KCET 2024 Mathematics Question Paper with Answer and Solution

(C) For the logarithm function $f(x) = \log_{b}(x)$ with base $b > 1$:
$1$. The domain is $(0, \infty)$,which is the set of all positive real numbers $R^{+}$.
$2$. The range is $(-\infty, \infty)$,which is the set of all real numbers $R$.
$3$. Since $\log_{b}(1) = 0$ for any base $b > 0, b \neq 1$,the point $(1, 0)$ always lies on the graph.
$4$. Since $b > 1$,the function is strictly increasing,not decreasing.
Therefore,the correct observation is that the point $(1, 0)$ is always on the graph of the logarithm function.

(C) Let $z = \frac{1-i \sin \alpha}{1+2 i \sin \alpha}$.
To make $z$ purely real,its imaginary part must be zero.
Multiply the numerator and denominator by the conjugate of the denominator $(1-2i \sin \alpha)$:
$z = \frac{(1-i \sin \alpha)(1-2i \sin \alpha)}{(1+2i \sin \alpha)(1-2i \sin \alpha)}$
$z = \frac{1 - 2i \sin \alpha - i \sin \alpha + 2i^2 \sin^2 \alpha}{1 + 4 \sin^2 \alpha}$
Since $i^2 = -1$,we have:
$z = \frac{1 - 3i \sin \alpha - 2 \sin^2 \alpha}{1 + 4 \sin^2 \alpha} = \frac{1 - 2 \sin^2 \alpha}{1 + 4 \sin^2 \alpha} - i \frac{3 \sin \alpha}{1 + 4 \sin^2 \alpha}$
For $z$ to be purely real,the imaginary part must be $0$:
$-\frac{3 \sin \alpha}{1 + 4 \sin^2 \alpha} = 0$
$\Rightarrow 3 \sin \alpha = 0$
$\Rightarrow \sin \alpha = 0$
$\Rightarrow \alpha = n \pi, n \in N$.

(A) We use the Pascal's identity: ${ }^{n} C_{r}+{ }^{n} C_{r+1}={ }^{n+1} C_{r+1}$.
Given expression: ${ }^{49} C_3+{ }^{48} C_3+{ }^{47} C_3+{ }^{46} C_3+{ }^{45} C_3+{ }^{45} C_4$.
Step $1$: Combine ${ }^{45} C_3+{ }^{45} C_4 = { }^{46} C_4$.
Step $2$: Combine ${ }^{46} C_3+{ }^{46} C_4 = { }^{47} C_4$.
Step $3$: Combine ${ }^{47} C_3+{ }^{47} C_4 = { }^{48} C_4$.
Step $4$: Combine ${ }^{48} C_3+{ }^{48} C_4 = { }^{49} C_4$.
Step $5$: Combine ${ }^{49} C_3+{ }^{49} C_4 = { }^{50} C_4$.
Thus,the final value is ${ }^{50} C_4$.

(B) The sum of $n$ terms of a $GP$ is given by $S_n = \frac{a(1 - r^n)}{1 - r}$.
Similarly,the sum of $2n$ terms is $S_{2n} = \frac{a(1 - r^{2n})}{1 - r}$.
Taking the ratio,we have $\frac{S_n}{S_{2n}} = \frac{a(1 - r^n)}{1 - r} \times \frac{1 - r}{a(1 - r^{2n})}$.
Using the identity $1 - r^{2n} = (1 - r^n)(1 + r^n)$,we get $\frac{S_n}{S_{2n}} = \frac{1 - r^n}{(1 - r^n)(1 + r^n)}$.
Simplifying this,we obtain $\frac{S_n}{S_{2n}} = \frac{1}{1 + r^n}$.

(D) Let $a$ and $b$ be the roots of the quadratic equation.
Then,the quadratic equation is given by $x^2-(a+b)x+ab=0$ ....$(i)$
It is given that $AM = 5$ and $GM = 4$.
Therefore,$\frac{a+b}{2} = 5 \Rightarrow a+b = 10$.
And $\sqrt{ab} = 4 \Rightarrow ab = 16$.
Substituting these values into equation $(i)$,we get $x^2-10x+16=0$.
Thus,the required quadratic equation is $x^2-10x+16=0$.

(A) We know that $C_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}$.
Thus,$\frac{C_r}{C_{r-1}} = \frac{n!}{r!(n-r)!} \times \frac{(r-1)!(n-r+1)!}{n!} = \frac{n-r+1}{r}$.
The given expression is $S = \sum_{r=1}^{n} r \frac{C_r}{C_{r-1}}$.
Substituting the ratio: $S = \sum_{r=1}^{n} r \left( \frac{n-r+1}{r} \right) = \sum_{r=1}^{n} (n-r+1)$.
Expanding the sum: $S = n + (n-1) + (n-2) + \ldots + 1$.
This is the sum of the first $n$ natural numbers,which is $\frac{n(n+1)}{2}$.

(C) In $\triangle ABC$,right-angled at $C$,the sides opposite to angles $A, B, C$ are $a, b, c$ respectively.
$\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{a}{b}$
$\tan B = \frac{\text{opposite}}{\text{adjacent}} = \frac{b}{a}$
$\tan A + \tan B = \frac{a}{b} + \frac{b}{a}$
$= \frac{a^2 + b^2}{ab}$
Since $\triangle ABC$ is a right-angled triangle,by Pythagoras theorem,$a^2 + b^2 = c^2$.
Therefore,$\tan A + \tan B = \frac{c^2}{ab}$.

(C) The slope of the line $x+y=3$ is $m_1 = -1$.
The slope of the line joining $(1,1)$ and $(-3,4)$ is $m_2 = \frac{4-1}{-3-1} = \frac{3}{-4} = -\frac{3}{4}$.
The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by $\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$.
Substituting the values: $\tan \theta = \left| \frac{-\frac{3}{4} - (-1)}{1 + (-1)(-\frac{3}{4})} \right| = \left| \frac{-\frac{3}{4} + 1}{1 + \frac{3}{4}} \right| = \left| \frac{\frac{1}{4}}{\frac{7}{4}} \right| = \frac{1}{7}$.
Therefore,$\theta = \tan ^{-1}\left(\frac{1}{7}\right)$.

