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

Given u (x, y) = \(\frac { { x }^{ 2 }+{ y }^{ 2 } }{ \sqrt { x+y } } \)
\(u({ \lambda }x,{ \lambda }y)=\frac { { \lambda }^{ 2 }{ x }^{ 2 }+{ { \lambda } }^{ 2 }{ y }^{ 2 } }{ \sqrt { { \lambda }x+{ \lambda }y } } \)
= \(\frac { { { \lambda } }^{ 2 }({ x }^{ 2 }+{ y }^{ 2 }) }{ \sqrt { { \lambda } } (\sqrt { x+y } ) } \)
= \({ { \lambda } }^{ 2-\frac { 1 }{ 2 } }u(x,y)\)
= \({ { \lambda } }^{ \frac { 3 }{ 2 } }u(x,y)\)
∴ u (x, y) is a homogeneous function of degree \(\frac32\)
∴ By Euler's theorem,
\(x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } \) = n.u ≍ \(x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } =\frac { 3 }{ 2 } u\)
Hence, proved.

The possible events of selection are
(i) both balls may be black, or
(ii) one white and one black or
(iii) both are white.
Therefore X is a random variable that take the values,
X (both are black balls) = Rs. 2(30) = Rs. 60
X (one black and one white ball) = Rs. 30 − Rs. 20 = Rs. 10
X (both are white balls) = Rs. 2( − 20) = - Rs. 40
Therefore X takes on the values 60,10, and − 40.

Let \(f(x)={ sin }^{ 2 }x\)
\(f(-x)={ (sin(-x)) }^{ 2 }={ sin }^{ 2 }x=f(x)\)
\(\therefore \int _{ -\frac { \pi }{ 4 } }^{ \frac { \pi }{ 4 } }{ { sin }^{ 2 }xdx } =2\int _{ 0 }^{ \frac { \pi }{ 4 } }{ { sin }^{ 2 }xdx } \)
\(cos\ 2x=1-2{ sin }^{ 2 }x\)
\(2{ sin }^{ 2 }x=1-cos2\)
\({ sin }^{ 2 }x=\frac { 1-cos\quad 2x }{ 2 } \)
\(=2\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \left( \frac { 1-cos2x }{ 2 } \right) dx } \)
\(=\frac { 2 }{ 2 } { \left[ x-\frac { sin2x }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 4 } }\)
\(=\frac { \pi }{ 4 } -\frac { sin\frac { \pi }{ 4 } }{ 2 } -0+\frac { sin0 }{ 2 } \)
\(\\ =\frac { \pi }{ 4 } -\frac { 1 }{ 2 } =\frac { \pi -2 }{ 4 } \)

\(Let\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }\left( \frac { sinx }{ 1+cosx } \right) dx } \)
\(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }\left( \frac { 1+sinx }{ { cos }^{ 2 }\frac { x }{ 2 } } \right) dx } \)
\(\left[ \because cosx=2{ cos }^{ 2 }x-1\Rightarrow 1+cos2x=2{ cos }^{ 2 }x \right] \)
\(=\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }\left( \frac { 1 }{ { cos }^{ 2 }\frac { x }{ 2 } } +\frac { sinx }{ { cos }^{ 2 }\frac { x }{ 2 } } \right) dx } \)
\(=\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }\left( { sec }^{ 2 }\frac { x }{ 2 } +\frac { 2sin\frac { x }{ 2 } cos\frac { x }{ 2 } }{ { cos }^{ 2 }\frac { x }{ 2 } } \right) dx } \)
\(\\ [\because sin2x=2sinx\quad cosx]\)
\(=\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }\left( { sec }^{ 2 }\frac { x }{ 2 } 2tan\frac { x }{ 2 } \right) dx } \)
\(=\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }(f'(x)+f(x)dx } \)
Where \(f(x)=2tan\frac { x }{ 2 } \)
\(=\frac { 1 }{ 2 } .{ e }^{ x }.f(x)={ \left[ \left( \frac { 1 }{ 2 } { e }^{ x }.2tan\frac { x }{ 2 } \right) \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\)
\(={ \left[ { e }^{ x }tan\frac { x }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\)
\(={ e }^{ \frac { \pi }{ 2 } }tan\frac { \pi }{ 4 } -{ e }^{ 0 }tan0={ e }^{ \frac { \pi }{ 2 } }(1)\)
\(\therefore I={ e }^{ \frac { \pi }{ 2 } }\)

