KCET 2014 Mathematics Question Paper with Answer and Solution

(A) Given that $\lim_{x \rightarrow 1} \frac{f(x)-2}{x^{2}-1} = \pi$.
Since the limit exists and the denominator $x^2-1$ approaches $0$ as $x \rightarrow 1$,the numerator $f(x)-2$ must also approach $0$ for the limit to be a finite value $\pi$.
Therefore,$\lim_{x \rightarrow 1} (f(x)-2) = 0$.
This implies $\lim_{x \rightarrow 1} f(x) = 2$.

(A) statement is a declarative sentence that is either true or false,but not both.
$(A)$ $ \sqrt{3} $ is a prime number: This is a false statement because $ \sqrt{3} \approx 1.732 $,which is not an integer,and prime numbers must be integers.
$(B)$ The sun is a star: This is a true statement.
$(C)$ Mathematics is interesting: This is an opinion,not a mathematical statement,but in the context of logic,it is often treated as subjective.
$(D)$ $ \sqrt{2} $ is an irrational number: This is a true statement.
Since we are looking for a statement that is not correct (i.e.,false),option $A$ is the correct answer.

(D) Let $n(U) = 60$ be the total number of students.
Let $C$ be the set of students who play cricket,$n(C) = 25$.
Let $T$ be the set of students who play tennis,$n(T) = 20$.
Given $n(C \cap T) = 10$.
The number of students who play at least one game is $n(C \cup T) = n(C) + n(T) - n(C \cap T)$.
$n(C \cup T) = 25 + 20 - 10 = 35$.
The number of students who play neither game is $n(U) - n(C \cup T)$.
$60 - 35 = 25$.

(D) Given the function $f(x) = b^{x}$ where $b > 1$.
For any exponential function of the form $f(x) = b^{x}$,the point $(0, 1)$ always lies on the graph because $f(0) = b^{0} = 1$.
The point $(1, b)$ lies on the graph because $f(1) = b^{1} = b$.
Therefore,the point $(1, 0)$ does not lie on the graph of the function $f(x) = b^{x}$.
Thus,option $D$ is not correct.

(A) The coordinates of the point $P$ are $(a, b, c)$.
To find the distance of point $P$ from the $x$-axis,we project the point onto the $x$-axis.
The projection of point $P(a, b, c)$ on the $x$-axis is the point $A(a, 0, 0)$.
The distance $d$ between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by the formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.
Substituting the coordinates of $P(a, b, c)$ and $A(a, 0, 0)$:
$d = \sqrt{(a-a)^2 + (0-b)^2 + (0-c)^2}$
$d = \sqrt{0^2 + (-b)^2 + (-c)^2}$
$d = \sqrt{b^2 + c^2}$.

(C) When two dice are thrown simultaneously,the total number of possible outcomes is $6 \times 6 = 36$.
The favorable outcomes for obtaining a total score of $5$ are the pairs $(1, 4), (4, 1), (2, 3), \text{ and } (3, 2)$.
The number of favorable outcomes is $4$.
The probability $P(E)$ is given by the ratio of the number of favorable outcomes to the total number of outcomes:
$P(E) = \frac{4}{36} = \frac{1}{9}$.

(A) Given that $A$ and $B$ are independent events.
We know that $P(A) = 1 - P(A^{\prime})$ and $P(B) = 1 - P(B^{\prime})$.
Given $P(A^{\prime}) = \frac{2}{3}$,so $P(A) = 1 - \frac{2}{3} = \frac{1}{3}$.
Given $P(B^{\prime}) = \frac{2}{7}$,so $P(B) = 1 - \frac{2}{7} = \frac{5}{7}$.
Since $A$ and $B$ are independent,$P(A \cap B) = P(A) \cdot P(B)$.
Therefore,$P(A \cap B) = \frac{1}{3} \cdot \frac{5}{7} = \frac{5}{21}$.

(C) The standard equation of a circle with center $(h, k)$ and radius $r$ is given by $(x-h)^{2} + (y-k)^{2} = r^{2}$.
Given center $(h, k) = (-a, -b)$ and radius $r = \sqrt{a^{2}-b^{2}}$.
Substituting these values into the formula:
$(x - (-a))^{2} + (y - (-b))^{2} = (\sqrt{a^{2}-b^{2}})^{2}$
$(x+a)^{2} + (y+b)^{2} = a^{2}-b^{2}$
Expanding the squares:
$x^{2} + 2ax + a^{2} + y^{2} + 2by + b^{2} = a^{2} - b^{2}$
Subtracting $a^{2}$ from both sides and adding $b^{2}$ to both sides:
$x^{2} + y^{2} + 2ax + 2by + b^{2} + b^{2} = 0$
$x^{2} + y^{2} + 2ax + 2by + 2b^{2} = 0$.

(B) The equation of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}$.
Substituting the points $(5,0)$ and $(0,3)$,we get $\frac{y-0}{x-5} = \frac{3-0}{0-5} = -\frac{3}{5}$.
This simplifies to $5y = -3(x-5)$,which is $3x + 5y - 15 = 0$.
The length of the perpendicular from a point $(x_0, y_0)$ to the line $ax + by + c = 0$ is given by $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$.
Substituting $(x_0, y_0) = (4,4)$ and the line equation $3x + 5y - 15 = 0$:
$d = \frac{|3(4) + 5(4) - 15|}{\sqrt{3^2 + 5^2}} = \frac{|12 + 20 - 15|}{\sqrt{9 + 25}} = \frac{17}{\sqrt{34}}$.
Rationalizing the denominator,we get $d = \frac{17\sqrt{34}}{34} = \frac{\sqrt{34}}{2} = \sqrt{\frac{34}{4}} = \sqrt{\frac{17}{2}}$.

(A) The given equation of the parabola is $x^{2}=12y$.
Comparing this with the standard form $x^{2}=4ay$,we get $4a=12$,which implies $a=3$.
The vertex of the parabola is at $(0,0)$.
The ends of the latus rectum are at $(2a, a)$ and $(-2a, a)$,which are $(6, 3)$ and $(-6, 3)$.
The area of the triangle formed by the vertex $(0,0)$ and the points $(6, 3)$ and $(-6, 3)$ is given by:
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
Here,the base is the length of the latus rectum,which is $4a = 12$.
The height is the distance from the vertex to the latus rectum,which is $a = 3$.
$\text{Area} = \frac{1}{2} \times 12 \times 3 = 18 \text{ sq. units}$.

