\(\int \frac{\sin \{(x+\alpha)-2 \alpha\}}{\sin (x+\alpha)} d x =\int \frac{\sin (x+\alpha) \cos 2 \alpha-\cos (x+\alpha) \sin 2 \alpha}{\sin (x+\alpha)} d x \)
\(=\cos 2 \alpha \int 1 \cdot d x-\sin 2 \alpha \int \cot (x+\alpha) d x \)
\(=\cos 2 \alpha \cdot x-\sin 2 \alpha \cdot \log |\sin (x+\alpha)|+C \)

log |x| + e ^-y/x = \(\frac{1}{e}\), as the particular solution.

Let \(I=\int \frac{2 \cos x}{(1-\sin x)\left(1+\sin ^{2} x\right)} d x\) 
Put sin x = t, then cos x dx = dt
\(\therefore I=\int \frac{2 d t}{(1-t)\left(1+t^{2}\right)}\) ...(i)
Now.Iet \(\frac{2}{(1-t)\left(1+t^{2}\right)}=\frac{A}{1-t}+\frac{B t+C}{1+t^{2}}\) 
\(\Rightarrow 2=\left(1+t^{2}\right) A+(1-t)(B t+C) \)
\(\Rightarrow 2=\left(1+t^{2}\right) A+\left(B t+C-B t^{2}-C t\right) \)
\(\Rightarrow 2=t^{2}(A-B)+t(B-\mid C)+(A+C) \) 
On comparing the coefficients of like powers of t, we get
\(A-B=0 ; B-C=0 \text { and } A+C=2\) 
\(\Rightarrow A=B ; B=C \text { and } A+C=2\)
\(\Rightarrow 2=t^{2}(A-B)+t(B-\mid C)+(A+C) \) 
\(\Rightarrow A=B=C=1\) 
\(\therefore \frac{2}{(1-t)\left(1+t^{2}\right)}=\frac{1}{1-t}+\frac{1+t}{1+t^{2}}\) 
Now, from Eq. (i), we get
\(I=\int\left(\frac{1}{1-t}+\frac{1+t}{1+t^{2}}\right) d t=\int \frac{d t}{1-t}+\int \frac{1}{1+t^{2}} d t+\frac{1}{2} \int \frac{2 t}{1+t^{2}} d t\) 
\(=\frac{\log |1-t|}{(-1)}+\tan ^{-1} t+\frac{1}{2} \log \left|1+t^{2}\right|+C\) 
\(=\frac{1}{2} \log \left|1+\sin ^{2} x\right|-\log |1-\sin x|+\tan ^{-1}(\sin x)+C\) 
\([\because t=\sin x]\) 
\(=\tan ^{-1}(\sin x)+\log \left|\frac{\sqrt{1+\sin ^{2} x}}{1-\sin x}\right|+C\) 
\(\left[\because \log m-\log n=\log \left(\frac{m}{n}\right) \text { and } n \log m=\log m^{n}\right]\)

