Given p : Jupiter is a planet and
q : India is an island.
(i) ¬p : Jupiter is not a planet.
(ii) p ∧ ¬q : Jupiter is a planet and India is not an island.
(iii) ¬p ∨ q : Jupiter is not a planet or India is an island.
(iv) p➝ ¬q : If Jupiter is a planet then India is not an island.
(v) p↔q : Jupiter is a planet if and only if India is an island.

We compute
F(\(\lambda \)x, \(\lambda \)y) = \(\frac { { \left( { \lambda x } \right) }^{ 2 }+5\left( \lambda x \right) \left( \lambda y \right) -10(\lambda y)^2 }{ 3\lambda x+7\lambda y } =\frac { { \lambda }^{ 2 } }{ \lambda } \left( \frac { { x }^{ 2 }+5xy-10{ y }^{ 2 } }{ 3x+7y } \right) =\lambda F(x,y)\)
for all \(\lambda \) ∈ R. So F is a homogeneous function of degree 1.

\( \because \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -ax }dx}=\frac { n! }{ { a }^{ n+1 } } \)
\(n=5,\quad a=3 \)
\(=\frac { 5! }{ { 3 }^{ 6 } } \)

Given n = 6, \(p=\frac { 1 }{ 3 } \), k = 3
\(P(X=k)=\left( \begin{matrix} n \\ k \end{matrix} \right) { p }^{ k }\left( 1-p \right) ^{ n-k },\) 
n = 0,1,2, ... n
\(\therefore P(X=k)=\left( \begin{matrix} n \\ k \end{matrix} \right) { p }^{ k }(1-p)^{ n-k }\) 
n = 0,1,2, ... n
\(P(X=3)=\left( \begin{matrix} 6 \\ 3 \end{matrix} \right) \left( \cfrac { 1 }{ 3 } \right) ^{ 3 }\left( 1-p \right) ^{ 6-3 }\) 
= \(\left( \begin{matrix} 6 \\ 3 \end{matrix} \right) \left( \cfrac { 1 }{ 3 } \right) ^{ 3 }\left( \cfrac { 2 }{ 3 } \right) ^{ 2 }\) 

\(P(X=3)=\frac { 160 }{ 729 } \) 

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

\(\begin{matrix} lim \\ (x,y)\rightarrow (0,0) \end{matrix}=\left| \frac { { y }^{ 2 }-xy }{ \sqrt { x } -\sqrt { y } } -0 \right| \)
= \(\left| \frac { { y }^{ 2 }-xy }{ \sqrt { x } -\sqrt { y } } \right| =\frac { \left| y \right| \left| y-x \right| }{ \left| \sqrt { x } -\sqrt { y } \right| } \)
= \(\frac { \left| y \right| \left| \sqrt { x } +\sqrt { y } \right| |\sqrt { y } -\sqrt { x } | }{ \left| -\sqrt { y } -\sqrt { x } \right| } \)
\(=\frac{|y||\sqrt{x}+\sqrt{y}| \sqrt{y}-\sqrt{\not x} \mid}{|\sqrt{y}-\sqrt{\not x}|}\)
= \(|y||\sqrt { x } +\sqrt { y } |\)
\(\therefore \begin{matrix} lim \\ (x,y)\rightarrow (0,0) \end{matrix}=\begin{matrix} lim \\ (x,y)\rightarrow (0,0) \end{matrix}|y||\sqrt { x } +\sqrt { y } |=0\)

If we directly substitute x = 0 we get an indeterminate form \(\frac{0}{0}\) and hence we apply the l’Hôpital’s rule to evaluate the limit as,
\(\underset{x\rightarrow 0^{+}}{lim}(\frac{sin \ x}{x^{2}})=\underset{x\rightarrow 0^{+}}{lim}(\frac{cos \ x}{2x})=\infty\)
\(\underset{x\rightarrow 0^{-}}{lim}(\frac{sin \ x}{x^{2}})=\underset{x\rightarrow 0^{-}}{lim}(\frac{cos \ x}{2x})=\infty\)
As the left limit and the right limit are not the same we conclude that the limit does not exist.
Remark
One may be tempted to use the l’Hôpital’s rule once again in \(\underset{x\rightarrow 0^{+}}{lim} (\frac{cos \ x}{2x})\) to conclude
\(\underset{x\rightarrow 0^{+}}{lim} (\frac{cos \ x}{2x})\)\(\underset{x\rightarrow 0^{+}}{lim} (\frac{-sin \ x}{2})\)=0
which is not true because it was not an indeterminate form.

