KCET 2022 Mathematics Question Paper with Answer and Solution

(B) Given $f(x) = \begin{cases} x^2-1, & 0 < x < 2 \\ 2x+3, & 2 \leq x < 3 \end{cases}$.
First,we find the left-hand limit at $x=2$:
$\lim_{x \rightarrow 2^{-}} f(x) = \lim_{x \rightarrow 2^{-}} (x^2-1) = 2^2 - 1 = 3$.
Next,we find the right-hand limit at $x=2$:
$\lim_{x \rightarrow 2^{+}} f(x) = \lim_{x \rightarrow 2^{+}} (2x+3) = 2(2) + 3 = 7$.
The roots of the required quadratic equation are $\alpha = 3$ and $\beta = 7$.
The quadratic equation is given by $x^2 - (\alpha + \beta)x + (\alpha \times \beta) = 0$.
Substituting the values: $x^2 - (3 + 7)x + (3 \times 7) = 0$.
Thus,the equation is $x^2 - 10x + 21 = 0$.

(C) Given the equation: $3x + i(4x - y) = 6 - i$
On comparing the real and imaginary parts of the $LHS$ and $RHS$,we get:
$3x = 6$ and $4x - y = -1$
From $3x = 6$,we find $x = 2$.
Substituting $x = 2$ into the equation $4x - y = -1$:
$4(2) - y = -1$
$8 - y = -1$
$y = 8 + 1 = 9$
Therefore,the values are $x = 2$ and $y = 9$.

(B) The letters of the word $MASK$ in alphabetical order are $A, K, M, S$.
Total number of permutations is $4! = 24$.
Words starting with $A$: $3! = 6$ words ($1$ to $6$).
Words starting with $K$: $3! = 6$ words ($7$ to $12$).
Words starting with $M$: $3! = 6$ words ($13$ to $18$).
Words starting with $S$:
$SAKM$ ($19^{th}$ word),
$SA MK$ ($20^{th}$ word),
$SKAM$ ($21^{st}$ word),
$SKMA$ ($22^{nd}$ word),
$SMAK$ ($23^{rd}$ word),
$SMKA$ ($24^{th}$ word).
Thus,the $19^{th}$ word is $SAKM$.

(B) Given that $a_1, a_2, a_3, \ldots, a_{10}$ is a geometric progression $(GP)$ with first term $a$ and common ratio $r$.
We are given $\frac{a_3}{a_1} = 25$.
Since $a_n = ar^{n-1}$,we have $a_3 = ar^2$ and $a_1 = a$.
Thus,$\frac{ar^2}{a} = 25$,which implies $r^2 = 25$.
We need to find the value of $\frac{a_9}{a_5}$.
Using the formula for the $n$-th term,$\frac{a_9}{a_5} = \frac{ar^8}{ar^4} = r^4$.
Since $r^2 = 25$,we have $r^4 = (r^2)^2 = (25)^2 = (5^2)^2 = 5^4$.

(C) To convert radians to degrees,we use the formula: $\text{Degree} = \text{Radian} \times \frac{180^{\circ}}{\pi}$.
Given the angle is $\frac{\pi}{32}$ radians.
$\text{Degree measure} = \frac{\pi}{32} \times \frac{180^{\circ}}{\pi} = \frac{180}{32}^{\circ} = \frac{45}{8}^{\circ} = 5 \frac{5}{8}^{\circ}$.
Now,convert the fractional part to minutes: $5^{\circ} + (\frac{5}{8} \times 60)^{\prime} = 5^{\circ} + (\frac{300}{8})^{\prime} = 5^{\circ} + 37.5^{\prime} = 5^{\circ} 37^{\prime} + 0.5^{\prime}$.
Convert the remaining fractional part to seconds: $0.5^{\prime} = (0.5 \times 60)^{\prime \prime} = 30^{\prime \prime}$.
Thus,the final measure is $5^{\circ} 37^{\prime} 30^{\prime \prime}$.

(D) Given,function $y = \tan x$.
In the $II$ quadrant,the angle $x$ ranges from $\frac{\pi}{2}$ to $\pi$.
As $x$ approaches $\frac{\pi}{2}^+$,$\tan x$ approaches $-\infty$.
As $x$ approaches $\pi^-$,$\tan x$ approaches $0$.
Since the function values move from $-\infty$ to $0$ as $x$ increases,the function $y = \tan x$ increases from $-\infty$ to $0$ in the $II$ quadrant.

(D) Given expression is $\sin \frac{5 \pi}{12} \sin \frac{\pi}{12}$.
We use the trigonometric identity $2 \sin A \sin B = \cos(A - B) - \cos(A + B)$.
Multiplying and dividing by $2$,we get:
$\frac{1}{2} [2 \sin \frac{5 \pi}{12} \sin \frac{\pi}{12}]$
$= \frac{1}{2} [\cos(\frac{5 \pi}{12} - \frac{\pi}{12}) - \cos(\frac{5 \pi}{12} + \frac{\pi}{12})]$
$= \frac{1}{2} [\cos(\frac{4 \pi}{12}) - \cos(\frac{6 \pi}{12})]$
$= \frac{1}{2} [\cos(\frac{\pi}{3}) - \cos(\frac{\pi}{2})]$
Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$ and $\cos(\frac{\pi}{2}) = 0$,
$= \frac{1}{2} [\frac{1}{2} - 0] = \frac{1}{4}$.

(B) Let $y = \sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 \theta}}}$.
We know that $1 + \cos 2A = 2 \cos^2 A$.
Using this identity,we simplify the expression step by step:
$y = \sqrt{2+\sqrt{2+\sqrt{2(1 + \cos 8 \theta)}}}$
$y = \sqrt{2+\sqrt{2+\sqrt{2(2 \cos^2 4 \theta)}}}$
$y = \sqrt{2+\sqrt{2+\sqrt{4 \cos^2 4 \theta}}}$
$y = \sqrt{2+\sqrt{2+2 \cos 4 \theta}}$
$y = \sqrt{2+\sqrt{2(1 + \cos 4 \theta)}}$
$y = \sqrt{2+\sqrt{2(2 \cos^2 2 \theta)}}$
$y = \sqrt{2+\sqrt{4 \cos^2 2 \theta}}$
$y = \sqrt{2+2 \cos 2 \theta}$
$y = \sqrt{2(1 + \cos 2 \theta)}$
$y = \sqrt{2(2 \cos^2 \theta)}$
$y = \sqrt{4 \cos^2 \theta} = 2 \cos \theta$.