f(x) = x³ - 5x + 12, x₀ = 2
f(x_o) = 2³ - 5(2) + 12
= 8 - 10 + 12 = 10
f'(x) = 3x² - 5
⇒ f'(x_o) = 3 (2²) - 5 = 7
∴ L(x) = f(x_o) +f'(x_o) (x - x_o)
= 10 + 7(x - 2)
= 10 + 7x - 14
L(x) = 7x- 4 

Let X be the random variable denotes the total C score is two throws of a die.
Sample space S

II	2	3	3	4	4	4
I
2	4	5	5	6	6	6
3	5	6	6	7	7	7
3	5	6	6	7	7	7
4	6	7	7	8	8	8
4	6	7	7	8	8	8
4	6	7	7	8	8	8

n (S) = 36
X = {4,5,6,7,8}
From the sample space

Values of random variable	4	5	6	7	8	Total
No of points in inverse image	1	4	10	12	9	36

This is an indeterminate of the form (0×∞). To evaluate this limit, we first simplify and bring it to the form \((\frac{\infty}{\infty})\) and apply L’Hôpital Rule.
\(\underset{x\rightarrow 0^{+}}{lim}xlog x= \underset{x\rightarrow 0^{+}}{(\frac{log x}{\frac{1}{x}})}\)
\(\underset{x\rightarrow 0^{+}}{lim}(\frac{\frac{1}{x}}{-\frac{1}{x^{2}}})=\underset{x\rightarrow 0^{+}}{lim}(-x)=0.\)

Given f (x) = (x − 2)(x − 7), x ∈ [3,11]
a) f(x) is continuous in [3, 11]
b) f(x) is differentiable in (3, 11)
c) f(11) = (11-2)(11-7)
= (9) (4) = 36
f(3) = (3 - 2)(3 - 7)
= (1)(-4) = - 4
∴ By Lagrange's mean value theorem, there exists c ∈ [3,11] such that f'(c) = \(\frac { f(b)-f(a) }{ b-a } \)
\(\left[ \begin{matrix} f(x)\begin{matrix} = & (x \end{matrix}- & 2) & \begin{matrix} (x & - \end{matrix}7) \\ \begin{matrix} = & { x }^{ 2 } \end{matrix}- & 7x & -2x\begin{matrix} + & 14 \end{matrix} \\ \begin{matrix} = & { x }^{ 2 } \end{matrix}- & 9x & +\begin{matrix} 14 & \end{matrix} \end{matrix} \right] \)
⇒ 2c - 9 = \(\frac{36+4}{11-3}\)
⇒ 3c - 9 = \(\frac{40}{8}\) = 5
⇒ 2c = 14
⇒ c = 7 ∈ [3 , 11]

General equation of a straight line in a plane is ax + by = 1  .....(1)
Since, the lines are non - horizontal, a ≠ 0 
Hence, differentiating with respect to y, equation (1)
a \(\\ \frac { dx }{ dy }+b =0\)
Differentiating again with respect to 'y'; we get,
\((a) \ \frac { { d }^{ 2 }x }{ d{ y }^{ 2 } } =0\Rightarrow \frac { { d }^{ 2 }x }{ d{ y }^{ 2 } } =0\quad [\because a\neq 0]\)
This is the differential equation of all non-horizontal lines in a plane.

p	q	r	~ q	~q V r	p➝(¬qVr)	~p	~pV(~qVr)
T	T	T	F	T	T	F	T
T	T	F	F	F	F	F	F
T	F	T	T	T	T	F	T
T	F	F	T	T	T	F	T
F	T	T	F	T	T	T	T
F	T	F	F	F	T	T	T
F	F	T	T	T	T	T	T
F	F	F	T	T	T	T	T

From the table, it is clear that the column of p➝(¬q V ~r) and ~pV(~q V r) are identical
∴  p➝(¬q V ~r) ≡ ~pV(~q V r) 
Hence proved.

