KCET 2021 Mathematics Question Paper with Answer and Solution

(A) Given,$C(x) = 20x + 4000$ and $R(x) = 60x + 2000$.
To earn a profit,the revenue must be greater than the cost,i.e.,$R(x) - C(x) > 0$.
Substituting the given functions:
$(60x + 2000) - (20x + 4000) > 0$
$60x + 2000 - 20x - 4000 > 0$
$40x - 2000 > 0$
$40x > 2000$
$x > \frac{2000}{40}$
$x > 50$.
Thus,the number of items $x$ must be greater than $50$ to earn a profit.

(D) Given,$\left(\frac{1+i}{1-i}\right)^{x}=1$
Rationalizing the base:
$\left[\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right]^{x}=1$
$\left[\frac{1+i^2+2i}{1^2-i^2}\right]^{x}=1$
Since $i^2 = -1$:
$\left[\frac{1-1+2i}{1+1}\right]^{x}=1$
$\left[\frac{2i}{2}\right]^{x}=1$
$i^x = 1$
We know that $i^k = 1$ if and only if $k$ is a multiple of $4$.
Therefore,$x = 4n$ for $n \in N$.

(C) Given,total questions in part $A = 6$ and total questions in part $B = 7$.
To choose $10$ questions with at least $4$ from each part,the possible combinations are:
$1$. $4$ from part $A$ and $6$ from part $B$
$2$. $5$ from part $A$ and $5$ from part $B$
$3$. $6$ from part $A$ and $4$ from part $B$
Total ways = $({ }^{6}C_{4} \times { }^{7}C_{6}) + ({ }^{6}C_{5} \times { }^{7}C_{5}) + ({ }^{6}C_{6} \times { }^{7}C_{4})$
$= (15 \times 7) + (6 \times 21) + (1 \times 35)$
$= 105 + 126 + 35 = 266$

(A) Given,number of terms,$n = 51$.
Since $n$ is odd,the middle term is the $\left(\frac{n+1}{2}\right)$-th term.
$\text{Middle term} = \left(\frac{51+1}{2}\right) = 26\text{-th term}$.
Thus,$T_{26} = a + 25d = 300$.
The sum of the first $n$ terms of an $AP$ is given by $S_n = \frac{n}{2}(a + l)$,where $l$ is the last term.
Here,$l = T_{51} = a + 50d$.
$S_{51} = \frac{51}{2}(a + a + 50d) = \frac{51}{2}(2a + 50d) = 51(a + 25d)$.
Substituting $a + 25d = 300$,we get:
$S_{51} = 51 \times 300 = 15300$.

(A) Given expression: $\cos 1200^{\circ} + \tan 1485^{\circ}$
Using the property $\cos(n \times 360^{\circ} + \theta) = \cos \theta$ and $\tan(n \times 360^{\circ} + \theta) = \tan \theta$:
$\cos 1200^{\circ} = \cos(3 \times 360^{\circ} + 120^{\circ}) = \cos 120^{\circ}$
$\tan 1485^{\circ} = \tan(4 \times 360^{\circ} + 45^{\circ}) = \tan 45^{\circ}$
Now,calculate the values:
$\cos 120^{\circ} = \cos(180^{\circ} - 60^{\circ}) = -\cos 60^{\circ} = -\frac{1}{2}$
$\tan 45^{\circ} = 1$
Therefore,$\cos 1200^{\circ} + \tan 1485^{\circ} = -\frac{1}{2} + 1 = \frac{1}{2}$

(B) We are given the expression: $\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}$.
Using the property $\tan(90^{\circ} - \theta) = \cot \theta$,we can rewrite the terms from $46^{\circ}$ to $89^{\circ}$ as:
$\tan 89^{\circ} = \tan(90^{\circ} - 1^{\circ}) = \cot 1^{\circ}$
$\tan 88^{\circ} = \tan(90^{\circ} - 2^{\circ}) = \cot 2^{\circ}$
$\dots$
$\tan 46^{\circ} = \tan(90^{\circ} - 44^{\circ}) = \cot 44^{\circ}$.
Substituting these into the expression:
$= (\tan 1^{\circ} \cot 1^{\circ}) \times (\tan 2^{\circ} \cot 2^{\circ}) \times \dots \times (\tan 44^{\circ} \cot 44^{\circ}) \times \tan 45^{\circ}$.
Since $\tan \theta \cot \theta = 1$ and $\tan 45^{\circ} = 1$,the expression becomes:
$= 1 \times 1 \times \dots \times 1 \times 1 = 1$.

(B) Given point $(x_{1}, y_{1}) = (a \cos^{3} \theta, a \sin^{3} \theta)$.
Given line: $x \sec \theta + y \operatorname{cosec} \theta = a$.
Rewriting the given line in slope-intercept form:
$y \operatorname{cosec} \theta = -x \sec \theta + a$
$y = -\frac{\sec \theta}{\operatorname{cosec} \theta} x + \frac{a}{\operatorname{cosec} \theta}$
$y = -\frac{\sin \theta}{\cos \theta} x + a \sin \theta$.
Thus,the slope $m_{1} = -\frac{\sin \theta}{\cos \theta}$.
The slope $m$ of the line perpendicular to this line is $m = -\frac{1}{m_{1}} = \frac{\cos \theta}{\sin \theta}$.
The equation of the required line is $(y - y_{1}) = m(x - x_{1})$:
$y - a \sin^{3} \theta = \frac{\cos \theta}{\sin \theta} (x - a \cos^{3} \theta)$
$y \sin \theta - a \sin^{4} \theta = x \cos \theta - a \cos^{4} \theta$
$x \cos \theta - y \sin \theta = a \cos^{4} \theta - a \sin^{4} \theta$
$x \cos \theta - y \sin \theta = a (\cos^{2} \theta - \sin^{2} \theta) (\cos^{2} \theta + \sin^{2} \theta)$
Since $\cos^{2} \theta + \sin^{2} \theta = 1$ and $\cos^{2} \theta - \sin^{2} \theta = \cos 2 \theta$,we get:
$x \cos \theta - y \sin \theta = a \cos 2 \theta$.

(C) Given the parabola equation: $y = \alpha x^{2} - 6x + \beta \dots (i)$
Since the parabola passes through the point $(0, 2)$,we substitute $x = 0$ and $y = 2$ into equation $(i)$:
$2 = \alpha(0)^{2} - 6(0) + \beta \implies \beta = 2$
Now,differentiate equation $(i)$ with respect to $x$:
$\frac{dy}{dx} = 2\alpha x - 6$
The tangent at $x = \frac{3}{2}$ is parallel to the $X$-axis,which means the slope $\frac{dy}{dx} = 0$ at $x = \frac{3}{2}$:
$\left(\frac{dy}{dx}\right)_{x = 3/2} = 2\alpha \left(\frac{3}{2}\right) - 6 = 0$
$3\alpha - 6 = 0 \implies 3\alpha = 6 \implies \alpha = 2$
Thus,the values are $\alpha = 2$ and $\beta = 2$.