(B) The slope of the line passing through $(7, 17)$ and $(15, \beta)$ is given by $m_1 = \frac{\beta - 17}{15 - 7} = \frac{\beta - 17}{8}$.
The given line is $2x - 3y + 17 = 0$,which can be written as $3y = 2x + 17$,or $y = \frac{2}{3}x + \frac{17}{3}$.
The slope of this line is $m_2 = \frac{2}{3}$.
Since the two lines are perpendicular,the product of their slopes must be $-1$,i.e.,$m_1 \cdot m_2 = -1$.
Substituting the values,we get $\left(\frac{\beta - 17}{8}\right) \cdot \left(\frac{2}{3}\right) = -1$.
$\frac{\beta - 17}{12} = -1$.
$\beta - 17 = -12$.
$\beta = 17 - 12 = 5$.

(A) Let $L = \lim _{y \rightarrow 0} \frac{\sqrt{3+y^3}-\sqrt{3}}{y^3}$
Rationalizing the numerator:
$L = \lim _{y}$ ${\rightarrow 0} \frac{(\sqrt{3+y^3}-\sqrt{3})}{y^3} \times \frac{(\sqrt{3+y^3}+\sqrt{3})}{(\sqrt{3+y^3}+\sqrt{3})}$
$L = \lim _{y \rightarrow 0} \frac{3+y^3-3}{y^3(\sqrt{3+y^3}+\sqrt{3})}$
$L = \lim _{y \rightarrow 0} \frac{y^3}{y^3(\sqrt{3+y^3}+\sqrt{3})}$
$L = \lim _{y \rightarrow 0} \frac{1}{\sqrt{3+y^3}+\sqrt{3}}$
Substituting $y = 0$:
$L = \frac{1}{\sqrt{3+0}+\sqrt{3}} = \frac{1}{\sqrt{3}+\sqrt{3}} = \frac{1}{2 \sqrt{3}}$

(D) Given,the numbers are $-1, 0, 1, k$.
Standard deviation,$\sigma = \sqrt{5}$.
We know that the variance $\sigma^2$ is given by $\sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2$.
Here,$n = 4$.
$\sigma^2 = 5 = \frac{(-1)^2 + 0^2 + 1^2 + k^2}{4} - \left(\frac{-1 + 0 + 1 + k}{4}\right)^2$.
$5 = \frac{2 + k^2}{4} - \left(\frac{k}{4}\right)^2$.
$5 = \frac{2 + k^2}{4} - \frac{k^2}{16}$.
Multiply the entire equation by $16$:
$80 = 4(2 + k^2) - k^2$.
$80 = 8 + 4k^2 - k^2$.
$72 = 3k^2$.
$k^2 = 24$.
Since $k > 0$,$k = \sqrt{24} = 2 \sqrt{6}$.

(A) Given,$n(A)=p, n(B)=q$ and $n(A \times B)=7$.
Since,$n(A \times B)=n(A) \times n(B)$,we have $p \times q = 7$.
Since $p$ and $q$ are the number of elements in sets,they must be positive integers. The factors of $7$ are $1$ and $7$.
Thus,the possible values for $(p, q)$ are $(7, 1)$ or $(1, 7)$.
In both cases,$p^2+q^2 = 7^2 + 1^2 = 49 + 1 = 50$.

(A) Given,$A = \{1, 2, 3, \ldots, 10\}$.
The set of odd numbers in $A$ is $S = \{1, 3, 5, 7, 9\}$.
The number of elements in $S$ is $n = 5$.
The total number of subsets of $S$ is $2^n = 2^5 = 32$.
Since the question asks for subsets containing only odd numbers,we exclude the empty set (which contains no elements and thus no odd numbers).
Therefore,the number of non-empty subsets is $2^5 - 1 = 32 - 1 = 31$.

(D) Given the relation $3 a+2 b=27$ where $a, b \in N$ (natural numbers).
$2 b = 27 - 3 a$
$b = \frac{3(9 - a)}{2}$
Since $b$ must be a natural number,$3(9 - a)$ must be even and positive.
For $a = 1, b = \frac{3(8)}{2} = 12$.
For $a = 3, b = \frac{3(6)}{2} = 9$.
For $a = 5, b = \frac{3(4)}{2} = 6$.
For $a = 7, b = \frac{3(2)}{2} = 3$.
For $a = 9, b = 0$ (not a natural number).
Thus,$R = \{(1, 12), (3, 9), (5, 6), (7, 3)\}$.

(B) Given $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$. Note that $A^2 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$.
We use the Binomial Theorem for matrices. Since $I$ and $A$ commute $(IA = AI = A)$,we have:
$(aI + bA)^n = \sum_{k=0}^{n} \binom{n}{k} (aI)^{n-k} (bA)^k$.
Since $A^k = O$ for all $k \ge 2$,only the terms for $k=0$ and $k=1$ are non-zero:
$(aI + bA)^n = \binom{n}{0} (aI)^n (bA)^0 + \binom{n}{1} (aI)^{n-1} (bA)^1$.
$(aI + bA)^n = 1 \cdot a^n I \cdot I + n \cdot a^{n-1} I \cdot bA$.
$(aI + bA)^n = a^n I + n a^{n-1} b A$.

(D) Given,$A^T = -A$.
Let $P = A^{2021}$.
Then,$P^T = (A^{2021})^T = (A^T)^{2021}$.
Substituting $A^T = -A$,we get $P^T = (-A)^{2021} = (-1)^{2021} A^{2021}$.
Since $2021$ is an odd number,$(-1)^{2021} = -1$.
Therefore,$P^T = -A^{2021} = -P$.
Since $P^T = -P$,$A^{2021}$ is a skew symmetric matrix.

(C) Given that $A$ is a $3 \times 3$ matrix,so $n = 3$.
We know that for any $n \times n$ matrix $M$,$|kM| = k^n |M|$.
Applying this to $|5 \times \text{adj}(A)| = 5$,we get $5^3 |\text{adj}(A)| = 5$.
This simplifies to $|\text{adj}(A)| = \frac{5}{5^3} = \frac{1}{5^2} = \frac{1}{25}$.
We also know the property $|\text{adj}(A)| = |A|^{n-1}$.
For $n = 3$,$|\text{adj}(A)| = |A|^{3-1} = |A|^2$.
Equating the two expressions for $|\text{adj}(A)|$,we have $|A|^2 = \frac{1}{25}$.
Taking the square root on both sides,we get $|A| = \pm \frac{1}{5}$.

(B) We know that for any invertible matrix $A$,the inverse of a power of a matrix is given by $(A^n)^{-1} = (A^{-1})^n$.
Applying this property to the given expression:
$(A^2)^{-1} = (A^{-1})^2$.
Thus,the correct option is $(A^{-1})^2$.