given A = {Q\{1}}
A is defined on A by x*y = x+y-xy
Let x,y ≠ 1
∴ x*y = x + y - xy
Now to prove that x + y - xy ≠ 1
Let us assume that x + y - xy = 1
x+y-xy-1 = 0
(x-1)-y(x-1) = 0
(x-1)(1-y) = 0
x =1 or y = 1 which is a false [∵x, y ≠ 1]
∴ is a binary operation on A.
Commutative property:
Let x,y ∈A ⇒ x, y≠1
∴x*y = x+y-xy
and y*x = y+x-yx
⇒x+y = y*x∀x, y∈A
A has commutative property under *
Associative property:
Let x,y,z ∊A ⇒x,y,z≠1
Consider (x*y)*z = (x+y-xy)*z
= x +y~xy +z- (x +y-xy)z
= x +y-xy+ z-xz- yz + xyz
= x +y+z-xy-yz-zx +xyz ...(1)
= x +y+z - yz - x (y + z - yz)
= x +y +z-yz-zy-xz +xyz ..(2)
From (1) & (2), (x*y)*z = x*(y*z)
A has associative property under *.

The lines 5x − 2y = 15, x + y + 4 = 0 intersect at (1, −5). The line 5x − 2y =15 meets the x-axis at (3, 0). The line x + y + 4 = 0 meets the x-axis at (−4, 0). The required area is shaded. It lies below the x-axis. It can be computed either by considering vertical strips or horizontal strips.
When we do by vertical strips, the region has to be divided into two sub-regions by the line x = 1. Then, we get

\(A=\left| \int _{ -4 }^{ 1 }{ ydx } \right| +\left| \int _{ 1 }^{ 3 }{ ydx } \right| \)
\(=\left| \int _{ -4 }^{ 1 }{ (-4-x)dx } \right| +\left| \int _{ 1 }^{ 3 }{ \left( \frac { 5x-15 }{ 2 } \right) dx } \right| \)
\(=\left| { \left( -4x-\frac { { x }^{ 2 } }{ 2 } \right) }_{ -4 }^{ 1 } \right| +\left| { \left( \frac { { 5x }^{ 2 } }{ 4 } -\frac { 15x }{ 2 } \right) }_{ 1 }^{ 3 } \right| \)
\(=\left| \left( -\frac { 9 }{ 2 } \right) -\left( 8 \right) \right| +\left| \left( -\frac { 45 }{ 4 } \right) -\left( -\frac { 25 }{ 4 } \right) \right| \)
\(=\frac { 25 }{ 2 } +5\)
\(=\frac { 35 }{ 2 } \)
When we do by horizontal strips, there is no need to subdivide the region. In this case, the area is bounded on the right by the line 5x − 2y = 15 and on the left by x + y + 4 = 0. So, we get
\(A=\int _{ -5 }^{ 0 }{ [{ x }_{ R }-{ x }_{ L }]dy } =\int _{ -5 }^{ 0 }{ \left[ \frac { 15+2y }{ 5 } -(-4-y) \right] dy } \)
\(=\int _{ -5 }^{ 0 }{ \left[ 7+\frac { 7y }{ 5 } \right] dy={ \left[ 7y+\frac { { 7y }^{ 2 } }{ 10 } \right] }_{ -5 }^{ 0 } } \)
\(\\ =0-\left[ -35+\frac { 35 }{ 2 } \right] =\frac { 35 }{ 2 } \)

