Given, ydx + (x - y²) dy = 0
\(\Rightarrow\) ydx + xdy - y²dy = 0
\(\Rightarrow\)  ydx +x dy = y²dy
\(\begin{array}{ll} \Rightarrow & y \frac{d x}{d y}+x=y^2 \end{array}\)
\(\begin{array}{ll} \Rightarrow & \frac{d x}{d y}+\frac{x}{y}=y \end{array}\)
which is of the form,
\(\frac{d x}{d y}+P x=Q\)
\(\therefore P=\frac{1}{y} \text { and } Q=y\)
Here, \(\mathrm{IF}=e^{\int P d y}=e^{\int \frac{1}{y} d y}=e^{\log y}=y\)
 Now, required solution is given by
\(\begin{aligned} x(\mathrm{IF}) & =\int Q(\mathrm{IF}) d y+C \end{aligned}\)
\(\begin{aligned} x y & =\int y^2 d y+C \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad x y & =\frac{y^3}{3}+C \end{aligned}\)

Given, \(x d y-y d x=\sqrt{x^2+y^2} d x\)
\(\begin{array}{ll} \Rightarrow & x d y=\sqrt{x^2+y^2} d x+y d x \end{array}\)
\(\begin{array}{ll} \Rightarrow & x d y=\left(\sqrt{x^2+y^2}+y\right) d x \end{array}\)
\(\begin{array}{ll} \Rightarrow & \frac{d y}{d x}=\frac{\sqrt{x^2+y^2}+y}{x} \end{array}\)    ....(i)
Now, putting y = vx
\(\Rightarrow \quad \frac{d y}{d x}=v+x \frac{d v}{d x}\)
From Eq.(i), we get
\(\begin{aligned} v+x \frac{d v}{d x} & =\frac{\sqrt{x^2+v^2 x^2}+v x}{x} \end{aligned}\)
\(\begin{aligned} \Rightarrow & & v+x \frac{d v}{d x} & =\frac{x \sqrt{1+v^2}+v x}{x} \end{aligned}\)
\(\Rightarrow v+x \frac{d v}{d x} =\sqrt{1+v^2}+v \)
\(\Rightarrow \frac{d v}{d x} =\sqrt{1+v^2} \)
\(\Rightarrow \sqrt{1+v^2} =\frac{d x}{x}\)         ....(ii)
On integrating both sides of Eq. (ii), we get
\(\int \frac{d v}{\sqrt{1+v^2}}=\int \frac{d x}{x}\)
\(\begin{aligned} \Rightarrow \quad \log \left|v+\sqrt{1+v^2}\right|=\log |x|+\log C, C>0 \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \log \left|\frac{y}{x}+\sqrt{1+\frac{y^2}{x^2}}\right|=\log |x|+\log C \quad\left[\because v=\frac{v}{x}\right] \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \log \left|y+\sqrt{x^2+y^2}\right|-\log |x|=\log |x|+\log C \end{aligned}\)
\(\begin{aligned} \Rightarrow \log \left|y+\sqrt{x^2+y^2}\right|=2 \log |x|+\log C=\log \left(x^2 C\right) \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad y+\sqrt{x^2+y^2}= \pm C x^2 \end{aligned}\)
\(\Rightarrow \quad y+\sqrt{x^2+y^2}=K x^2, \quad[\text { where, } K= \pm C]\)
which is the required general solution.

Let \(y=\sin ^2 x\) and \(z=e^{\cos x}\).
On differentiating y and z w.r.t. x, we get
\(\frac{d y}{d x}=2 \sin x \cos x\)
and \(\frac{d z}{d x}=e^{\cos x}(-\sin x)=-(\sin x) e^{\cos x}\)
Now, \(\frac{d y}{d z}=\frac{d y / d x}{d z / d x}=\frac{2 \sin x \cos x}{-(\sin x) e^{\cos x}}=\frac{-2 \cos x}{e^{\cos x}}\)

(9, 13, 15)

