Given, differential equation is \(\frac{d}{d x}\left(\frac{d y}{d x}\right)=5\)
\(\therefore \quad \frac{d^2 y}{d x^2}=5\)                ....(i)
Now, from Eq. (i),highest order derivative present in the given differential equation is 2.
\(\therefore\) Order = 2
and the exponent of highest order derivative is 1.
\(\therefore\) Degree = 1
So, the required sum is 2 + 1 = 3 

We have, \(\hat{i} \times(\hat{j}+\hat{k})+\hat{j} \times(\hat{k}+\hat{i})+\hat{k} \times(\hat{i}+\hat{j})\)
\(=\hat{i} \times \hat{j}+\hat{i} \times \hat{k}+\hat{j} \times \hat{k}+\hat{j} \times \hat{i}+\hat{k} \times \hat{i}+\hat{k} \times \hat{j}\)
[\(\because\) cross product is distributive over addition]
\(=\hat{k}-\hat{j}+\hat{i}-\hat{k}+\hat{j}-\hat{i}=0\)
[\(\because\) \(\hat{i} \times \hat{j}=\hat{k}, \hat{i} \times \hat{k}=-\hat{j}, \hat{j} \times \hat{k}=\hat{i}, \hat{j} \times \hat{i}=-\hat{k}, \hat{k} \times \hat{i}=\hat{j}, \hat{k} \times \hat{j}=-\hat{i}]\)

\(\int _{ 0 }^{ \pi /4 }{ sin2x } dx=-\frac { 1 }{ 2 } \left[ cos2x \right] ^{ \pi /4 }_{ 0 }\)
\(=-\frac { 1 }{ 2 } \left[ cos(\pi /2)-cos(0) \right] =\frac { 1 }{ 2 } \)

D-Co·smes af line are \(\frac { 1 }{ 2 } ,\frac { 1 }{ 2 } ,\frac { 1 }{ \sqrt { 2 } } \) 
Equation of line is :
\(\frac { x-2 }{ 1/2 } =\frac { y+3 }{ -1/2 } =\frac { z+4 }{ 1/2 } \)
2x-4=-2y-6
\(\quad \quad \quad \quad =\sqrt { 2 } (z-4)\)

\(\int \frac{1}{x(1+\log x)} d x =\int \frac{1}{t} d t \)
\(=\log |t|+C \)
= log |1 + log x| + c

The given equation is \(ydx-\left( x+2{ y }^{ 2 } \right) dy=0\)
\(\Rightarrow\)\(\frac { dx }{ dy } -\frac { x }{ y } =2y\)
Linear Equation in x
Computing with \(\frac { dx }{ dy } +Px=Q,\) 
we have: \('P'\quad =-\frac { 1 }{ y } and\quad 'Q'=2y.\)
\(\therefore\)  \(I.F.={ e }^{ \int { Pdy } }={ e }^{ \int { -\frac { 1 }{ y } dy } }={ e }^{ -log|y| }\)
\(={ e }^{ { log }^{ -1 } }={ y }^{ -1 }=\frac { 1 }{ y } .\)
Multiplying (1) by,\(\frac { 1 }{ y } \) we get
\(\frac { 1 }{ y } \frac { dx }{ dy } -\frac { x }{ { y }^{ 2 } } =2\Rightarrow \frac { d }{ dy } \left( x.\frac { 1 }{ y } \right) =2.\)
Integrating,\(x.\frac { 1 }{ y } =\int { 2.dy+c \Rightarrow } \frac { x }{ y } =2y+c\)
\(\Rightarrow \) \(x=2{ y }^{ 2 }+cy\)
Which is the reqd. general solution.

\(\text {Let } \vec{a}=2 \hat{i}+5 \hat{j}-3 \hat{k}, \vec{b}=-2 \hat{i}-3 \hat{j}+5 \hat{k}, \vec{c}=5 \hat{i}+3 \hat{j}-3 \hat{k}\)
Then the vector equation of the plane passing through \(\vec{a},\vec{b}\) and \(\vec{c}\) and is given by
\((\vec{r}-\vec{a}) \cdot(\overrightarrow{\mathrm{RS}} \times \overrightarrow{\mathrm{R} \mathrm{T}})=0 \quad \text { (Why?) } \)
\((\vec{r}-\vec{a}) \cdot[(\vec{b}-\vec{a}) \times(\vec{c}-\vec{a})]=0 \)
\({[\vec{r}-(2 \hat{i}+5 \hat{j}-3 \hat{k})] \cdot[(-4 \hat{i}-8 \hat{j}+8 \hat{k}) \times(3 \hat{i}-2 \hat{j})]=0} \)

