Given, differential equation is \(\frac{d y}{d x}=k(50-y) .\)
(i) Order of the given differential equation is 1.
(ii) (a) Method of solving differential equation \(\frac{d y}{d x}=k(50-y)\) is variable separable method.
(ii) (b) Given, differential equation,
\(\begin{aligned} \frac{d y}{d x} & =k(50-y) \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{d y}{50-y} & =k d x \end{aligned}\)
On integration, we get
-log |50 - y| = kx + C
(iv) (b) Now, -log(50 - y) = kx + C
On Puttin x = 0, y = 0 and k = 0.049, we get
=log(50 - 0) = 0 + C
\(\Rightarrow \quad C=-\log 50=\log (50)^{-1}=\log \frac{1}{50}\)
(v) (c) Solving differential equation, we get
-log_e (50 - y) = kx + C
\(\Rightarrow\) (50 - y) = e^{-kx - c} = e^-c.e^-kx = Ae^-kx
                                              [put e^-c = A]
\(\Rightarrow\) 50 - y = Ae^-kx \(\Rightarrow\) y = 50 - Ae^-kx
When x = 0, y = 0 then
0 = 50 - Ae⁰ \(\Rightarrow\) A = 50
\(\therefore \quad y=50-50 e^{-k x}=50\left(1-e^{-k x}\right)\)
\(\therefore\) The solution used to find the number of children who have been given the polio drops is
y = 50(1 - e^-kx).

Given, different equation,
\(\frac{d T}{d t} \propto(T-70) \Rightarrow \frac{d T}{T-70}=k d t\)
(i) Degree of the given differential equation is 1.
(ii) (a) Given differential equation is solved by variable separable method.
(iii) No; The time of death of Cat is 11:30 pm. Time of death is not depend on temperature
\(\therefore\) Time of death not change.
(iv) (a) We have, \(\frac{d T}{d t}=k(T-70)\)
\(\frac{d T}{T-70}=k d t\)
On integrating, we get
log |T - 70| = kt + C
(v) (d) Now, log |T - 70 | = kt + C
When t = 0,T = 72, we get
log(72 - 70) = 0 + C
\(\Rightarrow\) C = log 2

(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.

(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) (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) (b) : We have,\(\frac{d y}{d x}=\frac{a x+3}{2 y+f}\) 
\(\Rightarrow \quad(a x+3) d x=(2 y+f) d y\) 
\(\Rightarrow \quad a \frac{x^{2}}{2}+3 x=y^{2}+f y+c\) (Integrating)
\(\Rightarrow-\frac{a}{2} x^{2}+y^{2}-3 x+f y+C=0\)
This will represent a circle, if \(\frac{-a}{2}=1 \Rightarrow a=-2\) 
[\(\therefore\) In circle, coefficient of x² = coefficient of y²]
(ii) (c) : We have,\(\frac{y d y}{\sqrt{1-y^{2}}}=d x\) 
On integration, we get \(-\sqrt{1-y^{2}}=x+c\) 
\(\Rightarrow \quad 1-y^{2}=(x+c)^{2} \Rightarrow(x+c)^{2}+y^{2}=1\) ,which represents a circle with radius 1 and centre on the x-axis.
(iii) (c) :  \(y^{\prime}=y+1 \Rightarrow \frac{d y}{y+1}=d x\) 
\(\Rightarrow \ln (y+1)=x+c\) 
Now, \(y(0)=1 \Rightarrow c=\ln 2\) 
\(\therefore \quad \ln \left(\frac{y+1}{2}\right)=x \Rightarrow y+1=2 e^{x}\) 
So, y (In 2) = -1 + 2e^{1n 2} = -1 + 4 = 3
(iv) (c) : From the given differential equation, we have
\(\frac{d y}{d x}=\frac{e^{x}+x^{2}}{e^{y}} \Rightarrow e^{y} d y=\left(e^{x}+x^{2}\right) d x\) 
Integrating, we get \(y=e^{x}+\frac{x^{3}}{3}+c\) 
(v) (a) : We have, \(\frac{d y}{d x}=y \sin 2 x\) 
\(\Rightarrow \quad \frac{d y}{y}=\sin 2 x d x \Rightarrow \log y=-\frac{\cos 2 x}{2}+c\) 
Since x = 0, y = 1 therefore c = 1/2
Now, \(\log y=\frac{1}{2}(1-\cos 2 x)\) 
\(\Rightarrow \log y=\sin ^{2} x \Rightarrow y=e^{\sin ^{2} x}\)

