let $p$ be the statement: "$72$ is divisible by $2$ and $3$".
This can be written as $p = q \wedge r$,where $q$ is "$72$ is divisible by $2$" and $r$ is "$72$ is divisible by $3$".
The negation of a conjunction is given by De Morgan's Law: $\sim(q \wedge r) \equiv \sim q \vee \sim r$.
Here,$\sim q$ is "$72$ is not divisible by $2$" and $\sim r$ is "$72$ is not divisible by $3$".
Therefore,the negation is "$72$ is not divisible by $2$ or $72$ is not divisible by $3$".

(C) The given equation of the circle is $(x+1)^{2}+(y-3)^{2}=64$.
Comparing this with the standard form $(x-h)^{2}+(y-k)^{2}=r^{2}$,we get the radius $r = \sqrt{64} = 8$.
For a rectangle inscribed in a circle,the maximum area is achieved when the rectangle is a square.
The diagonal of this square is equal to the diameter of the circle,which is $d = 2r = 2 \times 8 = 16$.
Let the side of the square be $a$. Then,by Pythagoras theorem,$a^{2} + a^{2} = d^{2}$.
$2a^{2} = 16^{2} = 256$.
$a^{2} = 128$.
The area of the square is $a^{2} = 128 \text{ sq. units}$.

(C) Given: $P(A) = 0.5$ and $P(B) = 0.3$.
Since $A$ and $B$ are mutually exclusive events,$P(A \cap B) = 0$.
The probability of $A$ or $B$ occurring is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.5 + 0.3 - 0 = 0.8$.
The probability of neither $A$ nor $B$ is $P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)$.
Therefore,$P(A' \cap B') = 1 - 0.8 = 0.2$.

(C) When a pair of dice is thrown,the total number of possible outcomes is $6 \times 6 = 36$.
We need to find the probability of getting a sum greater than $7$.
The outcomes with a sum greater than $7$ are:
Sum $= 8$: $(2,6), (3,5), (4,4), (5,3), (6,2)$ ($5$ outcomes)
Sum $= 9$: $(3,6), (4,5), (5,4), (6,3)$ ($4$ outcomes)
Sum $= 10$: $(4,6), (5,5), (6,4)$ ($3$ outcomes)
Sum $= 11$: $(5,6), (6,5)$ ($2$ outcomes)
Sum $= 12$: $(6,6)$ ($1$ outcome)
Total favorable outcomes $= 5 + 4 + 3 + 2 + 1 = 15$.
Therefore,the required probability is $P = \frac{15}{36} = \frac{5}{12}$.

(A) Given that $A$ and $B$ are mutually exclusive events,the probability of $A$ or $B$ is given by the addition rule for mutually exclusive events:
$P(A \cup B) = P(A) + P(B)$
Substituting the given values:
$P(A \cup B) = \frac{3}{5} + \frac{1}{5} = \frac{4}{5}$
Converting the fraction to decimal:
$\frac{4}{5} = 0.8$

(A) Total number of batteries = $10$.
Number of dead batteries = $4$.
We need to select $3$ batteries without replacement.
The probability that the first battery is dead = $\frac{4}{10}$.
After selecting one dead battery,the remaining number of batteries is $9$ and the remaining number of dead batteries is $3$.
The probability that the second battery is dead = $\frac{3}{9}$.
After selecting two dead batteries,the remaining number of batteries is $8$ and the remaining number of dead batteries is $2$.
The probability that the third battery is dead = $\frac{2}{8}$.
Therefore,the probability that all $3$ batteries are dead = $\frac{4}{10} \times \frac{3}{9} \times \frac{2}{8} = \frac{24}{720} = \frac{1}{30}$.

(D) The given equation is $xy + yz = 0$.
Factoring out $y$,we get $y(x + z) = 0$.
This equation is satisfied if $y = 0$ or $x + z = 0$.
In $3D$ space,$y = 0$ represents the $xz$-plane and $x + z = 0$ represents a plane passing through the $y$-axis.
The normal vectors to these planes are $\vec{n_1} = (0, 1, 0)$ and $\vec{n_2} = (1, 0, 1)$.
The dot product of the normal vectors is $\vec{n_1} \cdot \vec{n_2} = (0)(1) + (1)(0) + (0)(1) = 0$.
Since the dot product of the normal vectors is $0$,the planes are perpendicular.
Therefore,the locus represents a pair of perpendicular planes.

(D) Given $A = \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix}$.
First,calculate $A^2$:
$A^2 = \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} = \begin{bmatrix} 4+4 & -4-4 \\ -4-4 & 4+4 \end{bmatrix} = \begin{bmatrix} 8 & -8 \\ -8 & 8 \end{bmatrix} = 4 \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} = 2^2 A$.
Next,calculate $A^3$:
$A^3 = A^2 \cdot A = (2^2 A) \cdot A = 2^2 A^2 = 2^2 (2^2 A) = 2^4 A$.
By observing the pattern:
$A^1 = 2^0 A$
$A^2 = 2^2 A$
$A^3 = 2^4 A$
$A^4 = 2^6 A$
In general,$A^n = 2^{2(n-1)} A$.
Comparing this with $A^n = 2^k A$,we get $k = 2(n-1)$.