Given differential equation is
\(\frac{d y}{d x}+2 y \tan x=\sin x\) 
which is a linear differential equation of the form \(\frac{d y}{d x}+P y=Q\) 
Here, P= 2tan x and Q sin x
Now, \(\mathrm{IF}=e^{\int P d x}=e^{\int 2 \tan x d x}\) 
\(=e^{2 \int \tan x d x}=e^{2 \log |\sec x|}\) 
\(=e^{\log \sec ^{2} x}=\sec ^{2} x \quad\left[\because e^{\log f(x)}=f(x)\right]\) 
Then, the solution of differential equation is given by
\(y \cdot(\mathrm{IF})=\int(\mathrm{IF}) Q d x+C\) 
\(\Rightarrow y \cdot \sec ^{2} \cdot x=\int \sec ^{2} x \cdot \sin x d x+C\) 
\(\Rightarrow y \cdot \sec ^{2} x=\int \frac{\sin x}{\cos ^{2} x} d x+C\) 
Put \(\cos x=t, \text { then }-\sin x d x=d t\) 
\(\therefore y \cdot \sec ^{2} x=-\int \frac{d t}{t^{2}}+C\) 
\(\Rightarrow y \cdot \sec ^{2} x=-\int t^{-2} d t+C\) 
\(\Rightarrow y \cdot \sec ^{2} x=-1 \frac{t^{-1}}{(-1)}+C\) 
\(\Rightarrow y \cdot \sec ^{2} x=\frac{1}{t}+C\) 
\(\Rightarrow y \cdot \sec ^{2} x=\frac{1}{\cos x}+C \quad[\because t=\cos x]\) 
\(\Rightarrow y \cdot \sec ^{2} x=\sec x+C\) 
\(\Rightarrow y=\frac{1}{\sec x}+\frac{C}{\sec ^{2} x}\) 
[dividing each term by sec² x]
\(\Rightarrow y=\cos x+C \cdot \cos ^{2} x\) 
Now, it given that \(y=0 \text { when } x=\frac{\pi}{3}\) 
\(\therefore 0=\cos \frac{\pi}{3}+C \cdot \cos ^{2} \frac{\pi}{3}\) 
\(\Rightarrow 0=\frac{1}{2}+C \cdot \frac{1}{4} \quad\left[\because \cos \frac{\pi}{3}=\frac{1}{2}\right]\) 
\(\Rightarrow \frac{-1}{2}=\frac{C}{4} \Rightarrow C=-2\) 
\(\therefore\) The required particular solution is
\(y=\cos x-2 \cos ^{2} x\)

Given points \((2,-3),(\lambda,-1) \text { and }(0,4)\) are collinear So, area of triangle formed by these three points will be zero.
\(\therefore \left|\begin{array}{rrr} 2 & -3 & 1 \\ \lambda & -1 & 1 \\ 0 & 4 & 1 \end{array}\right|=0\) 
Applying \(R_{2} \rightarrow R_{2}-R_{1}\) and \(R_{3} \rightarrow R_{3}-R_{1}\) we get
\(\left|\begin{array}{ccc} 2 & -3 & 1 \\ \lambda-2 & 2 & 0 \\ -2 & 7 & 0 \end{array}\right|=0\) 
Now, expanding along C₃, we get \(\left|\begin{array}{cc} \lambda-2 & 2 \\ -2 & 7 \end{array}\right|=0\) 
\( \Rightarrow 7(\lambda-2)-2(-2)=0 \)
\(\Rightarrow 7 \lambda-14+4=0 \)
\(\Rightarrow 7 \lambda-10=0 \Rightarrow \lambda \mid=\frac{10}{7} \) 
Hence, required value of \(\lambda \text { is } \frac{10}{7}\).

We have: \(xy={ e }^{ x-y }\) ....(1)
Diff.w.r.t.x, \(x.\frac { dy }{ dx } +y.1={ e }^{ x-y }.\frac { d }{ dx } (x-y)\)
\(x.\frac { dy }{ dx } +y={ e }^{ x-y }\left[ 1-\frac { dy }{ dx } \right] \)
\(\left( x+{ e }^{ x-y } \right) \frac { dy }{ dx } ={ e }^{ x-y }-y\)
\(\frac { dy }{ dx } =\frac { { e }^{ x-y }-y }{ x+{ e }^{ x-y } }\)
\(=\frac { xy-x }{ x+xy } \)
\(=\frac { y(x-1) }{ x(y+1) } \)
which is true.

\(\text { Here } \vec{b}_{1}=\hat{i}+2 \hat{j}+2 \hat{k} \text { and } \vec{b}_{2}=3 \hat{i}+2 \hat{j}+6 \hat{k}\)
\(\text { The angle } \theta \text { between the two lines is given by }\)
\(\cos \theta =\left|\frac{\vec{b}_{1} \cdot \vec{b}_{2}}{\left|\vec{b}_{1}\right|\left|\vec{b}_{2}\right|}\right|=\left|\frac{(\hat{i}+2 \hat{j}+2 \hat{k}) \cdot(3 \hat{i}+2 \hat{j}+6 \hat{k})}{\sqrt{1+4+4} \sqrt{9+4+36}}\right| \\ \)
\(=\left|\frac{3+4+12}{3 \times 7}\right|=\frac{19}{21} \)
\(\text { Hence }\theta=\cos ^{-1}\left(\frac{19}{21}\right) \)