(i) (b): We have, \(\frac{d y}{d x}=\frac{x^{2}+x y+y^{2}}{x^{2}}\)
Put y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
\(\therefore v+x \frac{d v}{d x}=\frac{x^{2}+x \cdot v x+v^{2} x^{2}}{x^{2}}=1+v+v^{2}\)
\(\Rightarrow x \frac{d v}{d x}=1+v^{2} \Rightarrow \int \frac{d v}{1+v^{2}}=\int \frac{d x}{x}+c\)
\(\Rightarrow \tan ^{-1} v=\log |x|+c \Rightarrow \tan ^{-1} \frac{y}{x}=\log |x|+c\)
(ii) (d): We have, \( 2 x y \frac{d y}{d x}=x^{2}+3 y^{2} \Rightarrow \frac{d y}{d x}=\frac{x^{2}+3 y^{2}}{2 x y}\)
Put y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
\(\therefore v+x \frac{d v}{d x}=\frac{x^{2}+3 v^{2} x^{2}}{2 v x^{2}} \Rightarrow x \frac{d v}{d x}=\frac{1+3 v^{2}}{2 v}-v\)
\(\Rightarrow x \frac{d v}{d x}=\frac{1+v^{2}}{2 v_{*}} \Rightarrow \int \frac{2 v}{1+v^{2}} d v=\int \frac{d x}{x}+\log c\)
\(\Rightarrow \log \left|1+v^{2}\right|=\log |x|+\log |c| \Rightarrow \log \left|v^{2}+1\right|=\log |x c|\)
\(\Rightarrow \quad v^{2}+1=x c \Rightarrow \frac{y^{2}}{2}+1=x c \Rightarrow x^{2}+y^{2}=x^{3} c\)
(iii) (d): We have, \(\left(x^{2}+3 x y+y^{2}\right) d x-x^{2} d y=0\)
\(\Rightarrow \frac{x^{2}+3 x y+y^{2}}{x^{2}}=\frac{d y}{d x}\)
Put y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
\(\therefore \frac{x^{2}+3 x^{2} v+x^{2} v^{2}}{x^{2}}=\left(v+x \frac{d v}{d x}\right)\)
\(\Rightarrow 1+3 v+v^{2}=v+x \frac{d v}{d x} \Rightarrow 1+2 v+v^{2}=x \frac{d v}{d x}\)
\(\Rightarrow \int \frac{d x}{x}-\int(v+1)^{-2} d v=c \Rightarrow \log x+\frac{1}{v+1}=c\)
\(\Rightarrow \log x+\frac{x}{x+y}=c\)
(iv) (c): We have, \(\frac{d y}{d x}=\frac{y}{x}\left\{\log \left(\frac{y}{x}\right)+1\right\}\)
Put y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
\(\therefore v+x \frac{d v}{d x}=v\{\log (v)+1\} \Rightarrow x \frac{d v}{d x}=v \log v\)
\(\Rightarrow \int \frac{d v}{v \log v}=\int \frac{d x}{x} \Rightarrow \log |\log v|=\log |x|+\log |c|\)
\(\Rightarrow \log \left(\frac{y}{x}\right)=c x\)
 (v) (a):  We have,\(\left(x \frac{d y}{d x}-y\right) e^{\frac{y}{x}}=x^{2} \cos x\)
\(\Rightarrow\left(\frac{d y}{d x}-\frac{y}{x}\right) e^{\frac{y}{x}}=x \cos x\)
\(\text { Put } y=v x \text { and } \frac{d y}{d x}=v+x \frac{d v}{d x}\)
\(\Rightarrow\left(v+x \frac{d v}{d x}-v\right) e^{v}=x \cos x \Rightarrow x e^{v} \frac{d v}{d x}=x \cos x\)
\(\Rightarrow \int e^{v} d v=\int \cos x d x \Rightarrow e^{v}=\sin x+c\)
\(\Rightarrow e^{\frac{y}{x}}-\sin x=c\)