(A) Given $A = \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix}$.
First,we calculate the determinant of $A$:
$|A| = (2)(-2) - (-1)(3) = -4 + 3 = -1$.
Since $|A| \neq 0$,$A$ is invertible.
Now,calculate $A^2$:
$A^2 = \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix} = \begin{bmatrix} 4-3 & -2+2 \\ 6-6 & -3+4 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$.
Since $A^2 = I$,we can find $A^3$:
$A^3 = A^2 \cdot A = I \cdot A = A$.
We need to find the inverse of $A^3$,which is $(A^3)^{-1}$.
Since $A^3 = A$,then $(A^3)^{-1} = A^{-1}$.
From $A^2 = I$,we know that $A \cdot A = I$,which implies $A^{-1} = A$.
Therefore,$(A^3)^{-1} = A$.

(A) Given the determinant equation: $\Delta=\left|\begin{array}{ccc}1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2a\end{array}\right|=86$
Expanding along the first row:
$1(a(2a) - (-1)(4)) - (-2)(2(2a) - (-1)(0)) + 5(2(4) - a(0)) = 86$
$1(2a^2 + 4) + 2(4a) + 5(8) = 86$
$2a^2 + 4 + 8a + 40 = 86$
$2a^2 + 8a + 44 = 86$
$2a^2 + 8a - 42 = 0$
Dividing by $2$:
$a^2 + 4a - 21 = 0$
This is a quadratic equation of the form $Aa^2 + Ba + C = 0$. The sum of the roots is given by $-\frac{B}{A}$.
Here,$A=1$ and $B=4$,so the sum of the values of '$a$' is $-\frac{4}{1} = -4$.

(B) Given vertices of the triangle are $(x_1, y_1) = (-2, 6)$,$(x_2, y_2) = (3, -6)$,and $(x_3, y_3) = (1, 5)$.
The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the determinant formula:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Substituting the values:
$\text{Area} = \frac{1}{2} |-2(-6 - 5) + 3(5 - 6) + 1(6 - (-6))|$
$\text{Area} = \frac{1}{2} |-2(-11) + 3(-1) + 1(12)|$
$\text{Area} = \frac{1}{2} |22 - 3 + 12|$
$\text{Area} = \frac{1}{2} |31| = 15.5 \text{ sq. units}$

(B) Given,$A_n = \begin{bmatrix} 1-n & n \\ n & 1-n \end{bmatrix}$.
The determinant $|A_n|$ is calculated as:
$|A_n| = (1-n)(1-n) - (n)(n)$
$|A_n| = 1 - 2n + n^2 - n^2 = 1 - 2n$.
Now,we need to find the sum $S = \sum_{n=1}^{2021} |A_n| = \sum_{n=1}^{2021} (1 - 2n)$.
This can be expanded as:
$S = (1-2) + (1-4) + (1-6) + \dots + (1 - 2 \times 2021)$
$S = (1 + 1 + \dots + 1) - 2(1 + 2 + 3 + \dots + 2021)$
There are $2021$ terms of $1$,so the first part is $2021$.
The second part is an arithmetic series sum: $2 \times \frac{2021(2021+1)}{2} = 2021 \times 2022$.
Thus,$S = 2021 - 2021 \times 2022$.
$S = 2021(1 - 2022) = 2021(-2021) = -(2021)^2$.

(D) For the function $f(x) = \frac{1}{\log_{10}(1-x)} + \sqrt{x+2}$ to be defined:
$1$. The term inside the square root must be non-negative: $x + 2 \geq 0 \implies x \geq -2$.
$2$. The argument of the logarithm must be positive: $1 - x > 0 \implies x < 1$.
$3$. The denominator cannot be zero: $\log_{10}(1-x) \neq 0 \implies 1 - x \neq 10^0 \implies 1 - x \neq 1 \implies x \neq 0$.
Combining these conditions: $x \geq -2$,$x < 1$,and $x \neq 0$.
Thus,the domain is $x \in [-2, 0) \cup (0, 1)$.

(D) The function is defined as $f(x) = \cos^{-1}[x]$.
We know that the domain of $\cos^{-1}(u)$ is $u \in [-1, 1]$.
Therefore,for $\cos^{-1}[x]$ to be defined,we must have $-1 \leq [x] \leq 1$.
Since $[x]$ is the greatest integer function,the possible integer values for $[x]$ are $-1, 0, 1$.
If $[x] = -1$,then $-1 \leq x < 0$.
If $[x] = 0$,then $0 \leq x < 1$.
If $[x] = 1$,then $1 \leq x < 2$.
Combining these intervals,we get $-1 \leq x < 2$.
Thus,the domain is $x \in [-1, 2)$.

(C) Given the 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 based on the given conditions:
$1$. For $f(-1)$,since $-1 \leq 1$,we use the third condition: $f(-1) = 3(-1) = -3$.
$2$. For $f(2)$,since $1 < 2 \leq 3$,we use the second condition: $f(2) = (2)^2 = 4$.
$3$. For $f(4)$,since $4 > 3$,we use the first condition: $f(4) = 2(4) = 8$.
Adding these values together: $f(-1) + f(2) + f(4) = -3 + 4 + 8 = 9$.

(A) bijection (one-to-one and onto function) exists between two sets if and only if they have the same number of elements (cardinality).
Since the number of elements in set $x$ is $7$ and the number of elements in set $y$ is $8$,the cardinalities are not equal.
Therefore,it is impossible to define a bijection from $x$ to $y$.
Thus,the number of bijections is $0$.

(A) Given,$x=e^{\theta} \sin \theta$ and $y=e^{\theta} \cos \theta$.
We differentiate both with respect to $\theta$:
$\frac{dx}{d\theta} = e^{\theta} \sin \theta + e^{\theta} \cos \theta = e^{\theta}(\sin \theta + \cos \theta)$
$\frac{dy}{d\theta} = e^{\theta} \cos \theta - e^{\theta} \sin \theta = e^{\theta}(\cos \theta - \sin \theta)$
Now,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{e^{\theta}(\cos \theta - \sin \theta)}{e^{\theta}(\cos \theta + \sin \theta)} = \frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta}$.
Dividing numerator and denominator by $\cos \theta$,we get $\frac{dy}{dx} = \frac{1 - \tan \theta}{1 + \tan \theta}$.
At $(x, y) = (1, 1)$,we have $x = e^{\theta} \sin \theta = 1$ and $y = e^{\theta} \cos \theta = 1$.
Dividing these,$\frac{y}{x} = \frac{e^{\theta} \cos \theta}{e^{\theta} \sin \theta} = \cot \theta = 1$,so $\tan \theta = 1$.
Substituting $\tan \theta = 1$ into the expression for $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0$.