\(At\quad x=0:\)
\(\lim _{ x\rightarrow { 0 }^{ - } }{ \frac { sinx }{ x } } =\lim _{ h\rightarrow 0 }{ \frac { sin(-h) }{ -h } } \)
\(=\lim _{ h\rightarrow 0 }{ \frac { -sin\quad h }{ -h } } \)
\(=\lim _{ h\rightarrow 0 }{ \frac { sin\quad h }{ h } } =1\)
\( \lim _{ x\rightarrow { 0 }^{ + } }{ f(x) } =\lim _{ x\rightarrow { 0 }^{ + } }{ (x+1) } =0+1=1\)
\( f(0)=0+1=1\)

Given, \(y=\sqrt{a x+b}\)
On differentiating both sides w.r.t. x, we get
\(\frac{d y}{d x}=\frac{1}{2 \sqrt{a x+b}} \frac{d}{d x}(a x+b) \)
\(\Rightarrow  \frac{d y}{d x}=\frac{a}{2 \sqrt{a x+b}}=\frac{a}{2 y} \)
\(\Rightarrow  y \frac{d y}{d x}=\frac{a}{2}\)
Again, differentiating both sides, w.r.t. x, we get
\(y \frac{d^2 y}{d x^2}+\frac{d y}{d x} \cdot \frac{d y}{d x}=0\)
\(\Rightarrow  y\left(\frac{d^2 y}{d x^2}\right)+\left(\frac{d y}{d x}\right)^2=0\)
Hence proved.

Given, \(A(\operatorname{adj} A)=\left[\begin{array}{ll}8 & 0 \\ 0 & 8\end{array}\right] \Rightarrow|A(\operatorname{adj} A)|=\left|\begin{array}{ll}8 & 0 \\ 0 & 8\end{array}\right|\)
\( \Rightarrow  |A||\operatorname{adj}(A)| =64-0\)
\(\Rightarrow |A||A|^{2-1}  =64\)  \(\left[\because|\operatorname{adj} A|=|A|^{n-1}\right]\)
\(\Rightarrow  |A|^2  =64\)
\(\Rightarrow  |A|  = \pm 8\)

We have, f(x) = |x - 5|
Test for differentiability at x =5
LHD = f'(5^-) \(=\lim _{h \rightarrow 0} \frac{f(5-h)-f(5)}{-h}\)
\(\begin{aligned} {\left[\because L f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h}\right]} \end{aligned}\)
\(\begin{aligned} =\lim _{h \rightarrow 0} \frac{|5-h-5|-|5-5|}{-h} \end{aligned}\)
\(\begin{aligned} =\lim _{h \rightarrow 0} \frac{|-h|}{-h}=\lim _{h \rightarrow 0} \frac{h}{-h}=-1 \end{aligned}\)
\([\because|-x|=x \text {, if } x>0]\)
RHD = f'(5⁺) \(=\lim _{h \rightarrow 0} \frac{f(5+h)-f(5)}{h}\)
\(\left[\because R f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}\right]\)
\(\begin{aligned} =\lim _{h \rightarrow 0} \frac{|5+h-5|-|5-5|}{h} \end{aligned}\)
\(\begin{aligned} =\lim _{h \rightarrow 0} \frac{|h|}{h}=\lim _{h \rightarrow 0} \frac{h}{h}=1 \quad[\because|x|=x \text {, if } x>0] \end{aligned}\)
Since, LHD \(≠\) RHD at x = 5
So, f is not differentiable.

We have,
\(\sin ^{-1}\left[k \tan \left(2 \cos ^{-1} \frac{\sqrt{3}}{2}\right)\right]=\frac{\pi}{3}\)
\(\Rightarrow \quad \sin ^{-1}\left[k \tan \left(2 \times \frac{\pi}{6}\right)\right]=\frac{\pi}{3} \quad\left[\because \cos ^{-1} \frac{\sqrt{3}}{2}=\frac{\pi}{6}\right]\)
\(\Rightarrow \quad \sin ^{-1}\left[k \tan \left(\frac{\pi}{3}\right)\right]=\frac{\pi}{3}\)
\(\Rightarrow \quad \sin ^{-1}(k \sqrt{3})=\frac{\pi}{3} \quad\left[\because \tan \frac{\pi}{3}=\sqrt{3}\right]\)
\(\begin{array}{lc} \Rightarrow & k \sqrt{3}=\sin \frac{\pi}{3} \end{array}\)
\(\begin{array}{lc} \Rightarrow & k \sqrt{3}=\frac{\sqrt{3}}{2} \end{array}\)
\(\begin{array}{lc} \Rightarrow & k=\frac{1}{2} \end{array}\)

