\( \int_{-\infty}^{\infty} f(x) \mathrm{dx}=1 \)
\(\int_{-\infty}^{0} 0 \mathrm{dx}+\int_{0}^{1} \mathrm{kx}(1-x)^{10} \mathrm{dx}+\int_{1}^{\infty} 0 \mathrm{dx}=1 \)
\(\int_{0}^{1} k x(1-x)^{10} \mathrm{dx}=1 \)
\(\int_{0}^{1}(1-x)(1-1+x)^{10} \mathrm{dx}=1 \)
\(k \int_{0}^{1}(1-x)(\mathrm{x})^{10} \mathrm{dx}=1 \)
\([\therefore \left.k \int_{0}^{1} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\int_{0}^{10} \mathrm{f}(\mathrm{a}-\mathrm{x}) \mathrm{dx}\right] \)
\( k \int_{0}^{1}(1-x)(x)^{10} d x =1 \)
\(k \int_{0}^{1}\left(x^{10}-x^{11}\right) d x =1 \)
\(k\left[\frac{x^{11}}{11}-\frac{x^{12}}{12}\right]_{0}^{1} =1 \)
\(k\left[\frac{1}{11}-\frac{1}{12}\right] =1 \)
\(k\left[\frac{1}{132}\right] =1\)
k = 132

Given  \(\frac{dy}{dx}=\) (4x + y + 1)²....(1)
put 4x + y + 1 = z
\(\Rightarrow 4+\frac { dy }{ dx } =\frac { dz }{ dx } \)
\(\Rightarrow \frac { dy }{ dx } =\frac { dz }{ dx } -4\)
Substituting these in (1) we get,
\(\frac { dz }{ dx } -4={ z }^{ 2 }\)
\(\Rightarrow \frac { dz }{ dx } ={ z }^{ 2 }+4\)
\(\Rightarrow \int { \frac { dz }{ { z }^{ 2 }+4 } } =\int { dx } \)
\(\\ \Rightarrow \frac { 1 }{ 2 } { tan }^{ -1 }\left( \frac { z }{ 2 } \right) =x+c\)
\(\\ \Rightarrow \frac { 1 }{ 2 } { tan }^{ -1 }\left( \frac { 4x+y+1 }{ 2 } \right) =x+c\)

Given E(X + 3) = 10
E(aX + b) = aE(X) + b
⇒ E(X) + 3 = 10
E(X) + 3 = 10
⇒ E(X) = 7
⇒μ = 7 ...(1)
E(X + 3)² = 116
E(X² + 6x + 9) 116
E(X²) + 6E(X) + 9 = 116 [ஃ E(9) = 9]
E(X²) + 6(7) + 9 = 116
E(X²) + 116 - 42 - 9 116 - 51
E(X²) = 65 ...(2)
Var(X) = E(X²) - [E(X)²]
65 - 7² = 65 - 49 = 16
ஃμ = 7 and σ² = 16.

Given \(f(x)=\begin{cases} \begin{matrix} kxe^{ -2x } & forx>0 \end{matrix} \\ \begin{matrix} 0 & for\quad x\le 0 \end{matrix} \end{cases}\)
Since the given function is a probability density function 
\(\int _{ -\infty }^{ \infty }{ f(x)dx } \) = 1 
\(\Rightarrow k\int _{ 0 }^{ \infty }{ { xe }^{ -2x }dx=1 } \)
\(\Rightarrow k \frac { 1! }{ \left( 2 \right) ^{ 2 } } =1\) 
[\(\int _{ 0 }^{ \infty }{ { x }^{ n }e^{ -ax } } =\frac { n! }{ { a }^{ +1 } } \), Here a = 2, n = 1]
\(\Rightarrow \frac { k }{ 4 } =1\\ \Rightarrow k=4\)

Given that \((1+{ x }^{ 2 })\frac { dy }{ dx } =1+{ y }^{ 2 }\) ..(1)
The given equation is written in the variables separable form 
\(\frac { dy }{ { 1+y }^{ 2 } } =\frac { dx }{ { 1+x }^{ x } } \) ...(2)
Integrating both sides of (2), we get tan⁻¹ tan⁻¹x +C.
But tan^-1 y - tan^-1 x = tan^-1 \(\left( \frac { y-x }{ 1+xy } \right) .\) ...(4)
Using (4) in (3) leads to tan^-1 \(\left( \frac { y-x }{ 1+xy } \right)\) = C, which implies \(\frac { y-x }{ 1+xy } \) = tan C = a (say).
Thus, y − x = a(1+ xy) gives the required solution

Given y = ax + \(\frac { b }{ x } \) .......(1)
Differentiating with respect to x
y' = ax - \(\frac { b }{ x ^2} \) ......(2)
Differentiating again with respect to x
\(y'' = \frac{-b(-2)}{x^3}= \frac{2b}{x^3} \)
\(Now, x^2y'' + xy'-y \)
\( = x^2 \times \frac{2b}{x^3}+x(a- \frac{b}{x^2})-(ax+\frac{b}{x}) \)
\(= 2\times (\frac{b}{x})+ax-(\frac{b}{x})-ax-(\frac{b}{x})\)
= 0
Hence, y = ax + b is the solution of the differential equation x²y"+xy'-y = 0.