(B) For Statement $1$:
$\lim _{x \rightarrow 1} \frac{a x^{2}+b x+c}{c x^{2}+b x+a} = \frac{a(1)^{2}+b(1)+c}{c(1)^{2}+b(1)+a} = \frac{a+b+c}{a+b+c} = 1$.
Since $a+b+c \neq 0$,the limit is $1$. Thus,Statement $1$ is true.
For Statement $2$:
$\lim _{x \rightarrow -2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2} = \lim _{x \rightarrow -2} \frac{\frac{2+x}{2x}}{x+2} = \lim _{x \rightarrow -2} \frac{2+x}{2x(x+2)} = \lim _{x \rightarrow -2} \frac{1}{2x} = \frac{1}{2(-2)} = -\frac{1}{4}$.
Since $-\frac{1}{4} \neq \frac{1}{4}$,Statement $2$ is false.

(C) The given numbers are $31, 32, 33, \ldots, 47$.
Subtracting $30$ from each term,we get the sequence $1, 2, 3, \ldots, 17$.
The standard deviation remains unchanged when a constant is subtracted from each term.
The formula for the standard deviation of the first $n$ natural numbers is $SD = \sqrt{\frac{n^{2}-1}{12}}$.
Here,$n = 17$.
$SD = \sqrt{\frac{17^{2}-1}{12}} = \sqrt{\frac{289-1}{12}} = \sqrt{\frac{288}{12}}$.
$SD = \sqrt{24} = 2 \sqrt{6}$.

(D) Given,$n(A \times B) = 35$ and $B \subset A$.
Since $n(A \times B) = n(A) \times n(B) = 35$,the possible factors of $35$ are $(35, 1)$ or $(7, 5)$.
Since $B \subset A$ and $A, B$ are non-singleton sets,$n(A) > n(B) > 1$.
Thus,$n(A) = 7$ and $n(B) = 5$.
Now,calculating ${}^{n(A)}C_{n(B)} = {}^{7}C_{5}$.
Using the property ${}^{n}C_{r} = {}^{n}C_{n-r}$,we get ${}^{7}C_{5} = {}^{7}C_{2}$.
${}^{7}C_{2} = \frac{7 \times 6}{2 \times 1} = 21$.

(A) The area of an ellipse given by the equation $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is calculated as $\pi |ab|$.
Given the equation $\frac{x^{2}}{25}+\frac{y^{2}}{\lambda^{2}}=1$,we identify $a^{2} = 25$ and $b^{2} = \lambda^{2}$.
Thus,$a = 5$ and $b = |\lambda|$.
The area is given as $20 \pi$ sq units.
Substituting these values into the area formula: $\pi \times 5 \times |\lambda| = 20 \pi$.
Dividing both sides by $5 \pi$,we get $|\lambda| = 4$.
Therefore,$\lambda = \pm 4$.

(B) The centroid of a triangle formed by joining the midpoints of the sides of another triangle is the same as the centroid of the original triangle.
Let the midpoints be $M_1 = (1, 5, -1)$,$M_2 = (0, 4, -2)$,and $M_3 = (2, 3, 4)$.
The centroid $G(x, y, z)$ of the triangle formed by these midpoints is given by the average of their coordinates:
$x = \frac{1 + 0 + 2}{3} = \frac{3}{3} = 1$
$y = \frac{5 + 4 + 3}{3} = \frac{12}{3} = 4$
$z = \frac{-1 - 2 + 4}{3} = \frac{1}{3}$
Therefore,the centroid is $(1, 4, 1/3)$.

(C) $1$. Analyze the line $l_1$ passing through $(0, 5)$ and $(4, 0)$. The equation of this line in intercept form is $\frac{x}{4} + \frac{y}{5} = 1$,which simplifies to $5x + 4y = 20$. Since the shaded region does not contain the origin $(0, 0)$,the inequality is $5x + 4y \geq 20$.
$2$. Analyze the horizontal line $l_2$. The shaded region lies below the line $y = 3$,so the inequality is $y \leq 3$.
$3$. Analyze the vertical line $l_3$. The shaded region lies to the left of the line $x = 6$,so the inequality is $x \leq 6$.
$4$. The region is in the first quadrant,so $x \geq 0$ and $y \geq 0$.
$5$. Combining these,the system of inequalities is $5x + 4y \geq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0$. This matches option $C$.

(C) Given,$P(A)=0.59, P(B)=0.30$ and $P(A \cap B)=0.21$.
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 of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the values: $P(A \cup B) = 0.59 + 0.30 - 0.21 = 0.89 - 0.21 = 0.68$.
Therefore,$P(A^{\prime} \cap B^{\prime}) = 1 - 0.68 = 0.32$.

(B) Let the total number of families be $x$.
Let $A$ be the set of families that own cell phones,so $n(A) = \frac{65}{100}x$.
Let $B$ be the set of families that own scooters,so $n(B) = 15000$.
The number of families that own both is $n(A \cap B) = \frac{15}{100}x$.
Since every family owns at least one,$n(A \cup B) = x$.
Using the formula $n(A \cup B) = n(A) + n(B) - n(A \cap B)$,we have:
$x = \frac{65x}{100} + 15000 - \frac{15x}{100}$
$x = \frac{50x}{100} + 15000$
$x = 0.5x + 15000$
$0.5x = 15000$
$x = \frac{15000}{0.5} = 30000$.
Thus,the total number of families in the town is $30000$.

(A) Given $f(x) = \left| \begin{array}{ccc} \cos x & 1 & 0 \\ 0 & 2 \cos x & 3 \\ 0 & 1 & 2 \cos x \end{array} \right|$.
Expanding the determinant along the first column $(C_1)$:
$f(x) = \cos x \cdot \left| \begin{array}{cc} 2 \cos x & 3 \\ 1 & 2 \cos x \end{array} \right| - 0 + 0$
$f(x) = \cos x \cdot ((2 \cos x)(2 \cos x) - (3)(1))$
$f(x) = \cos x (4 \cos^2 x - 3)$
$f(x) = 4 \cos^3 x - 3 \cos x$
Using the trigonometric identity $\cos 3x = 4 \cos^3 x - 3 \cos x$,we get:
$f(x) = \cos 3x$
Now,calculating the limit:
$\lim_{x \rightarrow \pi} f(x) = \lim_{x \rightarrow \pi} \cos 3x = \cos(3\pi)$
Since $\cos(3\pi) = -1$,the limit is $-1$.