Given differential equation is \((x-y) d y=(x+y) d x\) 
or \(\frac{d y}{d x}=\frac{x+y}{x-y}\)   ...(i)
Let \(F(x, y)=\frac{x+y}{x-y}\) 
On replacing x by \(\lambda x\) and y by \(\lambda y\) in Eq. (ii), we get
\(F(\lambda x, \lambda y)=\frac{\lambda x+\lambda y}{\lambda x-\lambda y}=\frac{\lambda(x+y)}{\lambda(x-y)}=\lambda^{0} F(x, y)\) 
Thus, F(x, y) is a homogeneous function of degree zero
So, put y = vx.
On differentiating both sides w.r.t. x, we get
\(\frac{d y}{d x}=v+x \frac{d v}{d x}\) 
On putting the values of y and \(\frac{d y}{d x}\) in Eq. (i), we get
\(v+x \frac{d v}{d x}=\frac{x+v x}{x-v x} \Rightarrow v+x \frac{d v}{d x}=\frac{1+v}{1-v}\) 
 \(\Rightarrow x \frac{d v}{d x}=\frac{1+v}{1-v}-v \Rightarrow \frac{x d v}{d x}=\frac{1+v-v+v^{2}}{1-v}\) 
\(\Rightarrow x \frac{d v}{d x}=\frac{1+v^{2}}{1-v}\) 
On separating the variables, we get
\(\frac{1-v}{1+v^{2}} d v=\frac{1}{x} d x\) 
On integrating both sides, we get \(\int \frac{1-v}{1+v^{2}} d v=\int \frac{1}{x} d x\) 
\(\Rightarrow \int \frac{1}{1+v^{2}} d v-\frac{1}{2} \int \frac{2 v}{1+v^{2}} d v=\int \frac{1}{x} d x\) 
\(\Rightarrow \tan ^{-1} v-\frac{1}{2} \log \left|\left(1+v^{2}\right)\right|=\log |x|+C\) 
\(\Rightarrow \tan ^{-1}\left(\frac{y}{x}\right)-\frac{1}{2} \log \left|\left(1+\frac{y^{2}}{x^{2}}\right)\right|=\log |x|+C\) \(\left[\text { put } v=\frac{y}{x}\right]\) 
\(\Rightarrow 2 \tan ^{-1}\left(\frac{y}{x}\right)-\log \left|\left(1+\frac{y^{2}}{x^{2}}\right)\right|=2(\log |x|+C)\) 
\(\Rightarrow 2 \tan ^{-1}\left(\frac{y}{x}\right)=\log \left|\frac{\left(x^{2}+y^{2}\right)}{x^{2}}\right|+\log |x|^{2}+2 C\) 
\(\Rightarrow 2 \tan ^{-1}\left(\frac{y}{x}\right)=\log \left[\left|\frac{x^{2}+y^{2}}{x^{2}}\right||x|^{2}\right]+2 C\) 
\([\because \log m+\log n=\log m n]\) 
\(\Rightarrow 2 \tan ^{-1}\left(\frac{y}{x}\right)-\log \left|x^{2}+y^{2}\right|=2 C\) 
\(\Rightarrow \tan ^{-1}\left(\frac{y}{x}\right)-\frac{1}{2} \log \left|x^{2}+y^{2}\right|=C\) 
which is the required solution.

Maximise Z = -3x - 5y
Subject to the constraints
\(\begin{aligned} -2 x+y \leq 4, x+y \geq 3, \end{aligned}\)
\(x-2 y \leq 2 \text { and } x \geq 0, y \geq 0 \)
Now, considering the inequations as equations, We get
            -2x + y = 4           ....(i)
               x + y = 3           ....(ii)
and        x - 2y = 2           ....(iii)
Table for line -2x + y = 4 is

So, it passes through the points (-2, 0) and (0, 4) On putting (0, 0) in the inequality -2x + y \(\leq\) 4, we have
0 \(\leq\) 4 (which is true)
So, the half plane is towards the origin.
Table for line x + y = 3 is

So, it passes through the points (3, 0) and (0, 3).
0 + 0 \(\geq\) 3 (which is false)
So, the half plane is away from the origin.
Table for line x - 2y = 2 is

So, it passes through the point (2, 0) and (0, -1).
On putting (0, 0) in the inequality x - 2 y \(\leq\) 2.
0 \(\leq\) 2 (which is true)
So, the half plane is towards the origin.
Also, x \(\geq\) 0, y \(\geq\) 0, so the feasible region lies in the Ist quadrant.
The point of intersection of Egs. (i) and (ii) is
\(\left(-\frac{1}{3}, \frac{10}{3}\right)\), Eqs. (ii) and (iii) is \(\left(\frac{8}{3}, \frac{1}{3}\right)\) and Eqs. (i) and (iii) is \(\left(\frac{-10}{3}, \frac{-8}{3}\right)\). The graphical representation of the above system of inequations is given below.

\(\therefore\) Clearly, feasible region is shaded.
The value of Z at corner points are as follows

Here, the feasible region is unbounded and the open half plane determined by -3x - 5y > \(\frac{-29}{3}\) has no point in common with the feasible region. Hence, \(Z=-\frac{29}{3}\) is maximum at \(A\left(\frac{8}{3}, \frac{1}{3}\right)\)