(D) Given the matrix equation: $\left[\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}2 \\ 4\end{array}\right]$.
Performing matrix multiplication on the left side,we get:
$\left[\begin{array}{c}1(x) + 1(y) \\ -1(x) + 1(y)\end{array}\right] = \left[\begin{array}{c}2 \\ 4\end{array}\right]$
$\left[\begin{array}{c}x+y \\ -x+y\end{array}\right] = \left[\begin{array}{c}2 \\ 4\end{array}\right]$
By equating the corresponding elements,we obtain the system of linear equations:
$x + y = 2$ (Equation $1$)
$-x + y = 4$ (Equation $2$)
Adding Equation $1$ and Equation $2$:
$(x + y) + (-x + y) = 2 + 4$
$2y = 6 \implies y = 3$
Substituting $y = 3$ into Equation $1$:
$x + 3 = 2 \implies x = -1$
Thus,the values are $x = -1$ and $y = 3$.

(A) Given that $ A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} $.
Then the transpose of $ A $ is $ A^{\prime} = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} $.
Now,calculating the product $ A A^{\prime} $:
$ A A^{\prime} = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} $
$ = \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & -\cos \alpha \sin \alpha + \sin \alpha \cos \alpha \\ -\sin \alpha \cos \alpha + \cos \alpha \sin \alpha & \sin^2 \alpha + \cos^2 \alpha \end{bmatrix} $
Using the trigonometric identity $ \cos^2 \alpha + \sin^2 \alpha = 1 $:
$ = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I $.

(D) Let the given determinant be $\Delta$.
Using the identity $(a+b)^2 - (a-b)^2 = 4ab$,we observe that for any term of the form $(n^x + n^{-x})^2 - (n^x - n^{-x})^2$,the result is $4(n^x)(n^{-x}) = 4(n^0) = 4$.
Applying the column operation $C_1 \rightarrow C_1 - C_2$,the first column becomes:
$C_1 = \begin{bmatrix} (5^x + 5^{-x})^2 - (5^x - 5^{-x})^2 \\ (6^x + 6^{-x})^2 - (6^x - 6^{-x})^2 \\ (7^x + 7^{-x})^2 - (7^x - 7^{-x})^2 \end{bmatrix} = \begin{bmatrix} 4 \\ 4 \\ 4 \end{bmatrix}$.
Now the determinant is $\left|\begin{array}{ccc} 4 & (5^x - 5^{-x})^2 & 1 \\ 4 & (6^x - 6^{-x})^2 & 1 \\ 4 & (7^x - 7^{-x})^2 & 1 \end{array}\right|$.
Since the first column consists of identical elements $(4)$,we can factor out $4$ from the determinant:
$4 \left|\begin{array}{ccc} 1 & (5^x - 5^{-x})^2 & 1 \\ 1 & (6^x - 6^{-x})^2 & 1 \\ 1 & (7^x - 7^{-x})^2 & 1 \end{array}\right|$.
Since column $1$ and column $3$ are identical,the value of the determinant is $4 \times 0 = 0$.

(C) Let $ \Delta = \left|\begin{array}{ccc}a-b & b+c & a \\ b-c & c+a & b \\ c-a & a+b & c\end{array}\right| $.
Applying the column operation $ C_{3} \rightarrow C_{3} + C_{2} $:
$ \Delta = \left|\begin{array}{ccc}a-b & b+c & a+b+c \\ b-c & c+a & a+b+c \\ c-a & a+b & a+b+c\end{array}\right| $
Taking $ (a+b+c) $ common from $ C_{3} $:
$ \Delta = (a+b+c) \left|\begin{array}{ccc}a-b & b+c & 1 \\ b-c & c+a & 1 \\ c-a & a+b & 1\end{array}\right| $
Applying $ R_{1} \rightarrow R_{1} - R_{2} $ and $ R_{2} \rightarrow R_{2} - R_{3} $:
$ \Delta = (a+b+c) \left|\begin{array}{ccc}a-2b+c & b-a & 0 \\ b-2c+a & c-b & 0 \\ c-a & a+b & 1\end{array}\right| $
Expanding along $ C_{3} $:
$ \Delta = (a+b+c) [ (a-2b+c)(c-b) - (b-a)(b-2c+a) ] $
$ \Delta = (a+b+c) [ (ac - ab - 2bc + 2b^{2} + c^{2} - bc) - (b^{2} - 2bc + ab - ab + 2ac - a^{2}) ] $
$ \Delta = (a+b+c) [ ac - ab - 3bc + 2b^{2} + c^{2} - b^{2} + 2bc - 2ac + a^{2} ] $
$ \Delta = (a+b+c) [ a^{2} + b^{2} + c^{2} - ab - bc - ac ] $
$ \Delta = a^{3} + b^{3} + c^{3} - 3abc $.

(C) The area of a triangle with vertices $(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3})$ is given by the formula:
$\frac{1}{2} \left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right| = k$
This implies that $\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right| = 2k$.
Now,consider the given determinant:
$D = \left|\begin{array}{lll}x_{1} & y_{1} & 4 \\ x_{2} & y_{2} & 4 \\ x_{3} & y_{3} & 4\end{array}\right|$
Taking the constant $4$ common from the third column,we get:
$D = 4 \left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right| = 4(2k) = 8k$.
Therefore,the square of the determinant is:
$D^{2} = (8k)^{2} = 64k^{2}$.

(B) We know that for any square matrix $A$ of order $n \times n$,the property of the determinant is given by $|kA| = k^n|A|$.
Here,the order of the matrix $A$ is $n = 3$ and the scalar constant is $k = 5$.
Substituting these values into the formula:
$|5A| = 5^3 |A|$
$|5A| = 125|A|$
Therefore,the correct option is $B$.