(A) Given the equation $x^{3}-2x^{2}-9x+18=0$.
Factoring the equation: $x^{2}(x-2)-9(x-2)=0 \Rightarrow (x^{2}-9)(x-2)=0 \Rightarrow (x-3)(x+3)(x-2)=0$.
Thus,the possible values for $x$ are $x=2, 3, -3$.
Now,evaluate the determinant $A = \left|\begin{array}{lll}1 & 2 & 3 \\ 4 & x & 6 \\ 7 & 8 & 9\end{array}\right|$.
Expanding along the first row: $A = 1(9x - 48) - 2(36 - 42) + 3(32 - 7x)$.
$A = 9x - 48 - 2(-6) + 96 - 21x$.
$A = 9x - 48 + 12 + 96 - 21x = -12x + 60$.
Now,substitute the values of $x$:
For $x=2$: $A = -12(2) + 60 = -24 + 60 = 36$.
For $x=3$: $A = -12(3) + 60 = -36 + 60 = 24$.
For $x=-3$: $A = -12(-3) + 60 = 36 + 60 = 96$.
Comparing the values $36, 24, 96$,the maximum value of $A$ is $96$.

(B) Given,$A=\left[\begin{array}{ccc}1 & -2 & 1 \\ 2 & 1 & 3\end{array}\right]$ and $B=\left[\begin{array}{cc}2 & 1 \\ 3 & 2 \\ 1 & 1\end{array}\right]$.
First,calculate the product $AB$:
$AB = \left[\begin{array}{ccc}1 & -2 & 1 \\ 2 & 1 & 3\end{array}\right] \left[\begin{array}{cc}2 & 1 \\ 3 & 2 \\ 1 & 1\end{array}\right]$
$AB = \left[\begin{array}{cc}(1)(2) + (-2)(3) + (1)(1) & (1)(1) + (-2)(2) + (1)(1) \\ (2)(2) + (1)(3) + (3)(1) & (2)(1) + (1)(2) + (3)(1)\end{array}\right]$
$AB = \left[\begin{array}{cc}2 - 6 + 1 & 1 - 4 + 1 \\ 4 + 3 + 3 & 2 + 2 + 3\end{array}\right]$
$AB = \left[\begin{array}{cc}-3 & -2 \\ 10 & 7\end{array}\right]$
Now,find the transpose $(AB)^{\prime}$ by interchanging rows and columns:
$(AB)^{\prime} = \left[\begin{array}{cc}-3 & 10 \\ -2 & 7\end{array}\right]$
Thus,the correct option is $B$.

(B) Given that $A$ and $B$ are square matrices of order $n=3$.
We are given $|A|=5$ and $|B|=3$.
Using the property of determinants,$|AB| = |A| \cdot |B| = 5 \times 3 = 15$.
Using the property $|kA| = k^n |A|$,where $k$ is a scalar and $n$ is the order of the square matrix,we have:
$|3AB| = 3^3 |AB|$.
Since $n=3$,$3^3 = 27$.
Therefore,$|3AB| = 27 \times 15 = 405$.

(D) Let $M$ be a symmetric matrix of the form $M = \begin{bmatrix} a & c \\ c & b \end{bmatrix}$,where $a, b, c \in \mathbb{Z}$.
For a matrix to be invertible,its determinant must be non-zero.
The determinant of $M$ is given by $|M| = ab - c^2$.
For $M$ to be invertible,we require $|M| \neq 0$,which implies $ab - c^2 \neq 0$,or $ab \neq c^2$.
In a symmetric matrix,the entries in the other diagonal are both $c$,so their product is $c^2$.
Thus,the condition for invertibility is that the product of the principal diagonal entries $(ab)$ is not equal to the product of the other diagonal entries $(c^2)$.

(D) For invertible matrices $A$ and $B$,the following properties hold:
$1$. The inverse of a product is the product of inverses in reverse order: $(AB)^{-1} = B^{-1} A^{-1}$.
$2$. The adjoint of a matrix is related to its inverse by: $\operatorname{adj} A = |A| A^{-1}$.
$3$. The determinant of an inverse matrix is the reciprocal of the determinant of the original matrix: $\operatorname{det}(A^{-1}) = [\operatorname{det}(A)]^{-1}$.
$4$. The inverse of a sum is generally not equal to the sum of the inverses: $(A+B)^{-1} \neq A^{-1} + B^{-1}$.
Therefore,the statement $(A+B)^{-1} = B^{-1} + A^{-1}$ is incorrect.

(D) Given expression: $\cos \left[\cot ^{-1}(-\sqrt{3})+\frac{\pi}{6}\right]$
We know that $\cot ^{-1}(-x) = \pi - \cot ^{-1}(x)$.
Therefore,$\cot ^{-1}(-\sqrt{3}) = \pi - \cot ^{-1}(\sqrt{3})$.
Since $\cot \left(\frac{\pi}{6}\right) = \sqrt{3}$,we have $\cot ^{-1}(\sqrt{3}) = \frac{\pi}{6}$.
Substituting this into the expression:
$\cos \left[\pi - \frac{\pi}{6} + \frac{\pi}{6}\right]$
$= \cos [\pi]$
$= -1$.

(C) Let the expression be $E = \tan ^{-1}\left[\frac{1}{\sqrt{3}} \sin \frac{5 \pi}{2}\right] + \sin ^{-1}\left[\cos \left(\sin ^{-1} \frac{\sqrt{3}}{2}\right)\right]$.
First,evaluate $\sin \frac{5 \pi}{2} = \sin \left(2 \pi + \frac{\pi}{2}\right) = \sin \frac{\pi}{2} = 1$.
Next,evaluate $\sin ^{-1} \frac{\sqrt{3}}{2} = \frac{\pi}{3}$.
Now substitute these values into the expression:
$E = \tan ^{-1}\left[\frac{1}{\sqrt{3}} \times 1\right] + \sin ^{-1}\left[\cos \frac{\pi}{3}\right]$.
Since $\cos \frac{\pi}{3} = \frac{1}{2}$,we have:
$E = \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right) + \sin ^{-1}\left(\frac{1}{2}\right)$.
We know that $\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$ and $\sin ^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$.
Therefore,$E = \frac{\pi}{6} + \frac{\pi}{6} = \frac{2 \pi}{6} = \frac{\pi}{3}$.

(D) Given function,$f(x) = \frac{x}{1-|x|}$.
For the function to be defined,the denominator must not be zero.
$1 - |x| \neq 0$
$|x| \neq 1$
$x \neq 1$ and $x \neq -1$.
Therefore,the domain is all real numbers except $1$ and $-1$,which is written as $R - \{-1, 1\}$.