Given circle equation be x² y² = r²
Let y = mx + c be the family of lines which touches the circle.
The condition for y = mx + c be all straight lines which towards the given circle x² y² = r² (1 + m²)
\(c=\sqrt { { 1+m }^{ 2 } } \)  
Hence, equation of tangent to the circle is .......(1)
y = mx + r\(\sqrt { { 1+m }^{ 2 } } \) .......(1) 
Differentiating with respect to x,
\(\frac { dy }{ dx } =m\quad ...(2)\)
Substituting 'm' in (1) we get,
\(y= \left( \frac { dy }{ dx } \right) \times x \pm r\sqrt { 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } } \)
\(y-x\left( \frac { dy }{ dx } \right) =\pm r\sqrt { 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } } \)
Squaring both sides we get,
\(\Rightarrow { \left[ y-x\left( \frac { dy }{ dx } \right) \right] }^{ 2 }={ r }^{ 2 }\left[ 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] \)
This is the differential equation of all straight lines touching the circle x² y² = r²

Equation of the parabola is y² = 4ax
Latus rectum LL′ passes through the focus (a, 0)  
Hence the point L is (a, y₁) 
Therefore y₁² = 4a²
Hence y₁ = ±2a
The end points of latus rectum are (a, 2a) and (a, -2a)  

Given that equation of the circle is
x² + y² = 16
⇒ a² = 16
and equation of the tangent is
y = \( 2\sqrt { 2 } \)x + c
⇒ m = \( 2\sqrt { 2 } \)  and c = c
[∵ y = mx + c is the tangent]
The condition for the line y = mx + c is a tangent to the circle x² +y² = a² is
c² = a² (1 + m²)
⇒ c² = 16(1 + \(\left( { \left( 2\sqrt { 2 } \right) }^{ 2 } \right) \)
⇒ c² = 16(1 + 8)
⇒  c² = 16(9)
⇒ c = ± 4(3)
⇒ ±12

Equation of the circle passing through the points of intersection of the chord and circle by
Theorem  is x² + y²−16+\(\lambda \)(3x + y + 5) = 0 .
The chord 3x + y + 5 = 0 is a diameter of this circle if the centre\(\left( \frac { -3\lambda }{ 2 } \frac { -\lambda }{ 2 } \right) \) lies on the chord.
So we have 3\(\left( \frac { -3\lambda }{ 2 } \right) \)-\(\frac { -\lambda }{ 2 } \)+5 = 0,
\(\frac { -9\lambda }{ 2 } \)-\(\frac { \lambda }{ 2 } \)+5 = 0,
−5λ + 5 = 0 ,
λ = 1.
Therefore, the equation of the required circle is x² + y²+3x + y −11 = 0.

Given = 10
\( \mathrm{P}(\mathrm{H})=\frac{1}{2} \Rightarrow p=\frac{1}{2} \)
\(\therefore q-1-p \)
\(=1-\frac{1}{2}=\frac{1}{2} \)
\(\mathrm{P}(\mathrm{X}=x)=n \mathrm{C}_{x} p^{x}(1-p)^{n-x} \)
\(x=0,1,2, \ldots n \)
\(\therefore \mathrm{P}(\mathrm{X}=5)=10 \mathrm{C}_{5} p^{5} q^{5} \)
\(=10 \mathrm{C}_{5}\left(\frac{1}{2}\right)^{5}\left(\frac{1}{2}\right)^{5} \)
\(=10 C_{5}\left(\frac{1}{2}\right)^{10}\)

This is a linear differential equation
Here P = 1, Q = e^-x
\(\therefore \int { p\ dx } =\int { 1.dx } =x\)
\(I.F={ e }^{ \int { pdx } }={ e }^{ x }\)
The solution is
\({ ye }^{ \int { pdx } }\int { { Qe }^{ \int { p\ dx } }dx+c } \)
\(\Rightarrow y{ e }^{ x }=\int { { e }^{ -x }.{ e }^{ x }dx+c=\int { dx+c } } \)
\(\int { y{ e }^{ x }=x+c } \)

\(S=\left|\begin{array}{l} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{array}\right|\)
Let X be a random variable whose values x are the number of fours.
The sample space S is given in the table.
It can also be written as
S = {(i, j)} , where i = 1, 2, 3, 6 and j = 1, 2, 3, 6
Therefore X takes on the values of 0, 1 and 2.
We observe that
(i) X = 0, if (i,  j) for i ≠ 4, j≠ 4,
(ii) X = 1, if (1, 4), (2, 4), (3, 4), (5, 4), (6, 4), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6)
(iii) X = 2, if (4, 4) ,
Therefore,

The probabilities are
\(f(0)=P(X=0)\cfrac { 25 }{ 36 } \) 
\(f(1)=P(X=1)=\cfrac { 10 }{ 36 } \) 
and \(f(20=P(X=2)=\cfrac { 1 }{ 36 } \)
Clearly the function f(x) satisfies the conditions
(i) f (x) ≥ 0, for x = 0, 1, 2 and
(ii) \(\underset { x }{ \Sigma } f(x)=\sum _{ x=0 }^{ x=-2 }{ f(x) } =f(0)+f(1)+f(2)=1\) 
\(=\frac{25}{36}+\frac{10}{36}+\frac{1}{36}=1\)
The probability mass function is presented as

