KCET 2015 Mathematics Question Paper with Answer and Solution

(B) The equation of the hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.
Since $2x+3y-6=0$ is a focal chord,it must pass through the focus $(ae, 0)$.
Substituting the coordinates of the focus into the equation of the chord:
$2(ae) + 3(0) - 6 = 0$
$2ae = 6$
$ae = 3$
Given the eccentricity $e = \frac{5}{4}$,we have:
$a \times \frac{5}{4} = 3$
$a = 3 \times \frac{4}{5} = \frac{12}{5}$
The length of the transverse axis is $2a$.
$2a = 2 \times \frac{12}{5} = \frac{24}{5}$.

(A) Given the quadratic expression $f(x) = x^{2}-8x+17$.
To find the minimum value,we can complete the square:
$f(x) = (x^{2}-8x+16) + 1$
$f(x) = (x-4)^{2} + 1$.
Since $(x-4)^{2} \ge 0$ for all real $x$,the minimum value occurs when $(x-4)^{2} = 0$,which is at $x = 4$.
Thus,the minimum value is $0 + 1 = 1$.

(A) Given that,$T_{2} = 24$ and $T_{5} = 3$.
In a $G$.$P$.,the $n^{\text{th}}$ term is given by $T_{n} = ar^{n-1}$.
So,$ar = 24$ $(1)$ and $ar^{4} = 3$ $(2)$.
Dividing equation $(2)$ by equation $(1)$,we get $\frac{ar^{4}}{ar} = \frac{3}{24}$ $\Rightarrow r^{3} = \frac{1}{8}$ $\Rightarrow r = \frac{1}{2}$.
Substituting $r = \frac{1}{2}$ in equation $(1)$,we get $a \times \frac{1}{2} = 24 \Rightarrow a = 48$.
The sum of the first $n$ terms of a $G$.$P$. is $S_{n} = \frac{a(1-r^{n})}{1-r}$.
For $n = 6$,$S_{6} = \frac{48(1-(1/2)^{6})}{1-1/2} = \frac{48(1-1/64)}{1/2} = 96 \times \frac{63}{64} = \frac{3 \times 63}{2} = \frac{189}{2}$.

(D) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
Let $S$ be the sum of the numbers on the two dice. We want to find the probability that $S > 5$.
It is easier to calculate the complement: the probability that $S \leq 5$.
The outcomes where $S \leq 5$ are:
$S=2: (1,1)$
$S=3: (1,2), (2,1)$
$S=4: (1,3), (2,2), (3,1)$
$S=5: (1,4), (2,3), (3,2), (4,1)$
Total outcomes where $S \leq 5$ is $1 + 2 + 3 + 4 = 10$.
Therefore,the number of outcomes where $S > 5$ is $36 - 10 = 26$.
The required probability is $P(S > 5) = \frac{26}{36} = \frac{13}{18}$.

(A) Given the expression: $ \sim[(\sim p) \wedge q] $.
By applying De Morgan's Law,which states that $ \sim(A \wedge B) = (\sim A) \vee (\sim B) $:
$ \sim[(\sim p) \wedge q] = \sim(\sim p) \vee (\sim q) $.
Since $ \sim(\sim p) = p $,the expression simplifies to:
$ p \vee (\sim q) $.
Therefore,the correct option is $ A $.

(D) Given equations are:
$\sin x + \sin y = \frac{1}{2} \quad (1)$
$\cos x + \cos y = 1 \quad (2)$
Using sum-to-product formulas:
$2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) = \frac{1}{2} \quad (3)$
$2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) = 1 \quad (4)$
Dividing equation $(3)$ by $(4)$:
$\frac{2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)} = \frac{1/2}{1}$
$\tan \left(\frac{x+y}{2}\right) = \frac{1}{2}$
Using the double angle formula $\tan(2\theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta}$ where $\theta = \frac{x+y}{2}$:
$\tan(x+y) = \frac{2 \tan \left(\frac{x+y}{2}\right)}{1 - \tan^2 \left(\frac{x+y}{2}\right)}$
Substituting $\tan \left(\frac{x+y}{2}\right) = \frac{1}{2}$:
$\tan(x+y) = \frac{2(1/2)}{1 - (1/2)^2} = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$

(B) The total number of red marbles is $6$ (numbered $1, 2, 3, 4, 5, 6$).
The total number of white marbles is $4$ (numbered $12, 13, 14, 15$).
The total number of marbles in the box is $6 + 4 = 10$.
We are looking for a marble that is both white and odd-numbered.
The white marbles are ${12, 13, 14, 15}$.
Among these,the odd-numbered marbles are $13$ and $15$.
Thus,there are $2$ favorable outcomes.
The probability $P$ is given by the ratio of favorable outcomes to the total number of outcomes:
$P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{10} = \frac{1}{5}$.

(C) Given $Z = \frac{(\sqrt{3} + i)^{3}(3i + 4)^{2}}{(8 + 6i)^{2}}$.
Taking the modulus on both sides,we get $|Z| = \frac{|\sqrt{3} + i|^{3} |3i + 4|^{2}}{|8 + 6i|^{2}}$.
Calculate the modulus of each complex number:
$|\sqrt{3} + i| = \sqrt{(\sqrt{3})^{2} + 1^{2}} = \sqrt{3 + 1} = \sqrt{4} = 2$.
$|3i + 4| = \sqrt{4^{2} + 3^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5$.
$|8 + 6i| = \sqrt{8^{2} + 6^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10$.
Substitute these values into the expression for $|Z|$:
$|Z| = \frac{2^{3} \times 5^{2}}{10^{2}} = \frac{8 \times 25}{100} = \frac{200}{100} = 2$.
Thus,$|Z| = 2$.

(A) Given the quadratic equation $x^{2}-ax+b^{2}=0$.
For a quadratic equation of the form $Ax^{2}+Bx+C=0$,the sum of the roots $\alpha+\beta = -B/A$ and the product of the roots $\alpha\beta = C/A$.
Here,$A=1$,$B=-a$,and $C=b^{2}$.
Therefore,$\alpha+\beta = -(-a)/1 = a$ and $\alpha\beta = b^{2}/1 = b^{2}$.
We need to find the value of $\alpha^{2}+\beta^{2}$.
Using the algebraic identity $\alpha^{2}+\beta^{2} = (\alpha+\beta)^{2}-2\alpha\beta$.
Substituting the values,we get $\alpha^{2}+\beta^{2} = (a)^{2}-2(b^{2}) = a^{2}-2b^{2}$.