(D) We are given the expression $a[b \cos C - c \cos B]$.
Using the projection formula for a triangle,we know that $a = b \cos C + c \cos B$.
Substituting this into the expression,we get $(b \cos C + c \cos B)(b \cos C - c \cos B)$.
This is in the form $(x+y)(x-y) = x^2 - y^2$,so it simplifies to $b^2 \cos^2 C - c^2 \cos^2 B$.
Alternatively,using the Law of Cosines,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ and $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.
Substituting these values: $a \left[ b \left( \frac{a^2 + b^2 - c^2}{2ab} \right) - c \left( \frac{a^2 + c^2 - b^2}{2ac} \right) \right]$.
$= a \left[ \frac{a^2 + b^2 - c^2}{2a} - \frac{a^2 + c^2 - b^2}{2a} \right]$.
$= a \left[ \frac{a^2 + b^2 - c^2 - a^2 - c^2 + b^2}{2a} \right]$.
$= a \left[ \frac{2b^2 - 2c^2}{2a} \right] = b^2 - c^2$.

(A) Let $ z = \frac{\beta-\alpha}{1-\bar{\alpha} \beta} $. We want to find $ |z| $.
Consider $ |z|^2 = z \cdot \bar{z} = \left( \frac{\beta-\alpha}{1-\bar{\alpha} \beta} \right) \left( \frac{\bar{\beta}-\bar{\alpha}}{1-\alpha \bar{\beta}} \right) $.
Expanding the numerator: $ (\beta-\alpha)(\bar{\beta}-\bar{\alpha}) = \beta \bar{\beta} - \beta \bar{\alpha} - \alpha \bar{\beta} + \alpha \bar{\alpha} = |\beta|^2 - \beta \bar{\alpha} - \alpha \bar{\beta} + |\alpha|^2 $.
Expanding the denominator: $ (1-\bar{\alpha} \beta)(1-\alpha \bar{\beta}) = 1 - \alpha \bar{\beta} - \bar{\alpha} \beta + \bar{\alpha} \alpha \beta \bar{\beta} = 1 - \alpha \bar{\beta} - \bar{\alpha} \beta + |\alpha|^2 |\beta|^2 $.
Since $ |\beta|=1 $,we have $ |\beta|^2 = 1 $.
Substituting $ |\beta|^2 = 1 $ into the expressions:
Numerator: $ 1 - \beta \bar{\alpha} - \alpha \bar{\beta} + |\alpha|^2 $.
Denominator: $ 1 - \alpha \bar{\beta} - \bar{\alpha} \beta + |\alpha|^2 (1) = 1 - \alpha \bar{\beta} - \bar{\alpha} \beta + |\alpha|^2 $.
Since the numerator and denominator are equal,$ |z|^2 = 1 $,which implies $ |z| = 1 $.

(A) Given $\sin \theta = \sin \alpha$.
Subtracting $\sin \alpha$ from both sides,we get $\sin \theta - \sin \alpha = 0$.
Using the trigonometric identity $\sin C - \sin D = 2 \cos \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)$,we have:
$2 \cos \left(\frac{\theta+\alpha}{2}\right) \sin \left(\frac{\theta-\alpha}{2}\right) = 0$.
This product is zero if either $\cos \left(\frac{\theta+\alpha}{2}\right) = 0$ or $\sin \left(\frac{\theta-\alpha}{2}\right) = 0$.
If $\cos \left(\frac{\theta+\alpha}{2}\right) = 0$,then $\frac{\theta+\alpha}{2} = (2n+1) \frac{\pi}{2}$,which means $\frac{\theta+\alpha}{2}$ is an odd multiple of $\frac{\pi}{2}$.
If $\sin \left(\frac{\theta-\alpha}{2}\right) = 0$,then $\frac{\theta-\alpha}{2} = n\pi$,which means $\frac{\theta-\alpha}{2}$ is a multiple of $\pi$.
Thus,the condition is that $\frac{\theta+\alpha}{2}$ is an odd multiple of $\frac{\pi}{2}$ $OR$ $\frac{\theta-\alpha}{2}$ is a multiple of $\pi$.

(C) Given,$\tan x = \frac{3}{4}$ and $\pi < x < \frac{3\pi}{2}$.
Since $\pi < x < \frac{3\pi}{2}$,dividing by $2$ gives $\frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$,which means $\frac{x}{2}$ lies in the second quadrant.
In the second quadrant,$\cos \theta$ is negative.
We know that $\sec^2 x = 1 + \tan^2 x = 1 + (\frac{3}{4})^2 = 1 + \frac{9}{16} = \frac{25}{16}$.
Therefore,$\sec x = \pm \frac{5}{4}$.
Since $x$ is in the third quadrant,$\cos x$ must be negative,so $\cos x = -\frac{4}{5}$.
Using the half-angle formula,$\cos^2 \frac{x}{2} = \frac{1 + \cos x}{2}$.
Substituting the value of $\cos x$,we get $\cos^2 \frac{x}{2} = \frac{1 - \frac{4}{5}}{2} = \frac{1/5}{2} = \frac{1}{10}$.
Since $\frac{x}{2}$ is in the second quadrant,$\cos \frac{x}{2}$ must be negative.
Thus,$\cos \frac{x}{2} = -\sqrt{\frac{1}{10}} = -\frac{1}{\sqrt{10}}$.

(D) We know that the formula for the coefficient of variation $(CV)$ is given by:
$CV = \frac{\sigma}{\mu} \times 100$
where $\sigma$ is the standard deviation and $\mu$ is the arithmetic mean.
Given: $CV = 60$ and $\sigma = 21$.
Substituting these values into the formula:
$60 = \frac{21}{\mu} \times 100$
Rearranging to solve for $\mu$:
$\mu = \frac{21 \times 100}{60}$
$\mu = \frac{2100}{60}$
$\mu = 35$
Therefore,the arithmetic mean of the distribution is $35$.