(or) 
\(f(x)=\begin{cases} \begin{matrix} \frac { 25 }{ 36 } & for \ x=0 \end{matrix} \\ \begin{matrix} \frac { 10 }{ 36 } & for \ x=1 \end{matrix} \\ \begin{matrix} \frac { 1 }{ 36 } & for \ x=2 \end{matrix} \end{cases}\)

\(\mathrm{n}=9, \mathrm{p}=\frac{1}{2}, \mathrm{k}=7 \)
\( \mathrm{P}(X=x)={ }^{n} C_{x} p^{x} q^{n-x}, x=0,1,2, \ldots, n \)
\(p =\frac{1}{2} \)
\(q =1-p=\frac{1}{2} \)
\(P(X=7) ={ }^{9} C_{7}\left(\frac{1}{2}\right)^{7}\left(\frac{1}{2}\right)^{2} \)
\( =\frac{9 \times 8}{2} \times \frac{1}{2^{9}} \)
\( =36 \times \frac{1}{512}=\frac{9}{128} \)

\(f(x)=\begin{cases} \begin{matrix} \frac { 1 }{ 2 } e^{ -\frac { x }{ 2 } } & for\quad x>0 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\)
\(\int _{ 0 }^{ \infty }{ x.f(x)dx } =\frac { 1 }{ 2 } \int _{ 0 }^{ \infty }{ x.{ e }^{ \frac { -x }{ 2 } } } dx\) 
\(\left[ \int _{ 0 }^{ \infty }{ { e }^{ -ax }.{ x }^{ n }dx=\cfrac { n! }{ { a }^{ n+1 } } } \right] \) 
= \(\frac { 1 }{ 2 } \times \frac { 1! }{ \left( \frac { 1 }{ 2 } \right) ^{ 2 } } =\frac { 1 }{ 2 } \times \frac { 1 }{ \frac { 1 }{ 4 } } \) 
= \(\frac { 1 }{ 2 } \times \frac { 4 }{ 1 } =2\) 
\(E({ X }^{ 2 })=\int _{ 0 }^{ \infty }{ { x }^{ 2 }.f(x) } dx\) 
= \(\int _{ 0 }^{ \infty }{ { x }^{ 2 }.\frac { 1 }{ 2 } { e }^{ -\frac { x }{ 2 } } } dx\) 
= \(\frac { 1 }{ 2 } \int { { x }^{ 2 }.{ e }^{ -\frac { x }{ 2 } }dx } \) 
= \(\frac { 1 }{ 2 } \times \frac { 2! }{ \left( \frac { 1 }{ 3 } \right) ^{ 3 } } =\frac { 1 }{ 2 } \times \frac { 2 }{ \frac { 1 }{ 8 } } \) 
= \(\frac { 1 }{ 2 } \times 2\times 8=8\) 
ஃVar(X)=E(X²) - [E(x)]²
= 8-2²
= 8 - 4 = 4

Given that \(\frac{dy}{dx}+2y\) = e^-x
This is a linear differential equation
Here P = 2 ; Q = e^−x.
\(\int { pdx } =\int { 2dx } =2x\)
Thus, I.F.\(={ e }^{ \int { pdx } }={ e }^{ 2x }\)
Hence the solution of (1) is \({ ye }^{ \int { pdx } }=\int { { Qe }^{ \int { Pdx } }dx+C } \)
That is, \({ ye }^{ 2x }=\int { { e }^{ -x }{ e }^{ 2x }dx+C } or\quad { ye }^{ 2x }={ e }^{ x }+C\quad or\quad y={ e }^{ -x }+{ Xe }^{ -2x }\) required solution

y = ae^x + be^−x; y − y = 0
Consider y = ae^x + be^−x  
Differentiating with respect to 'x' we get,
\(\frac { dy }{ dx } = ae^{x} -be^{-x} \)
Again differentiating, we get
\(\frac { d^2y }{ dx^2 } = ae^{x} -be^{-x} \)
\(\frac { d^2y }{ dx^2 } = y\)
\(\frac { d^2y }{ dx^2 } - y = 0\) is a required differential equation.
Thus y = ae^x + be^−x satisfies the diferential| equation y"- y = 0
Hence y = ae^x + be^−x is a solution of the differential equation y"- y = 0

\(\sqrt { \frac { dy }{ dx } } -4\frac { dy }{ dx } -7x=0\)
The given differential equation is 
\(\sqrt { \frac { dy }{ dx } } =4\frac { dy }{ dx } +7x\)
Squaring both sides,
\(\frac { dy }{ dx } =\quad { \left( 4\frac { dy }{ dx } +7x \right) }^{ 2 }\)
\(16{ \left( \frac { dy }{ dx } \right) }^{ 2 }+{ 49 }x^{ 2 }+56x{ \left( \frac { dy }{ dx } \right) }\)
The highest derivative is 1 and its maximum power is 2.
∴ Order 1, degree 2.

Given that 2a = 36.18, 2b = 9.12 , we get
e = \(\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\frac { \sqrt { { a }^{ 2 }-{ b }^{ 2 } } }{ a } \) = \(\frac { \sqrt { { \left( \frac { 36.18 }{ 2 } \right) }^{ 2 }{ \left( \frac { 9.12 }{ 2 } \right) }^{ 2 } } }{ \frac { 36.18 }{ 2 } } \)
\(\frac { \sqrt { { \left( 18.09 \right) }^{ 2 }-{ \left( 4.56 \right) }^{ 2 } } }{ \left( 8.09 \right) } \approx0.97\)

Centre of the circle is (3,-2) which lies on x + y − 1 = 0. So the line x + y − 1 = 0 passes through the centre and therefore the line x + y −1 = 0 is a diameter of the circle for all possible values of c .