Given \(f(x)=|\frac{1}{x}|, x\in [-1,1]\)
Rolle's theorem is not applicable since \(f(x)=(\frac{1}{x})\) is not continuous at x = 0 in [-1, 1] and not differentiable in (-1, 1)

Given = e^mx is the solution of
y' + 2y = 0 ...(1)
y = e^mx  ...... (2) 
\(\frac{dy}{dx} = e^{mx}. m\)
\(\frac{dy}{dx} = ym\)
\(\frac{dy}{dx} - my=0\)
⇒ y' - my = 0 ...(3)
Comparing equation (1) & (3),
we get m = -2

\(y\left( \frac { dy }{ dx } \right) =\frac { x }{ \left( \frac { dy }{ dx } \right) +{ \left( \frac { dy }{ dx } \right) }^{ 3 } } \)
is the given differential equation.
\(\Rightarrow y{ \left( \frac { dy }{ dx } \right) }^{ 2 }+{ \left( \frac { dy }{ dx } \right) }^{ 4 }=x\)
The highest derivative is 1 and its maximum power is 4.
∴ Order 1, degree 4.

p	q	q ➝ p	~p	~q	~q ➝ ~p
T	T	T	F	F	T
T	F	T	F	T	T
F	T	F	T	F	F
F	F	T	T	T	T

The entries in the columns corresponding q ➝ p and ~p ➝ ~q are identical and hence they are equivalent.
q ➝ p ≡ ~p ➝ ~q
Hence proved

(i) Since f (x) is a probability density function, f (x) ≥ 0 and \(\int _{ - }^{ \infty }{ f(x) } dx=1\) 
That is \(\int _{ -\infty }^{ 0 }{ 0dx } +\int _{ 0 }^{ \infty }{ k{ e }^{ -2x }dx } =1\) 
\(0+k\left( \frac { { e }^{ -2x } }{ -2 } \right) =1\Rightarrow k\left( \frac { { e }^{ -\infty }-{ e }^{ 0 } }{ -2 } \right) =1\Rightarrow k=2\) 
Therefore the probability density function is
\(f\left( x \right) =\begin{cases} \begin{matrix} 2{ e }^{ -2x } & forx>0 \end{matrix} \\ \begin{matrix} 0 & forx \end{matrix}\le 0 \end{cases}\)
 (ii) Distribution function
By definition the distribution function \(F(x)=P\left( x\le x \right) =\int _{ -\infty }^{ x }{ f(u) } du\) 
When x≤0 \(F(x)=\int _{ -\infty }^{ x }{ F(u) } du=\int _{ -\infty }^{ x }{ odu=0 } \) 
When x > 0 \(F(x)=\int _{ -\infty }^{ x }{ f(u) } du\int _{ -\infty }^{ x }{ 0du } +\int _{ 0 }^{ x }{ { 2e }^{ -2x }du\left( \frac { { e }^{ -2x } }{ -2 } \right) } =1-{ e }^{ 2x }\) 
This gives \(F(x)=\begin{cases} \begin{matrix} 0 & forx\le 0 \end{matrix} \\ \begin{matrix} 1-{ e }^{ 2x } & forx>0 \end{matrix} \end{cases}\) 
(iii) P(X < ) = P(X ≤2 ) = F(2 ) = 1-e^2\(\times\)2 (since F(x) is continuous)
(iv) The probability that X is at least equal to four unit of time is
P(X ≥ 4 ) = 1 - P(X < 4 ) = 1- F( 4) = 1 - ( 1-e^{-2\(\times\)4}) = e⁸
(v) In the continuous case, f (x) at x = a is not the probability that X takes the value a, that is f (x) at x = a is not equal to P( X ) a.  If X is continuous type, P(X = a) = 0 for a ∈ R. Therefore P(x = 3) = 0.