(B) Let the radii of the two circles be $r_1$ and $r_2$.
Given that the arc length $l$ is the same for both circles.
The formula for arc length is $l = r \theta$,where $\theta$ is in radians.
For the first circle,$l = r_1 \times (30^{\circ} \times \frac{\pi}{180^{\circ}}) = r_1 \times \frac{\pi}{6}$.
For the second circle,$l = r_2 \times (78^{\circ} \times \frac{\pi}{180^{\circ}}) = r_2 \times \frac{13\pi}{30}$.
Since the arc lengths are equal,$r_1 \times \frac{\pi}{6} = r_2 \times \frac{13\pi}{30}$.
Dividing both sides by $\pi$,we get $\frac{r_1}{6} = \frac{13r_2}{30}$.
Therefore,$\frac{r_1}{r_2} = \frac{13 \times 6}{30} = \frac{13}{5}$.

(A) The focus of the parabola is $(a, 0) = (6, 0)$,which implies $a = 6$.
The directrix of the parabola is $x = -a$,which is $x = -6$.
Since the focus lies on the $x$-axis and the directrix is a vertical line,the parabola is of the form $y^2 = 4ax$.
Substituting $a = 6$ into the equation,we get $y^2 = 4(6)x$.
Therefore,the equation of the parabola is $y^2 = 24x$.

(C) Given limit: $L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \cos x-1}{\cot x-1}$
Substituting $x = \frac{\pi}{4}$,we get the form $\frac{\sqrt{2}(\frac{1}{\sqrt{2}})-1}{1-1} = \frac{0}{0}$.
Applying $L'\text{Hospital rule}$ by differentiating the numerator and denominator with respect to $x$:
$L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{\frac{d}{dx}(\sqrt{2} \cos x - 1)}{\frac{d}{dx}(\cot x - 1)}$
$L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{-\sqrt{2} \sin x}{-\operatorname{cosec}^2 x} = \lim _{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \sin x}{\operatorname{cosec}^2 x}$
$L = \lim _{x \rightarrow \frac{\pi}{4}} \sqrt{2} \sin x \sin^2 x = \lim _{x \rightarrow \frac{\pi}{4}} \sqrt{2} \sin^3 x$
$L = \sqrt{2} \times (\frac{1}{\sqrt{2}})^3 = \sqrt{2} \times \frac{1}{2\sqrt{2}} = \frac{1}{2}$.

(C) The negation of a statement involving the universal quantifier "For every" is formed by replacing it with the existential quantifier "There exists at least one" and negating the predicate.
Given the statement: "For every real number $x$,$x^2+5$ is positive".
The negation is: "There exists at least one real number $x$ such that $x^2+5$ is not positive".

(D) The given observations are $a, b, c, d, e$ with mean $m$ and standard deviation $S$.
Standard deviation is defined as $S = \sqrt{\frac{\sum (x_i - m)^2}{n}}$.
When a constant $k$ is added to each observation,the new observations are $x_i' = x_i + k$.
The new mean $m'$ is $\frac{\sum (x_i + k)}{n} = \frac{\sum x_i}{n} + k = m + k$.
The new standard deviation $S'$ is $\sqrt{\frac{\sum (x_i' - m')^2}{n}}$.
Substituting the values,$S' = \sqrt{\frac{\sum ((x_i + k) - (m + k))^2}{n}} = \sqrt{\frac{\sum (x_i - m)^2}{n}}$.
Thus,$S' = S$.
Adding a constant to each observation does not change the standard deviation.

(C) Let set $A$ have $m$ elements and set $B$ have $n$ elements.
The number of subsets of set $A$ is $2^m$ and the number of subsets of set $B$ is $2^n$.
According to the problem,$2^m - 2^n = 56$.
We can write this as $2^n(2^{m-n} - 1) = 56$.
Factorizing $56$,we get $56 = 8 \times 7 = 2^3 \times (2^3 - 1)$.
Comparing $2^n(2^{m-n} - 1) = 2^3(2^3 - 1)$,we get $n = 3$ and $m - n = 3$.
Substituting $n = 3$ into $m - n = 3$,we get $m - 3 = 3$,which implies $m = 6$.
Thus,the values of $m$ and $n$ are $6$ and $3$ respectively.

(C) To find the pre-image of $17$,we set $f(x) = 17$:
$x^2 + 1 = 17$
$x^2 = 16$
$x = \pm 4$
So,the pre-image of $17$ is $\{4, -4\}$.
To find the pre-image of $-3$,we set $f(x) = -3$:
$x^2 + 1 = -3$
$x^2 = -4$
Since the square of a real number cannot be negative,there is no real value of $x$ such that $f(x) = -3$.
Thus,the pre-image of $-3$ is $\phi$ (the empty set).
Therefore,the pre-images are $\{4, -4\}$ and $\phi$.

(B) Given the equation $[x]^2-5[x]+6=0$.
Let $[x] = y$,then the equation becomes $y^2-5y+6=0$.
Factoring the quadratic equation: $(y-3)(y-2)=0$.
So,$[x]=2$ or $[x]=3$.
If $[x]=2$,then $2 \le x < 3$,which means $x \in [2, 3)$.
If $[x]=3$,then $3 \le x < 4$,which means $x \in [3, 4)$.
Combining these two intervals,we get $x \in [2, 3) \cup [3, 4) = [2, 4)$.
Therefore,the correct option is $B$.

(B) Let the breadth of the rectangle be $x \ cm$.
Then,the length of the rectangle is $5x \ cm$.
The perimeter of a rectangle is given by $P = 2(\text{length} + \text{breadth})$.
Substituting the values,$P = 2(5x + x) = 2(6x) = 12x$.
Given that the minimum perimeter is $180 \ cm$,we have $P \geq 180$.
Therefore,$12x \geq 180$.
Dividing both sides by $12$,we get $x \geq 15$.
Since $x$ represents the breadth,the breadth must be at least $15 \ cm$.

(B) Given $f(x) = x \cdot e^{x-x^2}$.
To find the intervals of increase or decrease,we compute the derivative $f'(x)$.
Using the product rule and chain rule: $f'(x) = 1 \cdot e^{x-x^2} + x \cdot e^{x-x^2} \cdot (1-2x)$.
$f'(x) = e^{x-x^2} [1 + x(1-2x)] = e^{x-x^2} [1 + x - 2x^2]$.
Factoring the quadratic expression: $1 + x - 2x^2 = -(2x^2 - x - 1) = -(2x+1)(x-1) = (2x+1)(1-x)$.
So,$f'(x) = e^{x-x^2} (2x+1)(1-x)$.
Since $e^{x-x^2} > 0$ for all $x \in R$,the sign of $f'(x)$ depends on $(2x+1)(1-x)$.
$f'(x) > 0$ when $(2x+1)(1-x) > 0$,which implies $-\frac{1}{2} < x < 1$.
Thus,$f(x)$ is increasing in the interval $\left(-\frac{1}{2}, 1\right)$.