Given 

(i) The probability density function. Differentiating F(x) with respect to 'x' at continuity points  of F(x), we get
\(F(x)=\begin{cases} \begin{matrix} 0 & x<0 \end{matrix} \\ \begin{matrix} \frac { 1 }{ 2 } ({ 2x }+1) & 0\le x<1 \end{matrix} \\ \begin{matrix} 1 & x\ge 1 \end{matrix} \end{cases}\) 
(ii) \(p(0.3\le X\le 0.6)=\int _{ 0.3 }^{ 0.6 }{ f(x)dx } \) 
= \(\int _{ 0.3 }^{ 0.6 }{ \frac { 1 }{ 2 } \left( 2x+1 \right) dx } =\frac { 1 }{ 2 } \left[ \frac { { 2x }^{ 2 } }{ 2 } +x \right] _{ 0.3 }^{ 0.6 }\) 
= \(\frac { 1 }{ 2 } \left( { x }^{ 2 }+x \right) _{ 0.3 }^{ 0.6 }=\frac { 1 }{ 2 } \left[ \left( { 0.6 }^{ 2 }+0.6 \right) -\left( { 0.3 }^{ 2 }+0.3 \right) \right] \) 
= \(\frac { 1 }{ 2 } \left[ \left( .36.6 \right) \right] -\left( .09+0.3 \right) ]\) 
= \(\frac { 1 }{ 2 } \left[ 0.96-.39 \right] =\frac { 0.57 }{ 2 } =0.285\) 
= 0.285

Given
\(f(x)=\begin{cases} \begin{matrix} { Ke }^{ \frac { -x }{ 3 } } & \begin{matrix} for & x>0 \end{matrix} \end{matrix} \\ \begin{matrix} 0 & \begin{matrix} for & x\le 0 \end{matrix} \end{matrix} \end{cases}\)
(i) Since f(x) is a probability density function
\(\int _{ -\infty }^{ \infty }{ f(x) } dx=1\) 
\(\Rightarrow \int _{ 0 }^{ \infty }{ K.{ e }^{ \frac { -x }{ 3 } } } dx=1\Rightarrow k\frac { \left[ { e }^{ \frac { -x }{ 3 } } \right] ^{ \infty } }{ -\frac { 1 }{ 3 } } \) 
\(\Rightarrow -3k\left[ { e }^{ -\infty }-{ e }^{ 0 } \right] =1\) [∵ e^∞ = 0, e⁰ = 1]
\(\Rightarrow 3k=1\Rightarrow k=\frac { 1 }{ 3 } \) 
\(\therefore k=\cfrac { 1 }{ 3 } \) 
(ii) The distribution function F(x) = \(\int _{ -\infty }^{ x }{ f(u)du } \)
Case 1: x < 0, 
\(F(x)=\int _{ -\infty }^{ x }{ f(x)dx=0 } \) 
Case 2: x > 0, 
\(f(x)=\int _{ -\infty }^{ x }{ f(x)dx } \) 
= \(\int _{ -\infty }^{ 0 }{ f(x)dx+\int _{ 0 }^{ x }{ f(x) } dx } \) 
= \(0+k\int _{ 0 }^{ x }{ { e }^{ \frac { -x }{ 3 } } } dx\) 
= \(\frac { 1 }{ 3 } \left[ \cfrac { { e }^{ \frac { -x }{ 3 } } }{ -\frac { 1 }{ 3 } } \right] =-\left[ { e }^{ -\frac { x }{ 3 } }-{ e }^{ o } \right] \) 
= \(-[{ e }^{ -\frac { x }{ 3 } }-1]\) 
= \(1-{ e }^{ -\frac { x }{ 3 } }\) 
\(\therefore F(x)=\begin{cases} \begin{matrix} 0 & x\le 0 \end{matrix} \\ \begin{matrix} 1-{ e }^{ \frac { -x }{ 3 } } & x>0 \end{matrix} \end{cases}\) 
(iii) p(X < 3)
= \(\int _{ 0 }^{ 3 }{ ke^{ -\frac { x }{ 3 } } } dx=\cfrac { 1 }{ 3 } \int _{ 0 }^{ 3 }{ { e }^{ -\frac { x }{ 3 } }dx } \)
= \(\cfrac { 1 }{ 3 } \left[ \frac { { e }^{ -\frac { x }{ 3 } } }{ \frac { -1 }{ 3 } } \right] \) 
= -[e^-1-e⁰] = -[e^-1-1]
(iv) \(p(5\le X)=p(X\ge 5)=\int _{ 5 }^{ \infty }{ f(x)dx } \) 
= \(\int _{ 5 }^{ \infty }{ ke^{ -\frac { x }{ 3 } } } dx=\cfrac { 1 }{ 3 } \cfrac { \left[ { e }^{ -\frac { x }{ 3 } } \right] _{ 5 }^{ \infty } }{ \frac { -1 }{ 3 } } \) 
= \(-\left[ { e }^{ -\infty }-e^{ \frac { -3 }{ 5 } } \right] =\left[ 0-{ e }^{ \frac { -5 }{ 3 } } \right] \) 
= \({ e }^{ \frac { -5 }{ 3 } }\) 
(v) \(p(X\le 4)=\int _{ -\infty }^{ 4 }{ f(x)dx } \) 
= \(\int _{ -\infty }^{ 0 }{ f(x)dx+\int _{ 0 }^{ 4 }{ f(x)dx } } \) 
= \(0+\int _{ 0 }^{ 4 }{ { ke }^{ -\frac { x }{ 3 } }dx } =k\left[ \frac { { e }^{ -\frac { x }{ 3 } } }{ -\frac { 1 }{ 3 } } \right] _{ 0 }^{ 4 }\) 
= \(\cfrac { 1 }{ 3 } \cfrac { \left[ { e }^{ -\frac { x }{ 3 } } \right] _{ 0 }^{ 4 } }{ -\frac { 1 }{ 3 } } =-\left[ { e }^{ \frac { -4 }{ 3 } }-{ e }^{ o } \right] \) 
= \(-\left[ { e }^{ \frac { -4 }{ 3 } }-1 \right] =1-{ e }^{ \frac { -4 }{ 3 } }\)