(B) Let $ \theta = \cos ^{-1} \left(\frac{2}{\sqrt{5}}\right) $. Then $ \cos \theta = \frac{2}{\sqrt{5}} $.
We need to find $ \tan \left(\frac{\theta}{2}\right) $.
Using the half-angle formula,$ \tan \left(\frac{\theta}{2}\right) = \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} $.
Substituting the value of $ \cos \theta $:
$ \tan \left(\frac{\theta}{2}\right) = \sqrt{\frac{1-\frac{2}{\sqrt{5}}}{1+\frac{2}{\sqrt{5}}}} = \sqrt{\frac{\sqrt{5}-2}{\sqrt{5}+2}} $.
Rationalizing the denominator:
$ \sqrt{\frac{(\sqrt{5}-2)(\sqrt{5}-2)}{(\sqrt{5}+2)(\sqrt{5}-2)}} = \sqrt{\frac{(\sqrt{5}-2)^2}{5-4}} = \sqrt{(\sqrt{5}-2)^2} = \sqrt{5}-2 $.
Thus,the value is $ \sqrt{5}-2 $.

(B) Given that,$\sin^{-1} x + \cos^{-1} y = \frac{2\pi}{5} \quad (1)$
We know the fundamental identities: $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ and $\sin^{-1} y + \cos^{-1} y = \frac{\pi}{2}$.
From these,we can write $\sin^{-1} x = \frac{\pi}{2} - \cos^{-1} x$ and $\cos^{-1} y = \frac{\pi}{2} - \sin^{-1} y$.
Substituting these into equation $(1)$:
$(\frac{\pi}{2} - \cos^{-1} x) + (\frac{\pi}{2} - \sin^{-1} y) = \frac{2\pi}{5}$
$\pi - (\cos^{-1} x + \sin^{-1} y) = \frac{2\pi}{5}$
$\cos^{-1} x + \sin^{-1} y = \pi - \frac{2\pi}{5}$
$\cos^{-1} x + \sin^{-1} y = \frac{3\pi}{5}$

(C) Given functions are $f(x) = |x| + x$ and $g(x) = |x| - x$.
For $x < 0$,we have $|x| = -x$.
Thus,$g(x) = -x - x = -2x$.
Now,we need to find $(f \circ g)(x) = f(g(x))$.
Substituting $g(x) = -2x$ into $f(x)$,we get $f(-2x) = |-2x| + (-2x)$.
Since $x < 0$,$-2x > 0$,so $|-2x| = -2x$.
Therefore,$f(-2x) = -2x - 2x = -4x$.

(A) Given the piecewise function:
$f(x) = \begin{cases} 2x; & x > 3 \\ x^2; & 1 < x \leq 3 \\ 3x; & x \leq 1 \end{cases}$
To find $f(-1) + f(2) + f(4)$,we evaluate each term individually:
$1$. For $f(-1)$: Since $-1 \leq 1$,we use the third condition $f(x) = 3x$. Thus,$f(-1) = 3(-1) = -3$.
$2$. For $f(2)$: Since $1 < 2 \leq 3$,we use the second condition $f(x) = x^2$. Thus,$f(2) = (2)^2 = 4$.
$3$. For $f(4)$: Since $4 > 3$,we use the first condition $f(x) = 2x$. Thus,$f(4) = 2(4) = 8$.
Adding these values together:
$f(-1) + f(2) + f(4) = -3 + 4 + 8 = 9$.

(C) Let $ A $ be a set with $ n = 6 $ elements.
Total number of functions from $ A $ to $ A $ is given by $ n^{n} = 6^{6} $.
$A$ function is a bijection if it is both one-to-one and onto. For a finite set $ A $ with $ n $ elements,the number of bijections from $ A $ to $ A $ is $ n ! = 6 ! $.
The number of functions that are not bijections is equal to the total number of functions minus the number of bijections.
Therefore,the required number of functions is $ 6^{6} - 6 ! $.

(B) For the function $f(x)$ to be continuous at $x=0$,the left-hand limit $(LHL)$ must equal the right-hand limit $(RHL)$ at $x=0$.
First,calculate the $LHL$:
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}$
Multiply the numerator and denominator by the conjugate $(\sqrt{1+kx} + \sqrt{1-kx})$:
$= \lim_{x \to 0^-} \frac{(1+kx) - (1-kx)}{x(\sqrt{1+kx} + \sqrt{1-kx})} = \lim_{x \to 0^-} \frac{2kx}{x(\sqrt{1+kx} + \sqrt{1-kx})} = \lim_{x \to 0^-} \frac{2k}{\sqrt{1+kx} + \sqrt{1-kx}} = \frac{2k}{1+1} = k$.
Next,calculate the $RHL$:
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{2x+1}{x-1} = \frac{2(0)+1}{0-1} = -1$.
Since the function is continuous at $x=0$,we set $LHL$ = $RHL$:
$k = -1$.

(A) For a function $f(x)$ to be continuous at $x=a$,the limit $\lim_{x \to a} f(x)$ must exist and be equal to $f(a)$.
Given $f(x) = \frac{\log_{e} x}{x-1}$ for $x \neq 1$ and $f(1) = k$.
We need to evaluate $\lim_{x \to 1} \frac{\log_{e} x}{x-1}$.
Since this is a $\frac{0}{0}$ indeterminate form,we apply $L$'$H$ôpital's Rule:
$\lim_{x \to 1} \frac{\frac{d}{dx}(\log_{e} x)}{\frac{d}{dx}(x-1)} = \lim_{x \to 1} \frac{1/x}{1} = \frac{1}{1} = 1$.
Since the function is continuous at $x=1$,we must have $k = \lim_{x \to 1} f(x) = 1$.