(A) The vertices of the shaded triangular region are $A(0, 14)$,$B(5, 0)$,and $C(19, 14)$.
$1$. The line passing through $A(0, 14)$ and $B(5, 0)$ has the equation $\frac{x}{5} + \frac{y}{14} = 1$,which simplifies to $14x + 5y = 70$. Since the shaded region lies above this line (e.g.,testing point $(5, 14)$ gives $14(5) + 5(14) = 70 + 70 = 140 > 70$),the inequality is $14x + 5y \geq 70$.
$2$. The line passing through $A(0, 14)$ and $C(19, 14)$ is the horizontal line $y = 14$. Since the shaded region lies below this line,the inequality is $y \leq 14$.
$3$. The line passing through $B(5, 0)$ and $C(19, 14)$ has a slope $m = \frac{14 - 0}{19 - 5} = \frac{14}{14} = 1$. Using the point-slope form $y - 0 = 1(x - 5)$,we get $y = x - 5$,or $x - y = 5$. Since the shaded region lies to the left of this line (e.g.,testing point $(5, 14)$ gives $5 - 14 = -9 < 5$),the inequality is $x - y \leq 5$.
Thus,the correct set of inequations is $14x + 5y \geq 70$,$y \leq 14$,and $x - y \leq 5$.

(B) The given expression is $ \left(\frac{10}{x}+\frac{x}{10}\right)^{10} $.
Here,the power $ n = 10 $,which is an even number.
Therefore,the number of terms in the expansion is $ n + 1 = 11 $.
The middle term is the $ \left(\frac{n}{2} + 1\right) $-th term,which is $ \left(\frac{10}{2} + 1\right) = 6 $-th term.
The general term of the expansion $ (a+b)^n $ is given by $ T_{r+1} = {}^{n}C_{r} a^{n-r} b^{r} $.
For the $ 6 $-th term,$ r = 5 $.
$ T_{6} = {}^{10}C_{5} \left(\frac{10}{x}\right)^{10-5} \left(\frac{x}{10}\right)^{5} $.
$ T_{6} = {}^{10}C_{5} \left(\frac{10}{x}\right)^{5} \left(\frac{x}{10}\right)^{5} $.
$ T_{6} = {}^{10}C_{5} \times 1 = {}^{10}C_{5} $.

(B) Given expression: $\tan \left(1^{\circ}\right)+\tan \left(89^{\circ}\right)$
Since $\tan \left(90^{\circ}-\theta\right)=\cot \theta$,we have $\tan \left(89^{\circ}\right)=\tan \left(90^{\circ}-1^{\circ}\right)=\cot \left(1^{\circ}\right)$
So,the expression becomes: $\tan \left(1^{\circ}\right)+\cot \left(1^{\circ}\right)$
$= \frac{\sin \left(1^{\circ}\right)}{\cos \left(1^{\circ}\right)}+\frac{\cos \left(1^{\circ}\right)}{\sin \left(1^{\circ}\right)}$
$= \frac{\sin^2 \left(1^{\circ}\right)+\cos^2 \left(1^{\circ}\right)}{\sin \left(1^{\circ}\right) \cos \left(1^{\circ}\right)}$
$= \frac{1}{\sin \left(1^{\circ}\right) \cos \left(1^{\circ}\right)}$
Multiply numerator and denominator by $2$:
$= \frac{2}{2 \sin \left(1^{\circ}\right) \cos \left(1^{\circ}\right)}$
Using the identity $\sin \left(2\theta\right)=2 \sin \theta \cos \theta$:
$= \frac{2}{\sin \left(2^{\circ}\right)}$

(D) Step $ 1 $: Calculate the mean $ \bar{x} $ of the given data.
$ \bar{x} = \frac{3 + 10 + 10 + 4 + 7 + 10 + 5}{7} = \frac{49}{7} = 7 $.
Step $ 2 $: Calculate the absolute deviations from the mean $ |x_i - \bar{x}| $.
$ |3 - 7| = 4 $
$ |10 - 7| = 3 $
$ |10 - 7| = 3 $
$ |4 - 7| = 3 $
$ |7 - 7| = 0 $
$ |10 - 7| = 3 $
$ |5 - 7| = 2 $
Step $ 3 $: Calculate the mean of these absolute deviations.
$ \text{Mean Deviation} = \frac{4 + 3 + 3 + 3 + 0 + 3 + 2}{7} = \frac{18}{7} \approx 2.57 $.

(C) Given line is $3x+y=3$.
Rewriting in slope-intercept form $y=mx+c$,we get $y=-3x+3$.
Thus,the slope of this line is $m_1 = -3$.
Since the required line is perpendicular to this line,its slope $m_2$ must satisfy $m_1 \times m_2 = -1$.
So,$-3 \times m_2 = -1 \Rightarrow m_2 = 1/3$.
The equation of a line passing through $(x_1, y_1)$ with slope $m$ is $(y-y_1) = m(x-x_1)$.
Substituting $(x_1, y_1) = (2,2)$ and $m = 1/3$:
$y-2 = 1/3(x-2)$
$y-2 = x/3 - 2/3$
$y = x/3 - 2/3 + 2$
$y = x/3 + 4/3$.
The $y$-intercept is the value of $y$ when $x=0$,which is $4/3$.

(A) We need to find the remainder of the sum $S = 1! + 2! + 3! + \dots + 11!$ when divided by $12$.
Note that for any $n \ge 4$,$n!$ contains the factors $4 \times 3 = 12$.
Therefore,$n!$ is divisible by $12$ for all $n \ge 4$.
This means $4!, 5!, 6!, \dots, 11!$ are all divisible by $12$,so their remainders when divided by $12$ are $0$.
The sum becomes $S \equiv 1! + 2! + 3! + 0 + \dots + 0 \pmod{12}$.
$S \equiv 1 + 2 + 6 \pmod{12}$.
$S \equiv 9 \pmod{12}$.
Thus,the remainder is $9$.