(A) Given,$y = (1 + x^2) \tan^{-1} x - x$.
Applying the product rule for the first term and the power rule for the remaining terms:
$\frac{dy}{dx} = \frac{d}{dx} [(1 + x^2) \tan^{-1} x] - \frac{d}{dx} (x)$
$\frac{dy}{dx} = [(1 + x^2) \cdot \frac{d}{dx}(\tan^{-1} x) + \tan^{-1} x \cdot \frac{d}{dx}(1 + x^2)] - 1$
$\frac{dy}{dx} = [(1 + x^2) \cdot \frac{1}{1 + x^2} + \tan^{-1} x \cdot (2x)] - 1$
$\frac{dy}{dx} = [1 + 2x \tan^{-1} x] - 1$
$\frac{dy}{dx} = 2x \tan^{-1} x$.

(C) Given,$y=x^{\sin x}+(\sin x)^x$.
Let $u=x^{\sin x}$ and $v=(\sin x)^x$.
Then $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$.
For $u=x^{\sin x}$,taking log on both sides: $\log u = \sin x \log x$.
Differentiating with respect to $x$: $\frac{1}{u} \frac{du}{dx} = \sin x \cdot \frac{1}{x} + \cos x \log x$.
So,$\frac{du}{dx} = x^{\sin x} \left( \frac{\sin x}{x} + \cos x \log x \right)$.
For $v=(\sin x)^x$,taking log on both sides: $\log v = x \log(\sin x)$.
Differentiating with respect to $x$: $\frac{1}{v} \frac{dv}{dx} = x \cdot \frac{\cos x}{\sin x} + \log(\sin x) = x \cot x + \log(\sin x)$.
So,$\frac{dv}{dx} = (\sin x)^x (x \cot x + \log(\sin x))$.
At $x = \frac{\pi}{2}$:
$\frac{du}{dx} = (\frac{\pi}{2})^{\sin(\pi/2)} (\frac{\sin(\pi/2)}{\pi/2} + \cos(\pi/2) \log(\pi/2)) = (\frac{\pi}{2})^1 (\frac{1}{\pi/2} + 0) = \frac{\pi}{2} \cdot \frac{2}{\pi} = 1$.
$\frac{dv}{dx} = (\sin(\pi/2))^{\pi/2} (\frac{\pi}{2} \cot(\pi/2) + \log(\sin(\pi/2))) = (1)^{\pi/2} (\frac{\pi}{2} \cdot 0 + \log(1)) = 1 \cdot (0 + 0) = 0$.
Therefore,$\frac{dy}{dx} = 1 + 0 = 1$.

(A) Given,$y=e^{\sqrt{x \sqrt{x \sqrt{x \ldots}}}}, x>1$.
Taking natural logarithm on both sides:
$\log _e y = \sqrt{x \sqrt{x \sqrt{x \ldots}}} \log _e e = \sqrt{x \sqrt{x \sqrt{x \ldots}}}$.
Since the expression under the square root repeats,we can write:
$\log _e y = \sqrt{x \log _e y}$.
Squaring both sides:
$(\log _e y)^2 = x \log _e y$.
Dividing by $\log _e y$ (since $x>1$,$y>1$,so $\log _e y \neq 0$):
$\log _e y = x$.
Now,differentiating with respect to $x$:
$\frac{1}{y} \frac{d y}{d x} = 1 \implies \frac{d y}{d x} = y$.
Differentiating again with respect to $x$:
$\frac{d^2 y}{d x^2} = \frac{d y}{d x} = y$.
At $x = \log _e 3$,we have $\log _e y = \log _e 3$,which implies $y = 3$.
Therefore,$\frac{d^2 y}{d x^2} = y = 3$.

(D) Given equation is $e^y + xy = e$.
At $x = 0$,$e^y + 0 = e \Rightarrow e^y = e \Rightarrow y = 1$.
Differentiating $e^y + xy = e$ with respect to $x$:
$e^y \frac{dy}{dx} + x \frac{dy}{dx} + y = 0$
$\Rightarrow \frac{dy}{dx} (e^y + x) = -y$
$\Rightarrow \frac{dy}{dx} = -\frac{y}{e^y + x}$.
At $x = 0$ and $y = 1$:
$\frac{dy}{dx} = -\frac{1}{e^1 + 0} = -\frac{1}{e}$.
Now,differentiate $\frac{dy}{dx} (e^y + x) = -y$ with respect to $x$:
$\frac{d^2y}{dx^2} (e^y + x) + \frac{dy}{dx} (e^y \frac{dy}{dx} + 1) = -\frac{dy}{dx}$
$\Rightarrow \frac{d^2y}{dx^2} (e^y + x) + e^y \left(\frac{dy}{dx}\right)^2 + \frac{dy}{dx} = -\frac{dy}{dx}$
$\Rightarrow \frac{d^2y}{dx^2} (e^y + x) + e^y \left(\frac{dy}{dx}\right)^2 + 2 \frac{dy}{dx} = 0$.
Substituting $x = 0, y = 1, \frac{dy}{dx} = -\frac{1}{e}$:
$\frac{d^2y}{dx^2} (e + 0) + e \left(-\frac{1}{e}\right)^2 + 2 \left(-\frac{1}{e}\right) = 0$
$e \frac{d^2y}{dx^2} + e \left(\frac{1}{e^2}\right) - \frac{2}{e} = 0$
$e \frac{d^2y}{dx^2} + \frac{1}{e} - \frac{2}{e} = 0$
$e \frac{d^2y}{dx^2} = \frac{1}{e}$
$\frac{d^2y}{dx^2} = \frac{1}{e^2}$.
Thus,the ordered pair is $\left(-\frac{1}{e}, \frac{1}{e^2}\right)$.

(B) Let $y = f(f(f(x))) + (f(x))^2$.
Applying the chain rule to differentiate with respect to $x$:
$\frac{dy}{dx} = f^{\prime}(f(f(x))) \cdot f^{\prime}(f(x)) \cdot f^{\prime}(x) + 2f(x) \cdot f^{\prime}(x)$.
At $x = 1$:
$\frac{dy}{dx} = f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot f^{\prime}(1) + 2f(1) \cdot f^{\prime}(1)$.
Given $f(1) = 1$ and $f^{\prime}(1) = 3$:
$\frac{dy}{dx} = f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot 3 + 2(1)(3)$.
Since $f(1) = 1$,then $f(f(1)) = f(1) = 1$.
$\frac{dy}{dx} = f^{\prime}(1) \cdot f^{\prime}(1) \cdot 3 + 6$.
Substituting $f^{\prime}(1) = 3$:
$\frac{dy}{dx} = 3 \cdot 3 \cdot 3 + 6 = 27 + 6 = 33$.