(C) Given the function $f(x) = x - \frac{1}{x}$.
To find the derivative $f^{\prime}(x)$,we differentiate with respect to $x$:
$f^{\prime}(x) = \frac{d}{dx}(x) - \frac{d}{dx}(x^{-1})$
$f^{\prime}(x) = 1 - (-1)x^{-2} = 1 + \frac{1}{x^{2}}$.
Now,substitute $x = -1$ into the derivative:
$f^{\prime}(-1) = 1 + \frac{1}{(-1)^{2}}$
$f^{\prime}(-1) = 1 + \frac{1}{1} = 1 + 1 = 2$.

(B) Given the equation: $ \cos y = x \cos (a+y) $
We can express $ x $ as: $ x = \frac{\cos y}{\cos (a+y)} $
Differentiating both sides with respect to $ x $ using the quotient rule: $ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2} $
$ 1 = \frac{\cos (a+y) \cdot (-\sin y \frac{dy}{dx}) - \cos y \cdot (-\sin (a+y) \frac{dy}{dx})}{\cos^2 (a+y)} $
$ 1 = \frac{dy}{dx} \left[ \frac{\sin (a+y) \cos y - \cos (a+y) \sin y}{\cos^2 (a+y)} \right] $
Using the trigonometric identity $ \sin (A-B) = \sin A \cos B - \cos A \sin B $:
$ 1 = \frac{dy}{dx} \left[ \frac{\sin (a+y-y)}{\cos^2 (a+y)} \right] $
$ 1 = \frac{dy}{dx} \left[ \frac{\sin a}{\cos^2 (a+y)} \right] $
Therefore,$ \frac{dy}{dx} = \frac{\cos^2 (a+y)}{\sin a} $

(A) Given the function $f(x) = |\cos x - \sin x|$.
In the neighborhood of $x = \frac{\pi}{6}$,we have $\cos x > \sin x$,so $f(x) = \cos x - \sin x$.
Now,differentiate $f(x)$ with respect to $x$:
$f^{\prime}(x) = \frac{d}{dx}(\cos x - \sin x) = -\sin x - \cos x$.
Substitute $x = \frac{\pi}{6}$ into the derivative:
$f^{\prime}\left(\frac{\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) - \cos\left(\frac{\pi}{6}\right)$.
Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ and $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$,we get:
$f^{\prime}\left(\frac{\pi}{6}\right) = -\frac{1}{2} - \frac{\sqrt{3}}{2} = -\frac{1}{2}(1 + \sqrt{3})$.

(D) Given that,$ y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}} $
Since the expression under the square root repeats infinitely,we can write:
$ y=\sqrt{x+y} $
Squaring both sides,we get:
$ y^{2}=x+y $
Differentiating both sides with respect to $ x $:
$ \frac{d}{dx}(y^{2}) = \frac{d}{dx}(x+y) $
$ 2y \frac{dy}{dx} = 1 + \frac{dy}{dx} $
Rearranging the terms to solve for $ \frac{dy}{dx} $:
$ 2y \frac{dy}{dx} - \frac{dy}{dx} = 1 $
$ \frac{dy}{dx}(2y - 1) = 1 $
$ \frac{dy}{dx} = \frac{1}{2y - 1} $

(A) Let the side of the cube be $x$ meters.
The volume of the cube is given by $V = x^{3}$.
Differentiating with respect to $x$,we get $\frac{dV}{dx} = 3x^{2}$,which implies $dV = 3x^{2} dx$.
Given that the side increases by $3\%$,we have $\frac{dx}{x} \times 100 = 3$,which means $\frac{dx}{x} = 0.03$,or $dx = 0.03x$.
Substituting $dx$ into the expression for $dV$:
$dV = 3x^{2} (0.03x) = 0.09x^{3} \text{ m}^{3}$.
Thus,the approximate change in volume is $0.09x^{3} \text{ m}^{3}$.

(B) Given the function $f(x) = x^{x}$.
Taking the natural logarithm on both sides,we get $\ln f(x) = x \ln x$.
Differentiating both sides with respect to $x$ using the chain rule and product rule:
$\frac{1}{f(x)} f'(x) = \ln x + x \cdot \frac{1}{x} = \ln x + 1$.
Thus,$f'(x) = f(x)(\ln x + 1) = x^{x}(\ln x + 1)$.
$A$ stationary point occurs where $f'(x) = 0$.
Since $x^{x} > 0$ for $x > 0$,we set $\ln x + 1 = 0$.
$\ln x = -1$.
Converting from logarithmic to exponential form,$x = e^{-1} = \frac{1}{e}$.

(C) To evaluate the integral $ I = \int \frac{1}{1+e^{x}} d x $:
Multiply the numerator and denominator by $ e^{-x} $:
$ I = \int \frac{e^{-x}}{e^{-x}(1+e^{x})} d x $
$ I = \int \frac{e^{-x}}{e^{-x}+1} d x $
Let $ u = e^{-x} + 1 $. Then $ du = -e^{-x} d x $,which implies $ e^{-x} d x = -du $.
Substituting these into the integral:
$ I = \int \frac{-du}{u} $
$ I = -\ln |u| + C $
$ I = -\ln |e^{-x} + 1| + C $
Simplify the expression inside the logarithm:
$ I = -\ln \left| \frac{1}{e^{x}} + 1 \right| + C $
$ I = -\ln \left| \frac{1+e^{x}}{e^{x}} \right| + C $
Using the property $ -\ln(a/b) = \ln(b/a) $:
$ I = \ln \left| \frac{e^{x}}{1+e^{x}} \right| + C $
Thus,the correct option is $ C $.