Given, differential equation is
\(\begin{aligned} x \frac{d y}{d x} & =y-x \sin \frac{y}{x} \end{aligned}\)
or \(\begin{aligned} \frac{d y}{d x} & =\frac{y}{x}-\sin \frac{y}{x} \end{aligned}\)                   ...(i)
Eq. (i) is a homogeneous differential equation.
On putting y = vx \(y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\) in Eq. (i), we get
\(\begin{array}{rlrl} v+x \frac{d v}{d x} =v-\sin v \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \quad \frac{d v}{\sin v} =-\frac{d x}{x} \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \quad \operatorname{cosec} v \quad d v =-\frac{d x}{x} \end{array}\)         ...(ii)
On integrating both sides of Eq. (ii), we get
\(\begin{aligned} \int \operatorname{cosec} v d v & =-\int \frac{d x}{x} \end{aligned}\)
\(\begin{aligned} \log |\operatorname{cosec} v-\cot v| & =-\log |x|+\log K, K>0 \end{aligned}\)
                                     [here, log K is an arbitrary constant]
\(\Rightarrow\) log| (cosecv - cot v)x| = log K
\(\Rightarrow\)    |(cosec v - cot v)x| = K
\(\Rightarrow \quad(\operatorname{cosec} v-\cot v) x= \pm K\)
\(\Rightarrow \quad\left(\operatorname{cosec} \frac{y}{x}-\cot \frac{y}{x}\right) x=C \quad\left[\because v=\frac{y}{x}\right]\)
which is the required general solution.

Let \(I =\int \frac{3-5 \sin x}{\cos ^2 x} d x\) 
\(=\int\left(\frac{3}{\cos ^2 x}-\frac{5 \sin x}{\cos ^2 x}\right) d x\)
\(=3 \int \sec ^2 x d x-5 \int \sec x \tan x d x\)
\(=3 \tan x-5 \sec x+C\)

We have, \(\int \frac{1}{\cos (x-a) \cos (x-b)} d x\)
\(=\frac{1}{\sin (b-a)} \int \frac{\sin \{(x-a)-(x-b)\}}{\cos (x-a) \cos (x-b)} d x\)
\(=\frac{1}{\sin (b-a)} \int \frac{-\cos (x-a) \sin (x-b)}{\cos (x-a) \cos (x-b)} d x\)
\(=\frac{1}{\sin (b-a)} \int[\tan (x-a)-\tan (x-b)] d x\)
\(=\frac{1}{\sin (b-a)}\left[\int \tan (x-a) d x-\int \tan (x-b) d x\right] \)
\(=\frac{1}{\sin (b-a)}\)  (- log |cos(x - a)] + log |cos(x - b)|]+C
\(=\frac{1}{\sin (b-a)}\left[\log \left|\frac{\cos (x-b)}{\cos (x-a)}\right|\right]+C\)

Let \(I=\int_{-1}^2\left|x^3-3 x^2+2 x\right| d x\)
Let \(f(x)=x^3-3 x^2+2 x \)
\(\because f(0)=0-0+0=0\)
\(\Rightarrow x-0=x\) is a factor of \(x^3-3 x^2+2 x\)
Now, \(\frac{x^3-3 x^2+2 x}{x}=x^2-3 x+2\)
\( \therefore \quad x^3-3 x^2+2 x=x\left(x^2-3 x+2\right)=x(x-1)(x-2)\)
\(\therefore I=\int_{-1}^2|x(x-1)(x-2)| d x\)
\(I=\int_{-1}^0|x(x-1)(x-2)| d x+\int_0^1|x(x-1)(x-2)| d x +\int_1^2|x(x-1)(x-2)| d x\)
[by property (iv)]
\(\therefore I=-\int_{-1}^0\left(x^3-3 x^2+2 x\right) d x+\int_0^1\left(x^3-3 x^2+2 x\right) d x  -\int_1^2\left(x^3-3 x^2+2 x\right) d x \)
\(=-\left[\frac{x^4}{4}-x^3+x^2\right]_{-1}^0+\left[\frac{x^4}{4}-x^3+x^2\right]_0^1  -\left[\frac{x^4}{4}-x^3+x^2\right]_1^2 \)
\(=-\left[0-\left(\frac{1}{4}+1+1\right)\right]+\left[\left(\frac{1}{4}-1+1\right)-(0)\right]\)
\(-\left[\left(\frac{16}{4}-8+4\right)-\left(\frac{1}{4}-1+1\right)\right] \)
\(=\frac{1}{4}+1+1+\frac{1}{4}-1+1+\frac{1}{4}=\frac{9}{4}+\frac{1}{4}+\frac{1}{4}=\frac{11}{4}\)

