(i) Write the given equation as \(\frac{d y}{d x}=2-\left(\frac{x}{y}\right)\) .
(ii) Substitute y = vx to solve it 
\(=\log |y-x|=C+\frac{x}{(y-x)}\)

\(\sin ^{2}(2 x+5)=\frac{1-\cos 2(2 x+5)}{2}=\frac{1-\cos (4 x+10)}{2} \)
\(\Rightarrow \int \sin ^{2}(2 x+5) d x=\int \frac{1-\cos (4 x+10)}{2} d x \)
\(=\frac{1}{2} \int 1 d x-\frac{1}{2} \int \cos (4 x+10) d x \)
\(=\frac{1}{2} x-\frac{1}{2}\left(\frac{\sin (4 x+10)}{4}\right)+\mathrm{C} \)
\(=\frac{1}{2} x-\frac{1}{8} \sin (4 x+10)+\mathrm{C} \)

Let \(x_{ 1 },x_{ 2 }\in R-\{ 3\} \)
Now \(f(x_{ 1 })=f(x_{ 2 })\)
\(\Rightarrow \) \(\frac { x_{ 1 }-2 }{ x_{ 1 }-3 } =\frac { x_{ 2 }-2 }{ x_{ 2 }-3 } \)
\(\Rightarrow \) \((​​x_{ 1 }-2)(x_{ 2 }-3)=(x_{ 2 }-2)(x_{ 1 }-3)\)
\(\Rightarrow \) \(x_{ 1 }x_{ 2 }-3x_{ 1 }-2x_{ 2 }+6=\) \(x_{ 2 }x_{ 1 }-3x_{ 2 }-2x_{ 1 }+6\)
\(\Rightarrow \) \(-3x_{ 1 }-2x_{ 2 }=3x_{ 2 }-2x_{ 1 }\)
\(\Rightarrow \) \(x_{ 1 }=x_{ 2 }\)
\(\Rightarrow \) \(f\) is one-one.
Let \(y\in R-\{ 1\} \).
Then  f(x)=y.
When \(\left( \frac { x-2 }{ x-3 } \right) =y,x\neq 3\)
\(\Rightarrow \) x - 2 = yx - 3y
\(\Rightarrow \) x - xy = 2 - 3y
\(\Rightarrow \) \(x=\frac { 2-3y }{ 1-y } \in A\)
\(\left[ \because \frac { 2-3y }{ 1-y } =3-\frac { 1 }{ 1-y } \neq 3\quad \right] \)
\(\therefore \) Corresponding to each \(y\in B\), there exists \(\frac { 2-3y }{ 1-y } \in A\)
such that \(f\left( \frac { 2-3y }{ 1-y } \right) =y\)
\(\Rightarrow \)f is onto.
Hence, 'f' is one-one and onto.

Let the events be:
E₁ : First bag is selected
E₂ : Second bag is selected and A : Ball drawn is red.
\(P({ { E }_{ 1 })=P({ E }_{ 2 })=\frac { 1 }{ 2 } }\)
and \(P(A/{ E }_{ 1 })=\frac { 4 }{ 8 } =\frac { 1 }{ 2 } ,P(A/{ E }_{ 2 })=\frac { 2 }{ 8 } =\frac { 1 }{ 4 } \)
By Bayes, Theorem
\(P({ E }_{ 1 })=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 }) } \)
\(=\frac { \frac { 1 }{ 2 } \frac { 1 }{ 2 } }{ \frac { 1 }{ 2 } \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \frac { 1 }{ 4 } } =\frac { 1 }{ 4 } \times \frac { 8 }{ 2+1 } =\frac { 2 }{ 3 } \)

(i) \(P(E)=\frac { 13 }{ 52 } =\frac { 1 }{ 4 } \)
\(P(F)=\frac { 4 }{ 52 } =\frac { 1 }{ 13 } \)
\(P(E\cap F)=\frac { 1 }{ 52 } =\frac { 1 }{ 4 } \times \frac { 1 }{ 13 } \)
= P(E).P(F)
Hence, the events E and F and independent.
(ii)  \(P(E)=\frac { 26 }{ 52 } =\frac { 1 }{ 2 } ,P(F)=\frac { 4 }{ 52 } =\frac { 1 }{ 13 } \)
\(P(E\cap F)=\frac { 2 }{ 52 } =\frac { 1 }{ 26 } =\frac { 1 }{ 2 } \times \frac { 1 }{ 13 } \)
= P(E).P(F)
(iii) \(P(E)=\frac { 8 }{ 52 } =\frac { 2 }{ 13 } ,P(F)=\frac { 8 }{ 52 } =\frac { 2 }{ 13 } \)
\(P(E\cap F)=\frac { 4 }{ 52 } =\frac { 1 }{ 13 } \)
and \(P(E)P(F)=\frac { 2 }{ 13 } \times \frac { 2 }{ 13 } =\frac { 4 }{ 169 } \)
Thus \(P(E\cap F)\neq P(E)P(F)\)
Hence, the events E and F are not independent.