(C) To evaluate the integral $ I = \int \frac{1}{\sqrt{3-6 x-9 x^{2}}} d x $,we first complete the square for the quadratic expression inside the square root.
$ 3-6 x-9 x^{2} = 3 - (9 x^{2} + 6 x) = 3 - ((3 x)^{2} + 2(3 x)(1) + 1^{2} - 1) = 3 - ((3 x+1)^{2} - 1) = 4 - (3 x+1)^{2} $.
Now,the integral becomes $ I = \int \frac{1}{\sqrt{4-(3 x+1)^{2}}} d x $.
Let $ u = 3x+1 $,then $ du = 3 dx $,which implies $ dx = \frac{du}{3} $.
Substituting these into the integral,we get $ I = \int \frac{1}{\sqrt{2^{2}-u^{2}}} \cdot \frac{du}{3} = \frac{1}{3} \int \frac{1}{\sqrt{2^{2}-u^{2}}} du $.
Using the standard formula $ \int \frac{1}{\sqrt{a^{2}-x^{2}}} dx = \sin^{-1}(\frac{x}{a}) + C $,we obtain $ I = \frac{1}{3} \sin^{-1}(\frac{u}{2}) + C $.
Substituting $ u = 3x+1 $ back,we get $ I = \frac{1}{3} \sin^{-1}(\frac{3x+1}{2}) + C $.

(A) Let $I = \int e^{\sin x} \cdot \left(\frac{\sin x+1}{\sec x}\right) d x$.
Since $\frac{1}{\sec x} = \cos x$,we can rewrite the integral as:
$I = \int e^{\sin x} (\sin x + 1) \cos x \, dx$.
Let $u = \sin x$,then $du = \cos x \, dx$.
Substituting these into the integral:
$I = \int e^u (u + 1) \, du$.
$I = \int (u e^u + e^u) \, du$.
Using the integration by parts formula $\int (f(u) + f'(u)) e^u \, du = e^u f(u) + C$,where $f(u) = u$ and $f'(u) = 1$:
$I = e^u \cdot u + C$.
Substituting $u = \sin x$ back:
$I = \sin x \cdot e^{\sin x} + C$.

(A) Let $ I = \int_{0}^{1/2} \frac{dx}{(1+x^{2}) \sqrt{1-x^{2}}} $.
Substitute $ x = \sin \theta $,then $ dx = \cos \theta d\theta $.
When $ x = 0, \theta = 0 $. When $ x = 1/2, \theta = \pi/6 $.
The integral becomes $ I = \int_{0}^{\pi/6} \frac{\cos \theta d\theta}{(1+\sin^{2} \theta) \sqrt{1-\sin^{2} \theta}} = \int_{0}^{\pi/6} \frac{\cos \theta d\theta}{(1+\sin^{2} \theta) \cos \theta} = \int_{0}^{\pi/6} \frac{d\theta}{1+\sin^{2} \theta} $.
Divide numerator and denominator by $ \cos^{2} \theta $: $ I = \int_{0}^{\pi/6} \frac{\sec^{2} \theta d\theta}{\sec^{2} \theta + \tan^{2} \theta} = \int_{0}^{\pi/6} \frac{\sec^{2} \theta d\theta}{1 + 2\tan^{2} \theta} $.
Let $ u = \sqrt{2} \tan \theta $,then $ du = \sqrt{2} \sec^{2} \theta d\theta $,so $ \sec^{2} \theta d\theta = \frac{du}{\sqrt{2}} $.
Limits: $ \theta = 0 \implies u = 0 $; $ \theta = \pi/6 \implies u = \sqrt{2} \tan(\pi/6) = \sqrt{2} \cdot \frac{1}{\sqrt{3}} = \sqrt{\frac{2}{3}} $.
Thus,$ I = \int_{0}^{\sqrt{2/3}} \frac{1}{1+u^{2}} \cdot \frac{du}{\sqrt{2}} = \frac{1}{\sqrt{2}} [\tan^{-1} u]_{0}^{\sqrt{2/3}} = \frac{1}{\sqrt{2}} \tan^{-1} \sqrt{\frac{2}{3}} $.

(B) To evaluate the integral $ I = \int_{0}^{1} \frac{dx}{e^{x}+e^{-x}} $,we first simplify the integrand by multiplying the numerator and denominator by $ e^{x} $:
$ I = \int_{0}^{1} \frac{e^{x}}{e^{2x}+1} dx $
Now,let $ u = e^{x} $. Then $ du = e^{x} dx $.
When $ x = 0 $,$ u = e^{0} = 1 $.
When $ x = 1 $,$ u = e^{1} = e $.
Substituting these into the integral,we get:
$ I = \int_{1}^{e} \frac{du}{u^{2}+1} $
The integral of $ \frac{1}{u^{2}+1} $ is $ \tan^{-1}(u) $.
Evaluating the definite integral:
$ I = [\tan^{-1}(u)]_{1}^{e} $
$ I = \tan^{-1}(e) - \tan^{-1}(1) $
Since $ \tan^{-1}(1) = \frac{\pi}{4} $,we have:
$ I = \tan^{-1}(e) - \frac{\pi}{4} $