(D) Let $\theta = \cot ^{-1} \sqrt{\frac{2+x}{2-x}}$. Then $\cot \theta = \sqrt{\frac{2+x}{2-x}}$.
From the right-angled triangle with base $\sqrt{2+x}$ and perpendicular $\sqrt{2-x}$,the hypotenuse is $\sqrt{(\sqrt{2+x})^2 + (\sqrt{2-x})^2} = \sqrt{2+x+2-x} = \sqrt{4} = 2$.
Thus,$\cos \theta = \frac{\text{base}}{\text{hypotenuse}} = \frac{\sqrt{2+x}}{2}$.
Now,the expression becomes $\frac{d}{d x}\left[\cos ^2 \theta\right] = \frac{d}{d x}\left[\left(\frac{\sqrt{2+x}}{2}\right)^2\right]$.
$= \frac{d}{d x}\left(\frac{2+x}{4}\right) = \frac{d}{d x}\left(\frac{1}{2} + \frac{x}{4}\right) = 0 + \frac{1}{4} = \frac{1}{4}$.

(A) Given $f(x+y)=f(x)f(y)$.
Putting $x=0, y=5$,we get $f(5)=f(0)f(5)$.
Since $f(5)=3 \neq 0$,we have $f(0)=1$.
Now,$f^{\prime}(5) = \lim_{h \to 0} \frac{f(5+h)-f(5)}{h}$.
Using the given functional equation,$f(5+h)=f(5)f(h)$.
So,$f^{\prime}(5) = \lim_{h \to 0} \frac{f(5)f(h)-f(5)}{h} = f(5) \lim_{h \to 0} \frac{f(h)-1}{h}$.
Since $f(0)=1$,this is $f(5) \lim_{h \to 0} \frac{f(h)-f(0)}{h} = f(5)f^{\prime}(0)$.
Substituting the values,$f^{\prime}(5) = 3 \times 2 = 6$.

(C) The given expression is $\lim _{n \rightarrow \infty} \sum_{r=1}^{2 n} \frac{n}{n^2+r^2}$.
Dividing the numerator and denominator by $n^2$,we get $\lim _{n \rightarrow \infty} \sum_{r=1}^{2 n} \frac{1}{n} \left(\frac{1}{1+(r/n)^2}\right)$.
Using the definition of a definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \sum_{r=1}^{kn} \frac{1}{n} f(r/n) = \int_0^k f(x) dx$.
Here,$f(x) = \frac{1}{1+x^2}$ and $k = 2$.
Thus,the integral is $\int_0^2 \frac{1}{1+x^2} dx$.
Evaluating the integral,we get $[\tan ^{-1}(x)]_0^2 = \tan ^{-1}(2) - \tan ^{-1}(0) = \tan ^{-1}(2)$.

(C) The equivalence class of $(3, 2)$ consists of all pairs $(x, y) \in A \times A$ such that $(x, y) R (3, 2)$.
This implies $2x = 3y$,or $\frac{x}{y} = \frac{3}{2}$.
Since $x, y \in \{2, 3, \ldots, 18\}$,we look for multiples of $(3, 2)$ such that both components are $\le 18$:
$(x, y) = (3, 2)$
$(x, y) = (6, 4)$
$(x, y) = (9, 6)$
$(x, y) = (12, 8)$
$(x, y) = (15, 10)$
$(x, y) = (18, 12)$
Counting these,we have $6$ such ordered pairs.

(C) Given that $A^2 = A$.
We need to expand $(I+A)^3$ using the binomial expansion formula $(I+A)^3 = I^3 + 3I^2A + 3IA^2 + A^3$.
Since $I$ is the identity matrix,$I^n = I$ and $IA = AI = A$.
Substituting these,we get:
$(I+A)^3 = I + 3A + 3A^2 + A^3$.
Since $A^2 = A$,it follows that $A^3 = A^2 \cdot A = A \cdot A = A^2 = A$.
Substituting $A^2 = A$ and $A^3 = A$ into the expression:
$(I+A)^3 = I + 3A + 3(A) + A$.
$(I+A)^3 = I + 3A + 3A + A$.
$(I+A)^3 = I + 7A$.

(B) Given $A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$.
First,calculate $A^2$:
$A^2 = A \times A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1+1 & 1+1 \\ 1+1 & 1+1 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} = 2 \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = 2A$.
Now,calculate $A^3$:
$A^3 = A^2 \times A = (2A) \times A = 2(A^2) = 2(2A) = 4A = 2^2 A$.
By observing the pattern,we can generalize for $A^n$:
$A^n = 2^{n-1} A$.
For $n = 10$:
$A^{10} = 2^{10-1} A = 2^9 A$.

(B) Given $f(x) = \left|\begin{array}{ccc} x-3 & 2(x^2-9) & 2x^3-81 \\ x-5 & 2(x^2-25) & 4(x^3-125) \\ 1 & 2 & 3 \end{array}\right|$.
For $x=5$,the second row becomes $5-5=0$,$2(25-25)=0$,and $4(125-125)=0$. Since a row is zero,$f(5)=0$.
Now calculate $f(1)$:
$f(1) = \left|\begin{array}{ccc} -2 & -16 & -79 \\ -4 & -48 & -496 \\ 1 & 2 & 3 \end{array}\right| = -2888$.
Now calculate $f(3)$:
$f(3) = \left|\begin{array}{ccc} 0 & 0 & -27 \\ -2 & -32 & -392 \\ 1 & 2 & 3 \end{array}\right| = -27(0 - (-32 + 32)) = 756$ (Correction: Calculating determinant: $-27(-4 - (-32)) = -27(28) = -756$).
Thus,$f(1) \cdot f(3) + f(3) \cdot f(5) + f(5) \cdot f(1) = (-2888 \times -756) + 0 + 0 = 2183328$.

(B) Given $f(x) = \left| \begin{array}{ccc} \cos x & x & 1 \\ 2 \sin x & x & 2x \\ \sin x & x & x \end{array} \right|$.
Expanding the determinant along the first row:
$f(x) = \cos x(x^2 - 2x^2) - x(2x \sin x - 2x \sin x) + 1(2x \sin x - x \sin x)$
$f(x) = \cos x(-x^2) - x(0) + x \sin x$
$f(x) = -x^2 \cos x + x \sin x$
Now,we need to evaluate $\lim_{x \rightarrow 0} \frac{f(x)}{x^2}$:
$\lim_{x \rightarrow 0} \frac{-x^2 \cos x + x \sin x}{x^2} = \lim_{x \rightarrow 0} \left( -\cos x + \frac{\sin x}{x} \right)$
Using the standard limit $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$ and $\cos(0) = 1$:
$\lim_{x \rightarrow 0} f(x) = -1 + 1 = 0$.