Given (x, y, z) = log (x³ + y³ + z³)
\(\frac { \partial U }{ \partial x } =\frac { 1 }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } { (3x }^{ 2 });\)
\(\frac { \partial U }{ \partial y } =\frac { { 3y }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \) and
\(\frac { \partial U }{ \partial z } =\frac { { 3z }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \)
\(\therefore \frac { \partial U }{ \partial x } +\frac { \partial U }{ \partial y } +\frac { \partial U }{ \partial z } =\frac { { 3x }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } +\frac { { 3y }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } +\frac { { 3z }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \)
\(=\frac { { 3({ x }^{ 2 }+y }^{ 2 }+{ z }^{ 2 }) }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \)

\(\int _{ b }^{ \infty }{ \frac { 1 }{ { a }^{ 2 }+{ x }^{ 2 } } dx } ={ \left[ \frac { 1 }{ a } { tan }^{ -1 }\frac { x }{ a } \right] }_{ b }^{ \infty }=\frac { 1 }{ a } { tan }^{ -1 }\infty -\frac { 1 }{ a } { tan }^{ -1 }\frac { b }{ a } =\frac { 1 }{ a } \left[ \frac { \pi }{ 2 } -{ tan }^{ -1 }\frac { b }{ a } \right] \)

Let us put I = \(\int ^{3}_{2} \frac{\sqrt {x}}{\sqrt {5-x}+\sqrt {x}}\) dx --- (1)
Applying the formula \(\int ^{a}_{b}\) f(x) dx =\(\int ^{a}_{b}\) f(a+b -x) dx, we get
I =  \(\int ^{3}_{2} \frac{\sqrt {(2+3-x)}}{\sqrt {5-(2+3-x)}+\sqrt { {(2+3-x)}}}\) dx = \(\int ^{3}_{2} \frac{\sqrt {(5-x)}}{\sqrt {x}+\sqrt { {(5-x}}}\) dx     -- (2)
Adding (1) and (2), we get 
2I =  \(\int ^{3}_{2} \frac{\sqrt {x} + \sqrt {5-x} }{\sqrt {x}+\sqrt { {5-x}}}\) dx =\(\int ^{3}_{2} \) dx = \([x]^{3}_{2}\) = 3 - 2 = 1
Hence, we get I = \(\frac{1}{2}\)

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]
\(\Rightarrow \frac { k }{ 4 } =1\\ \Rightarrow k=4\)

Let f(x) = \(f(x)={ x }^{ \frac { 2 }{ 3 } },{ x }_{ 0 }=125,\triangle x=-2\)
∴ (123)^\(\frac23\) = f(125) +1'(125) (-2) ... (1)
\(f(125)={ (125) }^{ \frac { 2 }{ 3 } }={ { (5 }^{ 3 }) }^{ \frac { 2 }{ 3 } }\) = 5² = 25
\({ f }^{ ' }(x)={ \frac { 2 }{ 3 } x }^{ \frac { 2 }{ 3 } -1 }={ \frac { 2 }{ 3 } x }^{ \frac { 1 }{ 3 } }=\frac { 2 }{ { 3x }^{ \frac { 1 }{ 3 } } } \)
\({ f }^{ ' }(125)=\frac { 2 }{ { 3(125)x }^{ \frac { 1 }{ 3 } } } =\frac { 2 }{ { 3{ (5 }^{ 3 } })^{ \frac { 1 }{ 3 } } } =\frac { 2 }{ 3(5) } =\frac { 2 }{ 15 } \)
∴ \({ (123) }^{ \frac { 2 }{ 3 } }=25+\frac { 2 }{ 15 } (-2)\)
\(=25-\frac { 4 }{ 15 } =25-0.27\)
\({ (123) }^{ \frac { 2 }{ 3 } }=24.73\)

This is an indeterminate of the form \((\frac{\infty}{\infty})\)
To evaluate this limit, we apply l’Hôpital Rule m times
\(\underset{x\rightarrow \infty}{lim}\frac{e^{x}}{x^{m}}=\underset{x\rightarrow \infty}{lim}\frac{e^{x}}{m!} = \infty\) 