By definition the cumulative distribution function for discrete random variable is
\(F(x)P\left( X\le x \right) =\underset { { x }_{ 1 }\le x }{ \Sigma } P(X={ x }_{ 1 })\) 
\(P(X<1)=0\) for -∞
\(F(1)=P\left( X\le 1 \right) =\underset { { x }_{ 1 }\le x }{ \Sigma } P(X={ x }_{ i })=\sum _{ -\infty }^{ 1 }{ P(X=x) } =P\left( X<1 \right) +P\left( X=1 \right) =0+\frac { 1 }{ 12 } =\frac { 1 }{ 12 } \)
\(F(2)=P\left( X\le 2 \right) =\sum _{ -\infty }^{ 2 }{ P\left( X=x \right) } =P\left( X\le 1 \right) +P\left( X=1 \right) +P\left( X=2 \right) \) 
= \(0+\frac { 1 }{ 12 } +\frac { 5 }{ 12 } =\frac { 1 }{ 2 } \) 
\(F(3)=P\left( X\le 3 \right) =\sum _{ -\infty }^{ 3 }{ P(X=x) } =P\left( X<1 \right) +P\left( X=1 \right) +P\left( X=2 \right) +P\left( X=3 \right) \)
= \(0+\frac { 1 }{ 2 } +\frac { 5 }{ 12 } +\frac { 5 }{ 12 } =\frac { 11 }{ 12 } \) 
\(F(4)=P\left( X\le 4 \right) =\sum _{ -\infty }^{ 4 }{ P\left( X=x \right) } =P\left( X=1 \right) +P\left( X=2 \right) +P\left( X=3 \right) +P(X=4)\)
= \(0+\frac { 1 }{ 12 } +\frac { 5 }{ 12 } +\frac { 5 }{ 12 } +\frac { 1 }{ 12 } =1\)
\(F(x)= \begin{cases}0, & -\infty

(ii) \(P(X\le 3)=F(3)\frac { 11 }{ 12 } \)
(iii) \(P(X\ge 2)=1-P\left( X<2 \right) =1-P(X\le 1)=1-F(1)=1-\frac { 1 }{ 12 } =\frac { 11 }{ 12 } \)

Let p be the probability of the useful life hours of a fluorescent light.
n = 12
P = 0.9
q = 1-p = 0.1
P(X= x)= ⁿC_x p^x q^n-x, x = 0, 1,2, .., n
(i) Exactly 10
P(X = 10) = ¹²C₁₀(0.9)¹⁰(1 - 0.9)²
= ¹²C₁₀(0.9)¹⁰ (0.1)²
(ii) Atleast 11
P(X≥11) = R(X = 11) + P(X = 12)
= ¹²C₁₁(0.9)¹¹ (0.1)¹ + ¹²C₁₂(0.9)¹²(0.1)⁶
= ¹²C₁ (0.9)¹¹ (0.1) + (0.9)¹²
= ¹²(0.9)¹¹ (0.1) + (0.9)¹²
= (0.9)¹¹ ((12)(0.1) + 0.9)
= (0.9)¹¹ (1.2 + 0.9)
= (0.9)¹¹ (2.1)
(ii) Atleast 2 will not have a useful
P(X,10) = 1 -P(X > 10)
= 1 - [P(X = 11) + P(X = 12)]
= 1-(2.1) (0.9)¹¹

Here again we shall use the tree diagram to calculate \(\frac { \partial g }{ \partial r } ,\frac { \partial g }{ \partial s } \) 
Hence we find \(\frac { \partial g }{ \partial x } \) = 2x, \(\frac { \partial g }{ \partial y } \) = 2, \(\frac { \partial x }{ \partial y } \) = 2, \(\frac { \partial x }{ \partial s } \) = -1, \(\frac { \partial y }{ \partial r } \) =2r, and \(\frac { \partial y }{ \partial s } \) = 2.
Now, \(\frac { \partial g }{ \partial r } \) = \(\frac { \partial g }{ \partial x } \frac { \partial x }{ \partial r } +\frac { \partial g }{ \partial y } \frac { \partial y }{ \partial r } \) = 2x(2) + 2(2r) 12r - 4s.
also,  \(\frac { \partial g }{ \partial s } \) = \(\frac { \partial g }{ \partial x } \frac { \partial x }{ \partial s } +\frac { \partial g }{ \partial y } \frac { \partial y }{ \partial s } \) = 2x(-1) +(2)2 = 2s - 4r + 4.