Let \(I=\int \frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x} d x  \)
\(=\int \frac{1-2 \sin ^{2} x+2 \sin ^{2} x \mid}{\cos ^{2} x} d x\left[\because \cos 2 A=1-2 \sin ^{2} A\right]  \)
\(=\int \frac{1}{\cos ^{2} x} d x=\int \sec ^{2} x d x \)
\(=\tan x+C\)

Let \(I=\int \frac{d x}{5-8 x-x^{2}}=\int \frac{d x}{5-2 \cdot 4 \cdot x-x^{2}-(4)^{2}+(4)^{2}}\) 
\( =\int \frac{d x}{5+16-\left[x^{2}+(4)^{2}+2 \cdot 4 \cdot x\right.} \)
\(=\int \frac{d x}{21-(x+4)^{2}} \)
\(=\int \frac{d x}{(\sqrt{21})^{2}-(x+4)^{2}} \)
\(=\frac{1}{2 \sqrt{21}} \log \left|\frac{\sqrt{21}+x+4}{\sqrt{21}-x-4}\right|+C \)
\({\left[\because \int \frac{d x}{a^{2}-x^{2}}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C\right]} \)

The given differential equation is
\(\frac{d y}{d x}=e^{x+y} \Rightarrow \frac{d y}{d x}=e^x \cdot e^y\)
\(\Rightarrow\) dy = e^x.e^y dx
\(\Rightarrow\) e^-ydy = e^xdx
On integrating both sides, we get
\(\Rightarrow \int e^{-y} d y=\int e^x d x \Rightarrow-e^{-y}=e^x+C\)
which is the required solution.

We have, ax² + by² = 1
On differentiating both sides w.r.t. x, we get
2ax + 2by y₁ = 0           [\(\because\) y₁ = dy/dx]
\(\Rightarrow\) 2(ax + by y₁) = 0
\(\Rightarrow\) ax + byy₁ = 0                        ...(i)
Again, on differentiating both sides w.r.t. x,we get
a + b(y₁y₁ + yy₂) = 0         [\(\because\) y₂ = d²y/dx²]
\(\begin{aligned} \Rightarrow \quad a+b\left(y_1^2+y y_2\right)=0 \end{aligned}\)
\(\Rightarrow \quad a=-b\left(y y_2+y_1^2\right)\)    ...(ii)
On putting a = - b(yy₂ + \(y_1^2\)) in Eq. (i), we get
\(\begin{array}{rlrl} -b\left(y y_2+y_1^2\right) x+b y y_1 =0 \end{array}\)
\(\begin{array}{rlrl} \Rightarrow b\left\{-\left(y y_2+y_1^2\right) x+y y_1\right\} =0 \end{array}\)
\(\begin{array}{rlrl} \Rightarrow x\left(y y_2+y_1^2\right) =y y_1 \end{array}\)
Hence Proved.

\(\text { Let } I=\int \frac{d x}{x\left(1+x^2\right)}\)
\( I=\int \frac{d x}{x^3\left(\frac{1}{x^2}+1\right)} \)
\( \text { Put } \frac{1}{x^2}+1=t \Rightarrow \frac{-2}{x^3} d x=d t\)
\(\therefore I=-\frac{1}{2} \int \frac{d t}{t}=-\frac{1}{2} \log |t|+C\)
\(\Rightarrow  I=-\frac{1}{2} \log \left|\frac{1}{x^2}+1\right|+C \)
\(\Rightarrow  I=-\frac{1}{2} \log \left|\frac{x^2+1}{x^2}\right|+C\)
\(\Rightarrow I=\frac{1}{2} \log \left|\frac{x^2}{x^2+1}\right|+C\)

Let \(I=\int \frac{x^2+1}{\left(x^2+2\right)\left(x^2+3\right)} d x\)
Let \(x^2=y\)
Then, \(\frac{x^2+1}{\left(x^2+2\right)\left(x^2+3\right)}=\frac{y+1}{(y+2)(y+3)}\)
Again, let \(\frac{y+1}{(y+2)(y+3)}=\frac{A}{y+2}+\frac{B}{y+3}\)
\(\Rightarrow \quad y+1=A(y+3)+B(y+2)\)
Putting y=-2 and y=-3 successively in Eq. (ii), we get
A=-1 and  B=2
On substituting the values of A and B in Eq. (i), we get
\(\frac{y+1}{(y+2)(y+3)}=\frac{-1}{y+2}+\frac{2}{y+3} \)
\(\therefore \int \frac{x^2+1}{\left(x^2+2\right)\left(x^2+3\right)} d x=\int \frac{-1}{x^2+2} d x+\int \frac{2}{x^2+3} d x\)
\(=-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x}{\sqrt{2}}\right)+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)+C\)