(C) Given the inequation $\frac{x^{2}+6x-7}{|x+4|} < 0$.
Since $|x+4|$ is always positive for $x \neq -4$,the expression is negative only when the numerator is negative.
Thus,$x^{2}+6x-7 < 0$ and $x \neq -4$.
Factoring the quadratic expression,we get $(x+7)(x-1) < 0$.
This inequality holds for $x \in (-7, 1)$.
Excluding the point where the denominator is zero $(x = -4)$,the solution set is $(-7, -4) \cup (-4, 1)$.

(C) Given the set $ A = \{-1, 1\} $.
We analyze the given options:
For option $ C $,the equation is $ x^{2} = 1 $.
Solving $ x^{2} = 1 $,we get $ x = \pm 1 $.
Thus,the set of roots is $ \{-1, 1\} $.
Therefore,the set builder form is $ A = \{x : x \text{ is a root of the equation } x^{2} = 1\} $.

(D) We have,$(1-\omega+\omega^{2})(1+\omega-\omega^{2}) \quad \dots(1)$
We know that,$1+\omega+\omega^{2}=0$.
From this,we can write:
$1+\omega^{2} = -\omega$,so $(1-\omega+\omega^{2}) = -\omega - \omega = -2\omega$.
$1+\omega = -\omega^{2}$,so $(1+\omega-\omega^{2}) = -\omega^{2} - \omega^{2} = -2\omega^{2}$.
Substituting these values into Eq. $(1)$:
$(-2\omega)(-2\omega^{2}) = 4\omega^{3}$.
Since $\omega^{3}=1$,the expression becomes $4(1) = 4$.

(C) Given the limit $ \lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}} $.
Substituting $ x = 0 $,we get the indeterminate form $ \left(\frac{0}{0}\right) $.
Applying $L$'$H$ôpital's rule by differentiating the numerator and denominator with respect to $ x $:
$ \lim _{x \rightarrow 0} \frac{\frac{d}{dx}(1-\cos x)}{\frac{d}{dx}(x^{2})} = \lim _{x \rightarrow 0} \frac{\sin x}{2x} $.
Using the standard limit $ \lim _{x \rightarrow 0} \frac{\sin x}{x} = 1 $,we get:
$ \frac{1}{2} \times \lim _{x \rightarrow 0} \frac{\sin x}{x} = \frac{1}{2} \times 1 = \frac{1}{2} $.
Thus,the value of the limit is $ \frac{1}{2} $.

(B) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\Delta = \frac{1}{2} \left| \begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array} \right|$.
Given the determinant $\left|\begin{array}{ccc}2 a & x_{1} & y_{1} \\ 2 b & x_{2} & y_{2} \\ 2 c & x_{3} & y_{3}\end{array}\right|=\frac{a b c}{2}$.
Taking $2$ common from the first column,we get $2 \left|\begin{array}{ccc} a & x_{1} & y_{1} \\ b & x_{2} & y_{2} \\ c & x_{3} & y_{3}\end{array}\right|=\frac{a b c}{2}$.
Dividing by $abc$,we get $2 \left|\begin{array}{ccc} 1 & x_{1}/a & y_{1}/a \\ 1 & x_{2}/b & y_{2}/b \\ 1 & x_{3}/c & y_{3}/c \end{array}\right|=\frac{1}{2}$.
Thus,$\left|\begin{array}{ccc} 1 & x_{1}/a & y_{1}/a \\ 1 & x_{2}/b & y_{2}/b \\ 1 & x_{3}/c & y_{3}/c \end{array}\right|=\frac{1}{4}$.
The area of the triangle is $\frac{1}{2} \left| \begin{array}{ccc} x_1/a & y_1/a & 1 \\ x_2/b & y_2/b & 1 \\ x_3/c & y_3/c & 1 \end{array} \right|$.
Since the determinant value is $\frac{1}{4}$,the area is $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$.

(C) Given $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$.
First,calculate the magnitude of $\vec{a}$:
$|\vec{a}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
We are given $|\vec{b}| = 5$ and the angle $\theta = \frac{\pi}{6}$.
The area of a triangle formed by two vectors $\vec{a}$ and $\vec{b}$ as adjacent sides is given by the formula:
$\text{Area} = \frac{1}{2} |\vec{a}| |\vec{b}| \sin(\theta)$.
Substituting the values:
$\text{Area} = \frac{1}{2} \times 3 \times 5 \times \sin\left(\frac{\pi}{6}\right)$.
Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$,we get:
$\text{Area} = \frac{1}{2} \times 15 \times \frac{1}{2} = \frac{15}{4}$.

(D) Given the function $f(x) = [x]$.
We know that the greatest integer function $[x]$ is discontinuous at all integer values of $x$.
It is continuous at all non-integer values.
Among the given options,$4$,$-2$,and $11$ are integers,while $1.5$ is a non-integer.
Therefore,$f(x)$ is continuous at $x = 1.5$.

(C) Given $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$.
We have $|\vec{a}|^2 = 1^2 + (-2)^2 + 3^2 = 1 + 4 + 9 = 14$.
Given $|\vec{a} - \vec{b}| = \sqrt{7}$,squaring both sides gives $|\vec{a} - \vec{b}|^2 = 7$.
Expanding the dot product,we get $|\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b}) = 7$.
Substitute $\vec{a} \cdot \vec{b} = |\vec{b}|^2$ into the equation:
$14 + |\vec{b}|^2 - 2|\vec{b}|^2 = 7$.
$14 - |\vec{b}|^2 = 7$.
$|\vec{b}|^2 = 14 - 7 = 7$.
Therefore,$|\vec{b}| = \sqrt{7}$.

(B) The direction cosines of a vector are given as $l = \frac{2}{3}$,$m = -\frac{1}{3}$,and $n = \frac{2}{3}$.
We know that a vector $\vec{V}$ with magnitude $|\vec{V}|$ and direction cosines $(l, m, n)$ is given by $\vec{V} = |\vec{V}|(l\hat{i} + m\hat{j} + n\hat{k})$.
Given the magnitude $|\vec{V}| = 3$.
Substituting the values,we get:
$\vec{V} = 3 \left( \frac{2}{3}\hat{i} - \frac{1}{3}\hat{j} + \frac{2}{3}\hat{k} \right)$
$\vec{V} = 2\hat{i} - \hat{j} + 2\hat{k}$.