(C) Given the set $A = \{x : |2x + 3| < 7\}$.
We know that the inequality $|f(x)| < a$ is equivalent to $-a < f(x) < a$.
Applying this to the given inequality:
$-7 < 2x + 3 < 7$
Subtracting $3$ from all parts:
$-7 - 3 < 2x < 7 - 3$
$-10 < 2x < 4$
Dividing by $2$:
$-5 < x < 2$
Now,let us check the condition for set $D = \{x : 0 < x + 5 < 7\}$:
$0 < x + 5 < 7$
Subtracting $5$ from all parts:
$0 - 5 < x < 7 - 5$
$-5 < x < 2$
Since the range of $x$ for set $A$ and set $D$ is the same,set $A$ is equal to set $D$.

(D) The general term in the expansion of $(1+x)^n$ is given by $T_{r+1} = {}^nC_r x^r$.
For the expansion of $(1+x)^{44}$,the $21^{\text{st}}$ term is $T_{21} = T_{20+1} = {}^{44}C_{20} x^{20}$.
The $22^{\text{nd}}$ term is $T_{22} = T_{21+1} = {}^{44}C_{21} x^{21}$.
Given that $T_{21} = T_{22}$,we have ${}^{44}C_{20} x^{20} = {}^{44}C_{21} x^{21}$.
Dividing both sides by $x^{20}$ (assuming $x \neq 0$),we get $x = \frac{{}^{44}C_{20}}{{}^{44}C_{21}}$.
Using the formula ${}^nC_r = \frac{n!}{r!(n-r)!}$,we have:
$x = \frac{44!}{20!24!} \times \frac{21!23!}{44!} = \frac{21!}{20!} \times \frac{23!}{24!} = \frac{21}{1} \times \frac{1}{24} = \frac{21}{24} = \frac{7}{8}$.

(A) $5$ digit telephone number starts with $67$.
Since the first two digits are fixed as $6$ and $7$,we need to fill the remaining $3$ positions.
The available digits are ${0, 1, 2, 3, 4, 5, 8, 9}$,which are $8$ distinct digits.
Since no digit can be repeated,we need to arrange $3$ digits out of these $8$ available digits.
The number of ways to arrange $3$ digits out of $8$ is given by the permutation formula $^nP_r = \frac{n!}{(n-r)!}$.
Here,$n = 8$ and $r = 3$.
$^8P_3 = \frac{8!}{(8-3)!} = \frac{8!}{5!} = 8 \times 7 \times 6 = 336$.

(B) The sum of an infinite geometric series is given by $ S = \frac{a}{1-r} $,where $ |r| < 1 $.
Given $ S = 4 $,we have $ \frac{a}{1-r} = 4 \Rightarrow a = 4(1-r) = 4 - 4r \Rightarrow a + 4r = 4 \dots(1) $.
The second term of a geometric series is $ t_2 = ar $.
Given $ t_2 = \frac{3}{4} $,we have $ ar = \frac{3}{4} \Rightarrow a = \frac{3}{4r} \dots(2) $.
Substituting $ a $ from $(2)$ into $(1)$:
$ \frac{3}{4r} + 4r = 4 $.
Multiplying by $ 4r $: $ 3 + 16r^2 = 16r \Rightarrow 16r^2 - 16r + 3 = 0 $.
Factoring the quadratic equation: $ 16r^2 - 12r - 4r + 3 = 0 \Rightarrow 4r(4r - 3) - 1(4r - 3) = 0 \Rightarrow (4r - 1)(4r - 3) = 0 $.
Thus,$ r = \frac{1}{4} $ or $ r = \frac{3}{4} $.
If $ r = \frac{1}{4} $,then $ a = 4(1 - \frac{1}{4}) = 4(\frac{3}{4}) = 3 $.
If $ r = \frac{3}{4} $,then $ a = 4(1 - \frac{3}{4}) = 4(\frac{1}{4}) = 1 $.
Comparing with the given options,the pair $ (a=3, r=\frac{1}{4}) $ matches option $ B $.

(D) Given $ y = (\tan^{-1} x)^2 $.
First,differentiate with respect to $ x $:
$ y_1 = \frac{dy}{dx} = 2(\tan^{-1} x) \cdot \frac{1}{1+x^2} $.
This implies $ (1+x^2) y_1 = 2 \tan^{-1} x $.
Differentiate both sides again with respect to $ x $ using the product rule:
$ (1+x^2) y_2 + y_1(2x) = 2 \cdot \frac{1}{1+x^2} $.
Multiply the entire equation by $ (1+x^2) $:
$ (1+x^2)^2 y_2 + 2x(1+x^2) y_1 = 2 $.
Thus,the value is $ 2 $.

(A) Given $y=(1+x)(1+x^{2})(1+x^{4})$.
We can write $y = \frac{(1-x)(1+x)(1+x^{2})(1+x^{4})}{1-x} = \frac{1-x^{8}}{1-x}$ for $x \neq 1$.
Using the quotient rule,$\frac{dy}{dx} = \frac{(1-x)(-8x^{7}) - (1-x^{8})(-1)}{(1-x)^{2}} = \frac{-8x^{7}+8x^{8}+1-x^{8}}{(1-x)^{2}} = \frac{7x^{8}-8x^{7}+1}{(1-x)^{2}}$.
Applying $L$'Hopital's rule or expanding the series,we evaluate the limit as $x \to 1$.
Alternatively,using logarithmic differentiation: $\ln y = \ln(1+x) + \ln(1+x^{2}) + \ln(1+x^{4})$.
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{1+x} + \frac{2x}{1+x^{2}} + \frac{4x^{3}}{1+x^{4}}$.
At $x=1$,$y = (2)(2)(2) = 8$.
$\frac{1}{8} \frac{dy}{dx} = \frac{1}{2} + \frac{2}{2} + \frac{4}{2} = \frac{1}{2} + 1 + 2 = 3.5$.
$\frac{dy}{dx} = 8 \times 3.5 = 28$.

(B) Given $f(x) = x^{3}$ and $g(x) = x^{3} - 4x$ on $[-2, 2]$.
Since $f(x)$ and $g(x)$ are polynomials,they are continuous on $[-2, 2]$ and differentiable on $(-2, 2)$. Thus,both satisfy the Mean Value Theorem.
For Rolle's Theorem,we check $f(a) = f(b)$:
$f(-2) = (-2)^{3} = -8$ and $f(2) = (2)^{3} = 8$. Since $f(-2) \neq f(2)$,$f(x)$ does not satisfy Rolle's Theorem.
$g(-2) = (-2)^{3} - 4(-2) = -8 + 8 = 0$ and $g(2) = (2)^{3} - 4(2) = 8 - 8 = 0$. Since $g(-2) = g(2)$,$g(x)$ satisfies Rolle's Theorem.
Therefore,statement $(a)$ and statement $(c)$ are correct.