We have, the equation of the ellipse
x² + 9y² = 36
\(\Rightarrow \frac{x^2}{36}+\frac{y^2}{4}=1 \Rightarrow \frac{x^2}{6^2}+\frac{y^2}{2^2}=1\)

\(\therefore\) Required area \(=4 \int_0^6 y d x\)
\(\begin{aligned} & =4 \int_0^6 \frac{2}{6} \sqrt{36-x^2} d x=\frac{4}{3} \int_0^6 \sqrt{36-x^2} d x \end{aligned}\)
\(\begin{aligned} =\frac{4}{3}\left[\frac{x}{2} \sqrt{36-x^2}+\frac{36}{2} \sin ^{-1} \frac{x}{6}\right]_0^6 \end{aligned}\)
\(\begin{aligned} & =\frac{4}{3}\left[0+\frac{36}{2} \sin ^{-1} 1-0\right]=\frac{4}{3} \times 18 \times \frac{\pi}{2} \end{aligned}\)
= 12 \(\pi\) sq units

Given, the circle \(x^{2}+y^{2}=32\) 
and the line y = x 
Let us find the point of intersection of Eqs. (i) and (ii
On substituting y = x in Eq. (i), we get
\(x^{2}+x^{2}=32\) 
\(\Rightarrow 2 x^{2}=32\) 
\(\Rightarrow x^{2}=16 \Rightarrow x=\pm 4\) 
Thus, the points of intersection are \(\begin{array}{c} (4,4) \text { and }(-4,-4) {[\because y=x]} \end{array}\) 
 
Clearly, the required area
= Area of shaded region OABO
\( =\int_{0}^{4} y(\text { line }) d x+\int_{4}^{4 \sqrt{2}} y(\text { circle }) d x\)
\(=\int_{0}^{4} x d x+\int_{4}^{4 \sqrt{2}} \sqrt{32-x^{2}} d x\)
\(\left[\because x^{2}+y^{2}=32 \Rightarrow y=\pm \sqrt{32-x^{2}} \text { and } y>0\right]\)
\(=\left[\frac{x^{2}}{2}\right]_{0}^{4}+\int_{4}^{4 \sqrt{2}} \sqrt{(4 \sqrt{2})^{2}-x^{2}} d x \) 
\(=\frac{1}{2}[16-0]+\frac{1}{2}\left[x \sqrt{(4 \sqrt{2})^{2}-x^{2}}+(4 \sqrt{2})^{2} \sin ^{-1}\left(\frac{x}{4 \sqrt{2}}\right)\right]_{4}^{4 \sqrt{2}}\) 
\(=8+\frac{1}{2}\left[\left(0+32 \sin ^{-1}(1)\right)-\left(4 \sqrt{32-16}+32 \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\right)\right]\) 
\(=8+\frac{1}{2}\left[32 \sin ^{-1}(1)-16-32 \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\right]\) 
\(=8+\frac{1}{2}\left[32 \cdot \frac{\pi}{2}-16-32 \cdot \frac{\pi}{4}\right]\)
\(=8+\frac{1}{2}[16 \pi-16-8 \pi]=8+\frac{1}{2}[8 \pi-16]=8+4 \pi-8\)
\(=4 \pi \text { sq units }\)

Given curve is  4x² + y² = 36 
\(\begin{array}{ll} \therefore & \frac{4 x^2}{36}+\frac{y^2}{36}=1 \end{array}\)
\(\begin{array}{ll} \therefore & \frac{x^2}{9}+\frac{y^2}{36}=1 \end{array}\)              ...(i)
We know that the standard equation of ellipse is
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)                      ...(ii)
On comparing Eqs. (i) and (ii), we get
a² = 9 and b² = 36
\(\Rightarrow\) a = 3 and b = 6
Here, we see that a < b, so the vertical ellipse will be formed.