(i) Probability mass function
For a discrete random variable we have
f(x) = p(X = x)
\(\therefore f(0)=F(0)=\frac { 1 }{ 2 } \) 
f(1) = F(1) - F(0)
= \(\frac { 3 }{ 5 } -\frac { 1 }{ 2 } =\frac { 6-5 }{ 10 } =\frac { 1 }{ 10 } \) 
f(2) = F(2)-F(1)
= \(\frac { 4 }{ 5 } -\frac { 3 }{ 5 } =\frac { 1 }{ 5 } \) 
f(3) = F(3) - F(2)
\(\frac { 9 }{ 10 } -\frac { 4 }{ 5 } =\frac { 9-8 }{ 10 } =\frac { 1 }{ 10 } \)
f(4) = F(4)-F(3)
= \(1-\frac { 9 }{ 10 } =\frac { 1 }{ 10 } \) 
ஃThe probability mass function is

(ii) p(x < 3) = p(x = 0) + p(x = 1) + p(x = 2)
= \(\frac { 1 }{ 2 } +\frac { 1 }{ 10 } +\frac { 1 }{ 5 } =\frac { 5+1+2 }{ 10 } =\frac { 8 }{ 10 } \) 
= \(\frac { 4 }{ 5 } \)
(iii) p(x≥2) = p(x = 2) + p(x = 3) + p(x = 4)
= \(\frac { 1 }{ 5 } +\frac { 1 }{ 10 } +\frac { 1 }{ 10 } =\frac { 2+1+1 }{ 10 } =\frac { 4 }{ 10 } \)
= \(\frac { 2 }{ 5 } \)

Given
\(\\ \\ \\ \\ \\ f(x)=\begin{cases} \begin{matrix} \frac { { x }^{ 2 }+1 }{ k } & forx=0,1,2 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\\ \\ \\ \\ \\ \\ \)
The random variable X take the values 0, 1, 2.
Probability mass function.

\(\sum _{ i=0 }^{ 2 }{ f(x_{ i })=1\Rightarrow f(0)+f(1)+f(2)=1 } \)
\(\Rightarrow \frac { 0+1 }{ k } +\frac { 1+1 }{ k } +\frac { 4+1 }{ k } \) 
\(\Rightarrow \frac { 1 }{ k } +\frac { 2 }{ k } +\frac { 5 }{ k } =1\) 
\(\Rightarrow \frac { 8 }{ k } =1\) 
\(\Rightarrow k=8\) 
(ii) Cumulative distribution function

\(F(0)=P(X<0)\\P(x=0)\\ =\frac { 1 }{ 8 } \) 
\(F(1)=P(X = 0)+ P(X = 1)\\ \frac { 1 }{ 8 } +\frac { 2 }{ 8 } =\frac { 3 }{ 8 } \)
\(F(2)=P(X= 0) + P(X = 1) + P(X = 2) = \frac { 1 }{ 8 } +\frac { 2 }{ 8 } +\frac { 5 }{ 8 } =1\) 
Cumulative distribution function is
\(f(x)=\begin{cases} \begin{matrix} \frac { 1 }{ 8 } & for & x\le 0 \end{matrix} \\ \begin{matrix} \frac { 2 }{ 8 } +\frac { 1 }{ 8 } & for & \frac { 3 }{ 8 } forx\le 1 \end{matrix} \\ \begin{matrix} \frac { 3 }{ 8 } +\frac { 5 }{ 8 } =1 & for & x\le 2 \end{matrix} \end{cases}\)
(iii) p(x ≥ 1) = p(x = 1) + p(x = 2)
= \(\frac { 2 }{ 8 } +\frac { 5 }{ 8 } \)
\(p(x\ge 1)=\frac { 7 }{ 8 } \)