(A) For a function to be continuous at $x = 0$,the limit of the function as $x \to 0$ must be equal to the value of the function at $x = 0$.
Given $f(x) = \frac{3 \sin(\pi x)}{5x}$ for $x \neq 0$ and $f(0) = 2K$.
We know that $\lim_{x \to 0} \frac{\sin(\theta x)}{x} = \theta$.
Therefore,$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{3 \sin(\pi x)}{5x} = \frac{3}{5} \lim_{x \to 0} \frac{\sin(\pi x)}{x} = \frac{3}{5} \times \pi = \frac{3\pi}{5}$.
Since the function is continuous at $x = 0$,we have $\lim_{x \to 0} f(x) = f(0)$.
So,$\frac{3\pi}{5} = 2K$.
Solving for $K$,we get $K = \frac{3\pi}{10}$.

(C) The graph represents the function $f(x) = |x-1|$.
At $x=1$,the graph is continuous because there is no break in the curve.
However,at $x=1$,there is a sharp corner (or cusp),which means the left-hand derivative and right-hand derivative are not equal.
Specifically,the left-hand derivative is $-1$ and the right-hand derivative is $1$.
Therefore,the function is continuous but not differentiable at $x=1$.

(C) Let $\phi$ be the angle made by the tangent with the positive $x$-axis. The slope of the tangent $m$ is given by $\tan \phi = \left. \frac{dy}{dx} \right|_{(1, 2)}$.
Given $y = x^{3} + 1$,we have $\frac{dy}{dx} = 3x^{2}$.
At $(1, 2)$,the slope $m = \tan \phi = 3(1)^{2} = 3$.
From the geometry of the line and the $y$-axis,the angle $\theta$ made with the $y$-axis satisfies $\theta + \phi = 90^{\circ}$ or $\theta = 90^{\circ} - \phi$ (depending on the orientation). Based on the provided diagram,the angle $\theta$ is the obtuse angle between the tangent and the positive $y$-axis,such that $\theta = 90^{\circ} + \phi$.
Then,$\tan \theta = \tan(90^{\circ} + \phi) = -\cot \phi$.
Since $\tan \phi = 3$,we have $\cot \phi = \frac{1}{3}$.
Therefore,$\tan \theta = -\frac{1}{3}$.

(D) Given the function $f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^2}{2} + x + 1$.
To find $f'(x)$,we differentiate $f(x)$ with respect to $x$ using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$:
$f'(x) = \frac{100x^{99}}{100} + \frac{99x^{98}}{99} + \dots + \frac{2x}{2} + 1$.
Simplifying this,we get $f'(x) = x^{99} + x^{98} + \dots + x + 1$.
Now,substituting $x = 0$ into the derivative:
$f'(0) = 0^{99} + 0^{98} + \dots + 0 + 1 = 1$.

(D) Let $I = \int_{e}^{\Pi} x f(x) dx$.
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$,we have:
$I = \int_{e}^{\Pi} (e + \Pi - x) f(e + \Pi - x) dx$.
Since $f(e + \Pi - x) = f(x)$,this becomes:
$I = \int_{e}^{\Pi} (e + \Pi - x) f(x) dx = (e + \Pi) \int_{e}^{\Pi} f(x) dx - \int_{e}^{\Pi} x f(x) dx$.
$I = (e + \Pi) \left( \frac{2}{e + \Pi} \right) - I$.
$2I = 2$.
$I = 1$.

(B) The function $f(x) = \sqrt{\cos x}$ is defined only when $\cos x \geq 0$.
In the interval $[0, 2\pi]$,$\cos x$ is non-negative in the first quadrant $[0, \frac{\pi}{2}]$ and the fourth quadrant $[\frac{3\pi}{2}, 2\pi]$.
Thus,the domain for $n=0$ is $[0, \frac{\pi}{2}] \cup [\frac{3\pi}{2}, 2\pi]$.

(C) Let $\Delta = \left|\begin{array}{ccc}x+2 & x+3 & x+a \\ x+4 & x+5 & x+b \\ x+6 & x+7 & x+c\end{array}\right|$.
Applying row operations $R_{1} \rightarrow R_{1}-R_{2}$ and $R_{2} \rightarrow R_{2}-R_{3}$:
$\Delta = \left|\begin{array}{ccc}(x+2)-(x+4) & (x+3)-(x+5) & (x+a)-(x+b) \\ (x+4)-(x+6) & (x+5)-(x+7) & (x+b)-(x+c) \\ x+6 & x+7 & x+c\end{array}\right|$
$\Delta = \left|\begin{array}{ccc}-2 & -2 & a-b \\ -2 & -2 & b-c \\ x+6 & x+7 & x+c\end{array}\right|$.
Since $a, b, c$ are in $A$.$P$.,we have $b-a = c-b$,which implies $a-b = b-c$.
Thus,the first two rows $R_{1}$ and $R_{2}$ are identical.
Since two rows are identical,the value of the determinant is $0$.

(D) Let $I = \int[(3x-1) \cos x + (1-2x) \sin x] dx$.
Using integration by parts,$\int u dv = uv - \int v du$:
For $\int (3x-1) \cos x dx$: Let $u = 3x-1, dv = \cos x dx \Rightarrow du = 3 dx, v = \sin x$.
$\int (3x-1) \cos x dx = (3x-1) \sin x - \int 3 \sin x dx = (3x-1) \sin x + 3 \cos x$.
For $\int (1-2x) \sin x dx$: Let $u = 1-2x, dv = \sin x dx \Rightarrow du = -2 dx, v = -\cos x$.
$\int (1-2x) \sin x dx = (1-2x)(-\cos x) - \int (-\cos x)(-2) dx = (2x-1) \cos x - 2 \sin x$.
Adding these: $I = (3x-1) \sin x + 3 \cos x + (2x-1) \cos x - 2 \sin x + C$.
$I = (3x-1-2) \sin x + (3+2x-1) \cos x + C = (3x-3) \sin x + (2x+2) \cos x + C$.
Comparing with $f(x) \cos x + g(x) \sin x + C$,we get $f(x) = 2x+2$ and $g(x) = 3x-3 = 3(x-1)$.