Now, required area = 4(Area of region OAB in first quadrant)
\(\begin{aligned} =4\left|\int_0^3 y d x\right| \end{aligned}\)
\(\begin{aligned} =\left|4 \int_0^3 2 \sqrt{9-x^2} d x\right| \end{aligned}\)
\(\begin{aligned} \left[\because \frac{y^2}{36}=1-\frac{x^2}{9} \Rightarrow y=2 \sqrt{9-x^2}\right] \end{aligned}\)
\(\begin{aligned} =8 \int_0^3 \sqrt{9-x^2} d x \end{aligned}\)
\(\begin{aligned} =8\left|\left[\frac{x}{2} \sqrt{9-x^2}+\frac{9}{2} \sin ^{-1}\left(\frac{x}{3}\right)\right]_0^3\right| \end{aligned}\)
\(\left[\because \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1}\left(\frac{x}{a}\right)\right]\)
\(=8\left\{\left[\frac{3}{2} \sqrt{9-9}+\frac{9}{2} \sin ^{-1} \frac{3}{3}-0-\frac{9}{2} \sin ^{-1} 0\right]\right\}\)
\(\begin{aligned} =8\left|\left[0+\frac{9}{2} \times \frac{\pi}{2}-0\right]\right| \end{aligned}\)
\(\begin{aligned} =8 \times \frac{9 \pi}{4} \end{aligned}\)
\(=18 \pi\)
Hence, the required area is 18\(\pi\) sq units.

We have, maximise, Z = 60x + 40y             ...(i)
Subject to the constraints, x + 2y \(\leq 12\)     ...(ii)
\(\begin{aligned} 2 x+y & \leq 12 \end{aligned}\)                                   ...(iii)
\(\begin{aligned} 4 x+5 y & \geq 20 \end{aligned}\)                                 ...(iv)
and \(\begin{aligned} x, y & \geq 0 \end{aligned}\)                                     ...(v)
Table for  line x + 2y = 12 is

So, the line x + 2y = 12 is passing through the points (0, 6) and (12, 0).
On putting (0,0) in the inequality \(x+2 y \leq 12\) we get \(0+2(0) \leq 12 \Rightarrow 0 \leq 12\), which is true
So,the half plane is towards the origin.
Table for line 2x + y = 12 is 

So, the line 2x + y = 12 is passing through the points (0, 12) and (6, 0).
On putting (0, 0) in the inequality \(2 x+y \leq 12\) we get
\(2(0)+0 \leq 12 \Rightarrow 0 \leq 12\) , which is true.
So, the half plane is towards the origin.
Table for line 4x + 5y = 20 is

So, the line 4x + 5y = 20 is passing through the points (0,4) and (5, 0).
On putting (0, 0) in the inequality \(4 x+5 y \geq 20\), we get 4(0) + 5(0) \(\geq\) 20 \(\Rightarrow\) 0 > 20 which is not true.
So, the half plane is away from the origin.
Also, x, y \(\geq\) 0
So, the region lies in Ist quadrant.

On solving Eqs. x + 2y = 12 and 2x + y = 12, we get D(4, 4)
Clearly, the feasible region is ABCDEA.
The corner points of the feasible region are A(0, 4), B(5, 0), C(6, 0), D(4, 4) and E(0, 6).
The value of Z at corner points are given below.

The maximum value of Z is 400 at D(4, 4).

Given that,
Minimise and Maximise Z = 3x + 9y       ...(i)
Subject to the constraints are
\(\begin{aligned} x+3 y & \leq 60 \end{aligned}\)        ...(ii)
\(\begin{aligned} x+y & \geq 10 \end{aligned}\)          ...(iii)
\(\begin{aligned} x & \leq y \end{aligned}\)                  ...(iv)
\(\begin{aligned} x \geq 0, y & \geq 0 \end{aligned}\)         ...(v)

First of all,let us plot the graph of the feasible region of the system of linear inequalities (ii) to (v). The feasible region ABCDA is shown in the figure.
Note That the region is bounded. The coordinates of the corner points A, B, C and D are (0, 10), (5, 5), (15, 15) and (0, 20), respectively.