(C) Given the curve equation is $\sqrt{x}+\sqrt{y}=6$.
Differentiating both sides with respect to $x$,we get:
$\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0$.
This implies $\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}$.
Since the tangent is equally inclined to the axes,its slope must be $\pm 1$.
Thus,$-\frac{\sqrt{y}}{\sqrt{x}} = \pm 1$,which gives $\sqrt{y} = \pm \sqrt{x}$.
Since $x$ and $y$ must be positive for the square roots to be defined,we have $\sqrt{y} = \sqrt{x}$,which implies $y = x$.
Substituting $y = x$ into the original equation:
$\sqrt{x} + \sqrt{x} = 6
2\sqrt{x} = 6
\sqrt{x} = 3
x = 9$.
Since $y = x$,we get $y = 9$.
Therefore,the required point is $(9, 9)$.

(C) Given,$f(x)=\log (1+x)-\frac{2 x}{2+x}$.
For the function to be defined,we must have $1+x > 0$,i.e.,$x > -1$.
Differentiating the function with respect to $x$,we get:
$f^{\prime}(x) = \frac{1}{1+x} - 2 \left[ \frac{(2+x)(1) - x(1)}{(2+x)^2} \right]$
$f^{\prime}(x) = \frac{1}{1+x} - 2 \left[ \frac{2+x-x}{(2+x)^2} \right] = \frac{1}{1+x} - \frac{4}{(2+x)^2}$
$f^{\prime}(x) = \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2} = \frac{4+x^2+4x-4-4x}{(1+x)(2+x)^2}$
$f^{\prime}(x) = \frac{x^2}{(1+x)(2+x)^2}$.
Since $x^2 \ge 0$ and $(2+x)^2 > 0$ for all $x > -1$,the sign of $f^{\prime}(x)$ depends on the term $\frac{1}{1+x}$.
For $f(x)$ to be increasing,we require $f^{\prime}(x) > 0$.
Since $x > -1$,$1+x > 0$,which implies $f^{\prime}(x) > 0$ for all $x \in (-1, \infty)$.
Thus,the function is increasing on $(-1, \infty)$.

(D) Given,$f(x)=4 \sin ^3 x-6 \sin ^2 x+12 \sin x+100$.
First,we find the derivative $f'(x)$:
$f'(x) = 12 \sin^2 x \cos x - 12 \sin x \cos x + 12 \cos x$.
Factoring out $12 \cos x$:
$f'(x) = 12 \cos x (\sin^2 x - \sin x + 1)$.
Consider the quadratic expression $g(t) = t^2 - t + 1$ where $t = \sin x$.
The discriminant $D = (-1)^2 - 4(1)(1) = 1 - 4 = -3 < 0$.
Since the coefficient of $t^2$ is positive and $D < 0$,the expression $\sin^2 x - \sin x + 1$ is always positive for all $x \in \mathbb{R}$.
Thus,the sign of $f'(x)$ depends only on $\cos x$.
$f'(x) > 0$ when $\cos x > 0$,which occurs in $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$.
$f'(x) < 0$ when $\cos x < 0$,which occurs in $x \in (\frac{\pi}{2}, \frac{3\pi}{2})$.
Therefore,$f(x)$ is strictly decreasing in the interval $(\frac{\pi}{2}, \pi)$.

(B) Given integral: $I = \int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} dx$
Using the identity $\cos 2\theta = 2\cos^2 \theta - 1$,we get:
$I = \int \frac{(2\cos^2 x - 1) - (2\cos^2 \alpha - 1)}{\cos x - \cos \alpha} dx$
$I = \int \frac{2\cos^2 x - 2\cos^2 \alpha}{\cos x - \cos \alpha} dx$
$I = 2 \int \frac{(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x - \cos \alpha} dx$
$I = 2 \int (\cos x + \cos \alpha) dx$
Integrating with respect to $x$:
$I = 2(\sin x + x \cos \alpha) + c$

(A) Given,$\int \frac{d x}{(x+2)\left(x^2+1\right)}=a \log \left|1+x^2\right|+b \tan ^{-1} x+\frac{1}{5} \log |x+2|+c$.
Let $\frac{1}{(x+2)\left(x^2+1\right)}=\frac{A}{(x+2)}+\frac{B x+C}{\left(x^2+1\right)}$.
Then $1=A(x^2+1)+(x+2)(Bx+C) = (A+B)x^2 + (2B+C)x + (A+2C)$.
Comparing coefficients,we get $A+B=0$,$2B+C=0$,and $A+2C=1$.
Solving these,we find $A=\frac{1}{5}$,$B=-\frac{1}{5}$,and $C=\frac{2}{5}$.
Thus,$\int \frac{d x}{(x+2)\left(x^2+1\right)} = \int \left( \frac{1}{5(x+2)} - \frac{x}{5(x^2+1)} + \frac{2}{5(x^2+1)} \right) dx$.
$= \frac{1}{5} \log |x+2| - \frac{1}{10} \log |x^2+1| + \frac{2}{5} \tan^{-1} x + c$.
Comparing this with the given expression,we get $a=-\frac{1}{10}$ and $b=\frac{2}{5}$.

(C) Let $I = \int_0^{\pi / 2} \sqrt{\sin \theta} \cos^3 \theta d \theta$.
Substitute $\sin \theta = t$,then $\cos \theta d \theta = dt$.
When $\theta = 0$,$t = 0$. When $\theta = \frac{\pi}{2}$,$t = 1$.
We can rewrite $\cos^3 \theta d \theta$ as $\cos^2 \theta (\cos \theta d \theta) = (1 - \sin^2 \theta) \cos \theta d \theta = (1 - t^2) dt$.
Thus,$I = \int_0^1 \sqrt{t} (1 - t^2) dt = \int_0^1 (t^{1/2} - t^{5/2}) dt$.
Integrating term by term: $I = \left[ \frac{t^{3/2}}{3/2} - \frac{t^{7/2}}{7/2} \right]_0^1$.
$I = \left[ \frac{2}{3} t^{3/2} - \frac{2}{7} t^{7/2} \right]_0^1$.
$I = \left( \frac{2}{3} - \frac{2}{7} \right) - (0 - 0) = \frac{14 - 6}{21} = \frac{8}{21}$.

(D) Let $I = \int_0^{\pi / 2} \frac{\cos x \sin x}{1+\sin x} d x$.
We can rewrite the numerator as $\cos x(1+\sin x - 1)$.
Then,$I = \int_0^{\pi / 2} \frac{\cos x(1+\sin x) - \cos x}{1+\sin x} d x$.
This simplifies to $I = \int_0^{\pi / 2} \cos x d x - \int_0^{\pi / 2} \frac{\cos x}{1+\sin x} d x$.
Evaluating the first integral: $\int_0^{\pi / 2} \cos x d x = [\sin x]_0^{\pi / 2} = \sin(\pi/2) - \sin(0) = 1 - 0 = 1$.
For the second integral,let $t = 1 + \sin x$,then $dt = \cos x dx$.
When $x = 0, t = 1$; when $x = \pi/2, t = 2$.
So,$\int_0^{\pi / 2} \frac{\cos x}{1+\sin x} d x = \int_1^2 \frac{dt}{t} = [\log |t|]_1^2 = \log 2 - \log 1 = \log 2$.
Therefore,$I = 1 - \log 2$.