(i) (c) : We have, \(2 \frac{d^{2} y}{d x^{2}}+3 \sqrt{1-\left(\frac{d y}{d x}\right)^{2}-y}=0\) 
\(\therefore \quad 2 \frac{d^{2} y}{d x^{2}}=-3 \sqrt{1-\left(\frac{d y}{d x}\right)^{2}-y}\) 
Squaring both sides, we get
\(4\left(\frac{d^{2} y}{d x^{2}}\right)^{2}=9\left[1-\left(\frac{d y}{d x}\right)^{2}-y\right]\) 
Here, highest order derivative is \(\frac{d^{2} y}{d x^{2}}\) and its power is 2. So, its degree is 2.
(ii) (d) : We have, \(y \frac{d y}{d x}=\frac{x}{\frac{d y}{d x}+\left(\frac{d y}{d x}\right)^{3}}\) 
\(\Rightarrow y\left(\frac{d y}{d x}\right)^{2}+y\left(\frac{d y}{d x}\right)^{4}=x\) 
\(\Rightarrow\) Here, highest order derivative is \(\frac{d y}{d x}\) is So , its order is 1 and degree is 4.
(iii) (a) : We have,\(y^{\prime \prime \prime}+y^{2}+e^{y^{\prime}}=0\) 
\(\frac{d^{3} y}{d x^{3}}+y^{2}+e^{(d y / d x)}=0\) 
Highest order derivative is \(\frac{d^{3} y}{d x^{3}}\) .So, its order is 3.
Also, the given differential cannot be expressed as a polynomial. So, its degree is not defined.
(iv) (c) : The given differential equation is,
\(\sqrt{a+x} \cdot\left(\frac{d y}{d x}\right)+x=0 \Rightarrow \frac{d y}{d x}=\frac{-x}{\sqrt{a+x}}\) 
Clearly, degree = 1
(v) (b) : We have \(y \frac{d y}{d x}=\frac{x}{\frac{d y}{d x}+\left(\frac{d y}{d x}\right)^{3}}\) 
\(\Rightarrow y\left(\frac{d y}{d x}\right)^{2}+y\left(\frac{d y}{d x}\right)^{4}=x\) 
\(\Rightarrow\) Here, highest order derivative is \(\frac{d y}{d x}\) ,So , its order is 1 and degree is 4.
(iii) (a) : We have, y'" +y² + e^y = 0
\(\frac{d^{3} y}{d x^{3}}+y^{2}+e^{(d y / d x)}=0\) 
Highest order derivative is \(\frac{d^{3} y}{d x^{3}}\) So, its order is 3.
Also, the given differential cannot be expressed as a polynomial. So, its degree is not defined
(iv) (c) :The given differential equation is,
\(\sqrt{a+x} \cdot\left(\frac{d y}{d x}\right)+x=0 \Rightarrow \frac{d y}{d x}=\frac{-x}{\sqrt{a+x}}\) 
Clearly, degree = 1.
(v) (b) : We have \(\left(1+\left(\frac{d y}{d x} \mid\right)^{3}\right)^{\frac{1}{3}}=7 \frac{d^{2} y}{d x^{2}}\) 
\(\therefore\) Order is 2 and degree is 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) : Given, T is the temperature of the body at any time t. Then, by Newton's law of cooling, we get \(\frac{d T}{d t}=k(T-50)\), where k is the constant of proportionality
(ii) (c) : From given information, we have
At 8 p.m. temperature is 70°F
\(\therefore\) At t = 0, T = 70°F
(iii) (b) : From given information, we have
At 10 p.m., temperature is 60°F.
\(\therefore\) At t = 2, T = 60° F
(iv) (c) : \(\frac{d T}{d t}=k(T-50) \Rightarrow \frac{d T}{T-50}=k d t\) 
On integrating both sides, we get
\(\log |T-50|=k t+\log C \Rightarrow T-50=C e^{\wedge t}\) 
Clearly, for t = 0, \(T=70^{\circ} \Rightarrow C=20\) 
Thus, T - 50 = 20e^kt
For \(t=2, T=60^{\circ} \Rightarrow 10=20 e^{2 k}\) 
\(\Rightarrow 2 k=\log \left(\frac{1}{2}\right) \Rightarrow k=\frac{1}{2} \log \left(\frac{1}{2}\right)\) 
Hence, \(T=50+20\left(\frac{1}{2}\right)^{\frac{t}{2}}\) 
(v) (b) : We have, \(T=50+20\left(\frac{1}{2}\right)^{\frac{2}{2}}\) 