(A) Given $A = \begin{vmatrix} x & 1 \\ 1 & x \end{vmatrix} = x^2 - 1$.
The determinant $B$ is given by $B = \begin{vmatrix} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{vmatrix}$.
Expanding $B$ along the first row:
$B = x(x^2 - 1) - 1(x - 1) + 1(1 - x)$
$B = x(x^2 - 1) - (x - 1) - (x - 1)$
$B = x(x^2 - 1) - 2(x - 1)$
$B = x(x - 1)(x + 1) - 2(x - 1)$
$B = (x - 1)[x(x + 1) - 2]$
$B = (x - 1)(x^2 + x - 2)$
$B = (x - 1)(x + 2)(x - 1) = (x - 1)^2(x + 2) = x^3 - 3x + 2$.
Now,differentiating $B$ with respect to $x$:
$\frac{dB}{dx} = \frac{d}{dx}(x^3 - 3x + 2) = 3x^2 - 3 = 3(x^2 - 1)$.
Since $A = x^2 - 1$,we have $\frac{dB}{dx} = 3A$.

(C) Given that $\cos ^{-1}(x)+\cos ^{-1}(y)+\cos ^{-1}(z)=3 \pi$.
We know that the range of $\cos ^{-1}(\theta)$ is $[0, \pi]$.
Therefore,the maximum value of each term $\cos ^{-1}(x)$,$\cos ^{-1}(y)$,and $\cos ^{-1}(z)$ is $\pi$.
For their sum to be $3 \pi$,each term must individually be equal to $\pi$.
Thus,$\cos ^{-1}(x)=\pi$,$\cos ^{-1}(y)=\pi$,and $\cos ^{-1}(z)=\pi$.
This implies $x = \cos(\pi) = -1$,$y = \cos(\pi) = -1$,and $z = \cos(\pi) = -1$.
Now,substitute these values into the expression $x(y+z)+y(z+x)+z(x+y)$:
$= (-1)(-1-1) + (-1)(-1-1) + (-1)(-1-1)$
$= (-1)(-2) + (-1)(-2) + (-1)(-2)$
$= 2 + 2 + 2 = 6$.

(B) Given that $2 \sin ^{-1} x - 3 \cos ^{-1} x = 4$.
We know that $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,so $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} x$.
Substituting this into the given equation:
$2 \sin ^{-1} x - 3(\frac{\pi}{2} - \sin ^{-1} x) = 4$
$2 \sin ^{-1} x - \frac{3 \pi}{2} + 3 \sin ^{-1} x = 4$
$5 \sin ^{-1} x = 4 + \frac{3 \pi}{2}$
$\sin ^{-1} x = \frac{8 + 3 \pi}{10}$
Now,we need to find the value of $2 \sin ^{-1} x + 3 \cos ^{-1} x$.
$2 \sin ^{-1} x + 3 \cos ^{-1} x = 2 \sin ^{-1} x + 3(\frac{\pi}{2} - \sin ^{-1} x)$
$= 2 \sin ^{-1} x + \frac{3 \pi}{2} - 3 \sin ^{-1} x$
$= \frac{3 \pi}{2} - \sin ^{-1} x$
$= \frac{3 \pi}{2} - \frac{8 + 3 \pi}{10}$
$= \frac{15 \pi - 8 - 3 \pi}{10}$
$= \frac{12 \pi - 8}{10} = \frac{6 \pi - 4}{5}$

(D) Given $(g \circ f)(x) = g(f(x)) = \sin x$ and $(f \circ g)(x) = f(g(x)) = (\sin \sqrt{x})^2$.
We test the options:
$(a)$ If $f(x) = \sin^2 x$ and $g(x) = x$,then $f(g(x)) = \sin^2 x$ and $g(f(x)) = \sin^2 x$. This does not match the given conditions.
$(b)$ If $f(x) = \sin \sqrt{x}$ and $g(x) = \sqrt{x}$,then $f(g(x)) = \sin \sqrt{\sqrt{x}} = \sin x^{1/4}$ and $g(f(x)) = \sqrt{\sin \sqrt{x}}$. This does not match.
$(c)$ If $f(x) = \sin^2 x$ and $g(x) = \sqrt{x}$,then $f(g(x)) = f(\sqrt{x}) = \sin^2 \sqrt{x} = (\sin \sqrt{x})^2$ and $g(f(x)) = g(\sin^2 x) = \sqrt{\sin^2 x} = |\sin x|$. This does not match the condition $g(f(x)) = \sin x$ exactly for all $x$.
$(d)$ If $f(x) = \sin \sqrt{x}$ and $g(x) = x^2$,then $f(g(x)) = f(x^2) = \sin \sqrt{x^2} = \sin |x|$ and $g(f(x)) = g(\sin \sqrt{x}) = (\sin \sqrt{x})^2$.
Note: The question implies finding the correct pair. Re-evaluating the provided options,none perfectly satisfy both conditions for all $x \in \mathbb{R}$. However,based on standard problem sets,there may be a typo in the question or options. Given the structure,option $(d)$ is the closest intended answer.

(B) Given $f(x) = \tan x$.
The inverse function $f^{-1}(y)$ is defined such that $f(f^{-1}(y)) = y$.
Here,we need to find $f^{-1}(1)$,so we set $f(x) = 1$.
$\tan x = 1$.
The general solution for $\tan x = \tan \alpha$ is $x = n \pi + \alpha$,where $n \in Z$.
Since $\tan(\frac{\pi}{4}) = 1$,the set of values for $x$ is $\{n \pi + \frac{\pi}{4} : n \in Z\}$.
Therefore,$f^{-1}(1) = \{n \pi + \frac{\pi}{4} : n \in Z\}$.

(B) The given function is $f(x) = |\cos x|$.
We know that the function $g(x) = \cos x$ is continuous and differentiable for all $x \in R$.
The modulus function $h(x) = |x|$ is continuous everywhere but not differentiable at $x = 0$.
Therefore,the composite function $f(x) = |\cos x|$ is continuous for all $x \in R$.
However,$f(x)$ will not be differentiable where the expression inside the modulus is zero,i.e.,at $\cos x = 0$.
This occurs at $x = (2n + 1) \frac{\pi}{2}$ for all $n \in Z$.
At these points,the graph of $f(x)$ has sharp corners (cusps),as shown in the graph.
Thus,$f(x)$ is everywhere continuous but not differentiable at odd multiples of $\frac{\pi}{2}$.