The given equation can be written as \(\frac { dx }{ dy } =\frac { \left( \frac { x }{ y } -1 \right) { 2e }^{ x/y } }{ 1+2{ e }^{ x/y } } =g\left( \frac { x }{ y } \right) ..(1)\)
The appearance of \(\frac{x}{y}\) in equation (1), suggests that the appropriate substitution is x = vy.
Put x = vy . Then, we have \(y\frac { dv }{ dy } =-\frac { 2{ e }^{ v }+v }{ 1+2{ e }^{ v } } \)
By separating the variables, we have \(-\frac { 1+2{ e }^{ v } }{ v+2{ e }^{ v } } dv=-\frac { dy }{ y } \)
On integration, we obtain
log |2e^v + v| = −log |y| + log |C| or log |2ye^v+vy| = log |C| or 2ye^v+ vy = ±C.
Replace v by \(\frac{x}{y}\) to get, 2ye^x/y+x = k, where k =土C, Which gives the required solution.

Separating the variables we get,
\(\frac { dy }{ \sqrt { 1-{ y }^{ 2 } } } \frac { dx }{ \sqrt { 1-{ x }^{ 2 } } } \)
Taking Integration on both sides, we get
\(\int \frac{d y}{\sqrt{1-y^{2}}}=\int \frac{d x}{\sqrt{1-x^{2}}}\)
sin^-1y = sin^-1 x + c

Distance moved in time 't' is s = \(\frac{t^{3}}{3}-t^{2}+3\) 
Velocity at time 't ' is V = \(\frac{ds}{dt}=t^{2}-2t\)
Acceleration at time 't ' is a(t) = \(\frac{dV}{dt}=2t-2\)
Therefore, the velocity is zero when t² − 2t = 0, that is t = 0, 2. The acceleration is zero when 2t − 2 = 0 . That is at time at time t = 1

Prove that p⟶(q⟶r) = (p ∧ q)⟶r without using truth table. From example we know that p⟶ q = ~p V q
Consider LHS = p⟶(q⟶r)
= p ⟶ (~q V r) [Implication Law]
= ~p V (~q V r) [Implication Law]
= (~p V ~q) V r [associative property]
= ~(p ∧ q) V r [using Demorgan's law]
= (p ∧ q) ⟶ r [Implication Law]
Hence proved.

(i) The output m+ n - mn is clearly an integer and hence∗ is a binary operation on Z.
(ii) m*n = m+ n − mn = n + m − nm = n*m, ∀m,n∈Z. So ∗ has commutative property.
(iii) Consider (m*n)*p = (m+ n −m n)* p= (m+ n −mn) + p − (m+ n −m n) p
= m+ n + p −mn −m p − n p + m n p ... (1)
Similarly m*(n*p) = m*(n + p − n p) = m+ (n + p − n p) −m (n + p − n p)
= m+ n + p − n p −m n −mp + m n p ... (2)
From (1) and (2), we see that m*(n*p) = (m*n)*p. Hence * has associative property.
(iv) An integer e is to be found such that 
m*e = e*m = m, ∀m∈Z ⇒ m + e - m e = m
⇒e(1-m) = 0 ⇒ e = 0 or m = 1. But m is an arbitrary integer and hence need not be equal to 1. So the only possibility is e = 0. Also m*0 = 0*m = m, ∀m∈Z. Hence 0 is the identity element and hence the existence of identity is assured.
(v) An element m'∈ Z is to be found such that m*m' = m' * m = e = 0, ∀m∈Z.
m*m' = 0 ⇒ m+m'-m m'= 0⇒ m m' = 0 ⇒ \(\frac{m}{m-1}\). when m = 1, m' is not defined.
When m = 2, m' is an integer. But except m = 2, m′ need not be an integer for all values of m. Hence inverse does not exist in Z.