Let T be the temperature of the coffee at time t
and T_m' the temperature of the kitchen.
By Newton's law of cooling
\(\frac { dT }{ dt } =K(T-{ T }_{ m })\)
\(\Rightarrow \frac { dT }{ dt } =K(T-70)\)
\(\Rightarrow \int { \frac { dT }{ T-70 } =K\int { dt } } \)
\(\Rightarrow log(T-70)=kt+logC\)
\(\Rightarrow log(T-70)-logC=Kt\)
\(\Rightarrow log\left( \frac { T-70 }{ C } \right) =Kt\)
\(\Rightarrow \frac { T-70 }{ C } ={ e }^{ Kt }\)
\(\Rightarrow T-70={ Ce }^{ Kt }...(1)\)
\(\\ When\ t=0,\ T={ 180 }^{ o }F\)
\(\therefore { 180 }^{ o }-{ 70 }^{ o }={ Ce }^{ 0 }\)
\(\Rightarrow C={ 11 }0^{ 0 }\)
\(\\ \therefore (1)\Rightarrow T-70=110{ e }^{ Kt } ..(2)\)
\(When\ t=0,T=160\)
\(\therefore 160-70=110{ e }^{ 10K }\)
\(90=110{ e }^{ 10K }\)
\(\Rightarrow { e }^{ 10K }=\frac { 9 }{ 11 } \)
\(\Rightarrow { e }^{ K }={ \left( \frac { 9 }{ 11 } \right) }^{ \frac { 1 }{ 10 } }...(3)\)
(i) when t = 15, (2) becomes,
\(\Rightarrow T-70=110{ \left( \frac { 9 }{ 11 } \right) }^{ \frac { 1 }{ 10 } \times 15 }\)
\(=110{ \left( \frac { 9 }{ 11 } \right) }^{ \frac { 3 }{ 2 } }\)
\(=110\times { \left( \frac { 9 }{ 11 } \right) }\left( \sqrt { \frac { 9 }{ 11 } } \right) \)
\(=110\times \frac { 9 }{ 11 } \times \frac { 3 }{ \sqrt { 11 } } \)
\(=\frac { 270 }{ \sqrt { 11 } } =\frac { 270 }{ 3.32 } =81.33\)
\(\Rightarrow\) T=81.33+70=151.3F
\(\therefore\) T = 151.3F
\(\therefore\) The temperature of the coffee at 10.15 am is 151.3F
(ii) when T = 130F, (2) becomes
T-70 = 110e^kt  ...(2)
\(\Rightarrow\) 130-70 = 110e^kt 
60 = 110e^kt 
e^kt  = \(\frac{6}{11}\)
\({ \left( \frac { 9 }{ 11 } \right) }^{ \frac { t }{ 10 } }=\frac { 6 }{ 11 } \)
\(\frac { t }{ 10 } =\frac { log\left( \frac { 6 }{ 11 } \right) }{ log\left( \frac { 9 }{ 11 } \right) } \)
\(=\frac { log(0.545) }{ log(0.818) } =\frac { -0.264 }{ -0.087 } \)
= 3.34
t = 30.34min
T = 140F (2)becomes
140-70 = 110e^kt ...(2)
\(\Rightarrow 70={ 110e }^{ kt }\)
\({ e }^{ kt }=\frac { 7 }{ 11 } \)
\({ \left( \frac { 9 }{ 11 } \right) }^{ \frac { t }{ 10 } }=\frac { 7 }{ 11 } \)
\(\frac { t }{ 10 } =\frac { log\left( \frac { 7 }{ 11 } \right) }{ log\left( \frac { 7 }{ 11 } \right) } =\frac { -0.197 }{ -0.087 } \)
= 2.26
t = 22.6min
\(\therefore\) Between 10.22 min to 10.30 min, the woman should have drunk the coffee.

Let T be the temperature of water at any time t.
Then, by Newton's law of cooling,
\(\frac { dT }{ dt } \infty (T-{ 25 }^{ o })\)
\(\Rightarrow \frac { dT }{ dt } =-\lambda (T-25)\)
\(\Rightarrow \int { \frac { dT }{ T-25 } } =-\lambda \int { dt } \)
\(\Rightarrow log(T-25)=-\lambda t+C\)  ...(1)
At t = 0, T = 100⁰e in (1) we get
\(\therefore(1)\Rightarrow\)log75 = 0+C
\(\Rightarrow\)C = log75
\(\therefore\)(1) becomes, log (T-25) = -\(\lambda\)t + log 75
\(\Rightarrow log\left( \frac { T-25 }{ 75 } \right) =-\lambda t...(2)\)
When t = 10, T = 80^oC
\(\therefore log\left( \frac { 80-25 }{ 75 } \right) =-10|\)
\(\Rightarrow log\left( \frac { 11 }{ 15 } \right) =-10\lambda \)
\(\Rightarrow \lambda =-\frac { 1 }{ 10 } log\left( \frac { 11 }{ 15 } \right) \)
Substituting \(\lambda\) in (2) we get
\(log\left( \frac { T-25 }{ 75 } \right) =\frac { 1 }{ 10 } log\left( \frac { 11 }{ 15 } \right) t ...(3)\)
\(\Rightarrow log\left( \frac { T-25 }{ 75 } \right) =\frac { 1 }{ 10 } log\left( \frac { 11 }{ 15 } \right) \times 20\)
\(={ \left( \frac { 11 }{ 15 } \right) }^{ 2 }\)
\(\Rightarrow \frac { T-25 }{ 75 } ={ \left( \frac { 11 }{ 15 } \right) }^{ 2 }\)
\(\Rightarrow T-25=\frac { 121 }{ 225 } \times 75=\frac { 121 }{ 3 } =40.33\)
\(\Rightarrow T=40.33+25\)
\(\\ =65{ .33 }^{ 0 }C\)
So the temperature of water after 20 minutes is 65.33°C
(ii) putting T = 40^oC in (3) we get
\(log\left( \frac { 40-25 }{ 75 } \right) =\frac { 1 }{ 10 } log\left( \frac { 11 }{ 15 } \right) t\)
\(\Rightarrow log\left( \frac { 1 }{ 5 } \right) =\frac { 1 }{ 10 } log\left( \frac { 11 }{ 15 } \right) t\)
\(t=\frac { 10log\left( \frac { 1 }{ 5 } \right) }{ log\left( \frac { 11 }{ 15 } \right) } =\frac { -10log5 }{ log\left( \frac { 11 }{ 15 } \right) } \)
\(=\frac { -10\times 1.6094 }{ -0.3101 } \)
\(\therefore\) t = 53.46 minutes