\(\Rightarrow \)  xy = \(\frac {{x}^{4}}{4} + c\)is the required solution.

\(={x\over log\ x}+c\)

Let P be the perimeter and A be the area of the rectangle of length x and width y.
Given \(\frac{d x}{d t}=-5 \mathrm{~cm} / \mathrm{min}\) 
[negative (-) sign for decreasing rate]
\(x=8 \mathrm{~cm}, \frac{d y}{d t}=4 \mathrm{~cm} / \mathrm{min} \text { and } y=6 \mathrm{~cm}\) 
(ii) Perimeter of the rectangle.\(P=2(x+y)\) 
On differentiating both sides w.r.t. t, we get
\( \frac{d P}{d t} =2\left(\frac{d x}{d t}+\frac{d y}{d t}\right) \)
\(=2(-5+4)=-2 \mathrm{~cm} / \mathrm{min}\) 
So, perimeter decreases at the rate of 2 cm/min.
(ii) Area of the rectangle, A = xy
On differentiating both sides w.r.t. t, we get
\( \frac{d A}{d t} =x \frac{d y}{d t}+y \frac{d x}{d t} \)
\(=8 \times 4+6 \times(-5)=32-30=2 \mathrm{~cm}^{2} / \mathrm{min} \) 
Hence, area increases at the rate of 2 cm ² /min.

Let \(I=\int \frac{1}{\cos ^2 x(1-\tan x)^2} d x=\int \frac{\sec ^2 x}{(1-\tan x)^2} d x\)
Now, put (1-tan x) = t
\(\Rightarrow -\sec ^2 x d x=d t \Rightarrow \sec ^2 x d x=-d t \)
\(\therefore  I=-\int \frac{d t}{t^2}=\frac{1}{t}+C=\frac{1}{1-\tan x}+C\)

Given, \(P(x)=72+42 x-x^2\)
On differentiating w.r.t, x, we get
\(P^{\prime}(x)=42-2 x\)
For maximum profit, we put \(P^{\prime}(x)=0\)
\(\Rightarrow 42-2 x  =0\)
\(\Rightarrow  x  =21\)
Therefore The maximum value of P(x) is at x = 21
\(\Rightarrow \quad P(21)  =72+42(21)-(21)^2 =72+882-441=513\)
Thus, the maximum profit is ₹ 513.

Write the given equation as \(\frac{d y}{d x}=(1-x)(1+y)\) and solve it.
\(=\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=C\)

Given differential equation is
\(\frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0  \)
\(\Rightarrow \frac{d y}{d x}=-\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}} \)
On separating the variables, we get
\(\frac{1}{\sqrt{1-y^{2}}} d y=-\frac{1}{\sqrt{1-x^{2}}} d x \)
On integrating both sides, we get
\(\int \frac{1}{\sqrt{1-y^{2}}} d y=-\int \frac{1}{\sqrt{1-x^{2}}} d x  \)
\(\Rightarrow \sin ^{-1} y=-\sin ^{-1} x+C \)
\(\Rightarrow \sin ^{-1} x+\sin ^{-1} y=C  \)
which is the required general solution.

Given, a function \(f:[1, \infty) \rightarrow(2,-)\) defined by \(f(x)=x+\frac{1}{x}\)
Now, let \(y=f(x)\)
\(\Rightarrow y=x+\frac{1}{x} \Rightarrow y=\frac{x^{2}+1}{x}\)
\(\Rightarrow x y=x^{2}+1 \rightarrow x^{2}-x y+1=0\)
\(\Rightarrow x=\frac{y+\sqrt{\gamma^{t}-4}}{2} \rightarrow x=\frac{x^{*} \sqrt{y^{2}-4}}{2}\)
\({\left[\text { neglecting } x=\frac{y-\sqrt{y^{2}-4}}{2} \because\right. \text { For } y=3 \in[2, \infty)} \)
\(x=\frac{3-\sqrt{9-4}}{2}=\frac{3-\sqrt{5}}{2}=\frac{3-223}{2}=\frac{0.77}{2} \notin[1, \infty) \) 
\(\Rightarrow f^{-1}(y)=\frac{y+\sqrt{y^{2}-4}}{2} \quad\left[\because y=f(x) \Rightarrow x=f^{-1}(y)\right]\)
\(\text { or } f^{-1}(x)=\frac{x+\sqrt{x^{2}-4}}{2}\)