Given w(x, y, z) = \(\left( \frac { { 5x }^{ 3 }{ y }^{ 4 }+7{ y }^{ 2 }{ xz }^{ 4 }-{ 75y }^{ 3 }{ z }^{ 4 } }{ { x }^{ 2 }+{ y }^{ 2 } } \right) \)
Let (x, y, z) = \(\frac { { 5x }^{ 3 }{ y }^{ 4 }+7{ y }^{ 2 }{ xz }^{ 4 }-{ 75y }^{ 3 }{ z }^{ 4 } }{ { x }^{ 2 }+{ y }^{ 2 } } \)
⇒ w = log f  ...(1)
⇒ e^w = f
f(λx, λy, λz) = \(\frac { { 5\lambda }^{ 3 }{ x }^{ 3 }{ \lambda }^{ 4 }{ y }^{ 4 }+7{ \lambda }^{ 2 }{ y }^{ 2 }\lambda x{ \lambda }^{ 4 }{ z }^{ 4 }-75{ \lambda }^{ 3 }{ y }^{ 3 }{ \lambda }^{ 4 }{ z }^{ 4 }{ }^{ } }{ { \lambda }^{ 2 }{ x }^{ 2 }+{ \lambda }^{ 2 }{ y }^{ 2 } } \)
= \(\frac { { \lambda }^{ 7 }(5{ x }^{ 3 }{ y }^{ 4 }+7{ y }^{ 2 }{ xz }^{ 4 }-{ 75 }y^{ 3 }{ z }^{ 4 } }{ { \lambda }^{ 2 }({ x }^{ 2 }+{ y }^{ 2 }) } ={ \lambda }^{ 5 }f(x,y,z)\)
∴ f(x, y, z) is a homogeneous function of degree 5.
∴ By Euler's theorem,
\(x.\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } +z\frac { \partial f }{ \partial z } =5.f\)
⇒ \(x.\frac { \partial }{ \partial x } ({ e }^{ w })+y.\frac { \partial }{ \partial y } ({ e }^{ w })+z.\frac { \partial }{ \partial z } ({ e }^{ w })=5.{ e }^{ w }\) [using (1)]
⇒ \(x.{ e }^{ w }\frac { \partial w }{ \partial x } +y.{ e }^{ w }\frac { \partial w }{ \partial y } +z.{ e }^{ w }\frac { \partial w }{ \partial z } ({ e }^{ w })=5{ e }^{ w }\)
⇒ \(x\frac { \partial w }{ \partial x } +y\frac { \partial w }{ \partial y } +z\frac { \partial w }{ \partial z } \)  [Divided by e^w]

Volume of the right circular conical frustum is obtained by revolving the line y = x between x = a and x = b around the x - axis

\(\therefore\) Height of the frustum h = b - a
\(\therefore\)Volume \(=\pi \int _{ a }^{ b }{ { x }^{ 2 }dx } =\pi { \left[ \frac { { x }^{ 3 } }{ 3 } \right] }_{ a }^{ b }\)
\(=\frac { \pi }{ 3 } [{ b }^{ 3 }-{ a }^{ 3 }]\)
\(=\frac { \pi }{ 3 } (b-a)({ b }^{ 2 }+ab+{ a }^{ 2 })\)
Now, substitute h = b - a, r = a and R = b we get Volume of the conical frustum
\(\frac { \pi }{ 3 } [h({ R }^{ 2 }+rR+{ r }^{ 2 })]\)
Given h = 2 m, r = 1 m, R = 2 m we get
Required volume \(=\frac { \pi }{ 3 } [2(4+2+1)]\)
\(=\frac { \pi }{ 3 } (14)\)
\(=\frac { 14\pi }{ 3 } \)

The ellipse is symmetric about both the axes. The major axis lies along x-axis. The region to be revolved is sketched
Hence, the required volume is given by
\(V=\pi \int _{ -a }^{ a }{ { y }^{ 2 }dx } =\pi \int _{ -a }^{ a }{ \frac { { b }^{ 2 } }{ { a }^{ 2 } } \left( { a }^{ 2 }-{ x }^{ 2 } \right) dx } \)
\(=\frac { 2\pi { b }^{ 2 } }{ { a }^{ 2 } } \int _{ 0 }^{ a }{ ({ a }^{ 2 }-{ x }^{ 2 })dx } \) since the integrand is an even function
\(=\frac { { 2\pi b }^{ 2 } }{ { a }^{ 2 } } { \left( { a }^{ 2 }x-\frac { { x }^{ 3 } }{ 3 } \right) }_{ 0 }^{ a }=\frac { 2\pi { b }^{ 2 } }{ 3 } ={ \left( { a }^{ 2 }-\frac { { a }^{ 3 } }{ 3 } \right) }_{ 0 }^{ a }=\frac { 2\pi { b }^{ 2 } }{ 3 } \left( \frac { { 2a }^{ 3 } }{ 3 } \right) =\frac { 4\pi { ab }^{ 2 } }{ 3 } \)