(B) The standard equation of a circle with a fixed radius '$a$' and variable center $(h, k)$ is given by:
$(x - h)^2 + (y - k)^2 = a^2$
Here,'$h$' and '$k$' are two arbitrary constants.
The order of a differential equation is equal to the number of independent arbitrary constants present in the general solution.
Since there are $2$ arbitrary constants,the order of the differential equation is $2$.

(D) Given the operation $ a \oplus b = a^{2} + b^{2} $.
First,calculate $ (2 \oplus 3) $:
$ 2 \oplus 3 = 2^{2} + 3^{2} = 4 + 9 = 13 $.
Now,substitute this result into the expression $ (2 \oplus 3) \oplus 4 $:
$ 13 \oplus 4 = 13^{2} + 4^{2} $.
$ 13^{2} + 4^{2} = 169 + 16 = 185 $.
Thus,the final result is $ 185 $.

(C) Given that $x=a \cos^{3} \theta$ and $y=a \sin^{3} \theta$.
First,find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = a(3 \cos^{2} \theta)(-\sin \theta) = -3a \cos^{2} \theta \sin \theta$.
$\frac{dy}{d\theta} = a(3 \sin^{2} \theta)(\cos \theta) = 3a \sin^{2} \theta \cos \theta$.
Now,calculate $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a \sin^{2} \theta \cos \theta}{-3a \cos^{2} \theta \sin \theta} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta$.
Finally,compute $1 + (\frac{dy}{dx})^{2}$:
$1 + (-\tan \theta)^{2} = 1 + \tan^{2} \theta = \sec^{2} \theta$.

(D) Given the equation: $\tan^{-1}\left(\frac{1-x}{1+x}\right) = \frac{1}{2} \tan^{-1} x$.
We know that $\tan^{-1}\left(\frac{1-x}{1+x}\right) = \tan^{-1}(1) - \tan^{-1}(x)$.
Since $\tan^{-1}(1) = \frac{\pi}{4}$,the equation becomes: $\frac{\pi}{4} - \tan^{-1} x = \frac{1}{2} \tan^{-1} x$.
Adding $\tan^{-1} x$ to both sides: $\frac{\pi}{4} = \frac{1}{2} \tan^{-1} x + \tan^{-1} x$.
$\frac{\pi}{4} = \frac{3}{2} \tan^{-1} x$.
$\tan^{-1} x = \frac{\pi}{4} \times \frac{2}{3} = \frac{\pi}{6}$.
Therefore,$x = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$.

(B) Let $\alpha = \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right)$. Then $\sin \alpha = \frac{2\sqrt{2}}{3}$.
Since $\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{8}{9} = \frac{1}{9}$,we have $\cos \alpha = \frac{1}{3}$.
Thus,$\alpha = \cos^{-1}\left(\frac{1}{3}\right)$.
Substituting this into the expression,we get $\cos^{-1}\left(\frac{1}{3}\right) + \sin^{-1}\left(\frac{1}{3}\right)$.
Using the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$,the value is $\frac{\pi}{2}$.

(A) The volume of a sphere is given by $V = \frac{4}{3} \Pi r^3$.
Differentiating with respect to time $t$,we get $\frac{dV}{dt} = 4 \Pi r^2 \frac{dr}{dt}$.
Given $\frac{dV}{dt} = 10 \text{ cm}^3 \text{s}^{-1}$ and $r = 15 \text{ cm}$.
Substituting these values: $10 = 4 \Pi (15)^2 \frac{dr}{dt}$.
$\frac{dr}{dt} = \frac{10}{4 \Pi (225)} = \frac{10}{900 \Pi} = \frac{1}{90 \Pi} \text{ cm s}^{-1}$.

(B) The given planes are $P_1: x - 2y - z + 5 = 0$ and $P_2: x + y + 3z = 6$.
Their normal vectors are $\vec{N}_1 = \hat{i} - 2\hat{j} - \hat{k}$ and $\vec{N}_2 = \hat{i} + \hat{j} + 3\hat{k}$.
The direction vector $\vec{b}$ of the line of intersection is given by the cross product of the normals:
$\vec{b} = \vec{N}_1 \times \vec{N}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & -1 \\ 1 & 1 & 3 \end{vmatrix} = \hat{i}(-6 - (-1)) - \hat{j}(3 - (-1)) + \hat{k}(1 - (-2)) = -5\hat{i} - 4\hat{j} + 3\hat{k}$.
The line passes through $(2, 3, 1)$ and is parallel to $\vec{b} = -5\hat{i} - 4\hat{j} + 3\hat{k}$.
Thus,the equation of the line is $\frac{x-2}{-5} = \frac{y-3}{-4} = \frac{z-1}{3}$.

(B) Given curves are $x^{3}-3xy^{2}+2=0$ $(1)$ and $3x^{2}y-y^{3}=2$ $(2)$.
Differentiating $(1)$ with respect to $x$: $3x^{2}-3(y^{2}+2xyy')=0 \Rightarrow x^{2}-y^{2}=2xyy' \Rightarrow y' = \frac{x^{2}-y^{2}}{2xy} = m_{1}$.
Differentiating $(2)$ with respect to $x$: $3(2xy+x^{2}y')-3y^{2}y'=0 \Rightarrow 2xy+x^{2}y'-y^{2}y'=0 \Rightarrow y'(x^{2}-y^{2}) = -2xy \Rightarrow y' = -\frac{2xy}{x^{2}-y^{2}} = m_{2}$.
Now,$m_{1} \cdot m_{2} = \left(\frac{x^{2}-y^{2}}{2xy}\right) \cdot \left(-\frac{2xy}{x^{2}-y^{2}}\right) = -1$.
Since the product of the slopes is $-1$,the curves cut at a right angle.