(B) Given $y = 2x^{3x}$.
Taking natural logarithm on both sides: $\ln y = \ln(2x^{3x}) = \ln 2 + 3x \ln x$.
Differentiating both sides with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = 0 + 3 \ln x + 3x \cdot \frac{1}{x} = 3 \ln x + 3$.
Thus,$\frac{dy}{dx} = y(3 \ln x + 3) = 2x^{3x}(3 \ln x + 3)$.
At $x = 1$,$\frac{dy}{dx} = 2(1)^{3(1)}(3 \ln 1 + 3)$.
Since $\ln 1 = 0$,we have $\frac{dy}{dx} = 2(1)(0 + 3) = 2 \times 3 = 6$.

(C) Let $f(x) = x^x$. Taking the natural logarithm on both sides,we get $\ln f(x) = x \ln x$.
Differentiating both sides with respect to $x$,we have $\frac{1}{f(x)} f'(x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$.
Thus,$f'(x) = x^x(1 + \ln x)$.
For the function to be strictly increasing,we require $f'(x) > 0$.
Since $x > 0$,$x^x$ is always positive.
Therefore,we must have $1 + \ln x > 0$,which implies $\ln x > -1$.
This is equivalent to $\ln x > \ln(\frac{1}{e})$.
Hence,$x > \frac{1}{e}$.

(B) Given the function $f(x)=x^3-6x^2+12x-3$. \\ First,find the first derivative: $f'(x)=3x^2-12x+12$. \\ Set $f'(x)=0$ to find critical points: $3(x^2-4x+4)=0 \Rightarrow 3(x-2)^2=0 \Rightarrow x=2$. \\ Next,find the second derivative: $f''(x)=6x-12$. \\ Evaluate the second derivative at $x=2$: $f''(2)=6(2)-12=0$. \\ Since $f''(2)=0$,we check the third derivative: $f'''(x)=6$. \\ Since $f'''(2)=6 \neq 0$,the concavity changes at $x=2$. \\ Therefore,$x=2$ is a point of inflexion.

(B) $\because$ Slant height of the cone,$L = 6$ units.
Let the radius be $r$ and height be $h$.
Volume $(V) = \frac{1}{3} \pi r^2 h$.
Since $L^2 = r^2 + h^2$,we have $r^2 = L^2 - h^2 = 36 - h^2$.
Substituting this into the volume formula: $V = \frac{1}{3} \pi (36 - h^2) h = \frac{1}{3} \pi (36h - h^3)$.
To find the maximum volume,differentiate $V$ with respect to $h$: $\frac{dV}{dh} = \frac{1}{3} \pi (36 - 3h^2)$.
Setting $\frac{dV}{dh} = 0$,we get $36 - 3h^2 = 0 \Rightarrow h^2 = 12 \Rightarrow h = 2 \sqrt{3}$ units.
Checking the second derivative: $\frac{d^2 V}{dh^2} = \frac{1}{3} \pi (-6h) = -2 \pi h$.
At $h = 2 \sqrt{3}$,$\frac{d^2 V}{dh^2} = -4 \sqrt{3} \pi < 0$,so the volume is maximum.
Maximum volume $V = \frac{1}{3} \pi (36 - (2 \sqrt{3})^2) (2 \sqrt{3}) = \frac{1}{3} \pi (36 - 12) (2 \sqrt{3}) = \frac{1}{3} \pi (24) (2 \sqrt{3}) = 16 \sqrt{3} \pi$ cu units.

(B) Given function $f(x) = x(x - 1)^2$ on the interval $[0, 2]$.
According to the Mean Value Theorem,there exists at least one $c \in (0, 2)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.
Here,$a = 0$ and $b = 2$.
$f(0) = 0(0 - 1)^2 = 0$.
$f(2) = 2(2 - 1)^2 = 2(1)^2 = 2$.
So,$f'(c) = \frac{2 - 0}{2 - 0} = 1$.
Now,find the derivative $f'(x)$:
$f(x) = x(x^2 - 2x + 1) = x^3 - 2x^2 + x$.
$f'(x) = 3x^2 - 4x + 1$.
Setting $f'(c) = 1$:
$3c^2 - 4c + 1 = 1$.
$3c^2 - 4c = 0$.
$c(3c - 4) = 0$.
This gives $c = 0$ or $c = 4/3$.
Since $c$ must be in the open interval $(0, 2)$,we choose $c = 4/3$.

(A) Let $I = \int \frac{\sin x}{3+4 \cos ^2 x} d x$.
Substitute $u = \cos x$,then $du = -\sin x d x$,or $\sin x d x = -du$.
Substituting these into the integral gives $I = \int \frac{-du}{3+4u^2} = -\int \frac{du}{3+(2u)^2}$.
Factor out $3$ from the denominator: $I = -\frac{1}{3} \int \frac{du}{1 + (\frac{2u}{\sqrt{3}})^2}$.
Let $t = \frac{2u}{\sqrt{3}}$,then $dt = \frac{2}{\sqrt{3}} du$,so $du = \frac{\sqrt{3}}{2} dt$.
Substituting $t$ into the integral: $I = -\frac{1}{3} \int \frac{\frac{\sqrt{3}}{2} dt}{1+t^2} = -\frac{\sqrt{3}}{6} \int \frac{dt}{1+t^2}$.
Since $\frac{\sqrt{3}}{6} = \frac{1}{2\sqrt{3}}$,we have $I = -\frac{1}{2\sqrt{3}} \tan^{-1}(t) + C$.
Substituting back $t = \frac{2 \cos x}{\sqrt{3}}$,we get $I = -\frac{1}{2\sqrt{3}} \tan^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right) + C$.