(A) For the function $f(x) = \frac{1}{\sqrt{[x]^2 - [x] - 6}}$ to be defined,the expression inside the square root must be strictly positive:
$[x]^2 - [x] - 6 > 0$
Factoring the quadratic expression:
$([x] - 3)([x] + 2) > 0$
This inequality holds when:
$[x] > 3$ or $[x] < -2$
If $[x] > 3$,then the smallest integer value for $[x]$ is $4$,which implies $x \geq 4$,or $x \in [4, \infty)$.
If $[x] < -2$,then the largest integer value for $[x]$ is $-3$,which implies $x < -2$,or $x \in (-\infty, -2)$.
Combining these,the domain is $x \in (-\infty, -2) \cup [4, \infty)$.

(A) Given the function $f(x) = \frac{2x}{x-1}$ where $A = \{x \in R : x \neq 1, 2, 3, \dots\}$.
To check for injectivity,we find the derivative $f'(x)$:
$f'(x) = \frac{(x-1)(2) - 2x(1)}{(x-1)^2} = \frac{2x - 2 - 2x}{(x-1)^2} = \frac{-2}{(x-1)^2}$.
Since $f'(x) < 0$ for all $x \in A$,the function is strictly decreasing,which implies $f$ is injective.
To check for surjectivity,let $y = \frac{2x}{x-1}$.
$y(x-1) = 2x \implies yx - y = 2x \implies x(y-2) = y \implies x = \frac{y}{y-2}$.
For $f$ to be surjective,for every $y \in R$,there must exist an $x \in A$. If $y = 2$,$x$ is undefined. Furthermore,we must ensure $x$ is not a positive integer. If $y = 0$,$x = 0$,which is allowed. However,if we choose $y$ such that $x$ becomes a positive integer (e.g.,if $x=2$,$y = \frac{2(2)}{2-1} = 4$),then $y=4$ has a pre-image $x=2$,but $2 \notin A$. Thus,$f$ is not surjective.

(A) Given function,$f(x) = \sqrt{3} \sin 2x - \cos 2x + 4$.
We can rewrite this as:
$f(x) = 2(\sin 2x \cdot \frac{\sqrt{3}}{2} - \cos 2x \cdot \frac{1}{2}) + 4$
$f(x) = 2(\sin 2x \cos \frac{\pi}{6} - \cos 2x \sin \frac{\pi}{6}) + 4$
$f(x) = 2 \sin(2x - \frac{\pi}{6}) + 4$.
$A$ function $f(x) = \sin(\theta)$ is one-one when the argument $\theta$ lies in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Therefore,for $f(x)$ to be one-one:
$-\frac{\pi}{2} \leq 2x - \frac{\pi}{6} \leq \frac{\pi}{2}$
Adding $\frac{\pi}{6}$ to all parts:
$-\frac{\pi}{2} + \frac{\pi}{6} \leq 2x \leq \frac{\pi}{2} + \frac{\pi}{6}$
$-\frac{2\pi}{6} \leq 2x \leq \frac{4\pi}{6}$
$-\frac{\pi}{3} \leq 2x \leq \frac{2\pi}{3}$
Dividing by $2$:
$-\frac{\pi}{6} \leq x \leq \frac{\pi}{3}$.
Thus,the function is one-one in the interval $[-\frac{\pi}{6}, \frac{\pi}{3}]$.

(D) Given the function $f(x) = \left\{\begin{array}{cc} 2x, & x > 3 \\ x^2, & 1 < x \leq 3 \\ 3x, & x \leq 1 \end{array}\right.$.
To find $f(-2) + f(3) + f(4)$,we calculate each term individually:
$1$. For $f(-2)$: Since $-2 \leq 1$,we use $f(x) = 3x$. Thus,$f(-2) = 3(-2) = -6$.
$2$. For $f(3)$: Since $1 < 3 \leq 3$,we use $f(x) = x^2$. Thus,$f(3) = (3)^2 = 9$.
$3$. For $f(4)$: Since $4 > 3$,we use $f(x) = 2x$. Thus,$f(4) = 2(4) = 8$.
Summing these values: $f(-2) + f(3) + f(4) = -6 + 9 + 8 = 11$.

(B) Given $f(x)=\begin{cases} x^{3}-1, & 1 < x < \infty \\ x-1, & -\infty < x \leq 1 \end{cases}$
First,we check the continuity at $x=1$.
$RHL = \lim_{x \rightarrow 1^{+}} f(x) = \lim_{x \rightarrow 1^{+}} (x^{3}-1) = 1^{3}-1 = 0$
$LHL = \lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{-}} (x-1) = 1-1 = 0$
$f(1) = 1-1 = 0$
Since $LHL = RHL = f(1)$,the function is continuous at $x=1$.
Next,we check the differentiability at $x=1$.
$f'(x) = \begin{cases} 3x^{2}, & 1 < x < \infty \\ 1, & -\infty < x < 1 \end{cases}$
$LHD = \lim_{x \rightarrow 1^{-}} f'(x) = 1$
$RHD = \lim_{x \rightarrow 1^{+}} f'(x) = 3(1)^{2} = 3$
Since $LHD \neq RHD$,the function is non-differentiable at $x=1$.
Therefore,the function is continuous and non-differentiable.

(B) Given the function is $f(x) = \frac{a}{x^{4}} - \frac{b}{x^{2}} + \cos x$.
To find the derivative with respect to $x$,we use the power rule $\frac{d}{dx}(x^{n}) = nx^{n-1}$ and the derivative of $\cos x$ is $-\sin x$.
$\frac{d}{dx} f(x) = \frac{d}{dx}(ax^{-4} - bx^{-2} + \cos x)$
$= a(-4x^{-5}) - b(-2x^{-3}) - \sin x$
$= -\frac{4a}{x^{5}} + \frac{2b}{x^{3}} - \sin x$
Comparing this with the expression $ma + nb - p$,we get:
$m = -\frac{4}{x^{5}}$,$n = \frac{2}{x^{3}}$,and $p = \sin x$.