(A) Let $I = \int_0^8 [x] dx$.
Since $[x]$ is the greatest integer function,it takes constant integer values on intervals $[n, n+1)$.
We can split the integral as:
$I = \int_0^1 0 dx + \int_1^2 1 dx + \int_2^3 2 dx + \int_3^4 3 dx + \int_4^5 4 dx + \int_5^6 5 dx + \int_6^7 6 dx + \int_7^8 7 dx$.
Evaluating each integral:
$I = 0 + 1(2-1) + 2(3-2) + 3(4-3) + 4(5-4) + 5(6-5) + 6(7-6) + 7(8-7)$.
$I = 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7$.
$I = \frac{7(7+1)}{2} = \frac{7 \times 8}{2} = 28$.

(D) We use the definition of the definite integral as the limit of a sum: $\int_a^b f(x) dx = \lim_{h \to 0} h \sum_{r=0}^{n-1} f(a+rh)$,where $nh = b-a$.
Here,$a=2$,$b=3$,and $f(x)=x^2$. Thus,$nh = 3-2 = 1$.
$I = \lim_{h \to 0} h \sum_{r=0}^{n-1} (2+rh)^2 = \lim_{h \to 0} h \sum_{r=0}^{n-1} (4 + 4rh + r^2h^2)$.
$I = \lim_{h \to 0} [4nh + 4h^2 \sum_{r=0}^{n-1} r + h^3 \sum_{r=0}^{n-1} r^2]$.
Using the formulas $\sum_{r=0}^{n-1} r = \frac{(n-1)n}{2}$ and $\sum_{r=0}^{n-1} r^2 = \frac{(n-1)n(2n-1)}{6}$:
$I = \lim_{h \to 0} [4nh + 4h^2 \frac{(n-1)n}{2} + h^3 \frac{(n-1)n(2n-1)}{6}]$.
Since $nh=1$,we have $I = \lim_{h \to 0} [4(1) + 2(nh-h)(nh) + \frac{(nh-h)(nh)(2nh-h)}{6}]$.
Substituting $nh=1$ and $h \to 0$:
$I = 4 + 2(1)(1) + \frac{(1)(1)(2)}{6} = 4 + 2 + \frac{1}{3} = 6 + \frac{1}{3} = \frac{19}{3}$.

(D) To evaluate $I = \int_0^1 \frac{x e^x}{(2+x)^3} d x$,we rewrite the integrand as follows:
$\frac{x}{(2+x)^3} = \frac{(x+2)-2}{(2+x)^3} = \frac{1}{(2+x)^2} - \frac{2}{(2+x)^3}$.
Let $f(x) = \frac{1}{(2+x)^2}$. Then $f'(x) = -2(2+x)^{-3} = -\frac{2}{(2+x)^3}$.
Thus,the integral becomes $\int_0^1 e^x [f(x) + f'(x)] d x$.
Using the standard result $\int e^x [f(x) + f'(x)] d x = e^x f(x) + C$,we get:
$I = [e^x \cdot \frac{1}{(2+x)^2}]_0^1$.
Evaluating at the limits:
$I = (e^1 \cdot \frac{1}{(2+1)^2}) - (e^0 \cdot \frac{1}{(2+0)^2}) = \frac{e}{9} - \frac{1}{4}$.

(B) The required area is given by the integral of the function $y = \tan x$ from $x = 0$ to $x = \frac{\pi}{3}$.
$\text{Required area} = \int_0^{\pi / 3} \tan x \, dx$
We know that the integral of $\tan x$ is $\log |\sec x|$.
$\text{Required area} = [\log |\sec x|]_0^{\pi / 3}$
Now,substitute the limits:
$= \log |\sec \frac{\pi}{3}| - \log |\sec 0|$
Since $\sec \frac{\pi}{3} = 2$ and $\sec 0 = 1$:
$= \log |2| - \log |1|$
$= \log 2 - 0 = \log 2 \text{ sq units}$.

(C) Given differential equation is $(1+y_1^2)^{2/3} = y_2$.
To find the degree,we need to eliminate the fractional exponent.
Cube both sides of the equation:
$((1+y_1^2)^{2/3})^3 = (y_2)^3$
$(1+y_1^2)^2 = y_2^3$.
Here,the highest order derivative is $y_2$,which represents the second derivative $\frac{d^2y}{dx^2}$.
Therefore,the order of the differential equation is $2$.
The degree is the power of the highest order derivative after the equation is made free from radicals and fractions.
The power of $y_2$ is $3$,so the degree is $3$.
The sum of the degree and order is $2 + 3 = 5$.

(A) Given the differential equation: $\frac{dy}{dx} = (x+y)^2$ $(i)$
Let $x+y = t$.
Differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dt}{dx}$,which implies $\frac{dy}{dx} = \frac{dt}{dx} - 1$.
Substituting this into equation $(i)$:
$\frac{dt}{dx} - 1 = t^2$
$\frac{dt}{dx} = t^2 + 1$
Separating the variables:
$\frac{dt}{t^2 + 1} = dx$
Integrating both sides:
$\int \frac{dt}{t^2 + 1} = \int dx$
$\tan^{-1}(t) = x + C$
Substituting $t = x+y$ back:
$\tan^{-1}(x+y) = x + C$

(B) The given differential equation is $\frac{dy}{dx} + \frac{y}{x} = x^2$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{1}{x}$ and $Q(x) = x^2$.
The integrating factor $(IF)$ is given by $IF = e^{\int P(x) dx} = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$.
The general solution is $y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$.
Substituting the values,we get $y \cdot x = \int x^2 \cdot x dx + C = \int x^3 dx + C = \frac{x^4}{4} + C$.
Thus,$y = \frac{x^3}{4} + \frac{C}{x}$.
Now,$y(2) = \frac{2^3}{4} + \frac{C}{2} = 2 + \frac{C}{2}$.
And $y(1) = \frac{1^3}{4} + \frac{C}{1} = \frac{1}{4} + C$.
Calculating $2y(2) - y(1) = 2(2 + \frac{C}{2}) - (\frac{1}{4} + C) = 4 + C - \frac{1}{4} - C = 4 - \frac{1}{4} = \frac{15}{4}$.