Given E = Ri + L \(\frac{di}{dt}\)
\(\frac { E }{ L } =\frac { Ri }{ L } +\frac { di }{ dt } \)
\(\Rightarrow \frac { Ri }{ L } +\frac { di }{ dt } =\frac { E }{ L } \)
This is a linear differential equation
\(Here\quad P=\frac { R }{ L } and\quad Q=\frac { E }{ L } \)
\(\therefore \int { pdt } =\int { \frac { R }{ L } dt } =\frac { R }{ L } t\)
\(\therefore I.F={ e }^{ \int { pdt } }={ e }^{ \frac { Rt }{ L } }\)
\(\therefore\) Solution is i\({ e }^{ \int { pdt } }=\int { Q{ e }^{ \int { pdt } }dt+C } \)
\(\Rightarrow i{ e }^{ \frac { Rt }{ L } }=\int { \frac { E }{ L } . } { e }^{ \frac { Rt }{ L } }dt+C\)
\(\therefore i{ e }^{ \frac { Rt }{ L } }=\frac { E }{ L } \frac { { e }^{ \frac { Rt }{ L } } }{ \frac { R }{ L } } dt+C\)
\(i=\frac { E }{ R } { e }^{ \frac { Rt }{ L } }+C\)
\(i=\frac { E }{ R } +c{ e }^{ -\frac { Rt }{ L } }\)
When E = 0,
\(i=0+c{ e }^{ -\frac { Rt }{ L } }\)
\(\Rightarrow i=c{ e }^{ -\frac { Rt }{ L } }\)

Let T be the temperature of the body at any time t and with time 0 taken to be 8 p.m. By Newton’s law of cooling \(\frac { dT }{ dt } =k(T-50)or\frac { dT }{ T-50 } =dt\).
Integrating on both sides, we get log |50 −T| = kt + logC or 50 −T = Ce^kt.
When t = 0, T = 70, and so C = −20
When t = 2,T = 60, we have −10 = −20 e^k2.
Thus, \(k=\frac { 1 }{ 2 } log\left( \frac { 1 }{ 2 } \right) \)
Hence, the solution is 50-T = -20e^{\(\frac{1}{2}\)tlog\((\frac{1}{2})\)} or T = 50 + 20\((\frac{1}{2})^\frac{t}{2}\)
Now, we would like to find the value of t, for which T(t) = 98.6 , and t = 2\(\left( \frac { log\left( \frac { 48.6 }{ 20 } \right) }{ log\left( \frac { 1 }{ 2 } \right) } \right) \approx -2.56\)
It appears that the person was murdered at about 5.30 p.m.

\(\frac { dy }{ dx } +\frac { 3y }{ x } =\frac { 1 }{ { x }^{ 2 } } \), given that y = 2 when x = 1 
This is a linear differential equation.
\(\therefore P=\frac { 3 }{ x } ;Q=\frac { 1 }{ { x }^{ 2 } } \)
\(\int { pdx } =3\int { \frac { 1 }{ x } dx=3logx=log{ x }^{ 3 } } \)
\(\therefore I.F.={ e }^{ \int { pdx } }={ e }^{ log{ x }^{ 3 } }={ x }^{ 3 }\)
\(\therefore\) The solution is \({ ye }^{ \int { pdx } }=\int { Q{ e }^{ \int { pdx } }dx+c } \)
\(\Rightarrow { yx }^{ 3 }=\int { \frac { 1 }{ { x }^{ 2 } } .{ x }^{ 3 } } dx+c\)
\(\Rightarrow { yx }^{ 3 }=\int { xdx+c } \)
\(\Rightarrow { yx }^{ 3 }=\frac { { x }^{ 2 } }{ 2 } +c...(1)\)
When x = 1, y = 2
\(\Rightarrow 2{ (1) }^{ 3 }=\frac { 1 }{ 2 } +c\Rightarrow 2-\frac { 1 }{ 2 } =\frac { 3 }{ 2 } \)
\({ yx }^{ 3 }=\frac { { x }^{ 2 } }{ 2 } +\frac { 3 }{ 2 } \)
\({ 2x }^{ 3 }y={ x }^{ 2 }+3\)