(B) Given the diagonal matrix $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix}$.
For a diagonal matrix $A = \text{diag}(a, b, c)$,the inverse is given by $A^{-1} = \text{diag}(\frac{1}{a}, \frac{1}{b}, \frac{1}{c})$.
Substituting the values $a=2, b=3, c=4$,we get:
$A^{-1} = \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{4} \end{bmatrix}$.
Thus,option $B$ is the correct answer.

(B) Given the equation of the line $y=mx$ and the boundaries $x=1$ and $x=2$.
The required area is given by the definite integral:
$\text{Area} = \int_{1}^{2} mx \, dx = 6$
$\Rightarrow m \left[ \frac{x^2}{2} \right]_{1}^{2} = 6$
$\Rightarrow \frac{m}{2} (2^2 - 1^2) = 6$
$\Rightarrow \frac{m}{2} (4 - 1) = 6$
$\Rightarrow \frac{3m}{2} = 6$
$\Rightarrow 3m = 12$
$\Rightarrow m = 4$

(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = 3x$.
First,we find the integrating factor $(IF)$:
$IF = e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = x$.
Multiplying the differential equation by the integrating factor $x$,we get:
$x \frac{dy}{dx} + y = 3x^2$.
This can be written as:
$\frac{d}{dx}(xy) = 3x^2$.
Integrating both sides with respect to $x$:
$\int \frac{d}{dx}(xy) dx = \int 3x^2 dx$.
$xy = x^3 + C$.
Dividing by $x$,we get the general solution:
$y = x^2 + \frac{C}{x}$.

(C) Consider a cube with side length $1$. Let the vertices be represented in a $3D$ coordinate system as shown in the figure.
Let the two diagonals be $\vec{OA}$ and $\vec{BC}$.
The coordinates are $O(0,0,0)$,$A(1,1,1)$,$B(1,0,0)$,and $C(0,1,1)$.
The vector $\vec{OA} = (1-0, 1-0, 1-0) = (1, 1, 1)$.
The vector $\vec{BC} = (0-1, 1-0, 1-0) = (-1, 1, 1)$.
The angle $\theta$ between two vectors $\vec{u}$ and $\vec{v}$ is given by $\cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}$.
$\vec{OA} \cdot \vec{BC} = (1)(-1) + (1)(1) + (1)(1) = -1 + 1 + 1 = 1$.
$|\vec{OA}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.
$|\vec{BC}| = \sqrt{(-1)^2 + 1^2 + 1^2} = \sqrt{3}$.
Therefore,$\cos \theta = \frac{1}{\sqrt{3} \cdot \sqrt{3}} = \frac{1}{3}$.
Thus,$\theta = \cos^{-1}\left(\frac{1}{3}\right)$.

(B) The given line is $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$.
Since the plane is perpendicular to this line,the direction ratios of the normal to the plane are the same as the direction ratios of the line,which are $(1, 2, 3)$.
The general equation of a plane with normal $(a, b, c)$ is $ax+by+cz+d=0$.
Substituting the direction ratios,we get $1x+2y+3z+d=0$.
Since the plane passes through the point $(2, 3, 4)$,we substitute these coordinates into the equation:
$1(2)+2(3)+3(4)+d=0$
$2+6+12+d=0$
$20+d=0$
$d=-20$.
Therefore,the equation of the plane is $x+2y+3z-20=0$,or $x+2y+3z=20$.

(D) The direction ratios of the given line are $(3, 4, 5)$.
For a line to be parallel to a plane,the normal vector of the plane must be perpendicular to the direction vector of the line.
If the plane equation is $ax+by+cz=d$,the normal vector is $(a, b, c)$.
The condition for perpendicularity is $a_1a_2 + b_1b_2 + c_1c_2 = 0$.
Checking option $D$: $2x+y-2z=0$,the normal vector is $(2, 1, -2)$.
Calculating the dot product: $3(2) + 4(1) + 5(-2) = 6 + 4 - 10 = 0$.
Since the dot product is $0$,the line is parallel to the plane $2x+y-2z=0$.

(A) The given lines are $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-K}$ and $\frac{x-1}{K}=\frac{y-4}{2}=\frac{z-5}{1}$.
For two lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ to be coplanar,the condition is $\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$.
Here,$(x_1, y_1, z_1) = (2, 3, 4)$ and $(x_2, y_2, z_2) = (1, 4, 5)$.
Also,$(a_1, b_1, c_1) = (1, 1, -K)$ and $(a_2, b_2, c_2) = (K, 2, 1)$.
Substituting these values into the determinant:
$\begin{vmatrix} 1-2 & 4-3 & 5-4 \\ 1 & 1 & -K \\ K & 2 & 1 \end{vmatrix} = 0$
$\begin{vmatrix} -1 & 1 & 1 \\ 1 & 1 & -K \\ K & 2 & 1 \end{vmatrix} = 0$
Expanding the determinant:
$-1(1 - (-2K)) - 1(1 - (-K^2)) + 1(2 - K) = 0$
$-1(1 + 2K) - 1(1 + K^2) + 2 - K = 0$
$-1 - 2K - 1 - K^2 + 2 - K = 0$
$-K^2 - 3K = 0$
$K(K + 3) = 0$
Thus,$K = 0$ or $K = -3$. Since $K = -3$ is not in the options,the correct value is $K = 0$.

(A) The given parabolas are $y=x^{2}$ and $x=y^{2}$.
First,we find the points of intersection by substituting $y=x^{2}$ into $x=y^{2}$:
$x=(x^{2})^{2} \implies x=x^{4} \implies x^{4}-x=0 \implies x(x^{3}-1)=0$.
This gives $x=0$ and $x=1$.
For $x=0$,$y=0$,and for $x=1$,$y=1$. The intersection points are $(0,0)$ and $(1,1)$.
The area $A$ is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=1$:
$A = \int_{0}^{1} (\sqrt{x} - x^{2}) dx$.
Evaluating the integral:
$A = [\frac{x^{3/2}}{3/2} - \frac{x^{3}}{3}]_{0}^{1} = [\frac{2}{3}x^{3/2} - \frac{1}{3}x^{3}]_{0}^{1}$.
$A = (\frac{2}{3} - \frac{1}{3}) - (0 - 0) = \frac{1}{3}$ square units.