(i) (b) : Here, P denotes the principal at any time t and the rate of interest be r% per annum compounded continuously, then according to the law given in the problem, we get
\(\frac{d P}{d t}=\frac{P r}{100}\) 
(ii) (a) : We have, \(\frac{d P}{d t}=\frac{\operatorname{Pr}}{100}\) 
\(\Rightarrow \frac{d P}{P}=\frac{r}{100} d t \Rightarrow \int \frac{1}{P} d P=\frac{r}{100} \int d t\) 
\(\Rightarrow \log P=\frac{r t}{100}+C\) 
At t = 0, P = P_o
\(\therefore \quad C=\log P_{0}\) 
So, \(\log P=\frac{r t}{100}+\log P_{0}\) 
\(\Rightarrow \log \left(\frac{P}{P_{0}}\right)=\frac{r t}{100}\) 
(iii) (c) : We have, r = 5, P_o = Rs. 100 and P = Rs. 200 = 2P_o
Substituting these values in (2), we get
\(\log 2=\frac{5}{100} t\) 
\(\Rightarrow\) t = 20 log_e 2 = 20 x 0.6931 years = 13.862 years
(iv) (d) : We have
P_o = Rs. 100, P = Rs. Rs. 200 = 2P_o and t = 10 years
Substituting these values in (2), we get
\(\log 2=\frac{10 r}{100} \Rightarrow r=10 \log 2=10 \times 0.6931=6.931\) 
(v) (a) : We have
P_o = Rs. 1000, r = 5 and t = 10
Substituting these values in (2), we get
\(\log \left(\frac{P}{1000}\right)=\frac{5 \times 10}{100}=\frac{1}{2}=0.5 \Rightarrow \frac{P}{1000}=e^{0.5}\) 
\(\Rightarrow\) P = 1000 x 1.648 = Rs. 1648

(i) (d) :Given, at any time t the thermometer reading be T^oF and the outside temperature be sop.
Then, by Newton's law of cooling, we have
\(\frac{d T}{d t} \propto(T-S) \Rightarrow \frac{d T}{d t}=-\lambda(T-S)\) 
(ii) (d) : Since, after 5 minutes, thermometer reads 60⁰F.
\(\therefore\) Value of T(5) = 60^0F
(iii) (a) : Clearly from given information, value of T( 10) is 50°F.
(iv) (b) : We have,\(\frac{d T}{d t}=-\lambda(T-S)\) 
\(\Rightarrow \frac{d T}{T-S}=-\lambda d t \Rightarrow \int \frac{1}{T-S} d T=-\lambda \int d t\) 
\(\Rightarrow \log (T-S)=-\lambda t+c\) 
(v) (c) : Since, at t = 0, T = 80°

(i) (d) : Since, maximum number of students in hostel is 1000.
\(\therefore\) Maximum value of n(t) is 1000.
(ii) (a) : Clearly, according to given information
\(\frac{d n}{d t}=\lambda n(1000-n)\) ,where \(\lambda\) is constant of proportionality.
(iii) (b) : Since, 50 students are infected after 4 days
\(\therefore\) n(4) = 50.
(iv) (c) : We have,\(\frac{d n}{d t}=\lambda n(100-n)\) 
\(\Rightarrow \int \frac{d n}{n(1000-n)}=\lambda \int d t\) 
\(\Rightarrow \frac{1}{1000} \int\left(\frac{1}{1000-n}+\frac{1}{n}\right) d n=\lambda \int d t\) 
\(\Rightarrow \quad \frac{1}{1000}\left[\frac{\log (1000-n)}{-1}+\log n\right]=\lambda t+C\) 
\(\Rightarrow \frac{1}{1000} \log \left(\frac{n}{1000-n}\right)=\lambda t+C\) 
(v) (a) : When, t = 0, n = 1
This condition is satisfied by option (a) only.