Given \(f(x)=\begin{cases} \begin{matrix} 16{ xe }^{ -4x } & foex>0 \end{matrix} \\ \begin{matrix} 0 & forx\le 0 \end{matrix} \end{cases}\) 
Mean :
= \(E(x)=\int _{ 0 }^{ \infty }{ x.f(x)dx } \) 
= \(\int _{ 0 }^{ \infty }{ x.16.xe^{ -4x }dx } \) 
= \(16\int _{ 0 }^{ \infty }{ { x }^{ 2 } } .{ e }^{ -4x }dx=16\times \frac { 2! }{ { 4 }^{ 3 } } \)  \(\left[ \because \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -ax }dx=\frac { n! }{ { a }^{ n+1 } } } \right] \) 
= \(16\times \frac { 2 }{ 64 } \)
\(=\frac { 1 }{ 2 } \) 
Variance :
\(E({ x }^{ 2 })=\int _{ 0 }^{ \infty }{ { x }^{ 2 }.f(x)dx } \) 
= \(\int _{ 0 }^{ \infty }{ { x }^{ 2 }.{ e }^{ -4x } } dx\) 
= \(16\int _{ 0 }^{ \infty }{ { x }^{ 3 }{ e }^{ -4x }dx } \) 

ஃ Var(X) = E(X²) - [E(x)]²
\( =\frac{3}{8}-\frac{1}{4} \)
\(=\frac{1}{8} \)
∴ Var(X) \(=\frac{1}{8} \)

Put x = 2a cos²\(\theta\).
Then, dx = -4a cos \(\theta\) sin \(\theta\)d\(\theta\)
when x = 0, 2a cos²\(\theta\) = 0 and so \(\theta\) = \(\frac{\pi}{2}\)
When x = 2a, 2a cos²\(\theta\) = 2a and so \(\theta\) = 0
Hence, we get
\(I=\int _{ 0 }^{ 2a }{ { x }^{ 2 }\sqrt { 2ax-{ x }^{ 2 } } dx } \)
\(\int _{ \frac { \pi }{ 2 } }^{ 0 }{ { 4a }^{ 2 }{ cos }^{ 2 }\theta \sqrt { { 4a }^{ 2 }{ cos }^{ 2 }\theta -4{ a }^{ 2 }{ cos }^{ 4 }\theta } } (-4a\ cos\theta sin\ \theta )d\theta \)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { 4a }^{ 2 }{ cos }^{ 2 }\theta\ 2a\ cos\ \theta sin\ \theta (4a\ cos\ \theta sin\theta )d\theta } \)
\(=32{ a }^{ 4 }\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { cos }^{ 4 }\theta { sin }^{ 2 }\theta d\theta } \)
\(=32{ a }^{ 4 }\times \frac { 1 }{ 6 } \times \frac { 3 }{ 4 } \times \frac { 1 }{ 2 } \times \frac { \pi }{ 2 } =\pi { a }^{ 4 }\) 