Total number of cards = 52
(a) Number of favourable cases = 26
\(\therefore\)  Required probability =\(\frac { 26 }{ 52 } =\frac { 1 }{ 2 } \)
(b) Number of favourable cases = 1
\(\therefore\)  Required probability = \(\frac { 1 }{ 52 } \)
(c) Number of favourable cases = 6
 \(\therefore\) Required probability =\(\frac { 6 }{ 52 } =\frac { 3 }{ 26 } \)
(d) Number of favourable cases = 4
\(\therefore\) Required probability = \(\frac { 4 }{ 52 } =\frac { 1 }{ 13 } \)

\( P(B / A)=\frac{P(A \cap B)}{P(A)} \)
\(\Rightarrow 0.6 \times 0.4=P(A \cap B) \)
\(\Rightarrow P(A \cap B)=0.24 \)
\( P(A \cup B)=P(A)+P(B)-P(A \cap B) \)
\(=0.4+0.7-0.24=0.86 \)

Let \(I=\int \frac{\cos \theta}{\left(4+\sin ^2 \theta\right)\left(5-4 \cos ^2 \theta\right)} d \theta\)
\(=\int \frac{\cos \theta}{\left(4+\sin ^2 \theta\right)\left[5-4\left(1-\sin ^2 \theta\right)\right]} d \theta \)
\(=\int \frac{\cos \theta}{\left(4+\sin ^2 \theta\right)\left(5-4+4 \sin ^2 \theta\right)} d \theta\)
\(=\int \frac{\cos \theta}{\left(4+\sin ^2 \theta\right)\left(1+4 \sin ^2 \theta\right)} d \theta\)
Let \(\sin \theta=t \Rightarrow \cos \theta d \theta=d t\)
Then, \(I=\int \frac{d t}{\left(4+t^2\right)\left(1+4 t^2\right)}\)
Again, let \(\frac{1}{\left(4+t^2\right)\left(1+4 t^2\right)}=\frac{A}{4+t^2}+\frac{B}{1+4 t^2}\)
[by partial fraction]
At \(t=0, \frac{A}{4}+\frac{B}{1}=\frac{1}{4 \times 1} \Rightarrow A+4 B=1\)
At \(t=1, \frac{A}{5}+\frac{B}{5}=\frac{1}{5 \times 5} \Rightarrow 5 A+5 B=1\)
On solving Eqs. (iii) and (iv), we get
\(A=-\frac{1}{15} \text { and } B=\frac{4}{15}\)
On putting \(A=-\frac{1}{15}\) and \(B=\frac{4}{15}\) in Eq. (ii), we get
\(\frac{1}{\left(4+t^2\right)\left(1+4 t^2\right)}=\frac{-\frac{1}{15}}{4+t^2}+\frac{\frac{4}{15}}{1+4 t^2}\)
\(\Rightarrow \quad  \frac{1}{\left(4+t^2\right)\left(1+4 t^2\right)}=\frac{-1}{15\left(4+t^2\right)}+\frac{4}{15\left(1+4 t^2\right)}\)
Now, \( I=\int \frac{1}{\left(4+t^2\right)\left(1+4 t^2\right)} d t \)
\(=\frac{-1}{15} \int \frac{1}{4+t^2} d t+\frac{4}{15} \int \frac{1}{1+4 t^2} d t\)
\(=\frac{-1}{15} \int \frac{1}{2^2+t^2}+\frac{4}{15 \times 4} \int \frac{1}{\left(\frac{1}{2}\right)^2+t^2} d t \)
\(=\frac{-1}{15} \cdot \frac{1}{2} \tan ^{-1} \frac{t}{2}+\frac{1}{15} \cdot \frac{1}{1 / 2} \tan ^{-1} \frac{t}{1 / 2}+C  \quad\left[\because \int \frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+C\right]\)
\(= \frac{-1}{30} \tan ^{-1} \frac{\sin \theta}{2}+\frac{2}{15} \tan ^{-1} 2 \sin \theta+C \quad[\because t=\sin \theta]\)

Write the given differential equation as
\(\frac{d x}{d y}=-\left(\frac{x-\sin y}{\tan y}\right)\) 
\(\Rightarrow \frac{d x}{d y}+(\cot y) x=\cos y\)  \(y=\sin ^{-1} 2 x\) 

Given, x² ≤ y..(i)
and y ≤ |x|...(ii)
Clearly, curve (i) is parabola directed upward with vertex at (0, 0) and symmetrical about y-axis.
Also \(y=|x|=\begin{cases} x\quad if\quad x\ge 0 \\ -x\quad ,if\quad x<\quad 0 \end{cases}\)
The lines y = x and y = - x both passes through origin and have slope of + 1& - 1 respectively.