(i) b : Curves y = cos x and y = x + 1 meet at point C(O, 1).

(ii) C : curve y=cosx meet the x axis at \(A^{\prime}\left(\frac{-\pi}{2}, 0\right)\) and \(A\left(\frac{\pi}{2}, 0\right)\) .
(iii) (a) : \(\int_{-1}^{0}(x+1) d x=\left[\frac{x^{2}}{2}+x\right]_{-1}^{0}=0-\left(\frac{1}{2}-1\right)=\frac{1}{2}\) 
(iv) (d) : \(\int_{0}^{\pi / 2} \cos x d x=[\sin x]_{0}^{\pi / 2}=\sin \frac{\pi}{2}-\sin 0=1\) 
(v) (b) : Required area \(\int_{-1}^{0}(x+1) d x+\int_{0}^{\pi / 2} \cos x d x\) 
\(=\frac{1}{2}+1=\frac{3}{2} \text { sq. units }\)

(i) (a) : Points (0, 2) and (3, 0) pass through both line and ellipse.
(ii) (b) :

(iii) (c) : \(\frac{2}{3} \int_{0}^{3} \sqrt{9-x^{2}} d x=\frac{2}{3} \int_{0}^{3} \sqrt{(3)^{2}-x^{2}} d x\) 
\(=\frac{2}{3}\left[\frac{1}{2} x \sqrt{9-x^{2}}+\frac{9}{2} \sin ^{-1}\left(\frac{x}{3}\right)\right]_{0}^{3}\) 
\(=\frac{2}{3}\left[\frac{3}{2} \sqrt{0}+\frac{9}{2} \sin ^{-1}(1)-\frac{1}{2}(0)-\frac{9}{2} \sin ^{-1}(0)\right]\) 
\(=\frac{2}{3}\left[\frac{9}{2} \cdot \frac{\pi}{2}\right]=\frac{3 \pi}{2}\) 
(iv) d : \(2 \int_{0}^{3}\left(1-\frac{x}{3}\right) d x=2\left[x-\frac{x^{2}}{6}\right]_{0}^{3}\) 
\(=2\left(3-\frac{9}{6}-0-0\right)=2 \times \frac{3}{2}=3\) 
(v) (d) : Area of smaller region bounded by the mirrorand scratch
\(=\frac{2}{3} \cdot \int_{0}^{3} \sqrt{9-x^{2}} d x-2 \int_{0}^{3}\left(1-\frac{x}{3}\right) d x\) 
\(=\frac{3 \pi}{2}-3=3\left(\frac{\pi}{2}-1\right) \text { sq. units }\)

(i) (b) : We have, (x - 1)²+y² = 1
\(\Rightarrow y=\sqrt{1-(x-1)^{2}}\)   ...(i) 
Also, \(x^{2}+y^{2}=1 \Rightarrow y=\sqrt{1-x^{2}}\) ...(ii)
From (i) and (ii), we get
\(\sqrt{1-(x-1)^{2}}=\sqrt{1-x^{2}}\) 
\(\Rightarrow(x-1)^{2}=x^{2} \Rightarrow 2 x=1 \Rightarrow x=\frac{1}{2}\) 
(ii) (c):
 