k = 4

Given,
\(\frac { 3-x }{ -3 } =\frac { y+2 }{ -2 } =\frac { z+2 }{ 6 } \) 
\(\Rightarrow \frac { x-3 }{ 3 } =\frac { y+2 }{ -2 } =\frac { z+2 }{ 6 } \)
Direction ratios are 3,-2,6
DRF parallel line 3,2,6
D.C's m,n :
\(I=\frac { a }{ \sqrt { a^{ 2 }+b^{ 2 }+c^{ 2 } } } \)
\(h=\frac { b }{ \sqrt { a^{ 2 }+b^{ 2 }+c^{ 2 } } } \)
\(n=\frac { c }{ \sqrt { a^{ 2 }+b^{ 2 }+c^{ 2 } } } \)
\(D.C.s=\frac { 3 }{ \sqrt { 9+4+36 } } ,\frac { -2 }{ \sqrt { 9+4+36 } } ,\frac { 6 }{ \sqrt { 9+4+36 } } \)
\(D.C's\quad are\quad \frac { 3 }{ 7 } ,\frac { -2 }{ 7 } ,\frac { 6 }{ 7 } \)

\(cot\left( \frac { \pi }{ 2 } -2cot^{ -1 }\sqrt { 3 } \right) =\sqrt { 3 } \)
Alternative Method :
\(cot\left( \frac { \pi }{ 2 } -2{ cot }^{ -1 }\sqrt { 3 } \right) =cot\left[ \frac { \pi }{ 2 } -2{ cot }^{ -1 }\left( cot\frac { \pi }{ 6 } \right) \right] \)
\(=cot\left[ \frac { \pi }{ 2 } -2\left( \frac { \pi }{ 6 } \right) \right] \)
\(\left[ \because { cot }^{ -1 }(cot\quad \theta )=\theta \ \forall \theta \ \in \ (0,\quad \pi ) \right] \)
\(=cot\left[ \frac { \pi }{ 2 } -\frac { \pi }{ 3 } \right] \)
\(=cot\left( \frac { \pi }{ 6 } \right) \)
\(=\sqrt { 3 } \)

Given, \(A=\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right]\)
Here, \(|A|=\left|\begin{array}{ccc}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right|=1(0+0)+1(9+2)+2(0-0)=11\)
Now, \(|A| I=11 I=\left[\begin{array}{ccc}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{array}\right]\)

Cofactors of |A| 
\(C_{11}=(-1)^{1+1}\left|\begin{array}{cc} 0 & -2 \\ 0 & 3 \end{array}\right|=0\)
\(C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 3 & -2 \\ 1 & 3 \end{array}\right|=-(9+2)=-11\)
\(C_{13}=(-1)^{1+3}\left|\begin{array}{ll} 3 & 0 \\ 1 & 0 \end{array}\right|=0\)
\(C_{21}=(-1)^{2+1}\left|\begin{array}{cc} -1 & 2 \\ 0 & 3 \end{array}\right|=-(-3-0)=3\)
\(C_{22}=(-1)^{2+2}\left|\begin{array}{ll} 1 & 2 \\ 1 & 3 \end{array}\right|=+(3-2)=1\)
\(C_{23}=(-1)^{2+3}\left|\begin{array}{cc} 1 & -1 \\ 1 & 0 \end{array}\right|=-(0+1)=-1\)
\(C_{31}=(-1)^{3+1}\left|\begin{array}{cc} -1 & 2 \\ 0 & -2 \end{array}\right|=+(2-0)=2\)
\(C_{32}=(-1)^{3+2}\left|\begin{array}{cc} 1 & 2 \\ 3 & -2 \end{array}\right|=-(-2-6)=8\)
\(C_{33}=(-1)^{3+3}\left|\begin{array}{cc} 1 & -1 \\ 3 & 0 \end{array}\right|=+(0+3)=3\)
\(\operatorname{adj} A=\left[\begin{array}{lll} C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33} \end{array}\right]=\left[\begin{array}{ccc} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{array}\right]\)
\(\text { Now, } A(\operatorname{adj} A)=\left[\begin{array}{ccc} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{array}\right]\left[\begin{array}{ccc} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{array}\right]\)
\(=\left[\begin{array}{lll} 0+11+0 & 3-1-2 & 2-8+6 \\ 0+0+0 & 9+0+2 & 6+0-6 \\ 0+0+0 & 3+0-3 & 2+0+9 \end{array}\right]\)
\(=\left[\begin{array}{ccc} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{array}\right]\)
and
\((\operatorname{adj} A) A  =\left[\begin{array}{ccc} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{array}\right]\left[\begin{array}{ccc} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{array}\right] \)
\(=\left[\begin{array}{ccc} 0+9+2 & 0+0+0 & 0-6+6 \\ -11+3+8 & 11+0+0 & -22-2+24 \\ 0-3+3 & 0+0+0 & 0+2+9 \end{array}\right] \)
\(=\left[\begin{array}{ccc} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{array}\right]\)
\(\text { Thus, } A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I \quad \text { Hence verified. }\)