(C) Given differential equation: $x \log x \frac{dy}{dx} + y = 2x \log x$.
Dividing by $x \log x$,we get: $\frac{dy}{dx} + \frac{1}{x \log x} y = 2$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x \log x}$ and $Q = 2$.
Integrating Factor $(IF)$ = $e^{\int P dx} = e^{\int \frac{1}{x \log x} dx} = e^{\log(\log x)} = \log x$.
The general solution is $y \cdot (IF) = \int Q \cdot (IF) dx + C$.
$y \log x = \int 2 \log x dx + C$.
$y \log x = 2(x \log x - x) + C$.
Since the domain of $\log x$ is $x > 0$ and the equation is defined for $x > 1$ (to avoid division by zero at $x=1$),we evaluate the constant. However,assuming the standard form holds,at $x=e$:
$y(e) \log e = 2(e \log e - e) + C$.
$y(e) = 2(e - e) + C = C$.
Assuming the initial condition $y(e)$ is well-defined,we find $y(e) = 2$.

(D) Given $\alpha=\hat{i}-3 \hat{j}$ and $\beta=\hat{i}+2 \hat{j}-\hat{k}$.
Since $\beta_1$ is parallel to $\alpha$,we can write $\beta_1 = \lambda \alpha = \lambda(\hat{i}-3 \hat{j}) = \lambda \hat{i} - 3\lambda \hat{j}$.
We have $\beta = \beta_1 + \beta_2$,so $\beta_2 = \beta - \beta_1 = (\hat{i} + 2 \hat{j} - \hat{k}) - (\lambda \hat{i} - 3\lambda \hat{j}) = (1-\lambda)\hat{i} + (2+3\lambda)\hat{j} - \hat{k}$.
Since $\beta_2$ is perpendicular to $\alpha$,their dot product is zero: $\beta_2 \cdot \alpha = 0$.
$((1-\lambda)\hat{i} + (2+3\lambda)\hat{j} - \hat{k}) \cdot (\hat{i} - 3\hat{j}) = 0$.
$(1-\lambda)(1) + (2+3\lambda)(-3) + (-1)(0) = 0$.
$1 - \lambda - 6 - 9\lambda = 0$.
$-5 - 10\lambda = 0 \Rightarrow \lambda = -\frac{1}{2}$.
Thus,$\beta_1 = -\frac{1}{2}(\hat{i} - 3\hat{j}) = -\frac{1}{2}\hat{i} + \frac{3}{2}\hat{j}$.
Comparing this with the given options,none of the options match the calculated result.

(NONE) We know that the dot product of two vectors is given by $a \cdot b = |a| |b| \cos \theta$.
Given $|a|=2$,$|b|=3$,and $\theta = 120^{\circ}$.
$a \cdot b = (2)(3) \cos 120^{\circ} = 6 \times (-1/2) = -3$.
Now,we need to find the magnitude $\left|\frac{a}{2} - \frac{b}{3}\right|$.
Let $X = \left|\frac{a}{2} - \frac{b}{3}\right|$. Then $X^2 = \left(\frac{a}{2} - \frac{b}{3}\right) \cdot \left(\frac{a}{2} - \frac{b}{3}\right)$.
$X^2 = \frac{1}{4}|a|^2 + \frac{1}{9}|b|^2 - 2 \left(\frac{1}{2} \cdot \frac{1}{3}\right) (a \cdot b)$.
$X^2 = \frac{1}{4}(2)^2 + \frac{1}{9}(3)^2 - \frac{1}{3} (a \cdot b)$.
$X^2 = \frac{1}{4}(4) + \frac{1}{9}(9) - \frac{1}{3}(-3)$.
$X^2 = 1 + 1 + 1 = 3$.
Therefore,$X = \sqrt{3}$.

(D) We are given the identity $|a \times b|^2 + |a \cdot b|^2 = |a|^2 |b|^2$.
Given that $|a \times b|^2 + |a \cdot b|^2 = 36$ and $|a| = 3$.
Substituting these values into the identity:
$|a|^2 |b|^2 = 36$
$(3)^2 |b|^2 = 36$
$9 |b|^2 = 36$
$|b|^2 = \frac{36}{9} = 4$
$|b| = \sqrt{4} = 2$.
Thus,the value of $|b|$ is $2$.

(A) The signs of the coordinates $(x, y, z)$ in the eight octants are given in the table below:
| Octant | $I$ | $II$ | $III$ | $IV$ | $V$ | $VI$ | $VII$ | $VIII$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| $x$ | $+$ | $-$ | $-$ | $+$ | $+$ | $-$ | $-$ | $+$ |
| $y$ | $+$ | $+$ | $-$ | $-$ | $+$ | $+$ | $-$ | $-$ |
| $z$ | $+$ | $+$ | $+$ | $+$ | $-$ | $-$ | $-$ | $-$ |
Given the point $(2, -4, -7)$,we have:
$x = 2$ (positive,$+$)
$y = -4$ (negative,$-$)
$z = -7$ (negative,$-$)
Looking at the table,the octant where $x$ is positive,$y$ is negative,and $z$ is negative is the $VIII$ octant.

(D) The direction ratios of the first line are $\vec{b_1} = \langle 3, 5, 4 \rangle$.
The direction ratios of the second line are $\vec{b_2} = \langle 1, 4, 2 \rangle$.
The angle $\theta$ between two lines with direction ratios $\langle a_1, b_1, c_1 \rangle$ and $\langle a_2, b_2, c_2 \rangle$ is given by $\cos \theta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \right|$.
Substituting the values:
$\cos \theta = \left| \frac{(3)(1) + (5)(4) + (4)(2)}{\sqrt{3^2 + 5^2 + 4^2} \sqrt{1^2 + 4^2 + 2^2}} \right|$
$\cos \theta = \left| \frac{3 + 20 + 8}{\sqrt{9 + 25 + 16} \sqrt{1 + 16 + 4}} \right|$
$\cos \theta = \left| \frac{31}{\sqrt{50} \sqrt{21}} \right| = \frac{31}{5 \sqrt{2} \sqrt{21}} = \frac{31}{5 \sqrt{42}}$.
Thus,$\theta = \cos^{-1} \left( \frac{31}{5 \sqrt{42}} \right)$.
Since this value is not present in the given options,the correct choice is $D$.