(B) We have $I = \int \frac{\sin \frac{5x}{2}}{\sin \frac{x}{2}} dx$.
Using the identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$,we multiply the numerator and denominator by $2 \cos \frac{x}{2}$:
$I = \int \frac{2 \sin \frac{5x}{2} \cos \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} dx = \int \frac{\sin(3x) + \sin(2x)}{\sin x} dx$.
Using the formulas $\sin 3x = 3 \sin x - 4 \sin^3 x$ and $\sin 2x = 2 \sin x \cos x$:
$I = \int \frac{3 \sin x - 4 \sin^3 x + 2 \sin x \cos x}{\sin x} dx = \int (3 - 4 \sin^2 x + 2 \cos x) dx$.
Using $4 \sin^2 x = 2(1 - \cos 2x)$:
$I = \int (3 - 2(1 - \cos 2x) + 2 \cos x) dx = \int (3 - 2 + 2 \cos 2x + 2 \cos x) dx$.
$I = \int (1 + 2 \cos 2x + 2 \cos x) dx = x + \sin 2x + 2 \sin x + C$.

(B) $I = \int \frac{1}{x[6(\log x)^2 + 7 \log x + 2]} dx$
Let $t = \log x$,then $dt = \frac{1}{x} dx$.
Substituting these into the integral,we get:
$I = \int \frac{dt}{6t^2 + 7t + 2}$
Factorizing the denominator: $6t^2 + 7t + 2 = 6t^2 + 4t + 3t + 2 = 2t(3t + 2) + 1(3t + 2) = (2t + 1)(3t + 2)$.
Using partial fractions: $\frac{1}{(3t + 2)(2t + 1)} = \frac{A}{3t + 2} + \frac{B}{2t + 1}$
$1 = A(2t + 1) + B(3t + 2)$
For $t = -\frac{1}{2}$,$1 = B(3(-\frac{1}{2}) + 2) = B(-\frac{3}{2} + 2) = B(\frac{1}{2}) \implies B = 2$.
For $t = -\frac{2}{3}$,$1 = A(2(-\frac{2}{3}) + 1) = A(-\frac{4}{3} + 1) = A(-\frac{1}{3}) \implies A = -3$.
Thus,$I = \int (\frac{-3}{3t + 2} + \frac{2}{2t + 1}) dt$
$I = -\frac{3 \log |3t + 2|}{3} + \frac{2 \log |2t + 1|}{2} + C$
$I = -\log |3t + 2| + \log |2t + 1| + C$
$I = \log |\frac{2t + 1}{3t + 2}| + C$
Substituting $t = \log x$ back,we get:
$I = \log |\frac{2 \log x + 1}{3 \log x + 2}| + C$

(D) Let $f(x) = (1-x^2) \sin x \cdot \cos^2 x$.
We check if the function is even or odd by evaluating $f(-x)$:
$f(-x) = (1-(-x)^2) \sin(-x) \cdot \cos^2(-x)$
Since $(-x)^2 = x^2$,$\sin(-x) = -\sin x$,and $\cos(-x) = \cos x$:
$f(-x) = (1-x^2) (-\sin x) \cdot (\cos x)^2$
$f(-x) = -(1-x^2) \sin x \cdot \cos^2 x = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.
According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) \, dx = 0$.
Therefore,$\int_{-\pi}^{\pi} (1-x^2) \sin x \cdot \cos^2 x \, dx = 0$.

(A) We need to evaluate the integral $I = \int_1^5 (|x-3| + |1-x|) \, dx$.
Since $x \in [1, 5]$,we have $|1-x| = x-1$.
Thus,$I = \int_1^5 |x-3| \, dx + \int_1^5 (x-1) \, dx$.
For the first part,$|x-3| = 3-x$ for $x \in [1, 3]$ and $|x-3| = x-3$ for $x \in [3, 5]$.
$\int_1^3 (3-x) \, dx = [3x - \frac{x^2}{2}]_1^3 = (9 - 4.5) - (3 - 0.5) = 4.5 - 2.5 = 2$.
$\int_3^5 (x-3) \, dx = [\frac{x^2}{2} - 3x]_3^5 = (12.5 - 15) - (4.5 - 9) = -2.5 - (-4.5) = 2$.
For the second part,$\int_1^5 (x-1) \, dx = [\frac{x^2}{2} - x]_1^5 = (12.5 - 5) - (0.5 - 1) = 7.5 - (-0.5) = 8$.
Adding these results: $I = 2 + 2 + 8 = 12$.

(B) Given equations are $y=3x$ and $y=x^2$.
To find the points of intersection,we set $3x = x^2$,which gives $x^2 - 3x = 0$,so $x(x-3) = 0$.
Thus,the points of intersection are $x=0$ and $x=3$.
The area of the region bounded by these curves is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=3$.
$\text{Area} = \int_{0}^{3} (3x - x^2) dx$
$= \left[ \frac{3x^2}{2} - \frac{x^3}{3} \right]_{0}^{3}$
$= \left( \frac{3(3)^2}{2} - \frac{(3)^3}{3} \right) - (0 - 0)$
$= \left( \frac{27}{2} - \frac{27}{3} \right)$
$= \frac{27}{2} - 9$
$= \frac{27 - 18}{2} = \frac{9}{2} \text{ square units}$.

(D) To find the area of the region bounded by $y=x$ and $y=x^3$,we first find the intersection points by setting $x^3 = x$.
This gives $x^3 - x = 0$,so $x(x^2 - 1) = 0$,which implies $x = -1, 0, 1$.
The area is symmetric about the origin.
Area $= 2 \int_0^1 (x - x^3) dx$
$= 2 \left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_0^1$
$= 2 \left( \frac{1}{2} - \frac{1}{4} \right)$
$= 2 \left( \frac{1}{4} \right) = 0.5 \text{ sq units}$.

(D) Given the differential equation $e^{dy/dx} = x+1$.
Taking the natural logarithm on both sides,we get $\frac{dy}{dx} = \log(x+1)$.
Integrating both sides with respect to $x$,we have $\int dy = \int \log(x+1) dx$.
Using integration by parts,$\int \log(x+1) dx = (x+1) \log(x+1) - (x+1) + C$.
Alternatively,$y = x \log(x+1) - \int \frac{x}{x+1} dx = x \log(x+1) - \int (1 - \frac{1}{x+1}) dx = x \log(x+1) - x + \log(x+1) + C$.
Thus,$y = (x+1) \log(x+1) - x + C$.
Given the initial condition $y(0) = 3$,we substitute $x=0$ and $y=3$: $3 = (0+1) \log(1) - 0 + C \Rightarrow 3 = 0 - 0 + C \Rightarrow C = 3$.
Substituting $C=3$ into the general solution,we get $y = (x+1) \log(x+1) - x + 3$.
Rearranging the terms,we obtain $y+x-3 = (x+1) \log(x+1)$.