We have : \(\overset { \rightarrow }{ |a| } =\overset { \rightarrow }{ |b| } =\overset { \rightarrow }{ |c| } =1\) ..(1)
Now   \(\overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } =\overrightarrow { 0 } \)..(2) 
Squaring,   \({ \left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \right) }^{ 2 }=0\)
\(\Rightarrow \) \(\overset { \rightarrow }{ { |a| }^{ 2 } } +\overset { \rightarrow }{ { |b| }^{ 2 } } +\overset { \rightarrow }{ { |c| }^{ 2 } } +2\left( \overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } +\overset { \rightarrow }{ b } .\overset { \rightarrow }{ c } +\overset { \rightarrow }{ c } .\overset { \rightarrow }{ a } \right) =0\)
\(\Rightarrow \) \({ (1) }^{ 2 }+{ (2) }^{ 2 }+{ (3) }^{ 2 }+2\left( \overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } +\overset { \rightarrow }{ b } .\overset { \rightarrow }{ c } +\overset { \rightarrow }{ c } .\overset { \rightarrow }{ a } \right) =0\) 
hence, \(\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } +\overset { \rightarrow }{ b } .\overset { \rightarrow }{ c } +\overset { \rightarrow }{ c } .\overset { \rightarrow }{ a } =-\frac { 3 }{ 2 } \) .

\(Let\ \vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k} \text { and } \vec{b}=-4 \hat{i}+6 \hat{j}-8 \hat{k} \text { . }\)
It is observed that \( \vec{b}=-4 \hat{i}+6 \hat{j}-8 \hat{k}=-2(2 \hat{i}-3 \hat{j}+4 \hat{k})=-2 \vec{a}\)
\(\therefore \vec{b}=\lambda \vec{a}\)
\(where, \lambda=-2 \)
 Hence, the given vectors are collinear.

\(\text {Let vector be } \vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)
According to given condition \( x-y+z=4 \text { ; }\)
\(2 x+y-3 z=0 ; x+y+z=2 \)
\(x-y+z=4 \)
\(2 x+y-3 z=0 \)
\(x+y+z=2 \)
From (ii), y = 3z − 2x
Substituting in (i) and (iii), we get
\(x-3 z+2 x+z=4 \Rightarrow 3 x-2 z=4 \)
\(x+3 z-2 x+z=2 \Rightarrow-x+4 z=2 \)
Solving (v) and (vi), we get
\(x=2, z=1\)
Substituting in (iv), we get
\(y=-1\)
\(\therefore \text { Vector is } 2 \hat{i}-\hat{j}+\hat{k} \)

\(\int \frac{x+2}{\sqrt{(x-2)(x-3)}}=\int \frac{x+2}{\sqrt{x^{2}-5 x+6}} d x\)
Let x + 2 = A (2x - 5) + B
Comparing coefficients of x and constant term, we get
\(2 A=1,-5 A+B=2 \Rightarrow A=\frac{1}{2}, B=\frac{9}{2} \)
\(\int \frac{x+2}{\sqrt{x^{2}-5 x+6}} d x=\frac{1}{2} \int \frac{2 x-5}{\sqrt{x^{2}-5 x+6}} d x+\frac{9}{2} \int \frac{1}{\sqrt{x^{2}-5 x+6}} d x\)
For first integral. substitute x² - 5x + 6 = I, for second integral make it a perfect square.
\(=\frac{1}{2} \int \frac{1}{t} d t+\frac{9}{2} \int \frac{1}{\sqrt{\left(x-\frac{5}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}} d x \)
\(=\frac{1}{2} \log |t|+\frac{9}{2} \log \left|\left(x-\frac{5}{2}\right)+\sqrt{x^{2}-5 x+6}\right|+C \)
\(=\log \left|\sqrt{x^{2}-5 x+6}\right|+\frac{9}{2} \log \left|\left(x-\frac{5}{2}\right)+\sqrt{x^{2}-5 x+6}\right|+C \)