(C) Given the differential equation: $ y = x \frac{dy}{dx} + \frac{2}{dy/dx} $.
Multiply both sides by $ \frac{dy}{dx} $ to eliminate the fraction:
$ y \left( \frac{dy}{dx} \right) = x \left( \frac{dy}{dx} \right)^2 + 2 $.
Rearranging the terms,we get: $ x \left( \frac{dy}{dx} \right)^2 - y \left( \frac{dy}{dx} \right) + 2 = 0 $.
The highest order derivative present is $ \frac{dy}{dx} $,so the order is $ 1 $.
The highest power of the highest order derivative is $ 2 $,so the degree is $ 2 $.
Therefore,the order and degree are $ 1 $ and $ 2 $ respectively.

(C) Given the rate of digging $\frac{dA}{dt} = \frac{2}{\sqrt{t}}$.
Integrating both sides with respect to $t$:
$A = \int \frac{2}{\sqrt{t}} dt = 2 \int t^{-1/2} dt = 2 \cdot \frac{t^{1/2}}{1/2} + C = 4\sqrt{t} + C$.
At $t = 0$,the area dug $A = 0$,so $C = 0$.
Thus,$A = 4\sqrt{t}$.
To find the time $t$ to dig $40$ square metres,set $A = 40$:
$40 = 4\sqrt{t} \Rightarrow \sqrt{t} = 10$.
Squaring both sides,$t = 10^2 = 100$ minutes.

(D) Given $a R b \Leftrightarrow |a-b| \leq 1$.
$1$. Reflexivity: For any $a \in S$,$|a-a| = 0 \leq 1$. Thus,$a R a$ holds. $R$ is reflexive.
$2$. Symmetry: If $a R b$,then $|a-b| \leq 1$. Since $|a-b| = |b-a|$,it follows that $|b-a| \leq 1$,so $b R a$ holds. $R$ is symmetric.
$3$. Transitivity: Consider $a = 1, b = 2, c = 3$.
$|a-b| = |1-2| = 1 \leq 1$ (True,so $a R b$).
$|b-c| = |2-3| = 1 \leq 1$ (True,so $b R c$).
However,$|a-c| = |1-3| = 2 > 1$ (False,so $a$ is not related to $c$).
Thus,$R$ is not transitive.
Conclusion: $R$ is reflexive and symmetric but not transitive.

(C) Given the function $f(x) = 3 + |x|$.
We know that the derivative of $|x|$ is given by:
$f'(x) = \frac{d}{dx}(3 + |x|) = \frac{d}{dx}(|x|) = \text{sgn}(x)$
Where $\text{sgn}(x)$ is the signum function defined as:
$f'(x) = \begin{cases} 1, & x > 0 \\ -1, & x < 0 \end{cases}$
The function $f'(x)$ is undefined at $x = 0$.
Looking at the values of $f'(x)$,it takes only two values: $1$ for $x > 0$ and $-1$ for $x < 0$.
The minimum value of this function $f'(x)$ is $-1$.

(B) Given that the rate of change of the radius is $ \frac{dr}{dt} = 5 \text{ cm s}^{-1} $.
The area of a circle is given by $ A = \pi r^2 $.
Differentiating both sides with respect to time $ t $,we get:
$ \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2\pi r \frac{dr}{dt} $.
Substituting the given values $ r = 8 \text{ cm} $ and $ \frac{dr}{dt} = 5 \text{ cm s}^{-1} $:
$ \frac{dA}{dt} = 2 \pi (8) (5) = 80 \pi \text{ cm}^2 \text{ s}^{-1} $.
Thus,the enclosed area is increasing at a rate of $ 80 \pi \text{ cm}^2 \text{ s}^{-1} $.

(A) Let $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \log (\sec \theta - \tan \theta) \, d\theta$.
Consider the function $f(\theta) = \log (\sec \theta - \tan \theta)$.
We check if $f(\theta)$ is an odd function by evaluating $f(-\theta)$:
$f(-\theta) = \log (\sec(-\theta) - \tan(-\theta)) = \log (\sec \theta + \tan \theta)$.
Since $\sec \theta + \tan \theta = \frac{1}{\sec \theta - \tan \theta}$,we have:
$f(-\theta) = \log \left( \frac{1}{\sec \theta - \tan \theta} \right) = -\log (\sec \theta - \tan \theta) = -f(\theta)$.
Since $f(\theta)$ is an odd function and the interval $[-\frac{\pi}{4}, \frac{\pi}{4}]$ is symmetric about the origin,the integral of an odd function over this interval is $0$.
Therefore,$I = 0$.

(A) Given that,$a * b = 1 + ab$ $\rightarrow (1)$
For commutativity,check if $a * b = b * a$:
$b * a = 1 + ba = 1 + ab = a * b$
Since $a * b = b * a$,the operation is commutative.
For associativity,check if $(a * b) * c = a * (b * c)$:
$(a * b) * c = (1 + ab) * c = 1 + (1 + ab)c = 1 + c + abc$
$a * (b * c) = a * (1 + bc) = 1 + a(1 + bc) = 1 + a + abc$
Since $1 + c + abc \neq 1 + a + abc$,the operation is not associative.
Therefore,the operation is commutative but not associative.

(B) Given that,$\sin \left(2 \sin ^{-1} 0.8\right)$.
Let $\sin ^{-1} 0.8 = \theta$,which implies $\sin \theta = 0.8$.
Since $\cos^2 \theta = 1 - \sin^2 \theta$,we have $\cos \theta = \sqrt{1 - (0.8)^2} = \sqrt{1 - 0.64} = \sqrt{0.36} = 0.6$.
Using the trigonometric identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we substitute the values:
$\sin 2\theta = 2 \times 0.8 \times 0.6$.
$\sin 2\theta = 1.6 \times 0.6 = 0.96$.
Thus,the value is $0.96$.