\( y e^{\left(\frac{1}{r}\right)} \cdot d x   =\left(x e^{\frac{1}{y}}+y\right) d y \)
\(\frac{d x}{d y}   =\frac{x \cdot e^{\left(\frac{6}{y}\right)}+y}{y e^{\left(\frac{x}{y}\right)}} \)
\(\frac{d x}{d y}   =\left(\frac{x}{y}\right)+\frac{1}{e^{\left(\frac{6}{y}\right)}}\)
Put x = \( \mathrm{vy}\) \( \Rightarrow\left(\frac{x}{y}\right)=\mathrm{v}\) and \(\frac{d x}{d y}=\cdot v+y \cdot \frac{d v}{d y} \)
\((1) \Rightarrow \ v+y \cdot \frac{d v}{d y}=v+\frac{1}{e^y} \)
\(\mathrm{e}^v \cdot \mathrm{dv}=\frac{d y}{y}\)
Integrating on both sides,
ie) \(\int e^v \cdot d v   =\int \frac{d y}{y} \)
\(e^v   =\log |y|+\log |c|  \) 
\(e^{\left(\frac{x}{y}\right)}   =\log |c y|\)

Let the of the parbola be x² = 4ay (1)

Since (30, 16) is a point on (1),
we get 30² = 4 \(\times\) a \(\times\) 16
⇒ a = \(\frac { 30\times 30 }{ 4\times 16 } =\frac { 225 }{ 16 } \)
∴ becomes, x² = \({ x }^{ 2 }=\frac { 4\times 225 }{ 16 } y=\frac { 225 }{ 4 } y\)
Let AC = h m and BD = lm
∴ A(6, h) is a point on the parabola [∵ OD = 6]
∴ \({ 6 }^{ 2 }=\frac { 225 }{ 4 } \times h\)
⇒ \(h=\frac { 36\times 4 }{ 225 } \Rightarrow h=0.52\)
∴ AD = 3 + h = 3 + 0.52 = 3.52 m
Also (12, 1) is a point on the parabola
[∵ ON = 6 + 6 = 12]
∴ \({ 12 }^{ 2 }=\frac { 225 }{ 4 } \times l\)
⇒ l = \(\frac { 12\times 12\times 4 }{ 225 } =\frac { 576 }{ 225 } =2.08\) = 5.08 m
Hence the length of first two vertical cables are 3.52 m and 5.08 m.

The parametric equation of tangent at 't₁' to the parabola y² = 4ax is yt₁ = x + at₁²  ...(1)
Also, the parametric equation of tangent at 't₂' to the parabola y = 4ax is yt₂ = x+ at₂² ...(2)
(1) ➝ yt₁ = x + at₁²
(2) ➝ yt₂ = x + at₂²
(1) - (2) y(t₁ -t₂) = a(t₁² - t₂²)

⇒ y = a(t₁ + t₂)
Substitutingy = a(t₁ + t₂) in (1) we get,
a(t₁ + t₂)t₁ = x + at₁²

⇒ x = at₁t₂
Hence, the point of intersection of two lengths is
[at₁t₂, a(t₁ + t₂)]

Equation of ellipse is
x²+ 4y² = 32
\(\frac { { x }^{ 2 } }{ 32 } + \frac { { y }^{ 2 } }{ 8 } =1\)
a² = 32, b² = 8
\(a=4\sqrt { 2 } ,b=2\sqrt { 2 } \)
Equation of tangent at \(\theta =\frac { \pi }{ 4 } \) is \(\frac { xcos\frac { \pi }{ 4 } }{ 4\sqrt { 2 } } \frac { ysin\frac { \pi }{ 4 } }{ 2\sqrt { 2 } } =1\) 
\(\frac { x }{ 8 } +\frac { y }{ 4 } =1\)
x+2y−8 = 0.
Equation of normal is \(\frac { 4\sqrt { 2X } }{ cos\frac { \pi }{ 4 } } -\frac { 2\sqrt { 2Y } }{ sin\frac { \pi }{ 4 } } =32-8\)
That is 8x-4y = 24 
2x-y-6 = 0
Aliter: 
At, \(\theta =\frac { \pi }{ 4 } \)
\((a\ cos \theta ,b\ sin \theta )=\left( 4\sqrt { 2 }\ cos\frac { \pi }{ 4 } ,2\sqrt { 2 }\ sin\frac { \pi }{ 4 } \right) \)
= (4, 2)
∴ Equation of tangent at \(\theta =\frac { \pi }{ 4 } \) is same at (4, 2) 
Equation of tangent in cartesian form is \(\frac { { xx }_{ 1 } }{ { a }^{ 2 } } +\frac { { yy }_{ 1 } }{ { b }^{ 2 } } =1\)
x+2y−8 = 0
Slope of tangent is -\(\frac { 1 }{ 2 } \)
Slope of normal is 2 Equation of normal is  y - 2 = 2(x -4) 
y−2x+6 = 0 

Let the equation of the circle be
x² + y² + 2gx + 2fy + c = 0 ........... (1)
(1) passes through (1, 0)
⇒ 1 + 0 + 2g(1) + 2f(0) + c = 0
⇒ 2g + c = -1 ................(2)
(1) passes through (-1, 0)
⇒ (-1)² + 0 + 2g(-1) + 2f(0)+ c = 0
⇒ -2g + c = -1 ..............(3)
Also (1) passes through (0, 1)
⇒ 0 + 1²+ 2g(0) +2f(1) + c = 0
⇒ 2f + c = -1 .................(4)
(2) + (3) ⇒ 2c = -2
⇒ c = -1
Substituting c = -1 in (2), we get
2g-1 = -1
⇒ 2g = 0
⇒ g = 0
Substituting c = -1 in (4) we get,
2f -1 = -1
⇒ 2f  = 0
⇒ f = 0
∴ The required equation of the circle 
x² + y² - 1 = 0