(D) Given the differential equation:
$x \frac{dy}{dx} + 2y = x^2$
Divide by $x$ to get the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$:
$\frac{dy}{dx} + \frac{2}{x}y = x$
Here,$P(x) = \frac{2}{x}$ and $Q(x) = x$.
The Integrating Factor ($I$.$F$.) is given by:
$I.F. = e^{\int P(x) dx} = e^{\int \frac{2}{x} dx} = e^{2 \ln|x|} = e^{\ln(x^2)} = x^2$
Multiplying the differential equation by the $I$.$F$. $(x^2)$:
$x^2 \frac{dy}{dx} + 2xy = x^3$
This can be written as:
$\frac{d}{dx}(y \cdot x^2) = x^3$
Integrating both sides with respect to $x$:
$y \cdot x^2 = \int x^3 dx$
$y \cdot x^2 = \frac{x^4}{4} + C$
Dividing by $x^2$:
$y = \frac{x^4 + 4C}{4x^2}$
Since $4C$ is an arbitrary constant,we can write it as $C$:
$y = \frac{x^4 + C}{4x^2}$

(D) Given the function $y = f(x^2 + 2)$.
Applying the chain rule to differentiate with respect to $x$,we get:
$\frac{dy}{dx} = f'(x^2 + 2) \cdot \frac{d}{dx}(x^2 + 2)$
$\frac{dy}{dx} = f'(x^2 + 2) \cdot (2x)$
Now,evaluate the derivative at $x = 1$:
$\left. \frac{dy}{dx} \right|_{x=1} = f'(1^2 + 2) \cdot (2 \cdot 1)$
$\left. \frac{dy}{dx} \right|_{x=1} = f'(3) \cdot 2$
Given that $f'(3) = 5$,substitute this value into the expression:
$\left. \frac{dy}{dx} \right|_{x=1} = 5 \cdot 2 = 10$

(A) Given $y = \log \left(\frac{1-x^{2}}{1+x^{2}}\right)$.
Using the chain rule and the quotient rule for differentiation:
$\frac{dy}{dx} = \frac{1}{\frac{1-x^{2}}{1+x^{2}}} \cdot \frac{d}{dx} \left(\frac{1-x^{2}}{1+x^{2}}\right)$
$\frac{dy}{dx} = \frac{1+x^{2}}{1-x^{2}} \cdot \left[ \frac{(-2x)(1+x^{2}) - (2x)(1-x^{2})}{(1+x^{2})^{2}} \right]$
$\frac{dy}{dx} = \frac{1}{1-x^{2}} \cdot \left[ \frac{-2x - 2x^{3} - 2x + 2x^{3}}{1+x^{2}} \right]$
$\frac{dy}{dx} = \frac{1}{1-x^{2}} \cdot \left[ \frac{-4x}{1+x^{2}} \right]$
$\frac{dy}{dx} = \frac{-4x}{(1-x^{2})(1+x^{2})}$
$\frac{dy}{dx} = \frac{-4x}{1-x^{4}}$

(A) Given the integral $ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{dx}{1+\cos 2x} $.
Using the trigonometric identity $ 1+\cos 2x = 2\cos^2 x $,we have:
$ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{dx}{2\cos^2 x} = \frac{1}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec^2 x \, dx $.
Since $ f(x) = \sec^2 x $ is an even function,we can use the property $ \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx $.
Thus,$ I = \frac{1}{2} \times 2 \int_{0}^{\frac{\pi}{4}} \sec^2 x \, dx = \int_{0}^{\frac{\pi}{4}} \sec^2 x \, dx $.
Evaluating the integral,we get $ [\tan x]_{0}^{\frac{\pi}{4}} = \tan(\frac{\pi}{4}) - \tan(0) = 1 - 0 = 1 $.

(B) Given the function: $g(x) = \frac{x^{200}}{200} + \frac{x^{199}}{199} + \dots + \frac{x^2}{2} + x + 5$.
To find $g'(x)$,we differentiate $g(x)$ with respect to $x$ using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$:
$g'(x) = \frac{200x^{199}}{200} + \frac{199x^{198}}{199} + \dots + \frac{2x}{2} + 1 + 0$.
Simplifying the expression,we get:
$g'(x) = x^{199} + x^{198} + \dots + x + 1$.
Now,substitute $x = 0$ into the derivative:
$g'(0) = 0^{199} + 0^{198} + \dots + 0 + 1$.
$g'(0) = 1$.

(B) Given equations are:
$x+y+z=6 \quad (1)$
$x+2y+3z=10 \quad (2)$
$x+2y+az=b \quad (3)$
Representing the system in matrix form $AX=B$,the augmented matrix $[A|B]$ is:
$\begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 1 & 2 & 3 & | & 10 \\ 1 & 2 & a & | & b \end{bmatrix}$
Applying row operations:
$R_2 \rightarrow R_2 - R_1$ gives $\begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 2 & | & 4 \\ 1 & 2 & a & | & b \end{bmatrix}$
$R_3 \rightarrow R_3 - R_1$ gives $\begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 2 & | & 4 \\ 0 & 1 & a-1 & | & b-6 \end{bmatrix}$
$R_3 \rightarrow R_3 - R_2$ gives $\begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 2 & | & 4 \\ 0 & 0 & a-3 & | & b-10 \end{bmatrix}$
For the system to have no solutions,the rank of the coefficient matrix must be less than the rank of the augmented matrix. This occurs when the last row represents an impossible equation,i.e.,$0x + 0y + 0z = k$ where $k \neq 0$.
Therefore,$a-3 = 0 \Rightarrow a=3$ and $b-10 \neq 0 \Rightarrow b \neq 10$.

(C) Given the matrix $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$.
We are given that $|A^3| = 27$.
Using the property of determinants,$|A^n| = |A|^n$,we have $|A|^3 = 27$.
Taking the cube root on both sides,we get $|A| = 3$.
Now,calculate the determinant of matrix $A$:
$|A| = (\alpha \times \alpha) - (2 \times 2) = \alpha^2 - 4$.
Equating this to $3$:
$\alpha^2 - 4 = 3$
$\alpha^2 = 7$
$\alpha = \pm \sqrt{7}$.

(D) Given that $ P=\left|\begin{array}{ll}x & 1 \\ 1 & x\end{array}\right| $ and $ Q=\left|\begin{array}{lll}x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x\end{array}\right| $.
Calculating the determinant $ P $:
$ P = x(x) - (1)(1) = x^{2}-1 $.
Calculating the determinant $ Q $:
$ Q = x(x^{2}-1) - 1(x-1) + 1(1-x) $.
$ Q = x^{3} - x - x + 1 + 1 - x $.
$ Q = x^{3} - 3x + 2 $.
Now,differentiating $ Q $ with respect to $ x $:
$ \frac{d Q}{d x} = \frac{d}{d x}(x^{3} - 3x + 2) = 3x^{2} - 3 $.
Factoring out $ 3 $:
$ \frac{d Q}{d x} = 3(x^{2} - 1) $.
Since $ P = x^{2} - 1 $,we have:
$ \frac{d Q}{d x} = 3P $.