We, now find the minimum and maximum value of Z. From the table, we find that the minimum value of Z is 60 at the point B(5, 5) of the feasible region.
The maximum value of Z on the feasible region occurs at the two corner points C (15, 15) and D (0, 20) and it is 180 in each case.
Remark Observe that in the above example, the problem has multiple optimal solutions at the corner points C and D, i.e. the both points produce same maximum value 180.
In such cases, you can see that every point on the line segment CD joining the two corner points C and D also give the same maximum value. Same is also true in the case, if the two points produce same minimum value.

Given, differential equation can be written as
\(\frac{d y}{d x}=\frac{y}{x}-\frac{1}{\sin \frac{y}{x}}\)       ....(i)
Let \(F(x, y)=\frac{y}{x}-\frac{1}{\sin \frac{y}{x}}\)
Now, \(F(\lambda x, \lambda y)=\frac{\lambda y}{\lambda x}-\frac{1}{\sin \left(\frac{\lambda y}{\lambda x}\right)}=\lambda^{\circ}\left(\frac{y}{x}-\frac{1}{\sin \frac{y}{x}}\right)\)
\(=\lambda^{\circ} F(x, y)\)
It is a homogeneous differential equation.
Now, on putting y = vx \(\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\)
\(\therefore\) From Eq. (i), we get \(v+x \frac{d v}{d x}=\frac{v x}{x}-\frac{1}{\sin \left(\frac{v x}{x}\right)}\)
\(\Rightarrow v+x \frac{d v}{d x}=v-\frac{1}{\sin v} \Rightarrow \sin v d v=-\frac{1}{x} d x\)
On integrating both sides, we get
- cos v = - log|x| - C
\(\Rightarrow-\cos \left(\frac{y}{x}\right)=-\log |x|-C\)
\(\Rightarrow \cos \left(\frac{y}{x}\right)=\log |x|+C\)        ...(ii)
Given that x = 1, when y = \(\frac{\pi}{2}\)
\(\therefore \quad \cos \left(\frac{\pi}{2}\right)=\log |1|+C \Rightarrow 0=0+C\)
\(\Rightarrow \quad C=0\)
On putting C = 0 in Eq. (ii), we get
\(\cos \left(\frac{y}{x}\right)=\log |x|+0\)

Let \( l=\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x\)
\( \Rightarrow l=\int_0^\pi \frac{(\pi-x) \sin (\pi-x)}{1+\cos ^2(\pi-x)} d x\)
\(\left[\because \int_0^a f(x) d x=\int_0^2 f(a-x) d x\right]\)
\(=\int_0^\pi \frac{(\pi-x) \sin x d x}{1+\cos ^2 x}\)
On adding Eqs. (i) and (ii), we get
\(=\pi \int_0^\pi \frac{\sin x d x}{2} \Rightarrow l=\frac{\pi}{2} \int_0^\pi \frac{\sin x}{1+c} d x\)
Using \(\int_0^\pi f(x) d x=2 \int_0^\pi f(x) d x\), if f(2 a-x)=f(x)
\(\therefore \quad I=\frac{\pi}{2} \times 2 \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x\)
\(\Rightarrow \quad l=\pi \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x\)
Put \(\cos x=t\), then -sin x d x=d t
Lower limit When x=0 then t=1
Upper limit When \(x=\frac{\pi}{2}\), then t=0
\(\therefore I=\pi \int_1^0 \frac{-d t}{1+t^2} \Rightarrow I=-\pi\left[\tan ^{-1} t\right]_1^0\)
\(\Rightarrow  I=-\pi\left[\tan ^{-1} 0-\tan ^{-1} 1\right]\)
\(\Rightarrow  I=-\pi\left[0-\frac{\pi}{4}\right]=\frac{\pi^2}{4}\)

(i) (c)
(ii) (b): Since (8, 12) satisfy all the inequalities therefore (8, .1.2) is the point in its feasible region.
(iii) (c) : At (0, 0), z = 0
At (16, 0), z = 352
At (8, 12), z = 392
At (0, 20), z = 360
It can be observed that max z occur at (8, 12). Thus, z will attain its optimal value at (8, 12).
(iv) (c) : We have, x + y = 20 ... (i)
and 3x + 2y = 48.. (ii)
On solving (i) and (ii), we get
X = 8, y = 12.
Thus, the coordinates of P are (8,12) and hence (8, 12) is one of its corner points.
(v) (b): The optimal solution occurs at every point on the line joining these two points.