Given w(x, y, z) = \(\frac { 1 }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } } }\)
= (x² + y² + z²) -^\(\frac12\)
\(\frac { \partial w }{ \partial x } =\frac { -1 }{ 2 } ({ x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ -\frac { 3 }{ 2 } }(2x)\)
= (-x² + y² + z²) -^\(\frac12\)
\(\frac { { \partial }^{ 2 }w }{ \partial { x }^{ 2 } } =\frac { \partial }{ \partial x } \left( \frac { \partial w }{ \partial x } \right) \)
= -[x\(\left( \frac { -3 }{ 2 } \right) \)( x² + y² +z²)^\(-\frac32\)
\((\not 2 x)+\left(x^{2}+y^{2}+z^{2}\right)^{\frac{3}{2}}\)
= (x² + y² + z²)^-\(\frac52\) [-3x² + x² +y + z²]
= - (x² + y² + z²)^-\(\frac52\) [y² + z² - 2x²] ....(1)
\(\frac { \partial w }{ \partial y } =\frac { -1 }{ 2 } ({ x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ -\frac { 3 }{ 2 } }(2y)\)
= -y(x² +y² + z²)-^\(\frac32\) 
\(\frac { { \partial }^{ 2 }w }{ \partial { y }^{ 2 } } =\frac { \partial }{ \partial y } \left( \frac { \partial w }{ \partial y } \right) \)
\(=-\left[y\left(\frac{-3}{\not 2}\right)\left(x^{2}+y^{2}+z^{2}\right)^{\frac{5}{2}}(\not 2 y)+\left(x^{2}+y^{2}+z^{2}\right)^{\frac{3}{2}}(1)\right]\)
= -(x² + y² + z²)-^{\(\frac52 \)} 
= -(x² + y² + z²)-^{\(\frac52 \)} [3y² + x² + y² + z²]
= -(x² + y² + z²)-^{\(\frac52 \)} [x² + y² + 2z²] ....(2)
Now \(\frac { \partial w }{ \partial z } =\frac { -1 }{ 2 } ({ x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ -\frac { 3 }{ 2 } }(2z)\)
= -z(x² + y² + z²)-^{\(\frac32 \) 
∴} \(\frac { { \partial }^{ 2 }w }{ \partial { z }^{ 2 } } \) = -(x² + y² + z²)-^{\(\frac52 \)} [x² + y² - 2z²] ...(3)
(1)+(2)+(3)⟶
^∴ \(\frac { { \partial }^{ 2 }w }{ \partial { x }^{ 2 } } +\frac { { \partial }^{ 2 }w }{ \partial { y }^{ 2 } } +\frac { { \partial }^{ 2 }w }{ \partial { y }^{ 2 } } \) ⁼ -(x² + y² + z²)-^{\(\frac52 \)} [y² + z² - 2x² + x² + z² - 2y + x²+ y-2z²]
= -(x² + y² + z²)-^{\(\frac52 \)}(0) = 0
Hence proved 

Let I = \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ (\sqrt { tan\ x } +\sqrt { cot\ x } )dx } \) Then we get
\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( \sqrt { \frac { sinx }{ cosx } } +\sqrt { \frac { cosx }{ sinx } } \right) dx } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { sinx+cosx }{ \sqrt { sinxcosx } } dx } =\sqrt { 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { sinx+cosx }{ \sqrt { 2sinxcosx } } dx } \)
\(=\sqrt { 2 } \int _{ 0 }^{ \frac { \pi }{2 } }{ \frac { (sinx+cosx)dx }{ \sqrt { 1-sinx-cosx{ ) }^{ 2 } } } } \)
Put u = sin x − cos x 
Then, du = (cos x + sin x)dx .
When x = 0, u = −1
When x =\(\frac{\pi}{2}\), u = 1
\(\therefore I=\sqrt { 2 } \int _{ -1 }^{ 1 }{ \frac { du }{ \sqrt { 1-{ u }^{ 2 } } } =\sqrt { 2 } { [{ sin }^{ -1 }u] }_{ -1 }^{ 1 }=\sqrt { 2 } \left[ { sin }^{ -1 }(1)-{ sin }^{ -1 }(-1)) \right] } =\pi \sqrt { 2 } \)