(C) $y = (\cos x^{2})^{2}$
Differentiating with respect to $x$ using the chain rule:
$\frac{dy}{dx} = 2(\cos x^{2}) \cdot \frac{d}{dx}(\cos x^{2})$
$\frac{dy}{dx} = 2(\cos x^{2}) \cdot (-\sin x^{2}) \cdot \frac{d}{dx}(x^{2})$
$\frac{dy}{dx} = 2(\cos x^{2}) \cdot (-\sin x^{2}) \cdot (2x)$
$\frac{dy}{dx} = -4x \sin x^{2} \cos x^{2}$
Using the trigonometric identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we have:
$\frac{dy}{dx} = -2x(2 \sin x^{2} \cos x^{2}) = -2x \sin 2x^{2}$

(B) We need to find the derivative of the sum: $\frac{d}{d x}\left(x^{x}+x^{a}+a^{x}+a^{a}\right)$.
Using the linearity property of the derivative,we have:
$\frac{d}{d x}\left(x^{x}\right)+\frac{d}{d x}\left(x^{a}\right)+\frac{d}{d x}\left(a^{x}\right)+\frac{d}{d x}\left(a^{a}\right)$.
$1$. For $\frac{d}{d x}(x^x)$: Let $y = x^x$. Taking $\log$ on both sides,$\log y = x \log x$. Differentiating with respect to $x$,$\frac{1}{y} \frac{dy}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$. Thus,$\frac{dy}{dx} = x^x(1 + \log x)$.
$2$. For $\frac{d}{d x}(x^a)$: Using the power rule,$\frac{d}{d x}(x^a) = a x^{a-1}$.
$3$. For $\frac{d}{d x}(a^x)$: Using the exponential derivative rule,$\frac{d}{d x}(a^x) = a^x \log a$.
$4$. For $\frac{d}{d x}(a^a)$: Since $a$ is a constant,$a^a$ is a constant,so its derivative is $0$.
Combining these,the result is $x^{x}(1+\log x)+a x^{a-1}+a^{x} \log a$.

(A) For Statement $1$:
$y = \log_{10} x + \log_{e} x$
Using the change of base formula,$\log_{a} b = \frac{\log_{e} b}{\log_{e} a}$,we have:
$y = \frac{\log_{e} x}{\log_{e} 10} + \log_{e} x$
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{x \log_{e} 10} + \frac{1}{x}$
Since $\frac{1}{\log_{e} 10} = \log_{10} e$,we get:
$\frac{dy}{dx} = \frac{\log_{10} e}{x} + \frac{1}{x}$.
Thus,Statement $1$ is true.
For Statement $2$:
$\frac{d}{dx}(\log_{10} x) = \frac{d}{dx} \left( \frac{\log_{e} x}{\log_{e} 10} \right) = \frac{1}{x \log_{e} 10}$.
The statement claims it is $\frac{\log x}{\log 10}$,which is incorrect.
Similarly,$\frac{d}{dx}(\log_{e} x) = \frac{1}{x}$.
The statement claims it is $\frac{\log x}{\log e}$,which is incorrect.
Thus,Statement $2$ is false.

(D) Given $y=(x-1)^{2}(x-2)^{3}(x-3)^{5}$.
Taking the natural logarithm on both sides:
$\log y = \log [(x-1)^{2}(x-2)^{3}(x-3)^{5}]$
Using logarithmic properties,$\log y = 2 \log (x-1) + 3 \log (x-2) + 5 \log (x-3)$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = \frac{2}{x-1} + \frac{3}{x-2} + \frac{5}{x-3}$.
Therefore,$\frac{dy}{dx} = y \left[ \frac{2}{x-1} + \frac{3}{x-2} + \frac{5}{x-3} \right]$.
Substituting $x=4$:
$y(4) = (4-1)^{2}(4-2)^{3}(4-3)^{5} = 3^{2} \times 2^{3} \times 1^{5} = 9 \times 8 \times 1 = 72$.
$\left( \frac{dy}{dx} \right)_{x=4} = 72 \left[ \frac{2}{4-1} + \frac{3}{4-2} + \frac{5}{4-3} \right] = 72 \left[ \frac{2}{3} + \frac{3}{2} + 5 \right]$.
$= 72 \left[ \frac{4 + 9 + 30}{6} \right] = 72 \times \frac{43}{6} = 12 \times 43 = 516$.

(C) The angular displacement is given by $\theta = \frac{t^{2}}{20} + \frac{t}{5}$.
Angular velocity $\omega$ is the rate of change of angular displacement with respect to time,given by $\omega = \frac{d\theta}{dt}$.
Differentiating $\theta$ with respect to $t$:
$\omega = \frac{d}{dt} \left( \frac{t^{2}}{20} + \frac{t}{5} \right) = \frac{2t}{20} + \frac{1}{5} = \frac{t}{10} + \frac{1}{5}$.
At $t = 4 \ s$,the angular velocity $k$ is:
$k = \left( \frac{4}{10} + \frac{1}{5} \right) = \frac{4}{10} + \frac{2}{10} = \frac{6}{10} = 0.6 \ rad/s$.
We need to find the value of $5k$:
$5k = 5 \times 0.6 = 3$.

(A) Given function is $f(x) = x^{2} - 2x$.
To find the interval where the function is strictly decreasing,we first find the derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(x^{2} - 2x) = 2x - 2$.
$A$ function is strictly decreasing when $f'(x) < 0$.
So,$2x - 2 < 0$.
$2(x - 1) < 0$.
$x - 1 < 0$.
$x < 1$.
Thus,the function $f(x)$ is strictly decreasing in the interval $(-\infty, 1)$.

(D) Given $x = \cos \theta + \log \tan \frac{\theta}{2}$.
On differentiating with respect to $\theta$,we get:
$\frac{dx}{d\theta} = -\sin \theta + \frac{1}{\tan \frac{\theta}{2}} \cdot \sec^2 \frac{\theta}{2} \cdot \frac{1}{2}$
$= -\sin \theta + \frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}} \cdot \frac{1}{\cos^2 \frac{\theta}{2}} \cdot \frac{1}{2}$
$= -\sin \theta + \frac{1}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}} = -\sin \theta + \frac{1}{\sin \theta}$
$= \frac{1 - \sin^2 \theta}{\sin \theta} = \frac{\cos^2 \theta}{\sin \theta} \quad ...(i)$
Now,$y = \sin \theta$. On differentiating with respect to $\theta$,we get:
$\frac{dy}{d\theta} = \cos \theta \quad ...(ii)$
Therefore,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{\cos \theta}{\frac{\cos^2 \theta}{\sin \theta}} = \tan \theta$.
For $\frac{dy}{dx} = 0$,we must have $\tan \theta = 0$.
This implies $\theta = n \pi$ for $n \in Z$.

(C) Given the curve $y = -x^{3} + 3x^{2} + 2x - 27$.
To find the slope of the curve,we differentiate with respect to $x$:
$\frac{dy}{dx} = -3x^{2} + 6x + 2$.
Let the slope be $m = -3x^{2} + 6x + 2$.
To find the maximum slope,we differentiate $m$ with respect to $x$:
$\frac{dm}{dx} = -6x + 6$.
Setting $\frac{dm}{dx} = 0$ for critical points:
$-6x + 6 = 0 \implies x = 1$.
Checking the second derivative:
$\frac{d^{2}m}{dx^{2}} = -6 < 0$.
Since the second derivative is negative,the slope is maximum at $x = 1$.
Substituting $x = 1$ into the expression for $m$:
$m_{\text{max}} = -3(1)^{2} + 6(1) + 2 = -3 + 6 + 2 = 5$.