(C) vector $\vec{v}$ in the plane of $\vec{a}$ and $\vec{b}$ is given by $\vec{v} = m\vec{a} + n\vec{b}$.
Substituting the given vectors: $\vec{v} = m(\hat{i} + \hat{j} + \hat{k}) + n(\hat{i} - \hat{j} + \hat{k}) = (m+n)\hat{i} + (m-n)\hat{j} + (m+n)\hat{k}$.
The projection of $\vec{v}$ on $\vec{c}$ is $\frac{\vec{v} \cdot \vec{c}}{|\vec{c}|} = \frac{1}{\sqrt{3}}$.
Calculating the dot product: $\vec{v} \cdot \vec{c} = (m+n)(1) + (m-n)(-1) + (m+n)(-1) = m+n - m+n - m-n = n-m$.
The magnitude $|\vec{c}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3}$.
Thus,$\frac{n-m}{\sqrt{3}} = \frac{1}{\sqrt{3}} \implies n-m = 1$,or $n = m+1$.
Substituting $n = m+1$ into $\vec{v}$: $\vec{v} = (2m+1)\hat{i} - \hat{j} + (2m+1)\hat{k}$.
For $m=0$,$\vec{v} = \hat{i} - \hat{j} + \hat{k}$. For $m=1$,$\vec{v} = 3\hat{i} - \hat{j} + 3\hat{k}$.
Comparing with the options,option $C$ is $3\hat{i} - \hat{j} + 3\hat{k}$.

(A) Given that $\frac{(x+1)^{2}}{x^{3}+x}=\frac{A}{x}+\frac{B x+C}{x^{2}+1}$.
Expanding the numerator on the left side: $\frac{x^{2}+2x+1}{x(x^{2}+1)} = \frac{x^{2}+1}{x(x^{2}+1)} + \frac{2x}{x(x^{2}+1)} = \frac{1}{x} + \frac{2}{x^{2}+1}$.
Comparing this with $\frac{A}{x} + \frac{Bx+C}{x^{2}+1}$,we get $A=1$,$B=0$,and $C=2$.
Now,we evaluate the expression $\operatorname{cosec}^{-1}\left(\frac{1}{A}\right)+\cot ^{-1}\left(\frac{1}{B}\right)+\sec ^{-1} C$.
Note that $\cot^{-1}(\frac{1}{0})$ is $\cot^{-1}(\infty) = 0$.
Substituting the values: $\operatorname{cosec}^{-1}(1) + \cot^{-1}(\infty) + \sec^{-1}(2) = \frac{\pi}{2} + 0 + \frac{\pi}{3} = \frac{5\pi}{6}$.

(C) Given equations are:
$y = x^{3} \implies x = y^{1/3}$
$y = 8$
$x = 0$
The area bounded by the curve $x = y^{1/3}$,the $y$-axis $(x=0)$,and the line $y=8$ is given by the integral with respect to $y$ from $y=0$ to $y=8$:
$\text{Area} = \int_{0}^{8} x \, dy$
$\text{Area} = \int_{0}^{8} y^{1/3} \, dy$
Evaluating the integral:
$\text{Area} = \left[ \frac{y^{(1/3) + 1}}{(1/3) + 1} \right]_{0}^{8} = \left[ \frac{3}{4} y^{4/3} \right]_{0}^{8}$
$\text{Area} = \frac{3}{4} (8^{4/3} - 0^{4/3})$
$\text{Area} = \frac{3}{4} ((2^{3})^{4/3}) = \frac{3}{4} (2^{4})$
$\text{Area} = \frac{3}{4} \times 16 = 3 \times 4 = 12 \text{ sq. units}$.

(A) Given the integral $ I = \int e^{x} \left( \frac{1+\sin x}{1+\cos x} \right) dx $.
Using trigonometric identities $ 1+\cos x = 2\cos^2 \left( \frac{x}{2} \right) $ and $ \sin x = 2\sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right) $:
$ I = \int e^{x} \left( \frac{1 + 2\sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right)}{2\cos^2 \left( \frac{x}{2} \right)} \right) dx $
$ I = \int e^{x} \left( \frac{1}{2\cos^2 \left( \frac{x}{2} \right)} + \frac{2\sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right)}{2\cos^2 \left( \frac{x}{2} \right)} \right) dx $
$ I = \int e^{x} \left( \frac{1}{2} \sec^2 \left( \frac{x}{2} \right) + \tan \left( \frac{x}{2} \right) \right) dx $
We know that $ \int e^{x} (f(x) + f'(x)) dx = e^{x} f(x) + C $.
Here,let $ f(x) = \tan \left( \frac{x}{2} \right) $. Then $ f'(x) = \sec^2 \left( \frac{x}{2} \right) \cdot \frac{1}{2} = \frac{1}{2} \sec^2 \left( \frac{x}{2} \right) $.
Thus,$ I = e^{x} \tan \left( \frac{x}{2} \right) + C $.

(D) Given the function $f: R \rightarrow R$ defined by $f(x) = \frac{1}{x}$.
For a function to be defined on the domain $R$,every element in $R$ must have a corresponding image in the codomain.
At $x = 0$,the expression $f(0) = \frac{1}{0}$ is undefined in the set of real numbers $R$.
Since $0 \in R$ and $f(0)$ does not exist,the function $f$ is not well-defined on the domain $R$.
Therefore,$f$ is not defined.

(A) We know that for any probability distribution,the sum of all probabilities must be equal to $1$.
Therefore,$\sum P(x) = 1$.
Substituting the given values from the table:
$0.2 + k + k + 2k = 1$
Combining the like terms:
$0.2 + 4k = 1$
Subtracting $0.2$ from both sides:
$4k = 1 - 0.2$
$4k = 0.8$
Dividing by $4$:
$k = \frac{0.8}{4} = 0.2$
Thus,the value of $k$ is $0.2$.