(A) Let $I = \int_{-2}^{2} |x \cos \pi x| \, dx$. Since $f(x) = |x \cos \pi x|$ is an even function,we have $I = 2 \int_{0}^{2} |x \cos \pi x| \, dx$.
The function $x \cos \pi x$ changes sign at $x = \frac{1}{2}$ and $x = \frac{3}{2}$ in the interval $[0, 2]$.
Thus,$I = 2 \left[ \int_{0}^{1/2} x \cos \pi x \, dx - \int_{1/2}^{3/2} x \cos \pi x \, dx + \int_{3/2}^{2} x \cos \pi x \, dx \right]$.
Using integration by parts,$\int x \cos \pi x \, dx = \frac{x \sin \pi x}{\pi} + \frac{\cos \pi x}{\pi^2}$.
Evaluating the integrals:
$1$. $\int_{0}^{1/2} x \cos \pi x \, dx = \left[ \frac{x \sin \pi x}{\pi} + \frac{\cos \pi x}{\pi^2} \right]_{0}^{1/2} = (\frac{1}{2\pi} + 0) - (0 + \frac{1}{\pi^2}) = \frac{1}{2\pi} - \frac{1}{\pi^2}$.
$2$. $\int_{1/2}^{3/2} x \cos \pi x \, dx = \left[ \frac{x \sin \pi x}{\pi} + \frac{\cos \pi x}{\pi^2} \right]_{1/2}^{3/2} = (-\frac{3}{2\pi} + 0) - (\frac{1}{2\pi} + 0) = -\frac{2}{\pi}$.
$3$. $\int_{3/2}^{2} x \cos \pi x \, dx = \left[ \frac{x \sin \pi x}{\pi} + \frac{\cos \pi x}{\pi^2} \right]_{3/2}^{2} = (0 + \frac{1}{\pi^2}) - (-\frac{3}{2\pi} + 0) = \frac{3}{2\pi} + \frac{1}{\pi^2}$.
Substituting these values: $I = 2 [(\frac{1}{2\pi} - \frac{1}{\pi^2}) - (-\frac{2}{\pi}) + (\frac{3}{2\pi} + \frac{1}{\pi^2})] = 2 [\frac{1}{2\pi} + \frac{2}{\pi} + \frac{3}{2\pi}] = 2 [\frac{4}{\pi}] = \frac{8}{\pi}$.

(C) The area bounded by the curve $y = \cos x$ from $x = 0$ to $x = \pi$ is given by the integral of the absolute value of the function:
$\text{Area} = \int_{0}^{\pi} |\cos x| \, dx$
Since $\cos x \ge 0$ for $x \in [0, \pi/2]$ and $\cos x \le 0$ for $x \in [\pi/2, \pi]$,we split the integral:
$\text{Area} = \int_{0}^{\pi/2} \cos x \, dx + \int_{\pi/2}^{\pi} (-\cos x) \, dx$
Evaluating the integrals:
$= [\sin x]_{0}^{\pi/2} + [-\sin x]_{\pi/2}^{\pi}$
$= (\sin(\pi/2) - \sin(0)) + (-(\sin(\pi) - \sin(\pi/2)))$
$= (1 - 0) + (-(0 - 1))$
$= 1 + 1 = 2 \text{ sq. units.}$

(B) The area bounded by the curve $y=x$,the $x$-axis,and the lines $x=-1$ and $x=2$ is given by the integral of the absolute value of $y$ with respect to $x$:
$Area = \int_{-1}^{2} |y| \, dx = \int_{-1}^{2} |x| \, dx$
We split the integral at $x=0$ because the function changes sign:
$Area = \int_{-1}^{0} |x| \, dx + \int_{0}^{2} |x| \, dx$
Since $|x| = -x$ for $x < 0$ and $|x| = x$ for $x \ge 0$:
$Area = \int_{-1}^{0} (-x) \, dx + \int_{0}^{2} (x) \, dx$
$Area = \left[ -\frac{x^2}{2} \right]_{-1}^{0} + \left[ \frac{x^2}{2} \right]_{0}^{2}$
$Area = (0 - (-\frac{(-1)^2}{2})) + (\frac{2^2}{2} - 0)$
$Area = \frac{1}{2} + \frac{4}{2} = \frac{1}{2} + 2 = \frac{5}{2} \text{ sq. units.}$

(A) Given differential equation is: $\frac{d^{2} y}{d x^{2}}=\sqrt[3]{1+\left(\frac{d y}{d x}\right)^{2}}$
To find the degree,we must eliminate the radical. Cube both sides of the equation:
$\left(\frac{d^{2} y}{d x^{2}}\right)^{3} = 1 + \left(\frac{d y}{d x}\right)^{2}$
The order of a differential equation is the highest derivative present. Here,the highest derivative is $\frac{d^{2} y}{d x^{2}}$,so the order is $2$.
The degree of a differential equation is the power of the highest derivative when the equation is expressed as a polynomial in derivatives. Here,the power of $\frac{d^{2} y}{d x^{2}}$ is $3$,so the degree is $3$.
Therefore,the degree is $3$ and the order is $2$.

(A) Given differential equation is $x \frac{dy}{dx} - y = 3$.
Rearranging the terms,we get $x \frac{dy}{dx} = y + 3$.
Separating the variables,we have $\frac{dy}{y + 3} = \frac{dx}{x}$.
Integrating both sides,we get $\int \frac{dy}{y + 3} = \int \frac{dx}{x}$.
This results in $\ln|y + 3| = \ln|x| + \ln|c|$,where $c$ is the constant of integration.
Taking the exponential of both sides,we get $y + 3 = cx$,or $y = cx - 3$.
This is the equation of a family of straight lines passing through the point $(0, -3)$.