(D) Let $I = \int \frac{x e^{x} d x}{(1+x)^{2}}$.
We can rewrite the numerator as $(x+1-1)$.
$I = \int \frac{e^{x}(x+1-1)}{(1+x)^{2}} d x$.
$I = \int e^{x} \left[ \frac{x+1}{(1+x)^{2}} - \frac{1}{(1+x)^{2}} \right] d x$.
$I = \int e^{x} \left[ \frac{1}{1+x} - \frac{1}{(1+x)^{2}} \right] d x$.
Let $f(x) = \frac{1}{1+x}$. Then $f'(x) = -\frac{1}{(1+x)^{2}}$.
Using the standard integration formula $\int e^{x} [f(x) + f'(x)] d x = e^{x} f(x) + C$,we get:
$I = e^{x} \left( \frac{1}{1+x} \right) + C = \frac{e^{x}}{1+x} + C$.

(A) Let $I = \int \frac{x^{3} \sin \left(\tan ^{-1}\left(x^{4}\right)\right)}{1+x^{8}} dx$.
Substitute $t = \tan ^{-1}\left(x^{4}\right)$.
Differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = \frac{1}{1+(x^{4})^{2}} \cdot 4x^{3} = \frac{4x^{3}}{1+x^{8}}$.
Thus,$\frac{x^{3}}{1+x^{8}} dx = \frac{1}{4} dt$.
Substituting these into the integral,we get $I = \int \sin(t) \cdot \frac{1}{4} dt$.
$I = \frac{1}{4} \int \sin(t) dt = \frac{1}{4} (-\cos(t)) + C$.
Substituting back $t = \tan ^{-1}\left(x^{4}\right)$,we get $I = -\frac{1}{4} \cos \left(\tan ^{-1}\left(x^{4}\right)\right) + C$.

(C) Let $I = \int \frac{x^{2}}{\sqrt{x^{6}+a^{6}}} dx$.
Substitute $x^{3} = t$.
Then,$3x^{2} dx = dt$,which implies $x^{2} dx = \frac{1}{3} dt$.
Substituting these into the integral,we get:
$I = \frac{1}{3} \int \frac{1}{\sqrt{t^{2} + (a^{3})^{2}}} dt$.
Using the standard integral formula $\int \frac{1}{\sqrt{x^{2} + a^{2}}} dx = \log |x + \sqrt{x^{2} + a^{2}}| + C$,we have:
$I = \frac{1}{3} \log |t + \sqrt{t^{2} + (a^{3})^{2}}| + C$.
Substituting $t = x^{3}$ back into the expression:
$I = \frac{1}{3} \log |x^{3} + \sqrt{x^{6} + a^{6}}| + C$.

(A) Let $I = \int e^{x} \left( \frac{1+\sin x}{1+\cos x} \right) dx$.
Using trigonometric identities,$1+\sin x = 1+2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1+\cos x = 2 \cos^2 \frac{x}{2}$.
Substituting these into the integral:
$I = \int e^{x} \left( \frac{1+2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \right) dx$
$I = \int e^{x} \left( \frac{1}{2 \cos^2 \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \right) dx$
$I = \int e^{x} \left( \frac{1}{2} \sec^2 \frac{x}{2} + \tan \frac{x}{2} \right) dx$.
Let $f(x) = \tan \frac{x}{2}$. Then $f'(x) = \sec^2 \frac{x}{2} \cdot \frac{1}{2} = \frac{1}{2} \sec^2 \frac{x}{2}$.
Since the integral is in the form $\int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + C$,
we get $I = e^{x} \tan \frac{x}{2} + C$.

(B) Let $I = \int_{0}^{4042} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{4042-x}} \, dx \quad \dots (i)$
Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we get:
$I = \int_{0}^{4042} \frac{\sqrt{4042-x}}{\sqrt{4042-x}+\sqrt{x}} \, dx \quad \dots (ii)$
Adding equations $(i)$ and $(ii)$:
$2I = \int_{0}^{4042} \frac{\sqrt{x}+\sqrt{4042-x}}{\sqrt{x}+\sqrt{4042-x}} \, dx$
$2I = \int_{0}^{4042} 1 \, dx$
$2I = [x]_{0}^{4042} = 4042$
$I = \frac{4042}{2} = 2021$

(D) $I_{n} = \int_{0}^{\frac{\pi}{4}} \tan^{n} x \, dx$ ... $(i)$
$I_{n+2} = \int_{0}^{\frac{\pi}{4}} \tan^{n+2} x \, dx$ ... $(ii)$
Adding equations $(i)$ and $(ii)$:
$I_{n} + I_{n+2} = \int_{0}^{\frac{\pi}{4}} \tan^{n} x (1 + \tan^{2} x) \, dx$
$I_{n} + I_{n+2} = \int_{0}^{\frac{\pi}{4}} \tan^{n} x \sec^{2} x \, dx$
Let $u = \tan x$,then $du = \sec^{2} x \, dx$.
When $x = 0, u = 0$ and when $x = \frac{\pi}{4}, u = 1$.
$I_{n} + I_{n+2} = \int_{0}^{1} u^{n} \, du = \left[ \frac{u^{n+1}}{n+1} \right]_{0}^{1} = \frac{1}{n+1}$.
For $n = 8$,we have $I_{8} + I_{10} = \frac{1}{8+1} = \frac{1}{9}$.

(A) The equation $y=-\sqrt{16-x^{2}}$ represents the lower semi-circle of the circle $x^{2}+y^{2}=16$,which has a radius $r=4$ and center at the origin $(0,0)$.
Since the area is bounded by this curve and the $X$-axis,we are looking for the area of the semi-circle below the $X$-axis.
The area of a full circle is $\pi r^{2} = \pi(4)^{2} = 16\pi$.
Therefore,the area of the semi-circle is $\frac{1}{2} \times 16\pi = 8\pi$ sq units.
Alternatively,using integration:
$\text{Area} = \left| \int_{-4}^{4} (-\sqrt{16-x^{2}}) dx \right|$
$= \left| \left[ \frac{x}{2} \sqrt{16-x^{2}} + \frac{16}{2} \sin^{-1} \frac{x}{4} \right]_{-4}^{4} \right|$
$= \left| [0 + 8 \sin^{-1}(1)] - [0 + 8 \sin^{-1}(-1)] \right|$
$= \left| 8(\frac{\pi}{2}) - 8(-\frac{\pi}{2}) \right| = |4\pi + 4\pi| = 8\pi$ sq units.

(C) Given the differential equation: $x dy - y dx = 0$
Rearranging the terms,we get: $x dy = y dx$
Dividing both sides by $xy$ (assuming $x, y \neq 0$): $\frac{dy}{y} = \frac{dx}{x}$
Integrating both sides: $\int \frac{1}{y} dy = \int \frac{1}{x} dx$
This gives: $\ln|y| = \ln|x| + C$
Taking the exponential of both sides: $y = cx$,where $c$ is an arbitrary constant.
This is the equation of a straight line passing through the origin.