Given

The random variable X take the values -1, 0, 1, 2, 3
For a discrete random variable X, we have
f(x) = p(X = x)
∴ f(-1) = p(X= -1) = F(-1) -F(0)
= 0.15-0 = 0.15
f(0) = p(X = 0) = F(0)-F(-1)
= 0.35-0.15 = 0.20
f(1) = p(X = 1) = F(1)-F(0)
= 0.60-0.35 = 0.25
f(2) = p(X=2) = F(2)-F(1)
= 0.85-0.60 = 0.25
f(3) = p(X = 3) = F = (3)-F(2)
= 1-0.85 = 0.15
(i) ஃThe probability mass function is

x	-1	0	1	2	3
f(x)	0.15	0.20	0.25	0.25	0.15

(ii) p(X<1)
= p(X = -1) + p(X = 0)
= 0.15 + 0.20 = 0.35
(iii) p(X ≥ 2)
= p(X = 2) + p(X = 3)
= 0.25 + 0.15
= 0.40

The given function is defined and is differentiable at all  \(x\in (-\infty, \infty) \). As
\(f(x)=\frac{1}{1+x^{2}}\).
We have \(f'(x)=-\frac{2x}{(1+x^{2})^{2}}\)
The stationary points are given by \(-\frac{2x}{(1+x^{2})^{2}}=0\) that is x = 0
Hence the intervals of monotonicity are \((-\infty,0)\) and \((0,\infty)\)
On the interval \((-\infty,0)\) the function strictly increases because f'(x) > 0 in that interval.
The function f(x) strictly decreases in the interval \((0,\infty)\) because f'(x) < 0 in that interval.
Since f′(x) changes from positive to negative when passing through x = 0, the first derivative test tells us there is local maximum at x = 0 and the local maximum value is f (0) = 1.

Let A be the number of bacteria at any time t
Given \(\frac { dA }{ A } \alpha A\)
\(\Rightarrow \frac { dA }{ dt } =kA\)
\(\Rightarrow \frac { dA }{ dt } =kdt\)
\(\Rightarrow \int { \frac { dA }{ dt } } =k\int { \frac { dt }{ t } } \)
\(\Rightarrow log\left( \frac { A }{ c } \right) =kt\)
\(\Rightarrow log\left( \frac { A }{ c } \right) =kt\)
\(\Rightarrow \frac { A }{ c } ={ e }^{ kt }\)
\(\Rightarrow A=c.{ e }^{ kt }\) ...(1)
Initially when t = 0, assume that A = A₀
\(\therefore\) becomes, = ce₀ \(\Rightarrow\) c = A₀ 
\(\therefore\) A = A0ekt  ...(2)
Given when t = 5, A = 3A₀
3A₀ = A₀e^5k \(\Rightarrow\) 3 = 35k 
when t = 10, 2 becomes,
A = A₀e^10k
= A₀(e^5k)2 = A₀(3)²
\(\Rightarrow\) A = 9A₀
Hence, the amount of bacteria will be present after 10 hours is 9 times the initial amount.

The given linear differential euation is of the form
\(\frac{d y}{d x}+P y=Q  \)
 where \(\mathrm{P}=\frac{1}{(1-x) \sqrt{x}} ; \mathrm{Q}=1-\sqrt{x}  \)
Take \(\sqrt{x}=\mathrm{t} \Rightarrow \mathrm{t}^2=x \)
\( x^{1 / 2}=\mathrm{t}  \)
\( \frac{1}{2} x^{1 / 2-1} \mathrm{~d} x=\mathrm{dt} \Rightarrow \frac{1}{2} x^{-1 / 2} d x=\mathrm{dt} \)
\( \frac{1}{2 x^{1 / 2}} d x=\mathrm{dt} \Rightarrow \frac{1}{2 \sqrt{x}} d x=\mathrm{dt}  \)
\( \frac{d x}{\sqrt{x}}=2 \mathrm{dt} \)
\( \text { I.F }=e^{\int P d x}=e^{\int \frac{1}{(1-x) \sqrt{x}} d x}  \)
\( =e^{\int \frac{1}{1-t^{t^2} 2 d t}}=e^{\int \frac{2 d t}{1-1^2}} \)
\(\because \int \frac{d x}{a^2-x^2}=\frac{1}{2 x} \log \left|\frac{a+x}{a-x}\right|  \) 
\(\text { Here } a=1 ; x=t\)
\(=e^{\log \left(\frac{1+1}{1-t}\right)} \)
\(\mathrm{I} . \mathrm{F}=\frac{1+t}{1-t} \)
\({ I.F }=\frac{1+\sqrt{x}}{1-\sqrt{x}} \)
The solution is y\(\times \mathrm{I} . \mathrm{F}=\int \mathrm{Q} \times \mathrm{I}.Fd x+c\)
\( y\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)=\int(1-\sqrt{x}) \frac{1+\sqrt{x}}{1-\sqrt{x}} d x+c  \)
\( =\int(1+\sqrt{x}) d x+c  \)
\(  y\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)=x+\frac{x^{3 / 2}}{3 / 2}+c \)
\( y\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)=x+\frac{2}{3} x^{3 / 2}+c \)
\( y\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)=x+\frac{2}{3} x \sqrt{x}+c \quad \because x^{3 / 2}=x \sqrt{x} \)
Which is the required solution