(C) Given the differential equation:
$\frac{dy}{dx} + y = \frac{1+y}{x}$
Rearranging the terms to the standard form $\frac{dy}{dx} + Py = Q$:
$\frac{dy}{dx} + y - \frac{y}{x} = \frac{1}{x}$
$\frac{dy}{dx} + y(1 - \frac{1}{x}) = \frac{1}{x}$
Here,$P = 1 - \frac{1}{x}$.
The integrating factor ($I$.$F$.) is given by:
$I.F. = e^{\int P dx} = e^{\int (1 - \frac{1}{x}) dx}$
$I.F. = e^{x - \ln|x|} = e^{x} \cdot e^{-\ln|x|} = e^{x} \cdot \frac{1}{x} = \frac{e^{x}}{x}$

(B) For three vectors to be coplanar,their scalar triple product must be zero.
Given the vectors $\vec{u} = a\hat{i}+\hat{j}+\hat{k}$,$\vec{v} = \hat{i}+b\hat{j}+\hat{k}$,and $\vec{w} = \hat{i}+\hat{j}+c\hat{k}$.
The condition for coplanarity is:
$\begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 0$
Expanding the determinant along the first row:
$a(bc - 1) - 1(c - 1) + 1(1 - b) = 0$
$abc - a - c + 1 + 1 - b = 0$
$abc - a - b - c + 2 = 0$
$abc - (a + b + c) = -2$

(A) Given vectors are $\vec{a}=\hat{i}+\lambda \hat{j}+2 \hat{k}$ and $\vec{b}=\mu \hat{i}+\hat{j}-\hat{k}$.
Since the vectors are orthogonal,their dot product is zero:
$\vec{a} \cdot \vec{b} = (1)(\mu) + (\lambda)(1) + (2)(-1) = 0$
$\mu + \lambda - 2 = 0 \Rightarrow \lambda + \mu = 2 \Rightarrow \mu = 2 - \lambda$ (Eq. $1$)
Given that $|\vec{a}| = |\vec{b}|$,we square both sides:
$|\vec{a}|^2 = |\vec{b}|^2$
$1^2 + \lambda^2 + 2^2 = \mu^2 + 1^2 + (-1)^2$
$1 + \lambda^2 + 4 = \mu^2 + 1 + 1$
$\lambda^2 + 3 = \mu^2$ (Eq. $2$)
Substitute $\mu = 2 - \lambda$ into Eq. $2$:
$\lambda^2 + 3 = (2 - \lambda)^2$
$\lambda^2 + 3 = 4 - 4\lambda + \lambda^2$
$3 = 4 - 4\lambda$
$4\lambda = 1 \Rightarrow \lambda = \frac{1}{4}$
Now,find $\mu$ using $\mu = 2 - \lambda$:
$\mu = 2 - \frac{1}{4} = \frac{7}{4}$
Thus,$(\lambda, \mu) = \left(\frac{1}{4}, \frac{7}{4}\right)$.

(C) Given that,$|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=144$ and $|\vec{a}|=4$.
We know the Lagrange's identity for vectors:
$|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=|\vec{a}|^{2} |\vec{b}|^{2} \sin^{2} \theta + |\vec{a}|^{2} |\vec{b}|^{2} \cos^{2} \theta$.
Factoring out $|\vec{a}|^{2} |\vec{b}|^{2}$,we get:
$|\vec{a}|^{2} |\vec{b}|^{2} (\sin^{2} \theta + \cos^{2} \theta) = |\vec{a}|^{2} |\vec{b}|^{2}$.
Therefore,$|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=|\vec{a}|^{2} |\vec{b}|^{2}$.
Substituting the given values:
$144 = (4)^{2} |\vec{b}|^{2}$.
$144 = 16 |\vec{b}|^{2}$.
$|\vec{b}|^{2} = \frac{144}{16} = 9$.
Taking the square root,we get $|\vec{b}| = 3$.

(A) Given that $\bar{a}$ and $\bar{b}$ are mutually perpendicular unit vectors.
This implies $|\bar{a}| = 1$,$|\bar{b}| = 1$,and $\bar{a} \cdot \bar{b} = 0$.
We need to evaluate the dot product:
$(3\bar{a} + 2\bar{b}) \cdot (5\bar{a} - 6\bar{b}) = 3\bar{a} \cdot (5\bar{a} - 6\bar{b}) + 2\bar{b} \cdot (5\bar{a} - 6\bar{b})$
$= 15(\bar{a} \cdot \bar{a}) - 18(\bar{a} \cdot \bar{b}) + 10(\bar{b} \cdot \bar{a}) - 12(\bar{b} \cdot \bar{b})$
$= 15|\bar{a}|^2 - 18(0) + 10(0) - 12|\bar{b}|^2$
$= 15(1)^2 - 12(1)^2$
$= 15 - 12 = 3$.

(A) Let the given line be $ \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} = \lambda $.
Any point $ R $ on the line is given by $ R(\lambda, 1+2\lambda, 2+3\lambda) $.
Let $ P $ be the point $ (1,6,3) $. The vector $ \vec{PR} $ is $ (\lambda-1, 2\lambda-5, 3\lambda-1) $.
Since $ PR $ is perpendicular to the line with direction ratios $ (1,2,3) $,their dot product is zero:
$ 1(\lambda-1) + 2(2\lambda-5) + 3(3\lambda-1) = 0 $
$ \lambda - 1 + 4\lambda - 10 + 9\lambda - 3 = 0 $
$ 14\lambda - 14 = 0 \Rightarrow \lambda = 1 $.
Substituting $ \lambda = 1 $ in $ R $,we get $ R(1, 3, 5) $.
Since $ R $ is the midpoint of $ PQ $,where $ Q(x_1, y_1, z_1) $ is the image of $ P $,
$ \frac{x_1+1}{2} = 1 \Rightarrow x_1 = 1 $
$ \frac{y_1+6}{2} = 3 \Rightarrow y_1 = 0 $
$ \frac{z_1+3}{2} = 5 \Rightarrow z_1 = 7 $.
Thus,the image $ Q $ is $ (1,0,7) $.