(D) Given the differential equation: $\frac{dy}{dx} = \frac{y+1}{x-1}$.
Separating the variables,we get: $\frac{1}{y+1} dy = \frac{1}{x-1} dx$.
Integrating both sides: $\int \frac{1}{y+1} dy = \int \frac{1}{x-1} dx$.
This gives: $\ln|y+1| = \ln|x-1| + \ln|C|$,which simplifies to $y+1 = C(x-1)$.
Using the initial condition $y(1) = 2$,we substitute $x=1$ and $y=2$ into the equation:
$2+1 = C(1-1) \Rightarrow 3 = C(0)$.
This implies $3 = 0$,which is a contradiction.
Since the initial condition $y(1) = 2$ is given at a point where the derivative $\frac{dy}{dx}$ is undefined (at $x=1$),the solution does not exist.
Therefore,the number of solutions is $0$.

(B) Let the equal acute angles of vector $a$ with the coordinate axes be $\alpha$.
Since the direction cosines are $l = \cos \alpha$,$m = \cos \alpha$,and $n = \cos \alpha$,we have the relation $l^2 + m^2 + n^2 = 1$.
Substituting the values,we get $3 \cos^2 \alpha = 1$,which implies $\cos \alpha = \frac{1}{\sqrt{3}}$.
Thus,the direction ratios of vector $a$ are proportional to $(1, 1, 1)$,so we can take $a = \hat{i} + \hat{j} + \hat{k}$.
The projection of vector $b$ on vector $a$ is given by the formula $\frac{b \cdot a}{|a|}$.
Calculating the dot product: $b \cdot a = (5\hat{i} + 7\hat{j} - \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = 5(1) + 7(1) - 1(1) = 5 + 7 - 1 = 11$.
The magnitude of vector $a$ is $|a| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.
Therefore,the projection is $\frac{11}{\sqrt{3}}$.

(A) Let $\vec{a}$ and $\vec{b}$ be the adjacent sides of the parallelogram,and $\vec{d_1}$ and $\vec{d_2}$ be the diagonals.
We know that $\vec{d_1} = \vec{a} + \vec{b}$ and $\vec{d_2} = \vec{a} - \vec{b}$ (or vice versa).
Thus,$\vec{a} = \frac{\vec{d_1} + \vec{d_2}}{2} = \frac{(3 \hat{i} + 6 \hat{j} - 2 \hat{k}) + (-\hat{i} - 2 \hat{j} - 8 \hat{k})}{2} = \frac{2 \hat{i} + 4 \hat{j} - 10 \hat{k}}{2} = \hat{i} + 2 \hat{j} - 5 \hat{k}$.
The length of side $\vec{a}$ is $|\vec{a}| = \sqrt{1^2 + 2^2 + (-5)^2} = \sqrt{1 + 4 + 25} = \sqrt{30}$.
Similarly,$\vec{b} = \frac{\vec{d_1} - \vec{d_2}}{2} = \frac{(3 \hat{i} + 6 \hat{j} - 2 \hat{k}) - (-\hat{i} - 2 \hat{j} - 8 \hat{k})}{2} = \frac{4 \hat{i} + 8 \hat{j} + 6 \hat{k}}{2} = 2 \hat{i} + 4 \hat{j} + 3 \hat{k}$.
The length of side $\vec{b}$ is $|\vec{b}| = \sqrt{2^2 + 4^2 + 3^2} = \sqrt{4 + 16 + 9} = \sqrt{29}$.
Comparing the lengths $\sqrt{30}$ and $\sqrt{29}$,the shorter side is $\sqrt{29}$.

(D) Given,$a \cdot b = 0$ and $(a + b)$ makes a $60^{\circ}$ angle with $a$.
Using the formula for the angle between two vectors:
$\cos 60^{\circ} = \frac{(a + b) \cdot a}{|a + b||a|}$
$\Rightarrow \frac{1}{2} = \frac{a \cdot a + b \cdot a}{|a + b||a|}$
Since $a \cdot b = 0$,we have $b \cdot a = 0$.
$\Rightarrow \frac{1}{2} = \frac{|a|^2}{|a + b||a|} = \frac{|a|}{|a + b|}$
$\Rightarrow |a + b| = 2|a|$
Squaring both sides:
$|a + b|^2 = 4|a|^2$
$|a|^2 + |b|^2 + 2(a \cdot b) = 4|a|^2$
Since $a \cdot b = 0$:
$|a|^2 + |b|^2 = 4|a|^2$
$|b|^2 = 3|a|^2$
$|b| = \sqrt{3}|a|$

(C) The area of a parallelogram with adjacent sides $a$ and $b$ is given by $|a \times b| = 15$ sq units.
Now,the area of the parallelogram with adjacent sides $(3a+2b)$ and $(a+3b)$ is given by the magnitude of their cross product:
Area $= |(3a+2b) \times (a+3b)|$
$= |3a \times a + 9a \times b + 2b \times a + 6b \times b|$
Since $a \times a = 0$ and $b \times b = 0$,the expression simplifies to:
$= |9(a \times b) + 2(b \times a)|$
Using the property $b \times a = -(a \times b)$:
$= |9(a \times b) - 2(a \times b)|$
$= |7(a \times b)|$
$= 7 |a \times b|$
Substituting the given value $|a \times b| = 15$:
$= 7 \times 15 = 105$ sq units.

(A) Given the vertices of the quadrilateral $ABCD$ are $A(0,4,1)$,$B(2,3,-1)$,$C(4,5,0)$,and $D(2,6,2)$.
We can divide the quadrilateral into two triangles,$\triangle ABD$ and $\triangle BCD$.
The area of the quadrilateral is the sum of the areas of these two triangles.
Area of $\triangle ABD = \frac{1}{2} |\vec{AB} \times \vec{AD}|$.
$\vec{AB} = (2-0)\hat{i} + (3-4)\hat{j} + (-1-1)\hat{k} = 2\hat{i} - \hat{j} - 2\hat{k}$.
$\vec{AD} = (2-0)\hat{i} + (6-4)\hat{j} + (2-1)\hat{k} = 2\hat{i} + 2\hat{j} + \hat{k}$.
$\vec{AB} \times \vec{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & -2 \\ 2 & 2 & 1 \end{vmatrix} = \hat{i}(-1+4) - \hat{j}(2+4) + \hat{k}(4+2) = 3\hat{i} - 6\hat{j} + 6\hat{k}$.
$|\vec{AB} \times \vec{AD}| = \sqrt{3^2 + (-6)^2 + 6^2} = \sqrt{9 + 36 + 36} = \sqrt{81} = 9$.
Area of $\triangle ABD = \frac{1}{2} \times 9 = 4.5 \text{ sq units}$.
Area of $\triangle BCD = \frac{1}{2} |\vec{BC} \times \vec{BD}|$.
$\vec{BC} = (4-2)\hat{i} + (5-3)\hat{j} + (0-(-1))\hat{k} = 2\hat{i} + 2\hat{j} + \hat{k}$.
$\vec{BD} = (2-2)\hat{i} + (6-3)\hat{j} + (2-(-1))\hat{k} = 0\hat{i} + 3\hat{j} + 3\hat{k}$.
$\vec{BC} \times \vec{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 1 \\ 0 & 3 & 3 \end{vmatrix} = \hat{i}(6-3) - \hat{j}(6-0) + \hat{k}(6-0) = 3\hat{i} - 6\hat{j} + 6\hat{k}$.
$|\vec{BC} \times \vec{BD}| = \sqrt{3^2 + (-6)^2 + 6^2} = 9$.
Area of $\triangle BCD = \frac{1}{2} \times 9 = 4.5 \text{ sq units}$.
Total Area = $4.5 + 4.5 = 9 \text{ sq units}$.