(C) Given the matrix $ A = \left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] $.
To find $ A^{2} $,we multiply the matrix $ A $ by itself:
$ A^{2} = A \cdot A = \left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] \left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] $
Performing matrix multiplication:
$ A^{2} = \left[\begin{array}{ll} (0 \times 0) + (1 \times 1) & (0 \times 1) + (1 \times 0) \\ (1 \times 0) + (0 \times 1) & (1 \times 1) + (0 \times 0) \end{array}\right] $
$ A^{2} = \left[\begin{array}{ll} 0 + 1 & 0 + 0 \\ 0 + 0 & 1 + 0 \end{array}\right] $
$ A^{2} = \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] $
This is the identity matrix $ I $ of order $ 2 \times 2 $.

(C) Given the expression $f(x) = 2 \sin^{-1} x + \cos^{-1} x$.
We can rewrite this as $f(x) = \sin^{-1} x + (\sin^{-1} x + \cos^{-1} x)$.
Since we know the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$,the expression becomes $f(x) = \sin^{-1} x + \frac{\pi}{2}$.
The range of $\sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Adding $\frac{\pi}{2}$ to all parts of the inequality,we get $-\frac{\pi}{2} + \frac{\pi}{2} \leq \sin^{-1} x + \frac{\pi}{2} \leq \frac{\pi}{2} + \frac{\pi}{2}$.
This simplifies to $0 \leq 2 \sin^{-1} x + \cos^{-1} x \leq \pi$.
Comparing this with $\alpha \leq 2 \sin^{-1} x + \cos^{-1} x \leq \beta$,we find $\alpha = 0$ and $\beta = \pi$.

(D) For a function $f(x)$ to be continuous at $x = 5$,the left-hand limit ($L$.$H$.$L$.) must equal the right-hand limit ($R$.$H$.$L$.) and the value of the function at $x = 5$.
First,calculate the value of the function at $x = 5$: $f(5) = 3(5) - 8 = 15 - 8 = 7$.
Next,calculate the $L$.$H$.$L$. as $x \to 5^-$: $\lim_{x \to 5^-} (3x - 8) = 3(5) - 8 = 7$.
Then,calculate the $R$.$H$.$L$. as $x \to 5^+$: $\lim_{x \to 5^+} (2k) = 2k$.
Since the function is continuous,we set $L$.$H$.$L$. = $R$.$H$.$L$.: $7 = 2k$.
Solving for $k$,we get $k = 7/2$.

(C) Given that,$f(x) = 2x^{2}$.
We need to evaluate the expression $\frac{f(3.8) - f(4)}{3.8 - 4}$.
Substitute the values into the function:
$\frac{f(3.8) - f(4)}{3.8 - 4} = \frac{2(3.8)^{2} - 2(4)^{2}}{3.8 - 4}$.
Factor out $2$ from the numerator:
$= \frac{2(3.8^{2} - 4^{2})}{3.8 - 4}$.
Using the algebraic identity $a^{2} - b^{2} = (a - b)(a + b)$,where $a = 3.8$ and $b = 4$:
$= \frac{2(3.8 - 4)(3.8 + 4)}{3.8 - 4}$.
Cancel the common term $(3.8 - 4)$ from the numerator and denominator:
$= 2(3.8 + 4)$.
$= 2(7.8) = 15.6$.

(A) The domain of $(f + g)$ is the intersection of the domains of $f(x)$ and $g(x)$.
For $f(x) = \frac{1}{2} - \tan^{-1}\left(\frac{\pi x}{2}\right)$,the given domain is $-1 < x < 1$.
For $g(x) = \sqrt{3 + 4x - 4x^2}$ to be defined,we must have $3 + 4x - 4x^2 \geq 0$.
Multiplying by $-1$,we get $4x^2 - 4x - 3 \leq 0$.
Factoring the quadratic: $(2x + 1)(2x - 3) \leq 0$.
This inequality holds for $x \in \left[-\frac{1}{2}, \frac{3}{2}\right]$.
The domain of $(f + g)$ is the intersection of $(-1, 1)$ and $\left[-\frac{1}{2}, \frac{3}{2}\right]$.
Intersection: $\left(-\frac{1}{2}, 1\right)$.

(A) Given the integral $ I = \int \frac{dx}{x^{2}(x^{4}+1)^{3/4}} $.
We can factor $ x^{4} $ out of the parenthesis:
$ I = \int \frac{dx}{x^{2}[x^{4}(1 + \frac{1}{x^{4}})]^{3/4}} $.
Simplifying the expression:
$ I = \int \frac{dx}{x^{2} \cdot x^{3}(1 + x^{-4})^{3/4}} = \int \frac{x^{-5}}{(1 + x^{-4})^{3/4}} dx $.
Let $ 1 + x^{-4} = t^{4} $.
Then,differentiating both sides with respect to $ x $,we get $ -4x^{-5} dx = 4t^{3} dt $,which implies $ x^{-5} dx = -t^{3} dt $.
Substituting these into the integral:
$ I = \int \frac{-t^{3} dt}{t^{3}} = -\int dt = -t + C $.
Substituting back $ t = (1 + x^{-4})^{1/4} $:
$ I = -(1 + x^{-4})^{1/4} + C = -(\frac{x^{4}+1}{x^{4}})^{1/4} + C = -\frac{(1+x^{4})^{1/4}}{x} + C $.

(A) Given the equation of the curve: $y = x^{2} - \frac{1}{x^{2}}$.
First,we find the derivative of the curve with respect to $x$ to find the slope of the tangent $(m_{1})$:
$y = x^{2} - x^{-2}$
$\frac{dy}{dx} = 2x - (-2)x^{-3} = 2x + \frac{2}{x^{3}}$.
Now,evaluate the slope of the tangent at the point $(-1, 0)$:
$m_{1} = \left. \frac{dy}{dx} \right|_{(-1, 0)} = 2(-1) + \frac{2}{(-1)^{3}} = -2 + \frac{2}{-1} = -2 - 2 = -4$.
The slope of the normal $(m_{2})$ is the negative reciprocal of the slope of the tangent:
$m_{1} \times m_{2} = -1$
$-4 \times m_{2} = -1$
$m_{2} = \frac{-1}{-4} = \frac{1}{4}$.
Therefore,the slope of the normal to the curve at $(-1, 0)$ is $1/4$.