(iii) (a) : \(\int_{0}^{1 / 2} \sqrt{1-(x-1)^{2}} d x\) 
\(=\left[\frac{x-1}{2} \sqrt{1-(x-1)^{2}}+\frac{1}{2} \sin ^{-1}\left(\frac{x-1}{1}\right)\right]_{0}^{1 / 2}\) 
\(\begin{array}{r} =\frac{1}{2}\left(\frac{1}{2}-1\right) \sqrt{1-\frac{1}{4}}+\frac{1}{2} \sin ^{-1}\left(-\frac{1}{2}\right)-\left(-\frac{1}{2}\right)(0) -\frac{1}{2} \sin ^{-1}(-1) \end{array}\) 
\(=\left[\frac{-1}{4} \cdot \frac{\sqrt{3}}{2}-\frac{1}{2} \cdot \frac{\pi}{6}+0+\frac{1}{2} \cdot \frac{\pi}{2}\right]=\frac{-\sqrt{3}}{8}-\frac{\pi}{12}+\frac{\pi}{4}\) 
\(=\frac{\pi}{6}-\frac{\sqrt{3}}{8}\) 
(iv) (c : \(\int_{1 / 2}^{1} \sqrt{1-x^{2}} d x=\left[\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x\right]_{1 / 2}^{1}\) 
\(=0+\frac{1}{2} \sin ^{-1}(1)-\frac{1}{4} \sqrt{1-\frac{1}{4}}-\frac{1}{2} \sin ^{-1}\left(\frac{1}{2}\right)\) 
\(=\frac{\pi}{4}-\frac{\sqrt{3}}{8}-\frac{\pi}{12}=\frac{\pi}{6}-\frac{\sqrt{3}}{8}\) 
(v) (d) : Required area
\(=2\left[\int_{0}^{1 / 2} \sqrt{1-(x-1)^{2}} d x+\int_{1 / 2}^{1} \sqrt{1-x^{2}} d x\right]\) 
\(=2\left[\frac{\pi}{6}-\frac{\sqrt{3}}{8}+\frac{\pi}{6}-\frac{\sqrt{3}}{8}\right]\) 
\(=2\left[\frac{\pi}{3}-\frac{\sqrt{3}}{4}\right]=\left(\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\right) \text { sq. units }\) 

(i) (a) : Equation of line AB is
\(y-0=\frac{3-0}{1+1}(x+1) \Rightarrow y=\frac{3}{2}(x+1)\) 
(ii) (c) : Equation of line BC is \(y-3=\frac{2-3}{3-1}(x-1)\) 
\(\Rightarrow y=-\frac{1}{2} x+\frac{1}{2}+3 \Rightarrow y=\frac{-1}{2} x+\frac{7}{2}\) 
(iii) (d) : Area of region ABCD
= Area of \(\triangle A B E\) + Area of region BCDE
\(=\int_{-1}^{1} \frac{3}{2}(x+1) d x+\int_{1}^{3}\left(\frac{-1}{2} x+\frac{7}{2}\right) d x\) 
\(=\frac{3}{2}\left[\frac{x^{2}}{2}+x\right]_{-1}^{1}+\left[\frac{-x^{2}}{4}+\frac{7}{2} x\right]_{1}^{3}\) 
\(=\frac{3}{2}\left[\frac{1}{2}+1-\frac{1}{2}+1\right]+\left[\frac{-9}{4}+\frac{21}{2}+\frac{1}{4}-\frac{7}{2}\right]\) 
= 3 + 5 = 8 sq. units
(iv) (a) : Equation of line AC is \(y-0=\frac{2-0}{3+1}(x+1)\) 
\(\Rightarrow y=\frac{1}{2}(x+1)\) 
\(\therefore \text { Area of } \Delta A D C=\int_{-1}^{3} \frac{1}{2}(x+1) d x=\left[\frac{x^{2}}{4}+\frac{1}{2} x\right]_{-1}^{3}\) 
\(=\frac{9}{4}+\frac{3}{2}-\frac{1}{4}+\frac{1}{2}=4 \text { sq. units }\) 
(v) (b) : Area of \(\Delta A B C\)= Area of region ABCD - Area of \(\Delta A C D=8-4=4 \mathrm{sq} . \text { units }\)

(i) (c) : For point of intersection, we have sin x = cos x
\(\Rightarrow \frac{\sin x}{\cos x}=1 \Rightarrow \tan x=1 \Rightarrow x=\frac{\pi}{4}\) 
(ii) (a) : \(\int_{0}^{\pi / 4} \sin x d x=[-\cos x]_{0}^{\pi / 4}=-\cos \frac{\pi}{4}+\cos 0\) \(=1-\frac{1}{\sqrt{2}}\) 
(iii) (b) : \(\int_{\pi / 4}^{\pi / 2} \cos x d x=[\sin x]_{\pi / 4}^{\pi / 2}=\sin \frac{\pi}{2}-\sin \frac{\pi}{4}\) \(=1-\frac{1}{\sqrt{2}}\) 
(iv) (c) : \(\int_{0}^{\pi} \sin x d x=[-\cos x]_{0}^{\pi}=[-\cos \pi+\cos 0]=2\) 
(v) (b) : \(\int_{0}^{\pi / 2} \sin x d x=[-\cos x]_{0}^{\pi / 2}=\left[-\cos \frac{\pi}{2}+\cos 0\right]\) = 0+1=1