(C) Given,direction cosines of line $1$ are $(l_{1}, m_{1}, n_{1}) = \left(\frac{\sqrt{3}}{4}, \frac{1}{4}, \frac{\sqrt{3}}{2}\right)$.
Direction cosines of line $2$ are $(l_{2}, m_{2}, n_{2}) = \left(\frac{\sqrt{3}}{4}, \frac{1}{4}, -\frac{\sqrt{3}}{2}\right)$.
The angle $\theta$ between two lines is given by $\cos \theta = |l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2}|$.
Substituting the values:
$\cos \theta = \left|\left(\frac{\sqrt{3}}{4} \times \frac{\sqrt{3}}{4}\right) + \left(\frac{1}{4} \times \frac{1}{4}\right) + \left(\frac{\sqrt{3}}{2} \times -\frac{\sqrt{3}}{2}\right)\right|$
$\cos \theta = \left|\frac{3}{16} + \frac{1}{16} - \frac{3}{4}\right|$
$\cos \theta = \left|\frac{3 + 1 - 12}{16}\right| = \left|-\frac{8}{16}\right| = \left|-\frac{1}{2}\right| = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,we have $\theta = \frac{\pi}{3}$.

(B) Given points are $A(-3, 4, 11)$ and $B(1, -2, 7)$.
Direction ratios of the line $AB$ are given by $(x_2 - x_1, y_2 - y_1, z_2 - z_1) = (1 - (-3), -2 - 4, 7 - 11) = (4, -6, -4)$.
Dividing by $-2$,we get the simplified direction ratios as $(-2, 3, 2)$.
The equation of a line passing through $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$.
Substituting the point $(-3, 4, 11)$ and direction ratios $(-2, 3, 2)$,we get $\frac{x - (-3)}{-2} = \frac{y - 4}{3} = \frac{z - 11}{2}$.
Thus,the equation is $\frac{x + 3}{-2} = \frac{y - 4}{3} = \frac{z - 11}{2}$.

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

(B) Given,$P(B) = \frac{3}{5}$,$P\left(\frac{A}{B}\right) = \frac{1}{2}$,and $P(A \cup B) = \frac{4}{5}$.
Since $P\left(\frac{A}{B}\right) = \frac{P(A \cap B)}{P(B)}$,we have:
$\frac{P(A \cap B)}{P(B)} = \frac{1}{2}$
$P(A \cap B) = \frac{1}{2} \times \frac{3}{5} = \frac{3}{10}$.
Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$:
$\frac{4}{5} = P(A) + \frac{3}{5} - \frac{3}{10}$
$P(A) = \frac{4}{5} - \frac{3}{5} + \frac{3}{10}$
$P(A) = \frac{1}{5} + \frac{3}{10} = \frac{2+3}{10} = \frac{5}{10} = \frac{1}{2}$.

(C) Let $E_{A}$ be the event that the sum of numbers on the dice is less than $6$.
The possible outcomes for $E_{A}$ are:
$E_{A} = \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)\}$
Thus,$n(E_{A}) = 10$.
Let $E_{B}$ be the event that the sum of numbers on the dice is $3$.
The possible outcomes for $E_{B}$ are:
$E_{B} = \{(1,2), (2,1)\}$
Thus,$n(E_{B}) = 2$.
The required conditional probability is $P(E_{B}|E_{A}) = \frac{n(E_{B} \cap E_{A})}{n(E_{A})} = \frac{2}{10} = \frac{1}{5}$.

(C) Given that $A, B$ and $C$ are independent events with $P(A)=P(B)=P(C)=P$.
The probability that at least two of $A, B$ and $C$ occur is the sum of the probabilities of the following mutually exclusive cases:
$1$. Exactly two events occur: $(A \cap B \cap C') \cup (A \cap B' \cap C) \cup (A' \cap B \cap C)$
$2$. All three events occur: $(A \cap B \cap C)$
Since the events are independent, $P(A \cap B \cap C') = P(A)P(B)P(C') = P \times P \times (1-P) = P^{2}(1-P)$.
Similarly, $P(A \cap B' \cap C) = P^{2}(1-P)$ and $P(A' \cap B \cap C) = P^{2}(1-P)$.
Also, $P(A \cap B \cap C) = P \times P \times P = P^{3}$.
Thus, $P(\text{at least two occur}) = 3 \times P^{2}(1-P) + P^{3}$.
$= 3P^{2} - 3P^{3} + P^{3} = 3P^{2} - 2P^{3}$.

(C) Let $E$ be the event that the car is of standard quality. Let $A_{1}$ be the event that the car is manufactured in plant $X$,and $A_{2}$ be the event that the car is manufactured in plant $Y$.
Given probabilities are:
$P(A_{1}) = \frac{70}{100} = \frac{7}{10}$
$P(A_{2}) = \frac{30}{100} = \frac{3}{10}$
$P(E|A_{1}) = \frac{80}{100} = \frac{8}{10}$
$P(E|A_{2}) = \frac{90}{100} = \frac{9}{10}$
Using Bayes' Theorem,the probability that the car came from plant $X$ given it is of standard quality is:
$P(A_{1}|E) = \frac{P(A_{1}) \times P(E|A_{1})}{P(A_{1}) \times P(E|A_{1}) + P(A_{2}) \times P(E|A_{2})}$
$P(A_{1}|E) = \frac{\frac{7}{10} \times \frac{8}{10}}{\frac{7}{10} \times \frac{8}{10} + \frac{3}{10} \times \frac{9}{10}}$
$P(A_{1}|E) = \frac{56/100}{56/100 + 27/100} = \frac{56}{56 + 27} = \frac{56}{83}$
Thus,the required probability is $\frac{56}{83}$.