Given slope curve
\(\Rightarrow \frac { dy }{ dx } =\frac { y-1 }{ { x }^{ 2 }+x } \) ..... (1)
\(\Rightarrow \frac { dy }{ y-1 } =\frac { dx }{ { x }^{ 2 }+x } \)
\( \Rightarrow \frac{d y}{y-1}=\frac{d x}{x^2+x}=\frac{d x}{x^2+2\left(\frac{x}{2}\right)+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2} \)
\(ie) \frac{d y}{y-1}=\frac{d x}{\left(x+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2} \)
Integrating on both sides, we get
 \( \int \frac{d y}{\log (y-1)}=\int \frac{d x}{\left(x+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2} \\  \text { ie) } \log (y-1)=\left(\frac{1}{2\left(\frac{1}{2}\right)}\right) \log \left[\frac{\left(x+\frac{1}{2}\right)-\left(\frac{1}{2}\right)}{\left(x+\frac{1}{2}\right)+\left(\frac{1}{2}\right)}\right]+\log C \\  \log (y-1)=\log \left(\frac{x}{x+1}\right)+\log C\\   ie)  (y-1)=\frac{C x}{x+1}  \) 
\(\Rightarrow log(y-1)=log\left( \frac { cx }{ x+1 } \right) \)
\(\Rightarrow y-1=\frac { cx }{ x+1 } \)
Since the curve passes through (1, 0) we get,
\(0-1=\frac { c }{ 2 } \Rightarrow c=-2\)
\(\Rightarrow y-1=\frac { -2x }{ x+1 } \)
\(\Rightarrow y=1-\frac { -2x }{ x+1 } \)
\(\Rightarrow y=\frac { x+1-2x }{ x+1 } =\frac { 1-x }{ x+1 } \)
\(\therefore y=\frac { 1-x }{ 1+x } \)

Let AB be the position of the ladder at any time t such that OA = x and OB = y
Then OA² + OB² = AB²
⇒ x² + y² = 17²
Given \(\frac { dx }{ dt } \) = 5 and x = 8
When x = 8, 8² + y² = 17²
 ⇒ y² = 289 - 64 = 225
 ⇒ y = 15
Differentiating (1) with respect to 't' we get,
\(2x\frac { dx }{ dt } +2y\frac { dy }{ dt } =0\)
⇒ 8(5) + 15 \(\frac { dy }{ dt } \) = 0 [∵ x = 8, \(\frac { dx }{ dt } \) = 15, y = 15]
⇒ 40 + 15\(\frac { dy }{ dt } \) = 0
⇒ \(\frac { dy }{ dt } =\frac { -40 }{ 15 } =\frac { -8 }{ 3 } \) m/sec
∴ The rate of top of the ladder moving down the wall is \(\frac{-8}{3}\) m/sec
(ii) The ladder, the wall and the floor forms a right angled triangle.
 Area = \(\frac12\)xy
Differentiating with respect to 't' we get,
\(\frac { dA }{ dt } =\frac { 1 }{ 2 } \left[ x\frac { dy }{ dx } +y\frac { dx }{ dt } \right] \)
\(=\frac { 1 }{ 2 } \left[ 8\left( -\frac { 8 }{ 3 } \right) +15(5) \right] \)
= \(\frac { 1 }{ 2 } \left[ \frac { -64 }{ 3 } +75 \right] =\frac { 1 }{ 2 } \)
= \(\frac { 1 }{ 2 } \left[ \frac { -64+225 }{ 3 } \right] =\frac { 1 }{ 2 } \left( \frac { 161 }{ 3 } \right) \)
= \(\frac { dA }{ dt } \) = 26.83 sq.m/sec