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, lines are
\(\vec{r}=6 \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})\)
or \(\frac{x-6}{1}=\frac{y-2}{-2}=\frac{z-2}{2}=\lambda\) (say)            ...(i)
and \(\vec{r}=-4 \hat{i}-\hat{k}+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k})\)
or \(\frac{x+4}{3}=\frac{y}{-2}=\frac{z+1}{-2}=\mu\)  (say)            ...(ii)
Let the insect crawling on line (i) should reach at point P and insect crawling on line (ii) should reach at point Q. Now, for distance between them to be shortest, line segment PQ must be at right angles to both the lines, then PQ is the shortest distance between them. Now, the position vector of point P lying on line (i) is \(\overrightarrow{O P}=(6+\lambda) \hat{i}+(2-2 \lambda) \hat{j}+(2+2 \lambda) \hat{k}\) for some \(\lambda\) and the position vector of point Q lying on line (ii) is \(\overrightarrow{O Q}=(-4+3 \mu) \hat{i}-2 \mu \hat{j}+(-1-2 \mu) \hat{k}\) for some \(\mu\) 
Now, \(\overrightarrow{P Q}=\overrightarrow{O Q}-\overrightarrow{O P}\)
\(\begin{aligned} \therefore \overrightarrow{P Q}=(-10+3 \mu-\lambda) \hat{i}+(-2 \mu & -2+2 \lambda) \hat{j}+(-3-2 \mu-2 \lambda) \hat{k} \end{aligned}\)
\(\because \overrightarrow{P Q}\) is perpendicular to both the lines.
\(\begin{aligned} \therefore(-10+3 \mu-\lambda)(1)+(-2 \mu-2+2 \lambda)(-2)+(-3-2 \mu-2 \lambda)(2)=0 \\ \end{aligned}\)
\(\begin{array}{ll} \Rightarrow & -10+3 \mu-\lambda+4 \mu+4-4 \lambda-6-4 \mu-4 \lambda=0 \end{array}\)
\(\begin{array}{ll} \Rightarrow & -12+3 \mu-9 \lambda=0 \end{array}\)
\(\begin{array}{ll} \Rightarrow & -4+\mu-3 \lambda=0 \end{array}\)
\(\Rightarrow \quad \mu-3 \lambda=4\)                           ...(iii)
and \(\begin{aligned} (-10+3 \mu-\lambda)(3)+ & (-2 \mu-2+2 \lambda)(-2) +(-3-2 \mu-2 \lambda)(-2)=0 \end{aligned}\)
\(\begin{aligned} \Rightarrow & -30+9 \mu-3 \lambda+4 \mu+4-4 \lambda+6+4 \mu+4 \lambda=0 \end{aligned}\)
\(\begin{aligned} \Rightarrow & & -20+17 \mu-3 \lambda & =0 \end{aligned}\)
\(\begin{aligned} \Rightarrow & & 17 \mu-3 \lambda & =20 \end{aligned}\)                             ....(iv)
Now, on solving Eqs. (iii) and (iv), we get
\(\mu=1 \text { and } \lambda=-1\)
\(\therefore P \equiv(5,4,0) \text { and } Q \equiv(-1,-2,-3)\)
\(\therefore\)  The insect on line (i) should reach at point P(5, 4, 0) and the insect on line (ii) should reach at point Q(-1-2,-3). so that the distance between, them is shortest.
Now, \(\begin{aligned} \overrightarrow{P Q}=(-1-5) \hat{i}+(-2-4) \hat{j}+(-3-0) \hat{k} \end{aligned}\)
\(\begin{aligned} \overrightarrow{P Q}=-6 \hat{i}-6 \hat{j}-3 \hat{k} \end{aligned}\)
\(\therefore\) Shortest possible distance between them is
\(|\overrightarrow{P Q}|=\sqrt{(-6)^2+(-6)^2+(-3)^2}\)
\(=\sqrt{36+36+9}=\sqrt{81}=9 \text { units }\)