(A) Let the point on the curve be $(x, y)$.
The equation of the tangent at $(x, y)$ is given by $Y - y = \frac{dy}{dx}(X - x)$,where $(X, Y)$ are the coordinates of any point on the tangent.
The $x$-intercept is found by setting $Y = 0$: $-y = \frac{dy}{dx}(X - x) \Rightarrow X = x - y \frac{dx}{dy}$.
The $y$-intercept is found by setting $X = 0$: $Y - y = \frac{dy}{dx}(-x) \Rightarrow Y = y - x \frac{dy}{dx}$.
According to the problem,the $x$-intercept is $2x$ and the $y$-intercept is $2y$.
So,$x - y \frac{dx}{dy} = 2x \Rightarrow -y \frac{dx}{dy} = x \Rightarrow -\frac{dx}{x} = \frac{dy}{y}$.
Integrating both sides: $-\int \frac{dx}{x} = \int \frac{dy}{y} \Rightarrow -\ln|x| = \ln|y| + \ln|C|$.
This simplifies to $\ln|y| + \ln|x| = \ln|C|$,which gives $xy = C$.

(C) Let $A$ be the origin $(0, 0, 0)$.
Then $\vec{AB}$ and $\vec{AC}$ represent the position vectors of vertices $B$ and $C$ respectively.
Let $M$ be the midpoint of side $BC$.
The position vector of $M$ is given by $\vec{AM} = \frac{\vec{AB} + \vec{AC}}{2}$.
Substituting the given vectors:
$\vec{AM} = \frac{(3\hat{i} + 4\hat{k}) + (5\hat{i} - 2\hat{j} + 4\hat{k})}{2}$
$\vec{AM} = \frac{(3+5)\hat{i} + (0-2)\hat{j} + (4+4)\hat{k}}{2}$
$\vec{AM} = \frac{8\hat{i} - 2\hat{j} + 8\hat{k}}{2} = 4\hat{i} - \hat{j} + 4\hat{k}$.
The length of the median $AM$ is the magnitude of vector $\vec{AM}$:
$|\vec{AM}| = \sqrt{4^2 + (-1)^2 + 4^2}$
$|\vec{AM}| = \sqrt{16 + 1 + 16} = \sqrt{33}$.

(C) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$.
Since $a+b$ is a unit vector,we have $|a+b| = 1$.
Squaring both sides,we get $|a+b|^2 = 1^2 = 1$.
Using the property $|a+b|^2 = |a|^2 + |b|^2 + 2(a \cdot b)$,we have $1 + 1 + 2(a \cdot b) = 1$.
This simplifies to $2 + 2(a \cdot b) = 1$,which gives $2(a \cdot b) = -1$,or $a \cdot b = -\frac{1}{2}$.
Since $a \cdot b = |a||b| \cos \theta$,we have $1 \times 1 \times \cos \theta = -\frac{1}{2}$.
Thus,$\cos \theta = -\frac{1}{2}$.
Since $\cos \theta = -\frac{1}{2}$,the angle $\theta$ is $\frac{2\pi}{3}$.

(B) The volume of a parallelopiped with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by the scalar triple product $|\vec{a} \cdot (\vec{b} \times \vec{c})|$, which is equivalent to the determinant of the matrix formed by the components of the vectors.
Given vectors are $\vec{a} = 0\hat{i} + 1\hat{j} + 1\hat{k}$, $\vec{b} = 1\hat{i} + 0\hat{j} + 1\hat{k}$, and $\vec{c} = 1\hat{i} + 1\hat{j} + 0\hat{k}$.
The volume $V = \left|\begin{array}{lll}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right|$.
Expanding the determinant along the first row:
$V = |0(0 - 1) - 1(0 - 1) + 1(1 - 0)|$.
$V = |0 + 1 + 1| = |2| = 2 \text{ cu units}$.

(D) Given $p=\frac{b \times c}{[a b c]}, q=\frac{c \times a}{[a b c]}, r=\frac{a \times b}{[a b c]}$.
We need to evaluate the expression $E = (a+b) \cdot p + (b+c) \cdot q + (c+a) \cdot r$.
Substituting the values of $p, q, r$:
$E = (a+b) \cdot \frac{b \times c}{[a b c]} + (b+c) \cdot \frac{c \times a}{[a b c]} + (c+a) \cdot \frac{a \times b}{[a b c]}$
$E = \frac{1}{[a b c]} [a \cdot (b \times c) + b \cdot (b \times c) + b \cdot (c \times a) + c \cdot (c \times a) + c \cdot (a \times b) + a \cdot (a \times b)]$
Using the property of scalar triple product $[x y z] = x \cdot (y \times z)$:
$E = \frac{1}{[a b c]} [[a b c] + 0 + [b c a] + 0 + [c a b] + 0]$
Since $[a b c] = [b c a] = [c a b]$:
$E = \frac{[a b c] + [a b c] + [a b c]}{[a b c]} = \frac{3[a b c]}{[a b c]} = 3$.

(A) The given lines are $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-5}{1}=\frac{z-6}{-5}$.
The direction ratios of the first line are $\vec{b_1} = -3\hat{i} + 2k\hat{j} + 2\hat{k}$.
The direction ratios of the second line are $\vec{b_2} = 3k\hat{i} + 1\hat{j} - 5\hat{k}$.
Since the lines are mutually perpendicular,their dot product must be zero,i.e.,$\vec{b_1} \cdot \vec{b_2} = 0$.
$(-3)(3k) + (2k)(1) + (2)(-5) = 0$.
$-9k + 2k - 10 = 0$.
$-7k - 10 = 0$.
$-7k = 10$.
$k = -\frac{10}{7}$.
Thus,the value of $k$ is $-\frac{10}{7}$.

(C) The equations of the planes are:
$2x + 3y + 4z = 4$ ....$(i)$
$4x + 6y + 8z = 12$
Dividing the second equation by $2$,we get:
$2x + 3y + 4z = 6$ ....$(ii)$
Since the normal vectors $(2, 3, 4)$ are the same,the planes are parallel.
The distance $D$ between two parallel planes $ax + by + cz = d_1$ and $ax + by + cz = d_2$ is given by the formula:
$D = \left| \frac{d_2 - d_1}{\sqrt{a^2 + b^2 + c^2}} \right|$
Substituting the values $a = 2, b = 3, c = 4, d_1 = 4$,and $d_2 = 6$:
$D = \left| \frac{6 - 4}{\sqrt{2^2 + 3^2 + 4^2}} \right| = \left| \frac{2}{\sqrt{4 + 9 + 16}} \right| = \frac{2}{\sqrt{29}}$ units.