(C) Given,the equation of the plane is $2x - 3y + 4z = 29$.
Since $OP$ is perpendicular to the plane,the direction ratios of $OP$ are the same as the normal vector to the plane,which is $\langle 2, -3, 4 \rangle$.
The equation of the line $OP$ passing through the origin $(0, 0, 0)$ with direction ratios $\langle 2, -3, 4 \rangle$ is given by:
$\frac{x - 0}{2} = \frac{y - 0}{-3} = \frac{z - 0}{4} = \lambda$
Thus,the general coordinates of any point $P$ on the line are $(2\lambda, -3\lambda, 4\lambda)$.
Since point $P$ lies on the plane $2x - 3y + 4z = 29$,we substitute these coordinates into the plane equation:
$2(2\lambda) - 3(-3\lambda) + 4(4\lambda) = 29$
$4\lambda + 9\lambda + 16\lambda = 29$
$29\lambda = 29$
$\lambda = 1$
Substituting $\lambda = 1$ back into the coordinates of $P$,we get:
$P = (2(1), -3(1), 4(1)) = (2, -3, 4)$.
Therefore,the coordinates of the foot of the perpendicular are $(2, -3, 4)$.

(A) The equation of the plane is given by $r \cdot(\hat{i}-2 \hat{j}+4 \hat{k})=4$.
Comparing this with the standard form $r \cdot n = d$,we have the normal vector $n = \hat{i}-2 \hat{j}+4 \hat{k}$ and $d = 4$.
The position vector of the point is $a = 2 \hat{i}+\hat{j}-\hat{k}$.
The perpendicular distance $D$ of a point $a$ from the plane $r \cdot n = d$ is given by the formula $D = \frac{|a \cdot n - d|}{|n|}$.
First,calculate the dot product $a \cdot n$:
$a \cdot n = (2 \hat{i}+\hat{j}-\hat{k}) \cdot (\hat{i}-2 \hat{j}+4 \hat{k}) = (2)(1) + (1)(-2) + (-1)(4) = 2 - 2 - 4 = -4$.
Next,calculate the magnitude of the normal vector $|n|$:
$|n| = \sqrt{1^2 + (-2)^2 + 4^2} = \sqrt{1 + 4 + 16} = \sqrt{21}$.
Now,substitute these values into the distance formula:
$D = \frac{|-4 - 4|}{\sqrt{21}} = \frac{|-8|}{\sqrt{21}} = \frac{8}{\sqrt{21}}$.
Thus,the distance is $\frac{8}{\sqrt{21}}$.

(B) To find the minimum value of $z = 4x + 6y$,we evaluate $z$ at each corner point:
At $(0,2): z = 4(0) + 6(2) = 12$
At $(3,0): z = 4(3) + 6(0) = 12$
At $(6,0): z = 4(6) + 6(0) = 24$
At $(6,8): z = 4(6) + 6(8) = 24 + 48 = 72$
At $(0,5): z = 4(0) + 6(5) = 30$
Since the minimum value $12$ occurs at two corner points $(0,2)$ and $(3,0)$,the minimum value of $z$ occurs at every point on the line segment joining these two points.
Since a line segment contains an infinite number of points,the correct option is $B$.

(A) Let $x$ and $y$ be the number of packets of food $X$ and $Y$ respectively. The constraints are given by:
$12x + 3y \geq 240 \Rightarrow 4x + y \geq 80$
$4x + 20y \geq 460 \Rightarrow x + 5y \geq 115$
$6x + 4y \leq 300 \Rightarrow 3x + 2y \leq 150$
$x \geq 0, y \geq 0$
To find the corner points,we find the intersection of these lines:
$1$. Intersection of $4x + y = 80$ and $x + 5y = 115$: Solving these,we get $x = 15, y = 20$.
$2$. Intersection of $4x + y = 80$ and $3x + 2y = 150$: Solving these,we get $x = 2, y = 72$.
$3$. Intersection of $x + 5y = 115$ and $3x + 2y = 150$: Solving these,we get $x = 40, y = 15$.
Thus,the corner points of the feasible region are $(2,72), (40,15), (15,20)$.

(D) Given,$P(A)=\frac{1}{2}$,$P(B)=\frac{1}{2}$ and $P(A \mid B)=\frac{1}{4}$.
Using the conditional probability formula,$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
Substituting the values,$\frac{1}{4} = \frac{P(A \cap B)}{1/2}$.
Therefore,$P(A \cap B) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$.
By De Morgan's Law,$P(A^{\prime} \cap B^{\prime}) = P((A \cup B)^{\prime}) = 1 - P(A \cup B)$.
Using the addition theorem,$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{2} + \frac{1}{2} - \frac{1}{8} = 1 - \frac{1}{8} = \frac{7}{8}$.
Thus,$P(A^{\prime} \cap B^{\prime}) = 1 - \frac{7}{8} = \frac{1}{8}$.

(B) Given,$P(\bar{A})=0.75$,$P(A \cup B)=0.65$ and $P(B)=x$.
First,find $P(A)$: $P(A) = 1 - P(\bar{A}) = 1 - 0.75 = 0.25$.
Since $A$ and $B$ are independent events,$P(A \cap B) = P(A) \cdot P(B) = 0.25x$.
Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$:
$0.65 = 0.25 + x - 0.25x$.
$0.65 - 0.25 = x(1 - 0.25)$.
$0.40 = 0.75x$.
$x = \frac{0.40}{0.75} = \frac{40}{75} = \frac{8}{15}$.

(A) Let event $E_1$ be the event that there is a lockdown and $E_2$ be the event that there is no lockdown. Let $A$ be the event that the pandemic is controlled in one month.
Given probabilities are:
$P(E_1) = 0.7$
$P(E_2) = 1 - P(E_1) = 1 - 0.7 = 0.3$
$P(A \mid E_1) = 0.8$
$P(A \mid E_2) = 0.3$
Using the Law of Total Probability:
$P(A) = P(E_1) \times P(A \mid E_1) + P(E_2) \times P(A \mid E_2)$
$P(A) = (0.7 \times 0.8) + (0.3 \times 0.3)$
$P(A) = 0.56 + 0.09 = 0.65$
Thus,the probability that the pandemic will be controlled in one month is $0.65$.

(A) Given that three coins are tossed.
The sample space $S$ is given by:
$S = \{TTT, TTH, THT, HTT, HHT, HTH, THH, HHH\}$
Therefore,$n(S) = 8$.
Let $X$ represent the number of heads.
Then $X$ can take values $0, 1, 2, 3$.
The probability distribution of $X$ is:
| $X$ | $0$ | $1$ | $2$ | $3$ |
|---|---|---|---|---|
| $P(X)$ | $1/8$ | $3/8$ | $3/8$ | $1/8$ |
The mean $\mu$ is given by $\Sigma X_i P(X_i)$:
$\mu = 0 \times \left(\frac{1}{8}\right) + 1 \times \left(\frac{3}{8}\right) + 2 \times \left(\frac{3}{8}\right) + 3 \times \left(\frac{1}{8}\right)$
$\mu = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8}$
$\mu = \frac{12}{8} = 1.5$