(C) The given lines are $2x = 3y = -z$ and $6x = -y = -4z$.
First,we write the equations in the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
For the first line $2x = 3y = -z$,dividing by $6$,we get $\frac{x}{3} = \frac{y}{2} = \frac{z}{-6}$. Thus,the direction ratios are $\vec{v_1} = (3, 2, -6)$.
For the second line $6x = -y = -4z$,dividing by $12$,we get $\frac{x}{2} = \frac{y}{-12} = \frac{z}{-3}$. Thus,the direction ratios are $\vec{v_2} = (2, -12, -3)$.
Two lines are perpendicular if the dot product of their direction vectors is zero,i.e.,$a_1a_2 + b_1b_2 + c_1c_2 = 0$.
Calculating the dot product: $(3)(2) + (2)(-12) + (-6)(-3) = 6 - 24 + 18 = 0$.
Since the dot product is $0$,the lines are perpendicular to each other.
Therefore,the angle between the lines is $90^{\circ}$.

(D) The given line is $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$.
Since the line lies on the plane $2x-4y+z=7$,any point on the line must satisfy the equation of the plane.
$A$ point on the line is $(4, 2, k)$.
Substituting this point into the plane equation $2x-4y+z=7$:
$2(4) - 4(2) + k = 7$
$8 - 8 + k = 7$
$k = 7$
Additionally,the direction vector of the line $(1, 1, 2)$ must be perpendicular to the normal vector of the plane $(2, -4, 1)$.
Checking the dot product: $(1)(2) + (1)(-4) + (2)(1) = 2 - 4 + 2 = 0$.
Since the dot product is $0$,the line is parallel to the plane. Since the point $(4, 2, 7)$ satisfies the plane equation,the entire line lies on the plane.

(A) The feasible region is a polygon with vertices at $(5, 5)$,$(0, 10)$,$(0, 20)$,and $(15, 15)$.
We evaluate the objective function $z = 3x + 9y$ at each vertex:
At $(5, 5)$: $z = 3(5) + 9(5) = 15 + 45 = 60$
At $(0, 10)$: $z = 3(0) + 9(10) = 0 + 90 = 90$
At $(0, 20)$: $z = 3(0) + 9(20) = 0 + 180 = 180$
At $(15, 15)$: $z = 3(15) + 9(15) = 45 + 135 = 180$
Comparing these values,the minimum value of $z$ is $60$,which occurs at the point $(5, 5)$.

(D) Given the objective function $z=x+4y$ and the constraints:
$1) x+2y \leq 2$
$2) x+2y \geq 8$
$3) x, y \geq 0$
Analyzing the constraints:
Constraint $(1)$ represents the region on or below the line $x+2y=2$,which passes through $(2,0)$ and $(0,1)$. Since $0+2(0) \leq 2$ is true,the region includes the origin.
Constraint $(2)$ represents the region on or above the line $x+2y=8$,which passes through $(8,0)$ and $(0,4)$. Since $0+2(0) \geq 8$ is false,the region does not include the origin.
Constraint $(3)$ restricts the solution to the first quadrant.
Comparing the regions defined by $(1)$ and $(2)$,we observe that the region satisfying $x+2y \leq 2$ and the region satisfying $x+2y \geq 8$ are disjoint. There is no point $(x, y)$ that satisfies both inequalities simultaneously.
Therefore,the $LPP$ has no feasible solution.

(A) The total number of tickets is $ 17 $.
The even-numbered tickets are $ 2, 4, 6, 8, 10, 12, 14, 16 $.
Thus,the total number of even-numbered tickets is $ 8 $.
Let $ E_1 $ be the event that the first ticket drawn is even,and $ E_2 $ be the event that the second ticket drawn is even.
The probability of drawing an even ticket first is $ P(E_1) = \frac{8}{17} $.
Since the ticket is not replaced,the remaining number of tickets is $ 16 $,and the number of remaining even tickets is $ 7 $.
The conditional probability of drawing an even ticket second is $ P(E_2|E_1) = \frac{7}{16} $.
The probability that both tickets show even numbers is $ P(E_1 \cap E_2) = P(E_1) \times P(E_2|E_1) = \frac{8}{17} \times \frac{7}{16} = \frac{7}{34} $.

(B) The mean $(\mu)$ of the probability distribution is given by $\mu = \sum x_{i} P_{i}$.
$\mu = (0 \times \frac{25}{36}) + (1 \times \frac{5}{18}) + (2 \times \frac{1}{36}) = 0 + \frac{10}{36} + \frac{2}{36} = \frac{12}{36} = \frac{1}{3}$.
The variance $(\sigma^2)$ is given by $\sigma^2 = \sum x_{i}^2 P_{i} - \mu^2$.
First,calculate $\sum x_{i}^2 P_{i} = (0^2 \times \frac{25}{36}) + (1^2 \times \frac{5}{18}) + (2^2 \times \frac{1}{36}) = 0 + \frac{10}{36} + \frac{4}{36} = \frac{14}{36} = \frac{7}{18}$.
Now,$\sigma^2 = \frac{7}{18} - (\frac{1}{3})^2 = \frac{7}{18} - \frac{1}{9} = \frac{7-2}{18} = \frac{5}{18}$.
Therefore,the standard deviation $\sigma = \sqrt{\frac{5}{18}} = \sqrt{\frac{5 \times 2}{18 \times 2}} = \sqrt{\frac{10}{36}} = \frac{\sqrt{10}}{6}$.
Alternatively,$\sigma = \sqrt{\frac{5}{18}} = \frac{1}{3} \sqrt{\frac{5}{2}}$.
Thus,option $B$ is correct.