(C) Given the equation $xy = 0$.
This implies that either $x = 0$ or $y = 0$.
In three-dimensional space,the equation $x = 0$ represents the $YZ$-plane.
The equation $y = 0$ represents the $ZX$-plane.
Since the $YZ$-plane and $ZX$-plane are perpendicular to each other,the equation $xy = 0$ represents a pair of planes at right angles.

(NONE) The equation of the line is $\frac{x-2}{2}=\frac{y-3}{4}=\frac{z-4}{-2}$.
This line is parallel to the vector $\vec{b} = 2\hat{i} + 4\hat{j} - 2\hat{k}$.
The equation of the plane is $2x - 2y + z = 5$.
The normal vector to the plane is $\vec{n} = 2\hat{i} - 2\hat{j} + \hat{k}$.
The sine of the angle $\theta$ between the line and the plane is given by $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.
Calculating the dot product: $\vec{b} \cdot \vec{n} = (2)(2) + (4)(-2) + (-2)(1) = 4 - 8 - 2 = -6$.
So,$|\vec{b} \cdot \vec{n}| = |-6| = 6$.
Calculating the magnitudes: $|\vec{b}| = \sqrt{2^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6}$.
$|\vec{n}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
Thus,$\sin \theta = \frac{6}{(2\sqrt{6})(3)} = \frac{6}{6\sqrt{6}} = \frac{1}{\sqrt{6}}$.

(A) The plane contains the point $P(3,2,0)$ and the line $\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}$.
Any point on the line is $Q(3,6,4)$.
The vector along the line is $\vec{v} = \hat{i} + 5\hat{j} + 4\hat{k}$.
The vector joining $P(3,2,0)$ and $Q(3,6,4)$ is $\vec{PQ} = (3-3)\hat{i} + (6-2)\hat{j} + (4-0)\hat{k} = 4\hat{j} + 4\hat{k}$.
The normal vector $\vec{n}$ to the plane is given by the cross product $\vec{v} \times \vec{PQ}$:
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 5 & 4 \\ 0 & 4 & 4 \end{vmatrix} = \hat{i}(20-16) - \hat{j}(4-0) + \hat{k}(4-0) = 4\hat{i} - 4\hat{j} + 4\hat{k}$.
Dividing by $4$,we can take the normal vector as $\vec{n}' = \hat{i} - \hat{j} + \hat{k}$.
The equation of the plane passing through $(3,2,0)$ is $1(x-3) - 1(y-2) + 1(z-0) = 0$.
$x - 3 - y + 2 + z = 0$.
$x - y + z = 1$.

(D) To find the minimum value of the objective function $Z = 4x + 6y$,we evaluate $Z$ at each corner point:
$1$. At $(0,2): Z = 4(0) + 6(2) = 12$
$2$. At $(3,0): Z = 4(3) + 6(0) = 12$
$3$. At $(6,0): Z = 4(6) + 6(0) = 24$
$4$. At $(6,8): Z = 4(6) + 6(8) = 24 + 48 = 72$
$5$. At $(0,5): Z = 4(0) + 6(5) = 30$
Since the minimum value of $Z$ is $12$,which occurs at both corner points $(0,2)$ and $(3,0)$,the minimum value of $Z$ occurs at every point on the line segment joining these two points.

(C) Given,$n=10$.
Probability of getting an odd number in a single throw,$p = \frac{3}{6} = \frac{1}{2}$.
Probability of not getting an odd number,$q = 1 - p = \frac{1}{2}$.
We need to find the probability that an odd number appears at least once,which is $P(X \geq 1)$.
Using the complement rule,$P(X \geq 1) = 1 - P(X=0)$.
Using the binomial distribution formula $P(X=k) = {}^{n}C_{k} p^{k} q^{n-k}$,we have:
$P(X=0) = {}^{10}C_{0} (\frac{1}{2})^{0} (\frac{1}{2})^{10} = 1 \times 1 \times \frac{1}{1024} = \frac{1}{1024}$.
Therefore,$P(X \geq 1) = 1 - \frac{1}{1024} = \frac{1023}{1024}$.

(B) We know that the sum of probabilities in a probability distribution is $1$. Therefore,$\frac{25}{36} + k + \frac{1}{36} = 1 \Rightarrow k + \frac{26}{36} = 1 \Rightarrow k = 1 - \frac{13}{18} = \frac{5}{18}$.
The mean of a random variable $X$ is given by $E(X) = \sum x_i P(x_i)$.
Given $E(X) = \frac{1}{3}$,we have:
$E(X) = 0 \times \frac{25}{36} + 1 \times k + 2 \times \frac{1}{36} = k + \frac{2}{36} = k + \frac{1}{18}$.
Since $E(X) = \frac{1}{3}$,we have $k + \frac{1}{18} = \frac{1}{3} \Rightarrow k = \frac{1}{3} - \frac{1}{18} = \frac{6-1}{18} = \frac{5}{18}$.
The variance of $X$ is given by $Var(X) = E(X^2) - [E(X)]^2$.
First,calculate $E(X^2) = \sum x_i^2 P(x_i)$:
$E(X^2) = 0^2 \times \frac{25}{36} + 1^2 \times k + 2^2 \times \frac{1}{36} = 0 + k + \frac{4}{36} = k + \frac{1}{9}$.
Substituting $k = \frac{5}{18}$:
$E(X^2) = \frac{5}{18} + \frac{1}{9} = \frac{5+2}{18} = \frac{7}{18}$.
Now,$Var(X) = \frac{7}{18} - (\frac{1}{3})^2 = \frac{7}{18} - \frac{1}{9} = \frac{7-2}{18} = \frac{5}{18}$.

(B) Given that the random variable $X$ follows a binomial distribution with parameters $n=5$ and $p$. The probability mass function is given by $P(X=k) = { }^n C_k p^k (1-p)^{n-k}$.
Given $P(X=2) = 9 P(X=3)$.
Substituting the values: ${ }^5 C_2 p^2 (1-p)^{5-2} = 9 \times { }^5 C_3 p^3 (1-p)^{5-3}$.
Since ${ }^5 C_2 = \frac{5 \times 4}{2 \times 1} = 10$ and ${ }^5 C_3 = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$,we have:
$10 p^2 (1-p)^3 = 9 \times 10 p^3 (1-p)^2$.
Dividing both sides by $10 p^2 (1-p)^2$ (assuming $p \neq 0, 1$):
$(1-p) = 9p$.
$1 = 10p$.
$p = 1/10$.