(C) Given the equations $x = ct$ and $y = \frac{c}{t}$.
We need to find $\frac{dy}{dt}$ at $t = 2$.
Since $y = \frac{c}{t} = c \cdot t^{-1}$,we differentiate with respect to $t$:
$\frac{dy}{dt} = c \cdot (-1) \cdot t^{-2} = -\frac{c}{t^2}$.
At $t = 2$,we substitute the value into the derivative:
$\left. \frac{dy}{dt} \right|_{t=2} = -\frac{c}{(2)^2} = -\frac{c}{4}$.

(B) Given that $f(x) = \frac{x}{x^{2}+1}$.
First,calculate $f(2)$ by substituting $x = 2$ into the function:
$f(2) = \frac{2}{2^{2}+1} = \frac{2}{4+1} = \frac{2}{5}$.
Now,calculate $f(f(2))$ by substituting $f(2) = \frac{2}{5}$ back into the function:
$f(f(2)) = f\left(\frac{2}{5}\right) = \frac{\frac{2}{5}}{\left(\frac{2}{5}\right)^{2}+1}$.
Simplify the denominator:
$\left(\frac{2}{5}\right)^{2} + 1 = \frac{4}{25} + 1 = \frac{4+25}{25} = \frac{29}{25}$.
Therefore,$f(f(2)) = \frac{\frac{2}{5}}{\frac{29}{25}} = \frac{2}{5} \times \frac{25}{29} = \frac{2 \times 5}{29} = \frac{10}{29}$.

(B) To evaluate the determinant $\left|\begin{array}{cc}\cos 15^{\circ} & \sin 15^{\circ} \\ \sin 75^{\circ} & \cos 75^{\circ}\end{array}\right|$,we use the formula for a $2 \times 2$ determinant: $\left|\begin{array}{cc}a & b \\ c & d\end{array}\right| = ad - bc$.
Applying this to the given matrix:
$\cos 15^{\circ} \cos 75^{\circ} - \sin 15^{\circ} \sin 75^{\circ}$
Using the trigonometric identity $\cos(A + B) = \cos A \cos B - \sin A \sin B$,where $A = 15^{\circ}$ and $B = 75^{\circ}$:
$\cos(15^{\circ} + 75^{\circ}) = \cos(90^{\circ})$
Since $\cos(90^{\circ}) = 0$,the value of the determinant is $0$.

(B) Let $p$ be the probability of taking a step forward,so $p = 0.4$.
Let $q$ be the probability of taking a step backward,so $q = 0.6$.
Total steps $n = 11$.
Let $x$ be the number of forward steps and $y$ be the number of backward steps.
We have $x + y = 11$.
For the man to be one step away from the starting point,the net displacement must be $x - y = 1$ or $x - y = -1$.
Case $1$: $x - y = 1$. Adding the equations,$2x = 12 \implies x = 6$,so $y = 5$.
The probability for this case is $^{11}C_{6} \times p^{6} \times q^{5} = ^{11}C_{6} \times (0.4)^{6} \times (0.6)^{5}$.
Case $2$: $x - y = -1$. Adding the equations,$2x = 10 \implies x = 5$,so $y = 6$.
The probability for this case is $^{11}C_{5} \times p^{5} \times q^{6} = ^{11}C_{5} \times (0.4)^{5} \times (0.6)^{6}$.
Since $^{11}C_{6} = ^{11}C_{5}$,the total probability is $^{11}C_{6} \times (0.4)^{5} \times (0.6)^{5} \times (0.4 + 0.6) = ^{11}C_{6} \times (0.24)^{5} \times 1 = ^{11}C_{6} \times (0.24)^{5}$.

(B) The equation of the plane is $2x - 3y + 4z = 29$.
The normal vector to the plane is $\vec{n} = 2\hat{i} - 3\hat{j} + 4\hat{k}$.
The line passing through the origin $(0, 0, 0)$ and perpendicular to the plane has the direction ratios of the normal vector. Thus,the equation of the line is $\frac{x}{2} = \frac{y}{-3} = \frac{z}{4} = \lambda$.
Any point on this line is given by $(2\lambda, -3\lambda, 4\lambda)$.
Since this point lies on the plane,we substitute these coordinates into the plane equation: $2(2\lambda) - 3(-3\lambda) + 4(4\lambda) = 29$.
$4\lambda + 9\lambda + 16\lambda = 29 \Rightarrow 29\lambda = 29 \Rightarrow \lambda = 1$.
Substituting $\lambda = 1$ back into the point coordinates,we get $(2(1), -3(1), 4(1)) = (2, -3, 4)$.

(C) Let $I = \int_{0}^{\pi / 4} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x$.
Simplifying the integrand,we get $I = \int_{0}^{\pi / 4} \log (1+\tan x) d x$.
Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,we have:
$I = \int_{0}^{\pi / 4} \log \left(1+\tan \left(\frac{\pi}{4}-x\right)\right) d x$.
Since $\tan(\frac{\pi}{4}-x) = \frac{1-\tan x}{1+\tan x}$,the integral becomes:
$I = \int_{0}^{\pi / 4} \log \left(1+\frac{1-\tan x}{1+\tan x}\right) d x = \int_{0}^{\pi / 4} \log \left(\frac{2}{1+\tan x}\right) d x$.
Using the property $\log(\frac{a}{b}) = \log a - \log b$:
$I = \int_{0}^{\pi / 4} \log 2 d x - \int_{0}^{\pi / 4} \log (1+\tan x) d x$.
$I = \log 2 [x]_{0}^{\pi / 4} - I$.
$2I = \frac{\pi}{4} \log 2$.
Therefore,$I = \frac{\pi}{8} \log 2$.

(B) Given the integral $ I = \int \frac{\sin ^{2} x}{1+\cos x} d x $.
Using the trigonometric identity $ \sin ^{2} x = 1 - \cos ^{2} x $,we get:
$ I = \int \frac{1 - \cos ^{2} x}{1 + \cos x} d x $.
Since $ 1 - \cos ^{2} x = (1 - \cos x)(1 + \cos x) $,the expression becomes:
$ I = \int \frac{(1 - \cos x)(1 + \cos x)}{1 + \cos x} d x $.
Canceling the common term $ (1 + \cos x) $,we have:
$ I = \int (1 - \cos x) d x $.
Integrating term by term,we get:
$ I = x - \sin x + C $.