Given * on A is commutative
Given b * a = c ⇒ a * b = c
Given c * a = a ⇒ a * c = a
Given b * c = a ⇒ c * b = a
Hence

*	a	b	c
a	b	c	a
b	c	b	a
c	a	a	c

In this problem a ∗ b is in the quotient form. Since the division by 0 is undefined, the denominator b -1 must be nonzero.
It is clear that b −1 = 0 if b = 1. As 1∈Q, ∗ is not a binary operation on the whole of Q. However it can be found that by omitting 1 from Q, the output a ∗b exists in Q\{1}. Hence ∗ is a binary operation on Q\{1}.

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
\(P(X=3)=\left( \begin{matrix} 6 \\ 3 \end{matrix} \right) \left( \cfrac { 1 }{ 3 } \right) ^{ 3 }\left( 1-p \right) ^{ 6-3 }\) 
= \(\left( \begin{matrix} 6 \\ 3 \end{matrix} \right) \left( \cfrac { 1 }{ 3 } \right) ^{ 3 }\left( \cfrac { 2 }{ 3 } \right) ^{ 2 }\) 

\(P(X=3)=\frac { 160 }{ 729 } \) 

Given \(f(x)=\begin{cases} \begin{matrix} { 3e }^{ -3x } & x>0 \end{matrix} \\ \begin{matrix} 0 & elsewhere \end{matrix} \end{cases}\)
\(E(X)=\int _{ 0 }^{ \infty }{ x.f(x)dx } =\int _{ 0 }^{ \infty }{ x.3.{ e }^{ -3x }dx } \) 
= \(3\int _{ 0 }^{ \infty }{ x.{ e }^{ -3x }dx } \left[ \therefore \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -ax }dx=\cfrac { n! }{ { a }^{ n+1 } } } \right] \) 
= \(3\times \frac { 1! }{ { 3 }^{ 2 } } =\frac { 3 }{ 9 } =\frac { 1 }{ 3 } \)
ஃ Expected life of the electronic equipment is = \(\frac { 1 }{ 3 } \) 

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

Let f(x) = e^-|-x| = e^-|x| = f(x)
So f (x) is an even function.
Hence, \(\int ^{log 2}_{-log 2} e ^{-|x|}\)dx = 2\(\int ^{log 2}_{0} e ^{-|x|}\)dx = 2\(\int ^{log 2}_{0} e ^{-x}\) dx
= 2(-e^-x)\(^{log2}_{0}\) = 2 (-e^-log2 + e⁰) = 2 \((-e ^{log \frac{1}{2}} + 1)\)
= 2\((-\frac {1}{2}+1)=1\).

x = 2 and dx = 0.1
Taking differentials,
df = (2x + 3) dx
Whenx = 2, dx = 0.1
df = (2(2) + 3)(0.1) = 7(0.1) = 0.7

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

Given y = x⁴ + 2x² - x
\(\frac { dy }{ dx } \) = 4x³ + 4x - 1
Slope of the tangent at x = 1 is
m = \(\left( \frac { dy }{ dx } \right) \)(x = 1)
= 4(1)³+ 4 (1) - 1
= 4+4-1 = 7
∴ m = 7

The given function is y mx +\(\frac{7}{m}\), where m is an arbitrary constant ....(1)
Differentiating both sides of equation (1) with respect to x, we get y' = m.
Substituting the values of y' and y in the given differential equation
we get xy'\(\frac{1}{y'}\)-y = xm +\(\frac{7}{m}\)- mx -\(\frac{7}{m}\) = 0
Therefore, the given function is a solution of the differential equation xy' + 7\(\frac{1}{y'}\) - y = 0