(i) (b) : The given differential equation can be written as \(\frac{d y}{d x}+\frac{\cos x}{1+\sin x} y=\frac{-x}{1+\sin x}\) 
Compare it with \(\frac{d y}{d x}+P y=Q\) ,we get
\(P=\frac{\cos x}{1+\sin x}\) and \(Q=\frac{-x}{1+\sin x}\) 
(ii) (c) : \(\text { I.F. }=e^{\int P d x}=e^{\int \frac{\cos x}{1+\sin x} d x}\) 
Put \(1+\sin x=t \Rightarrow \cos x d x=d t\) 
\(\therefore \quad \text { I.F. }=e^{\int \frac{1}{t} d t}=e^{\log t}=t=1+\sin x\) 
(iii) (d) : Solution of given differential equation is given by \(y \cdot(\mathrm{I} . \mathrm{F} .)=\int Q(\mathrm{I.F.}) d x+c\) 
\(\Rightarrow y(1+\sin x)=\int \frac{-x}{1+\sin x} \cdot(1+\sin x) d x+c\) 
\(\Rightarrow \quad y(1+\sin x)=\frac{-x^{2}}{2}+c\) 
(iv) (a) : We have, y(0) = 1 i.e., x = 0, y = 1
\(\therefore \quad 1(1+\sin 0)=c \Rightarrow c=1\) 
\(\therefore \quad y(1+\sin x)=\frac{-x^{2}}{2}+1=\frac{2-x^{2}}{2}\) 
\(\therefore \quad y=\frac{2-x^{2}}{2(1+\sin x)}\) 
(v) (b) : We have, \(y=\frac{2-x^{2}}{2(1+\sin x)}\)
\(\therefore \quad y\left(\frac{\pi}{2}\right)=\frac{2-\left(\frac{\pi}{2}\right)^{2}}{2\left(1+\sin \frac{\pi}{2}\right)}=\frac{2-\frac{\pi^{2}}{4}}{4}=\frac{8-\pi^{2}}{16}\)

(i) Given, differential equation is (x² - y²)dx + 2xydy = 0
\(\begin{aligned} \Rightarrow \frac{d y}{d x} & =\frac{-\left(x^2-y^2\right)}{2 x y}=\frac{y^2-x^2}{2 x y} \end{aligned}\)
\(\begin{aligned} =\frac{x^2\left(\frac{y^2}{x^2}-1\right)}{2 x y}=\frac{\left(\frac{y}{x}\right)^2-1}{2\left(\frac{y}{x}\right)} \end{aligned}\)
\(\therefore\)  In RHS, degree of numerator and denominator is same
\(\therefore\)  It is a homogeneous differential equation and can be written as
\(\frac{d y}{d x}=g\left(\frac{y}{x}\right)\)
(ii) Given, differential equation is (x² - y²)dx + 2xy dy = 0
\(\Rightarrow \quad \frac{d y}{d x}=-\frac{\left(x^2-y^2\right)}{2 x y}=\frac{y^2-x^2}{2 x y}\) ...(i)
This is a homogeneous differential equation
On putting y = vx \(\Rightarrow \frac{d y}{d x}=v+x \cdot \frac{d v}{d x}\)
\(\therefore\) From Eq (i), we get
\(\begin{aligned} v+x \cdot \frac{d v}{d x} & =\frac{v^2-1}{2 v} \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad x \frac{d v}{d x} & =\frac{v^2-1}{2 v}-v \end{aligned}\)
\(\begin{aligned} =\frac{v^2-1-2 v^2}{2 v}=\frac{-v^2-1}{2 v} \end{aligned}\)
\(\Rightarrow \frac{2 v}{v^2+1} d v=\frac{-d x}{x}\)
on integrating both sides, we get
log |v² + 1| = -log x + log c
\(\begin{aligned} & \Rightarrow \quad \log \left|\frac{y^2}{x^2}+1\right|=-\log x+\log c \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad \log \left|\frac{y^2+x^2}{x^2} \cdot x\right|=\log c \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad \frac{y^2+x^2}{x}=c \end{aligned}\)
\(\Rightarrow\)   y² + x² = cx
which is the required solution.

(c) Assertion is correct, Reason is incorrect

(c) Assertion is correct, Reason is incorrect