Given U(x, y, z) = \(\frac { { x }^{ 2 }+{ y }^{ 2 } }{ xy } +3{ z }^{ 2 }y\)
\(\frac { \partial U }{ \partial x } =\frac { xy(2x)-({ x }^{ 2 }+{ y }^{ 2 })(y) }{ { x }^{ 2 }{ y }^{ 2 } } +0\)
\(=\frac { 2{ x }^{ 2 }y-{ x }^{ 2 }y-{ y }^{ 3 } }{ { x }^{ 2 }{ y }^{ 2 } } =\frac { { x }^{ 2 }y-{ y }^{ 3 } }{ { x }^{ 2 }{ y }^{ 2 } } \)
\(=\frac { y({ x }^{ 2 }-{ y }^{ 2 }) }{ { x }^{ 2 }{ y }^{ 2 } } =\frac { { x }^{ 2 }-{ y }^{ 2 } }{ { x }^{ 2 }y } \)
\(\frac { \partial U }{ \partial y } =\frac { xy(2y)-({ x }^{ 2 }+{ y }^{ 2 })(x) }{ { x }^{ 2 }{ y }^{ 2 } } +3{ z }^{ 2 }\)
= \(\frac { { 2xy }^{ 2 }-{ x }^{ 3 }-{ xy }^{ 2 } }{ { x }^{ 2 }{ y }^{ 2 } } +3{ z }^{ 2 }\)
\(=\frac { { xy }^{ 2 }-{ x }^{ 3} }{ { x }^{ 2 }{ y }^{ 2 } } +3{ z }^{ 2 }\)
\(=\frac { { y }^{ 2 }-{ x }^{ 2 } }{ x{ y }^{ 2 } } +3{ z }^{ 2 }\)
\(\frac { \partial U }{ \partial z } =0+3y(2z)=6yz\)

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) \(p(X\le 4)=\int _{ -\infty }^{ 4 }{ f(x)dx } \) 
= \(\int _{ -\infty }^{ 0 }{ f(x)dx+\int _{ 0 }^{ 4 }{ f(x)dx } } \) 
= \(0+\int _{ 0 }^{ 4 }{ { ke }^{ -\frac { x }{ 3 } }dx } =k\left[ \frac { { e }^{ -\frac { x }{ 3 } } }{ -\frac { 1 }{ 3 } } \right] _{ 0 }^{ 4 }\) 
= \(\cfrac { 1 }{ 3 } \cfrac { \left[ { e }^{ -\frac { x }{ 3 } } \right] _{ 0 }^{ 4 } }{ -\frac { 1 }{ 3 } } =-\left[ { e }^{ \frac { -4 }{ 3 } }-{ e }^{ o } \right] \) 
= \(-\left[ { e }^{ \frac { -4 }{ 3 } }-1 \right] =1-{ e }^{ \frac { -4 }{ 3 } }\)

Let T represent the temperature of the boiling water and Tm represents the temperature of the kitchen.
By Newton's law of cooling
\(\Rightarrow \int { \frac { dT }{ T-{ T }_{ m } } =K\int { dt } } \)
\(\Rightarrow log(T-{ T }_{ m })=Kt+logC\)
\(\Rightarrow log(T-{ T }_{ m })-logC=Kt\)
\(\Rightarrow log\left( \frac { T-{ T }_{ m } }{ C } \right) =Kt\)
\(\Rightarrow T-{ T }_{ m }={ Ce }^{ Kt } ...(1)\)
when t=0,T=100
\(\therefore 100-{ T }_{ m }={ Ce }^{ 0 }\)
\(\Rightarrow C=100-{ T }_{ m }\)
\(\Rightarrow becomes,\ T-{ T }_{ m }=(100-{ T }_{ m }){ e }^{ Kt }\)
Also when t = 5, T = 80
\(\therefore 80-{ T }_{ m }=(100-{ T }_{ m }){ e }^{ 5K }\)
\(\Rightarrow { e }^{ 5K }=\frac { 80-{ T }_{ m } }{ 100-{ T }_{ m } } ..(2)\)
When t = 10, T = 65
(2) \(\Rightarrow\) 65 - T = (100-T_m)e^10K
= (100-T_m)(e^5K)²
\(=(100-{ T }_{ m }){ \left( \frac { 80-{ T }_{ m } }{ 100-{ T }_{ m } } \right) }^{ 2 }\)
[using(2)]
\(\Rightarrow 65-{ T }_{ m }=\frac { { (80-{ T }_{ m } })^{ 2 } }{ 100-{ T }_{ m } } \)
\(\Rightarrow\) 6500-65T_m-100T_m+T_m² = 6400+T_m²-160T_m
\(\Rightarrow\) 6500-6400 = 165Tm-160Tm
\(\Rightarrow\) 100 = 5Tm
\(\\ \Rightarrow { T }_{ m }=\frac { 100 }{ 5 } ={ 20 }^{ o }C\)
Hence the temperature of the kitchen is 20^oC