(i) (a) : Here, teacher explained about cosine curve.
(ii) (c) : Required area \(\int_{0}^{\pi / 2} \cos x d x\) 
\(=[\sin x]_{0}^{\pi / 2}=\sin \frac{\pi}{2}-\sin 0=1-0=1 \text { sq. unit }\) 
(iii) (b) : Required area = \(\left|\int_{\pi / 2}^{3 \pi / 2} \cos x d x\right|=\left|[\sin x]_{\pi / 2}^{3 \pi / 2}\right|\) 
\(=\left|\sin \frac{3 \pi}{2}-\sin \frac{\pi}{2}\right|=|-1-1|=|-2|\) 
= 2 sq. units [Since, area can't be negative]
(iv) (a) : Required area = \(\int_{3 \pi / 2}^{2 \pi} \cos x d x=[\sin x]_{3 \pi / 2}^{2 \pi}\) 
\(=\sin 2 \pi-\sin \frac{3 \pi}{2}=0-(-1)=1 \text { sq. unit }\) 
(v) (d) : Required area
\(=\int_{0}^{\pi / 2} \cos x d x+\left|\int_{\pi / 2}^{3 \pi / 2} \cos x d x\right|+\int_{3 \pi / 2}^{2 \pi} \cos x d x\) 
= 1 + 2 + 1 = 4 sq. units.

(i) (d) : Line x + y = 3 cuts the x-axis and y-axis at (3, 0) and (0, 3) respectively
[Since, at x-axis, y = 0 and at y-axis, x = 0]
(ii) (c) : We have, y² = 4x and x + y = 3
From (i) and (ii), we have y² =  4(3 - y)
\(\Rightarrow y^{2}+4 y-12=0 \Rightarrow y^{2}+6 y-2 y-12=0\) 
\(\Rightarrow y(y+6)-2(y+6)=0\) 
\(\Rightarrow (y+6)(y-2)=0 \Rightarrow y=2, y=-6\) 
From (ii), x = 3 - 2 = 1 or x = 3 + 6 = 9
\(\therefore\) Required points of intersection are (1, 2), (9, - 6)
(iii) (b) :

(iv) (d): \(\int_{-6}^{2}(3-y) d y=\left[3 y-\frac{y^{2}}{2}\right]_{-6}^{2}\) 
\(=\left[6-\frac{4}{2}-\left[3(-6)-\frac{(-6)^{2}}{2}\right]\right]=4+36=40\) 
(v) (c): Required area = \(\int_{-6}^{2}(3-y) d y-\int_{-6}^{2} \frac{y^{2}}{4} d y\) 
\(=40-\frac{1}{4}\left[\frac{y^{3}}{3}\right]_{-6}^{2}=40-\frac{1}{4}\left[\frac{8}{3}-\frac{(-6)^{3}}{3}\right]\) 
\(=40-\frac{2}{3}-\frac{216}{12}=\frac{480-8-216}{12}=\frac{256}{12}=\frac{64}{3} \mathrm{sq} . \text { units }\)

(i) (b) : We have, y² + 1 = 4  ...(i)
and \(x=\sqrt{3} y\)   ...(ii)
From (i) and (ii), we get
\(3 y^{2}+y^{2}=4 \Rightarrow 4 y^{2}=4 \Rightarrow y^{2}=1 \Rightarrow y=\pm 1\) 
From (ii), \(x=\sqrt{3},-\sqrt[4]{3}\) 
\(\therefore\) Points of intersection of pizza and edge of knife are \((\sqrt{3}, 1),(-\sqrt{3},-1)\) .
(ii) (a) :