Let \(I=\int \frac{2 x}{\left(x^2+1\right)\left(x^4+4\right)} d x\)
Put \(x^2=t \Rightarrow 2 x d x=d t\)
\(\therefore \quad I=\int \frac{d t}{(t+1)\left(t^2+4\right)}\)
Now, \(\frac{1}{(t+1)\left(t^2+4\right)}=\frac{A}{t+1}+\frac{B t+C}{t^2+4}\)
\(\Rightarrow  1=A\left(t^2+4\right)+(B t+C)(t+1)\)
\(\Rightarrow  1=A\left(t^2+4\right)+\left(B t^2+B t+C t+C\right)\)
\(\Rightarrow  1=t^2(A+B)+t(B+C)+(4 A+C)\)
On comparing the coefficients of \(t^2, t\) and constant terms from both sides, we get
A+B=0 ...(i)
B+C=0 ...(ii)
4 A+C=1 ...(iii)
From Eqs. (i) and (ii), we get
A-C=0 ..(iv)
From Eqs. (iii) and (iv), we get
\(5A=1 \Rightarrow A=\frac{1}{5}\)
Then, \(C=\frac{1}{5}\) and \(B=-\frac{1}{5}\)
Now, \(=\frac{1}{5} \int \frac{d t}{t+1}-\frac{1}{5} \int \frac{t-1}{t^2+4} d t \)
\(=\frac{1}{5} \log |t+1|-\frac{1}{5}\left[\int \frac{t}{t^2+4} d t-\int \frac{1}{t^2+4} d t\right]\)
\(=\frac{1}{5} \log |t+1|-\frac{1}{5}\left[\frac{1}{2} \log \left|t^2+4\right|-\frac{1}{2} \tan ^{-1}\left(\frac{t}{2}\right)\right]+C\)
\(=\frac{1}{5} \log \left|x^2+1\right|-\frac{1}{5}\left[\frac{1}{2} \log \left|x^4+4\right|\right.\left.\quad-\frac{1}{2} \tan ^{-1}\left(\frac{x^2}{2}\right)\right]+C \quad\left[\because t=x^2\right]\)

The vector equations of given lines are
\(\begin{aligned} \vec{r}=3 \hat{i}+2 \hat{j}-4 \hat{k}+\lambda(\hat{i}+2 \hat{j}+2 \hat{k}) \end{aligned}\)
and \(\begin{aligned} \vec{r}=5 \hat{i}-2 \hat{j}+\mu(3 \hat{i}+2 \hat{j}+6 \hat{k}) \end{aligned}\)
On comparing them with \(\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}\) and \(\vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}\), we get
\(\overrightarrow{a_1}=3 \hat{i}+2 \hat{j}-4 \hat{k}, \overrightarrow{a_2}=5 \hat{i}-2 \hat{j}, \vec{b}_1=\hat{i}+2 \hat{j}+2 \hat{k}\)
and \(\vec{b}_2=3 \hat{i}+2 \hat{j}+6 \hat{k}\)
\(\therefore \quad \overrightarrow{a_2}-\overrightarrow{a_1}=(5 \hat{i}-2 \hat{j})-(3 \hat{i}+2 \hat{j}-4 \hat{k})\)
\(=2 \hat{i}-4 \hat{j}+4 \hat{k}\)
\(\begin{aligned} \therefore \quad \vec{b}_1 \times \vec{b}_2 & =\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 3 & 2 & 6 \end{array}\right| \end{aligned}\)
\(\begin{aligned} =\hat{i}(12-4)-\hat{j}(6-6)+\hat{k}(2-6) \end{aligned}\)
\(\begin{aligned} =8 \hat{i}-4 \hat{k} \end{aligned}\)
\(\therefore\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)=(2 \hat{i}-4 \hat{j}+4 \hat{k}) \cdot(8 \hat{i}-4 \hat{k})\)
= 16 + 0 - 16 = 0
\(\therefore\) The lines are intersecting and the shortest distance between the lines is 0.
Now, the position vectors of arbitrary points on the given lines are \((3+\lambda) \hat{i}+(2+2 \lambda) \hat{j}+(-4+2 \lambda) \hat{k}\) and \((5+3 \mu) \hat{i}+(-2+2 \mu) \hat{j}+6 \mu \hat{k}\), respectively.
Since, lines intersect then they have a common point.
\(\therefore \begin{aligned} 3+\lambda & =5+3 \mu \end{aligned}\)         ...(i)
\(\begin{aligned} 2+2 \lambda & =-2+2 \mu \end{aligned}\)        ...(ii)
\(\begin{aligned} -4+2 \lambda & =6 \mu \end{aligned}\)               ...(iii)
On solving Eqs. (i) and (ii), we get
\(\lambda=-4 \text { and } \mu=-2\)
\(\therefore\) Point of intersection is (3 - 4, 2 - 8, -4 - 8)
i.e. (-1, -6, -12). 