Given,
\(f(x)=\left\{\begin{array}{c} \frac{\sqrt{1+k x}-\sqrt{1-k x}}{x}, \text { if }-1 \leq x<0 \\ \frac{2 x+1}{x-1}, \quad \text { if } 0 \leq x<1 \end{array}\right.\) 
is continuous at x = 0.
Now, \(f(0)=\frac{2 \cdot 0+1}{0-1}=\frac{1}{-1}=-1\) 
 \(L H L =\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} \frac{\sqrt{1-k h}-\sqrt{1+k h}}{-h} \)
\(=\lim _{h \rightarrow 0} \frac{\sqrt{1-k h}-\sqrt{1+k h} \times(\sqrt{1-k h}+\sqrt{1+k h})}{-h(\sqrt{1-k h}+\sqrt{1+k h})} \)
\(=\lim _{h \rightarrow 0} \frac{(1-k h)-(1+k h)}{-h(\sqrt{1-k h}+\sqrt{1+k h})} \) 
\(\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]\) 
\(=\lim _{h \rightarrow 0} \frac{-2 k h}{-h(\sqrt{1-k h}+\sqrt{1+k h})} \)
\(=\lim _{h \rightarrow 0} \frac{2 k}{\sqrt{1-k h}+\sqrt{1+k h}}=\frac{2 k}{1+1}=\frac{2 k}{2}=k \)
\(\because f(x)\) is continuous at x = 0
\(\therefore f(0)=\mathrm{LHL} \Rightarrow-1=k\)
So, \( k=-1\)

The given differential equation is :
\(x\frac { dy }{ dx } \sin { \left( \frac { y }{ x } \right) } +x-y\sin { \left( \frac { y }{ x } \right) } =0\)
\(\Rightarrow y\sin { \left( \frac { y }{ x } \right) } -x=\frac { xdy }{ dx } \sin { \left( \frac { y }{ x } \right) } \)
\(\Rightarrow \frac { dy }{ dx } =\frac { y\sin { \left( \frac { y }{ x } \right) } -x }{ x\sin { \left( \frac { y }{ x } \right) } } \) ... (i)
\(f\left( x,y \right) =\frac { y }{ x } -cosec\left( \frac { y }{ x } \right) \)
Now, put \(x=\lambda x,y=\lambda y\), then
\(f\left( \lambda x,\lambda y \right) =\frac { \lambda y }{ \lambda x } -cosec\left( \frac { \lambda y }{ \lambda x } \right) \)
\(=\frac { y }{ x } -cosec\left( \frac { y }{ x } \right) \)
= f (x, y)
So, it is homogeneous
Now, put y = vx in equation (i).
On differentiating both sides w.r.t. x, we get
\(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
Now from (i), \(v+x\frac { dv }{ dx } =\frac { vx }{ x } -\frac { 1 }{ \sin { \left( \frac { vx }{ x } \right) } } \)
\(\Rightarrow x\frac { dv }{ dx } =-\frac { 1 }{ \sin { v } } \)
\(\Rightarrow -\int { \sin { v } dv } =\int { \frac { dx }{ x } } \)
\(\Rightarrow \cos { v } =\log { \left| x \right| } +C\)
\(\Rightarrow \cos { \left( \frac { y }{ x } \right) } =\log { \left| x \right| } +C\)
It is given that x = 1 and y = \(y={ \pi }/{ 2 }\)
So, \(\cos { \frac { \left( \frac { \pi }{ 2 } \right) }{ 1 } } =C+\log { \left| 1 \right| } \)
\(\Rightarrow C=0\)
\(\Rightarrow \cos { \left( \frac { y }{ x } \right) } =\log { \left| x \right| } \)
Hence, required solution is
\(\cos { \left( \frac { y }{ x } \right) } =\log { \left| x \right| } \)

Given, (cos x)^y = (cos y)^x
Taking log on both sides, we get
log(cos x)^y = log (cos y)^x
\(\therefore\) y log (cos x) = x log(cos y)
On differentiating both sides w.r.t. x, we get
\(y \cdot \frac{d}{d x} \log (\cos x)+\log (\cos x) \cdot \frac{d}{d x}(y)\)
\(\begin{aligned} =x \frac{d}{d x} \log (\cos y)+\log (\cos y) \frac{d}{d x}(x) \end{aligned}\)
\(\begin{aligned} \left[\because \frac{d}{d x}(u v)=u \frac{d v}{d x}+v \frac{d u}{d x}\right] \end{aligned}\)
\(\begin{aligned} \Rightarrow y \cdot \frac{1}{\cos x} \frac{d}{d x}(\cos x)+\log (\cos x) \frac{d y}{d x} \end{aligned}\)
\(\begin{aligned} =x \cdot \frac{1}{\cos y} \frac{d}{d x}(\cos y)+\log (\cos y) \cdot 1 \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad y \cdot \frac{1}{\cos x}(-\sin x)+\log (\cos x) \cdot \frac{d y}{d x} \end{aligned}\)
\(\begin{aligned} =x \frac{1}{\cos y}(-\sin y) \cdot \frac{d y}{d x}+\log (\cos y) \cdot 1 \end{aligned}\)
\(\begin{aligned} \Rightarrow-y \tan x+\log (\cos x) \frac{d y}{d x}=-x \tan y \frac{d y}{d x}+\log (\cos y) \end{aligned}\)
\(\begin{aligned} \Rightarrow[x \tan y+\log (\cos x)] \frac{d y}{d x}=\log (\cos y)+y \tan x \end{aligned}\)
\(\begin{aligned} \therefore \quad \frac{d y}{d x}=\frac{\log (\cos y)+y \tan x}{x \tan y+\log (\cos x)} \end{aligned}\)