(C) Let the given expression be $E = \tan \left[\sin ^{-1}\left\{\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^{2}}}{\sqrt{2}}\right\}-\sin ^{-1} x\right]$.
Substitute $x = \sin \theta$,where $\theta = \sin^{-1} x$.
Since $0 \leq x \leq \frac{1}{2}$,we have $0 \leq \theta \leq \frac{\pi}{6}$.
The expression inside the $\sin^{-1}$ becomes $\frac{\sin \theta}{\sqrt{2}} + \frac{\cos \theta}{\sqrt{2}} = \sin \theta \cos \frac{\pi}{4} + \cos \theta \sin \frac{\pi}{4} = \sin \left(\theta + \frac{\pi}{4}\right)$.
Now,the expression becomes $E = \tan \left[\sin ^{-1}\left\{\sin \left(\theta + \frac{\pi}{4}\right)\right\} - \theta\right]$.
Since $0 \leq \theta \leq \frac{\pi}{6}$,then $\frac{\pi}{4} \leq \theta + \frac{\pi}{4} \leq \frac{\pi}{6} + \frac{\pi}{4} = \frac{5\pi}{12}$.
Since $\frac{\pi}{4} \leq \theta + \frac{\pi}{4} \leq \frac{5\pi}{12}$,the value lies within the principal range of $\sin^{-1}$,so $\sin^{-1}(\sin(\theta + \frac{\pi}{4})) = \theta + \frac{\pi}{4}$.
Thus,$E = \tan \left(\theta + \frac{\pi}{4} - \theta\right) = \tan \frac{\pi}{4} = 1$.

(B) We are given that $P(A) \neq 0$. The conditional probability is defined as $P(B \mid A) = \frac{P(A \cap B)}{P(A)}$.
$(i)$ If $A \subset B$,then $A \cap B = A$. Therefore,$P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{P(A)}{P(A)} = 1$.
(ii) If $A \cap B = \phi$,then $P(A \cap B) = P(\phi) = 0$. Therefore,$P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{0}{P(A)} = 0$.
Thus,the values are $1$ and $0$ respectively.

(B) Given that $f(x)=(x+1)^{2}$ for $x \geq -1$.
Since $g(x)$ is the reflection of the graph of $f(x)$ in the line $y=x$,$g(x)$ is the inverse function of $f(x)$,denoted as $f^{-1}(x)$.
To find the inverse,let $y = (x+1)^{2}$.
Since $x \geq -1$,we have $x+1 \geq 0$,so taking the square root of both sides gives $\sqrt{y} = x+1$.
Solving for $x$,we get $x = \sqrt{y} - 1$.
By interchanging $x$ and $y$,we obtain $f^{-1}(x) = \sqrt{x} - 1$.
Thus,$g(x) = \sqrt{x} - 1$.

(D) Statement $(a)$ is a fundamental property of determinants: if two rows or columns are identical,the determinant is $0$.
Statement $(b)$ refers to the property that the determinant of a matrix $A$ is equal to the determinant of its transpose $A^T$,i.e.,$|A| = |A^T|$. Thus,interchanging rows and columns does not change the value.
Statement $(c)$ is a property stating that swapping any two rows or columns multiplies the determinant by $-1$,thus changing its sign.
Therefore,all statements $(a)$,$(b)$,and $(c)$ are correct.

(B) The symmetric part of a square matrix $A$ is given by the formula $\frac{1}{2}(A + A^T)$.
Given $A = \begin{bmatrix} 1 & 2 & 4 \\ 6 & 8 & 2 \\ 2 & -2 & 7 \end{bmatrix}$,its transpose $A^T$ is $\begin{bmatrix} 1 & 6 & 2 \\ 2 & 8 & -2 \\ 4 & 2 & 7 \end{bmatrix}$.
Now,$A + A^T = \begin{bmatrix} 1+1 & 2+6 & 4+2 \\ 6+2 & 8+8 & 2-2 \\ 2+4 & -2+2 & 7+7 \end{bmatrix} = \begin{bmatrix} 2 & 8 & 6 \\ 8 & 16 & 0 \\ 6 & 0 & 14 \end{bmatrix}$.
Finally,$\frac{1}{2}(A + A^T) = \frac{1}{2} \begin{bmatrix} 2 & 8 & 6 \\ 8 & 16 & 0 \\ 6 & 0 & 14 \end{bmatrix} = \begin{bmatrix} 1 & 4 & 3 \\ 4 & 8 & 0 \\ 3 & 0 & 7 \end{bmatrix}$.

(A) Given that $A$ is a $3 \times 4$ matrix.
Then,the transpose matrix $A^{\prime}$ is a $4 \times 3$ matrix.
For the product $A^{\prime}B$ to be defined,the number of columns in $A^{\prime}$ must equal the number of rows in $B$. Since $A^{\prime}$ is $4 \times 3$,$B$ must have $3$ rows.
For the product $BA^{\prime}$ to be defined,the number of columns in $B$ must equal the number of rows in $A^{\prime}$. Since $A^{\prime}$ has $4$ rows,$B$ must have $4$ columns.
Therefore,$B$ must be a $3 \times 4$ matrix.

(D) We are given the property $A(\operatorname{adj} A) = |A|I$.
Comparing this with the given equation $A(\operatorname{adj} A) = 10I$,we get $|A| = 10$.
We know the property for the determinant of the adjoint of a matrix: $|\operatorname{adj} A| = |A|^{n-1}$,where $n$ is the order of the matrix.
Here,$n = 3$,so $|\operatorname{adj} A| = |A|^{3-1} = |A|^2$.
Substituting the value of $|A|$,we get $|\operatorname{adj} A| = 10^2 = 100$.

(D) Total number of bulbs $N = 100$.
Number of defective bulbs $D = 10$.
Number of non-defective bulbs $G = 100 - 10 = 90$.
We are drawing a sample of $n = 5$ bulbs without replacement.
The probability that none is defective is given by the hypergeometric distribution formula:
$P(X = 0) = \frac{\binom{90}{5} \binom{10}{0}}{\binom{100}{5}}$
$P(X = 0) = \frac{\frac{90 \times 89 \times 88 \times 87 \times 86}{5 \times 4 \times 3 \times 2 \times 1}}{\frac{100 \times 99 \times 98 \times 97 \times 96}{5 \times 4 \times 3 \times 2 \times 1}}$
$P(X = 0) = \frac{90 \times 89 \times 88 \times 87 \times 86}{100 \times 99 \times 98 \times 97 \times 96} \approx (0.9)^{5}$.
Given the options provided,the intended model is binomial approximation (sampling with replacement),where $p = \frac{90}{100} = 0.9$ is the probability of picking a non-defective bulb.
Thus,$P(X = 0) = (0.9)^{5} = (\frac{9}{10})^{5}$.