⇒ Required Area = 2x Standard Area on a side
= \(-\left[ \left( \frac { 1 }{ 2 } -1 \right) -\left( \frac { 16 }{ 2 } -4 \right) \right] \)
= \(2{ \left[ \frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 2 } \right] }_{ 0 }^{ 1 }\)
= \(2\left[ \frac { 1 }{ 2 } -\frac { 1 }{ 3 } \right] =2\times \frac { 1 }{ 6 } \)
= \(\frac { 1 }{ 3 } \) sq.units

Let the events be:
E₁ = Choosing 1^st coin
E₂ = Choosing 2^nd coin
E₃ = Choosing 3^rd coin
A: Getting Heads
\(P({ E }_{ 1 })=P({ E }_{ 2 })=P(E_{ 3 })=\frac { 1 }{ 3 } \)
\(P(A/{ E }_{ 1 })=\frac { 40 }{ 100 } ,P(A/{ E }_{ 2 })=\frac { 75 }{ 100 } ,\)
\(P(A/{ E }_{ 3 })=\frac { 1 }{ 2 } \)
\(P({ E }_{ 1 }/A)\)
\(=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 })+P({ E }_{ 3 })P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 1 }{ 3 } .\frac { 40 }{ 100 } }{ \frac { 1 }{ 3 } .\frac { 40 }{ 100 } +\frac { 1 }{ 3 } .\frac { 75 }{ 100 } +\frac { 1 }{ 3 } .\frac { 1 }{ 2 } } =\frac { 8 }{ 33 } \)

Let the events defined are :
E₁: Person chosen is a cyclist
E_2: Person chosen is a scooter driver
E₃: Person chosen is a motorbike driver
A:Person meets with an accident
P(E₁_= 1/6, P(E₂) = 1/3, P(E₃) = 1/2
P(A/E₁) = 0.01, P(A/E₂) = 0.03,
P(A/E₃) = 0.15
\(P{ (E }_{ 2 }/A)=\frac { P({ E }_{ 2 }).P(A/{ E }_{ 2 }) }{ P({ E }_{ 1 }).P(A/{ E }_{ 1 })+P({ E }_{ 2 }).P(A/{ E }_{ 2 })+P({ E }_{ 3 }).P(A/{ E }_{ 3) } } \)
\(=\frac { \frac { 1 }{ 3 } \times 0.03 }{ \frac { 1 }{ 6 } \times 0.01+\frac { 1 }{ 3 } \times 0.03+\frac { 1 }{ 2 } \times 0.15 } \)
\(=\frac { 0.01 }{ 0.086 } \)
= 0.11627
Suggestion: Cycle should be promoted as it is:
(i) Good for health
(ii) Pollution free
(iii) Saves energy (no petrol).  

As differential equation is
1+ \(3\frac { dy }{ dx } +3{ \left( \frac { dy }{ dx } \right) }^{ 2 }+{ \left( \frac { dy }{ dx } \right) }^{ 3 }={ \left( \frac { dy }{ dx } \right) }^{ 2 }\)
Exponent of highest order derivative is 3.

As \(\frac { dx }{ dt } =2\) cm/s, x is side of equiolatral triangle.
A = \(\frac { \sqrt { 3 } }{ 4 } { x }^{ 2 }\)
\(\Rightarrow \frac { dA }{ dx } =\frac { \sqrt { 3 } }{ 2 } { x }\frac { dx }{ dt } \)
= \(\frac { \sqrt { 3 } }{ 2 } x2=\sqrt { 3 } x\)
∴\(|\frac{dA}{dx}|\)x=10 = 10\(\sqrt3\) cm²/s