\(A=\left[ \begin{matrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{matrix} \right] \)
|A| = 2 x 0-(-3) x -2+5 x 1
= -6 + 5 = -1
a₁₁ = (-4 + 4) = 0
a₁₂ = -(-6 + 4) = 2
a₁₃ = (3 - 2) = 1
a₂₁ = -(6 - 5) = -1
a₂₂ = (-4 - 5) = -9
a₂₃ = -(2 + 3) = -5
a₃₁ = (12 - 10) = 2
a₃₂ = -(-8 - 15) = 23
a₃₃ = (4 + 9) = 13
\({ A }^{ -1 }=\frac { 1 }{ -1 } \left[ \begin{matrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{matrix} \right] \)
Given set of equations may be written as
\(AX=B,\)
where \(B=\left[ \begin{matrix} 11 \\ -5 \\ -3 \end{matrix} \right] \)
\(X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \)
\(\Rightarrow X={ A }^{ -1 }B\)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{matrix} \right] \left[ \begin{matrix} 11 \\ -5 \\ -3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 0-5+6 \\ -22-45+69 \\ 11-25+39 \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \right] \)
Hence x = 1, y = 2 and x = 3.

Let the price of a pen, bag and instrument box are ₹ x, ₹y and ₹z respectively, then
\(5 x+3 y+z=160\)
(for Gautam)
\(2 x+y+3 z=190\)
(for Vikram)
\(\text { and } x+2 y+4 z=250\)
(for Ankur)
(i) This system of equation can be written as AX=B, where
\(A=\left[\begin{array}{lll} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{array}\right]\)
\(X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text { and } B=\left[\begin{array}{l} 160 \\ 190 \\ 250 \end{array}\right]\)
(ii) |A|  =5(4-6)-3(8-3)+1(4-1)
\(=5(-2)-3(5)+1(3) =-10-15+3=-22 \neq 0\)
(iii) Cofactors of |A| are
\(A_{11}=(-1)^{1+1}\left|\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right|=(4-6)=-2\)
\(A_{12}=(-1)^{1+2}\left|\begin{array}{ll} 2 & 3 \\ 1 & 4 \end{array}\right|=-(8-3)=-5\)
\(A_{13}=(-1)^{1+3}\left|\begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array}\right|=4-1=3\)
\(A_{21}=(-1)^{2+1}\left|\begin{array}{ll} 3 & 1 \\ 2 & 4 \end{array}\right|=-(12-2)=-10 \)
\( A_{22}=(-1)^{2+2}\left|\begin{array}{ll} 5 & 1 \\ 1 & 4 \end{array}\right|=20-1=19\)
\(A_{23}=(-1)^{2+3}\left|\begin{array}{ll} 5 & 3 \\ 1 & 2 \end{array}\right|=-(10-3)=-7\)
\( A_{31}=(-1)^{3+1}\left|\begin{array}{ll} 3 & 1 \\ 1 & 3 \end{array}\right|=(9-1)=8 \)
\(A_{32}=(-1)^{3+2}\left|\begin{array}{ll} 5 & 1 \\ 2 & 3 \end{array}\right|=-(15-2)=-13\)
\(A_{33}=(-1)^{3+3}\left|\begin{array}{ll} 5 & 3 \\ 2 & 1 \end{array}\right|^2=(5-6)=-1 \)
\(\therefore \operatorname{adj} A =\left[\begin{array}{lll} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{array}\right]^\tau\)
\(=\left[\begin{array}{ccc} -2 & -5 & 3 \\ -10 & 19 & -7 \\ 8 & -13 & -1 \end{array}\right]^T\)
\( =\left[\begin{array}{ccc} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{array}\right]^2\)
\( \therefore A^{-1}=\frac{\operatorname{adj} A}{|A|}=-\frac{1}{22}\left[\begin{array}{ccc} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{array}\right] \)
Or 
(iii) P =A²-5 A
\(=\left[\begin{array}{ccc} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{lll} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{array}\right]-5\left[\begin{array}{lll} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{array}\right]\)
\( =\left[\begin{array}{ccc} 25+6+1 & 15+3+2 & 5+9+4 \\ 10+2+3 & 6+1+6 & 2+3+12 \\ 5+4+4 & 3+2+8 & 1+6+16 \end{array}\right] \)
\(=\left[\begin{array}{lll} 32 & 20 & 18 \\ 15 & 13 & 17 \\ 13 & 13 & 23 \end{array}\right]-\left[\begin{array}{ccc} 25 & 15 & 5 \\ 10 & 5 & 15 \\ 5 & 10 & 20 \end{array}\right] \)
\(=\left[\begin{array}{lll} 7 & 5 & 13 \\ 5 & 8 & 2 \\ 8 & 3 & 3 \end{array}\right]\)