(B) Given the adjacent sides of the parallelogram are $\vec{a} = \hat{i} + \hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} + \hat{k}$.
The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by the magnitude of their cross product: $|\vec{a} \times \vec{b}|$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 1 \\ 2 & 1 & 1 \end{vmatrix}$
$= \hat{i}(0(1) - 1(1)) - \hat{j}(1(1) - 1(2)) + \hat{k}(1(1) - 0(2))$
$= \hat{i}(-1) - \hat{j}(-1) + \hat{k}(1)$
$= -\hat{i} + \hat{j} + \hat{k}$.
Now,calculate the magnitude of the resulting vector:
$|\vec{a} \times \vec{b}| = \sqrt{(-1)^2 + (1)^2 + (1)^2}$
$= \sqrt{1 + 1 + 1} = \sqrt{3}$.
Thus,the area of the parallelogram is $\sqrt{3}$ square units.

(A) Given that $|\vec{a}| = |\vec{b}| = 1$ and the angle $\theta = \frac{\pi}{3}$.
Using the property of the magnitude of the sum of two vectors:
$|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b})$
$|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}| \cos \theta$
Substituting the given values:
$|\vec{a} + \vec{b}|^2 = 1^2 + 1^2 + 2(1)(1) \cos \frac{\pi}{3}$
$|\vec{a} + \vec{b}|^2 = 1 + 1 + 2 \times \frac{1}{2}$
$|\vec{a} + \vec{b}|^2 = 1 + 1 + 1 = 3$
Taking the square root on both sides:
$|\vec{a} + \vec{b}| = \sqrt{3}$
Since $\sqrt{3} \approx 1.732$,which is greater than $1$,the value of $|\vec{a} + \vec{b}|$ is greater than $1$.

(C) Given the function $f: N \rightarrow N$ defined by $f(n) = \begin{cases} \frac{n+1}{2} & n \text{ is odd} \\ \frac{n}{2} & n \text{ is even} \end{cases}$.
To check for one-one: Consider $n=1$ (odd) and $n=2$ (even).
$f(1) = \frac{1+1}{2} = 1$.
$f(2) = \frac{2}{2} = 1$.
Since $f(1) = f(2)$ but $1 \neq 2$,the function $f$ is not one-one.
To check for onto: For any $y \in N$,we need to find $n \in N$ such that $f(n) = y$.
If $y$ is any natural number,we can choose $n = 2y$ (which is even). Then $f(2y) = \frac{2y}{2} = y$.
Since for every $y \in N$,there exists an $n \in N$ such that $f(n) = y$,the function $f$ is onto.
Thus,$f$ is onto but not one-one.

(C) The scalar triple product is defined as $[\vec{x} \quad \vec{y} \quad \vec{z}] = \vec{x} \cdot (\vec{y} \times \vec{z})$.
Given expression: $[\vec{a}-\vec{b} \quad \vec{b}-\vec{c} \quad \vec{c}-\vec{a}] = (\vec{a}-\vec{b}) \cdot ((\vec{b}-\vec{c}) \times (\vec{c}-\vec{a}))$.
First,calculate the cross product: $(\vec{b}-\vec{c}) \times (\vec{c}-\vec{a}) = \vec{b} \times \vec{c} - \vec{b} \times \vec{a} - \vec{c} \times \vec{c} + \vec{c} \times \vec{a}$.
Since $\vec{c} \times \vec{c} = 0$,this simplifies to $\vec{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a}$.
Now,take the dot product with $(\vec{a}-\vec{b})$:
$(\vec{a}-\vec{b}) \cdot (\vec{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a})$
$= \vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{a} \cdot (\vec{a} \times \vec{b}) + \vec{a} \cdot (\vec{c} \times \vec{a}) - \vec{b} \cdot (\vec{b} \times \vec{c}) - \vec{b} \cdot (\vec{a} \times \vec{b}) - \vec{b} \cdot (\vec{c} \times \vec{a})$.
Using the property that the scalar triple product is zero if any two vectors are identical:
$= [\vec{a} \vec{b} \vec{c}] + 0 + 0 - 0 - 0 - [\vec{b} \vec{c} \vec{a}]$.
Since $[\vec{a} \vec{b} \vec{c}] = [\vec{b} \vec{c} \vec{a}]$,the expression becomes $[\vec{a} \vec{b} \vec{c}] - [\vec{a} \vec{b} \vec{c}] = 0$.

(D) The given constraints are $x+y \leq 2$,$x \geq 0$,and $y \geq 0$.
These constraints define a feasible region in the first quadrant with corner points $(0,0)$,$(2,0)$,and $(0,2)$.
We evaluate the objective function $Z = 3x+2y$ at these corner points:
$1$. At $(0,0)$: $Z = 3(0) + 2(0) = 0$.
$2$. At $(2,0)$: $Z = 3(2) + 2(0) = 6$.
$3$. At $(0,2)$: $Z = 3(0) + 2(2) = 4$.
Comparing these values,the maximum value is $6$,which occurs at the point $(2,0)$.

(D) Let $I = \int \frac{\sin(2x)}{\sin^2(x) + 2\cos^2(x)} dx$.
Using the identity $\sin^2(x) = 1 - \cos^2(x)$,the denominator becomes $1 - \cos^2(x) + 2\cos^2(x) = 1 + \cos^2(x)$.
So,$I = \int \frac{\sin(2x)}{1 + \cos^2(x)} dx$.
Let $t = 1 + \cos^2(x)$.
Then $dt = 2\cos(x)(-\sin(x)) dx = -\sin(2x) dx$.
This implies $\sin(2x) dx = -dt$.
Substituting these into the integral:
$I = \int \frac{-dt}{t} = -\log |t| + c$.
Substituting back $t = 1 + \cos^2(x)$,we get $I = -\log |1 + \cos^2(x)| + c$.
Thus,option $D$ is correct.