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,
\(i=0+c{ e }^{ -\frac { Rt }{ L } }\)
\(\Rightarrow i=c{ e }^{ -\frac { Rt }{ L } }\)

The given differential cquation may be written as
\(\left(1+x+x y^2\right) \frac{d y}{d x}+\left(y+y^3\right)=0  \)
\(\left(1+x+x y^2\right) \frac{d y}{d x}=-\left(y+y^3\right)  \)
\(\left(1+x+x y^2\right)=-1\left(y^2+1\right) \frac{d x}{d y}  \)
\(y\left(y^2+1\right) \frac{d x}{d y}+1+x\left(y^2+1\right)=0\)
Divided by \( y\left(y^2+1\right) ,\) 
\(\frac{d x}{d y}+\frac{1}{y\left(y^2+1\right)}+\frac{x\left(y^2+1\right)}{y\left(y^2+1\right)}   =0  \)
\(\frac{d x}{d y}+\frac{x}{y}  =-\frac{1}{y\left(y^2+1\right)}\)
This is the form of \( \frac{d x}{d y}+\mathrm{Px}=\mathrm{Q}  \) where \( \mathrm{P}=\frac{1}{y} and  \mathrm{Q}=\frac{-1}{y\left(1+y^2\right)}\)
\(\text { I.F }=e^{\int P d y}=e^{\int \frac{1}{y} d y}=e^{\log y}=y\)
So, the solution of the equation is given by
\(x \times \mathrm{I} . \mathrm{F}  =\int(Q \times I . F) d y+c  \)
\(x \times \mathrm{y}  =\int \frac{-1}{y\left(1+y^2\right)} \times y \times d y+c  \)
\(x \mathrm{y}  =\int \frac{-1}{1+y^2} d y+c=-\int \frac{1}{1+y^2} d y+c\)
xy = -tan^-1y + c
xy + tan^-1y = c
Which is the required solution.

Let x and y be the length and breadth of the rectangle.
ஃ Perimeter P = 2x + 2y
\(\Rightarrow 2y=P-2x\Rightarrow y=\frac { P-2x }{ 2 } \) 
Let \(f(x)=Area=xy=x\left( \frac { P-2x }{ 2 } \right) \)
\(f(x)=\frac { Px-{ 2x }^{ 2 } }{ 2 } \)
\(f'(x)=\frac { 1 }{ 2 } \left[ P-4x \right] \)
f'(x) = 0
\(\Rightarrow \frac { 1 }{ 2 } \left[ P-4x \right] =0\)
\(\Rightarrow P=4x\) 
\(\Rightarrow x=\frac { P }{ 4 } \)
∴The critical number is \(\frac { P }{ 4 } \) 
 Now,\(f''\left( \frac { P }{ 4 } \right) =-2<0\)
ஃf(x) is maximum when \(x=\frac { P }{ 4 } \)
When \(x=\frac { P }{ 4 } \)
\(\Rightarrow y=\frac { P-2\left( \frac { P }{ 4 } \right) }{ 2 } =\frac { P-\frac { P }{ 2 } }{ 2 } =\frac { P }{ 4 } \)
\(\therefore x=y=\frac { P }{ 4 } \)
ஃThe rectangle is a square when the area is maximum for a given perimeter.

Given that s(t) = t³ − 6t² + 9t + 1. On differentiating, we get v(t) = 3t² -12t + 9 and a(t) = 6t −12.
(i) The particle is at rest when v(t) = 0 . Therefore, v(t) = 3(t −1)(t − 3) = 0 gives t = 1 and t = 3.
(ii) The particle changes direction when v (t) changes its sign. Now.
if 0 ≤ t < 1 then both (t −1) and (t − 3) < 0 and hence, v(t) > 0.
If 1< t < 3 then (t −1) > 0 and (t − 3) < 0 and hence, v(t) < 0.
If t > 3 then both (t −1) and (t − 3) > 0 and hence, v(t) > 0.
Therefore, the particle changes direction when t = 1 and t = 3.
(iii) The total distance travelled by the particle from time t = 0 to t = 2 is given by,
|s(0) − s(1)| + |s(1) − s(2)| = |1− 5 | + | 5 − 3| = 6 metres.