 (i) (c) : The given differential equation can be written as \( \frac{d y}{d x}+2 y \cot x=\operatorname{cosec} x\)
\(\therefore \text { I.F. }=e^{\int 2 \cot x d x}=e^{2 \log |\sin x|}=(\sin x)^{2} \)
\(\therefore \quad \lambda=2\)
(ii) (c) : We have, \(\left(1-x^{2}\right) \frac{d y}{d x}-x y=1\)
\(\Rightarrow \frac{d y}{d x}-\frac{x}{1-x^{2}} \cdot y=\frac{1}{1-x^{2}} \)
\(\therefore \text { I.F. }=e^{-\int \frac{x}{1-x^{2}} d x}=e^{\frac{1}{2} \int \frac{-2 x}{1-x^{2}} d x}\)
\(=e^{\frac{1}{2} \log \left(1-x^{2}\right)}=e^{\log \left(1-x^{2}\right)^{\frac{1}{2}}}=\sqrt{1-x^{2}}\)
(iii) (b) : We have, \(\frac{d y}{d x}+y=e^{-x} \)
 It is a linear differential equation with I.F. = \(e^{\int d x}=e^{x}\)
Now, solution is \(y \cdot e^{x}=\int e^{x} \cdot e^{-x} d x+c\)
\(\Rightarrow y e^{x}=\int d x+c \Rightarrow y e^{x}=x+c \Rightarrow y=x e^{-x}+c e^{-x}\)
\(\because y(0)=0 \Rightarrow c=0 \quad \therefore y=x e^{-x}\)
(iv) (a) : We have,\( \frac{d y}{d x}+y \tan x=\sec x\)
It is a linear differential equation with I.F. = \(e^{\int \tan x d x}=e^{\log |\sec x|}=\sec x\)
Now, solution is \(y \sec x=\int \sec ^{2} x d x+c\)
\(\Rightarrow y \sec x=\tan x+c\)
(v) (c) : We have, \(\frac{d y}{d x}-3 y=\sin 2 x\)
It is a linear differential equation with \(\text { I.F. }=e^{\int-3 d x}=e^{-3 x}\)

(i) (a) : Let \(\begin{equation} y=x^{x} \Rightarrow \log y=x \log x \end{equation}\) 
\(\begin{equation} \Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{d}{d x}(x \log x) \Rightarrow \frac{d y}{d x}=x^{x}\left[1 \times \log x+x \times \frac{1}{x}\right] \end{equation}\) 
(ii) (d)
(iii) (d) : Given \(\begin{equation} x=e^{x / y} \Rightarrow \log x=\frac{x}{v} \log e \Rightarrow y \log x=x \end{equation}\) 
\(\begin{equation} \Rightarrow y \frac{1}{x}+(\log x) \frac{d y}{d x}=1 \end{equation}\) 
\(\begin{equation} \Rightarrow \frac{d y}{d x}=\left(1-\frac{y}{x}\right) \frac{1}{\log x} \Rightarrow \frac{1}{x \log x}(x-y) \end{equation}\) 
(iv) (c) : y = (2 - x)³ (3 + 2x)⁵ '
\(\begin{equation} \Rightarrow \log y=\log (2-x)^{3}+\log (3+2 x)^{5} \end{equation}\) 
= 3 log (2 - x) + 5log (3 + 2x)
\(\begin{equation} \Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{3 \times(-1)}{2-x}+\frac{5}{3+2 x} \times(2) \end{equation}\) 
\(\begin{equation} \Rightarrow \frac{d y}{d x}=(2-x)^{3}(3+2 x)^{5}\left[\frac{10}{3+2 x}-\frac{3}{2-x}\right] \end{equation}\) 
(v) (d) : y = x^x.e^{(2x + 5)}
\(\begin{equation} \Rightarrow \log y=x \log x+(2 x+5) \end{equation}\) 
\(\begin{equation} \Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}=\left(x \cdot \frac{1}{x}+\log x\right)+2 \end{equation}\) 
\(\begin{equation} \Rightarrow \frac{d y}{d x}=x^{x} \cdot e^{2 x+5} \cdot(3+\log x) \end{equation}\)

(b) Assertion is correct, Reason is correct; Reason is not a correct explanation for Assertion

(c) Assertion is correct, Reason is incorrect