(iii) (b) : Required area = \(\int_{0}^{\sqrt{3}} \frac{x}{\sqrt{3}} d x+\int_{\sqrt{3}}^{2} \sqrt{4-x^{2}} d x\) 
\(=\frac{1}{\sqrt{3}}\left[\frac{x^{2}}{2}\right]_{0}^{\sqrt{3}}+\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1}\left(\frac{x}{2}\right)\right]_{\sqrt{3}}^{2}\) 
 \(=\frac{1}{\sqrt{3}}\left[\frac{3}{2}-0\right]+\left[2 \sin ^{-1}(1)-\left(\frac{\sqrt{3}}{2}+2 \sin ^{-1} \frac{\sqrt{3}}{2}\right)\right]\) 
\(=\frac{\sqrt{3}}{2}+\frac{2 \pi}{2}-\frac{\sqrt{3}}{2}-\frac{2 \pi}{3}=\frac{\pi}{3} \text { sq. units }\) 
(iv) (a) : We have,x²+y²=4
\(\Rightarrow(x-0)^{2}+(y-0)^{2}=(2)^{2}\) 
\(\therefore\) Radius = 2
Area of \(\frac{1}{4} \text { th slice of pizza }=\frac{1}{4} \pi(2)^{2}=\pi \mathrm{sq} . \text { units }\) 
(v) (d) : Area of whole pizza = \(\pi(2)^{2}=4 \pi \mathrm{sq} . \text { units }\) 

(i) (c) : We have, x² +y²= 16  ..(i)
and y = x ...(ii)
From (i) and (ii), \(2 x^{2}=16 \Rightarrow x^{2}=8 \Rightarrow x=2 \sqrt{2}\)  (\(\therefore\) x lies in first quadrant)
\(\therefore\) Point of intersection of (i) and (ii) in first quadrant is \((2 \sqrt{2}, 2 \sqrt{2})\) .
(ii) (b) : The shaded region which represent the areabounded by two given curves in first quadrant is shown below.
 
(iii)( d) : \(\int_{0}^{2 \sqrt{2}} x d x=\left[\frac{x^{2}}{2}\right]_{0}^{2 \sqrt{2}}=\frac{(2 \sqrt{2})^{2}}{2}=\frac{8}{2}=4\) 
(iv) (a) : \(\int_{2 \sqrt{2}}^{4} \sqrt{16-x^{2}} d x=\left[\frac{x}{2} \sqrt{16-x^{2}}+\frac{16}{2} \cdot \sin ^{-1}\left(\frac{x}{4}\right)\right]_{2 \sqrt{2}}^{4}\) 
 \(=8 \sin ^{-1}(1)-4-8 \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\) 
\(=8\left(\frac{\pi}{2}\right)-4-8\left(\frac{\pi}{4}\right)=4 \pi-4-2 \pi=2 \pi-4=2(\pi-2)\) 
(v) (d) : Required area = Area (OLA) + Area (BAL)
\(=\int_{0}^{2 \sqrt{2}} x d x+\int_{2 \sqrt{2}}^{4} \sqrt{16-x^{2}} d x\) 
\(=4+2(\pi-2)=2 \pi \text { sq. units. }\)

(i) (b) : We have, x² = y ...(i) and x = y. , ...(ii)
From (i) and (ii), x²= x \(\Rightarrow\)x² - x = 0
\(\Rightarrow\) x(x - 1) = 0 ~ x = 0, 1
From (ii), y = 0, 1
Required points of intersection are (0, 0), (1, 0)
(ii) (a) :
 
(iii) (c) : \(\int_{0}^{1} x d x=\left[\frac{x^{2}}{2}\right]_{0}^{1}=\frac{1}{2}-0=\frac{1}{2}\) 
(iv) (b) : \(\int_{0}^{1} x^{2} d x=\left[\frac{x^{3}}{3}\right]_{0}^{1}=\frac{1}{3}-0=\frac{1}{3}\) 
(v) (a) : Required area \(=\int_{0}^{1} x d x-\int_{0}^{1} x^{2} d x\) 
\(=\frac{1}{2}-\frac{1}{3}=\frac{1}{6} \text { sq. units. }\)