Given differential equation is
\(\left(1+x^{2}\right) d y+2 x y d x=\cot x d x \quad[\because x \neq 0]\) 
Above equation can be rewritten as,
\(\left(1+x^{2}\right) d y+(2 x y-\cot x) d x=0\) 
\(\Rightarrow\left(1+x^{2}\right) d y=(\cot x-2 x y) d x\) 
On dividing both sides by \(1+x^{2}\) ,we get
\(d y=\frac{\cot x-2 x y}{1+x^{2}} d x\) 
\(\Rightarrow \frac{d y}{d x}=\frac{\cot x}{1+x^{2}}-\frac{2 x y}{1+x^{2}}\) 
\(\Rightarrow \frac{d y}{d x}+\frac{2 x}{1+x^{2}} y=\frac{\cot x}{1+x^{2}}\) 
which is a linear differential equation of the form of
\(\frac{d y}{d x}+P y=Q\) 
Here, \(P=\frac{2 x}{1+x^{2}} \text { and } Q=\frac{\cot x}{1+x^{2}}\) 
Now,\(\mathrm{IF}=e^{\int P d x}=e^{\frac{1}{1+x^{2}} d x}=e^{\log \left|1+x^{2}\right|}=1+x^{2}\) 
\(\left[\because I_{1}=\int \frac{2 x}{1+x^{2}} d x,\right. \text { put } 1+x^{2}=t \Rightarrow 2 x d x=d t\) 
\(\left.\Rightarrow I_{1}=\int \frac{d t}{t}=\log |t|=\log \left|1+x^{2}\right|\right]\)
and the solution of linear differential equation is given by
\(y \times I F=\int(Q \times I F) d x+C\) 
\(\Rightarrow y\left(1+x^{2}\right)=\int \frac{\cot x}{1+x^{2}} \times\left(1+x^{2}\right) d x+C\) 
\(\Rightarrow y\left(1+x^{2}\right)=\int \cot x d x+C\) 
\(\Rightarrow y\left(1+x^{2}\right)=\log |\sin x|+C\) 
\(\Rightarrow y=\frac{\log |\sin x|}{1+x^{2}}+\frac{C}{1+x^{2}}\) 
which is the required solution.

As shown in the Figure, the line y = 3x + 2 meets x-axis at x\(=\frac{-2}{3}\) and its graph lies below x-axis for \(x \in\left(-1, \frac{-2}{3}\right)\) and above x-axis for \(x \in\left(\frac{-2}{3}, 1\right)\)
The required area = Area of the region ACBA + Area of the region ADEA
\(=\left|\int_{-1}^{\frac{-2}{3}}(3 x+2) d x\right|+\int_{\frac{-2}{3}}^1(3 x+2) d x\)
\(=\left|\left[\frac{3 x^2}{2}+2 x\right]_{-1}^{\frac{-2}{3}}\right|+\left[\frac{3 x^2}{2}+2 x\right]_{\frac{-2}{3}}^1=\frac{1}{6}+\frac{25}{6}=\frac{13}{3}\)

(i) (c) : Since, size of population is 5000.
\(\therefore\) Maximum value of y(t) is 5000.
(ii) (d) : Clearly, according to given information
\(\frac{d y}{d t}=k y(5000-y)\) ,where k is the constant of proportionality.
(iii) (a): Since, rumour is initiated with 100 people.
\(\therefore\) When t = 0, then y = 100
Thus y(O) = 100
(iv) (b) : Since, rumour is spread in 500 people, after
2 days.
\(\therefore\) When t = 2, then y = 500.
Thus, y(2) = 500
(v) (c) : We know that, when t = 0, then y = 100
This condition is satisfied by option (c) only.

(i) (b) : D.R:s of OA are < 1-0, 0-0, 0-0 >, i.e., < 1, 0, 0 >.
(ii) (a) : Equation of diagonal OB' is \(\frac{x-0}{1}=\frac{y-0}{2}=\frac{z-0}{3} \text { i.e., } \frac{x}{1}=\frac{y}{2}=\frac{z}{3}\)
(iii) (c) : OABC is xy-plane, therefore its equation is z = 0.
(iv) (c) : Plane O'A'B'C is parallel to xy-plane passing through (0, 0, 3), therefore its equation is z = 3.
(v) (a) : Plane ABB' A' is parallel to yz-plane passing through (1, 0, 0), therefore its equation is x = 1.

Construct the following table of values of objective function

(i) (d): Minimum value of Z is -48 which occurs at (0,8).
(ii) (a): Maximumvalue of Z is 20, which occurs at (5,0).
(iii) (b): Maximum of Z - Minimum of Z
= 20 - (-48) = 20 + 48 = 68
(iv) (c): The corner points of the feasible region are O(0,0), A(3, 0), B(3, 2), C(2, 3), D(0, 3).
(v) (d)