\(\text { Given, } h(t)=-\frac{7}{2} t^2+\frac{13}{2} t+1\)
(i) Yes,
Since, h(t) is a polynomial function.
So, it is everywhere continuous.
(ii) \( h^{\prime}(t)=-\frac{7}{2}(2 t)+\frac{13}{2} \)
\(\Rightarrow \quad h^{\prime}(t)=-7 t+\frac{13}{2}\)
Also, \(h^{\prime \prime}(t)=-7\)
For maxima or minima, put \(h^{\prime}(t)=0\)
\(\Rightarrow  -7 t+\frac{13}{2}=0 \)
\(\Rightarrow  -7 t=-\frac{13}{2}\)
\(\Rightarrow  t=\frac{13}{14}\)
At \(t=\frac{13}{14}, h^{\prime \prime}(t)<0\)
So, \(t=\frac{13}{14}\) is point of maxima.
Thus, at \(t=\frac{13}{14}\) the height of the ball is maximum.

Given, 70% customers pay their first month bill in time, therefore, the customers who do not pay bill in time is 30%.
(i) Let E₁ and E₂, denote the event of customer paying or not paying the fìrst month bill in time, respectively.
Then, P(E₁) = 70% = 0.7
and P(E₂) = 30% = 0.3
(ii) Given that the probability of customers who pays the bill in time next month is 0.8 and who does not pay in time is 0.4.
Let A denotes the event of customer paying second month's bill in time.
\(\therefore P\left(\frac{A}{E_1}\right)=P\) (customer paying second month bill in time when they pay first month bill)
= 0.8
and \(P\left(\frac{A}{E_2}\right)=P\) (customer paying second month bill in time when they do not pay first month bill in time)
= 0.4
(ii) The probability of customer paying second month's bill in time = 0.7 \(\times\) 0.8 = 0.56
Or
The probability of customer paying first month's bill in time, if it is found that customer has paid the second month's bill in time is \(P\left(\frac{E_1}{A}\right)\)
\(\begin{aligned} P\left(\frac{E_1}{A}\right) & =\frac{P\left(E_1\right) \cdot P\left(\frac{A}{E_1}\right)}{P\left(E_1\right) \cdot P\left(\frac{A}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{A}{E_2}\right)} \end{aligned}\)
\(\begin{aligned} =\frac{0.7 \times 0.8}{0.7 \times 0.8+0.3 \times 0.4} \end{aligned}\)
\(\begin{aligned} =\frac{0.56}{0.56+0.12}=\frac{0.56}{0.68} \end{aligned}\)
= 0.824

We have, R = {(l₁,l₂) :l₁ is parallel to l₂}
(i) If l₁ is parallel to l₂, then l₂ is parallel to l_1.
So, if (l₁, l₂) \(\in R,\) then (l₂, l₁) \(\in R\)
\(\therefore\) R is symmetric.
(ii) If l₁ is parallel to l₂ and l₂ is parallel to l₃, then l₁ is parallel to l₃.
So, if \(\left(l_1, l_2\right) \in R,\left(l_2, l_3\right) \in R\), then (l₁,l₃)\(\in R\)
\(\therefore\) R is transitive.
(iii) Let equation of line parallel to y = 3x + 2 be y = mx + c, where m is the slope of line. Since, y = 3x + 2 and y = mx + c are parallel. Slope of (y =3x + 2) = Slope of(y = mx + c)
\(\Rightarrow\) 3 = m i.e. m = 3
Hence, the required line is
y = 3x + c, where c \(\in R\)
Or
We have, S = {(I₁, I₂) : l₁ is perpendicular to l₂}
For Symmetric If I₁, is perpendicular to I₂, then l₂ is perpendicular to l₁.
So, if (l₁,l₂) \(\in S\), then (l₂, l₁) \(\in S\)
\(\therefore\) S is symmetric.
For Transitive If I₁, is perpendicular to l₂, and l₂, is perpendicular to l₃, then I₁, is not perpendicular to l₃, it is parallel to I₃.
So, if \(\left(l_1, l_2\right) \in S,\left(l_2, l_3\right) \in S \text {, then }\left(l_1, l_3\right) \notin S \text {. }\)
\(\therefore\) S is not transitive.

(b) Both A and R are correct; R is not the correct explanation of A

(b) We have, f(x) = x² - 4x + 6 or f'(x) = 2x - 4

Therefore, f'(x) = 0 gives x = 2. Now, the point x = 2 divides the real line into two disjoint intervals namely, (- \(\infty\), 2) and (2, \(\infty\)). In the interval ( -\(\infty\), 2), f'(x) = 2x - 4 < 0.
Therefore,f is strictly decreasing in this interval. Also, in the interval (2, \(\infty\)), f'(x) > 0 and so the function f is strictly increasing in this interval.
Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.