\( \mathrm{u} =\sin ^{-1}\left(\frac{x-y}{\sqrt{x}+\sqrt{y}}\right) \)
\(\sin \mathrm{u} =\left(\frac{x-y}{\sqrt{x}+\sqrt{y}}\right)=\mathrm{f}(\mathrm{x}, \mathrm{y}) \)
\( f(\lambda x, \lambda y) =\frac{\lambda x-\lambda y}{\sqrt{\lambda x}+\sqrt{\lambda y}} \)
\( =\frac{\lambda(x-y)}{\sqrt{\lambda}(\sqrt{x}+\sqrt{y})} \)
\( =\lambda^{1 / 2} \frac{(x-y)}{(\sqrt{x}+\sqrt{y})} \)
f is a homogeneous function of degree \(\frac{1}{2}\) By Eulers Theorem
\( \mathrm{x} \frac{\partial f}{\partial x}+\mathrm{y} \frac{\partial u}{\partial y} =\frac{1}{2} \mathrm{f} \)
\(\mathrm{x} \frac{\partial}{\partial x} \sin u+\mathrm{y} \frac{\partial}{\partial y} \sin \mathrm{u} =\frac{1}{2} \sin u \)
\(\mathrm{x} \cos \mathrm{u} \frac{\partial u}{\partial x}+\mathrm{y} \cos \mathrm{u} \frac{\partial u}{\partial y} =\frac{1}{2} \sin \mathrm{u} \)
\(\mathrm{x} \frac{\partial u}{\partial x}+\mathrm{y} \frac{\partial u}{\partial y} =\frac{1}{2} \frac{\sin u}{\cos u} \)
\(\mathrm{x} \frac{\partial u}{\partial x}+\mathrm{y} \frac{\partial u}{\partial y} =\frac{1}{2} \tan \mathrm{u}\)

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.

It is known that Z₅ = {[0], [1], [2], [3], [4]}. The table corresponding to addition modulo 5 is as follows: We take reminders {0,1,2,3,4} to represent the classes {[0], [1], [2], [3], [4]}.

+5	0	1	2	3	4
0	0	1	2	3	4
1	1	2	3	4	0
2	2	3	4	0	1
3	3	4	0	1	2
4	4	0	1	2	3

(i) Since each box in the table is filled by exactly one element of Z₅, the output a +₅ b is unique and hence +₅ is a binary operation.
(ii) The entries are symmetrically placed with respect to the main diagonal. So +₅ has commutative property
(iii) The table cannot be used directly for the verification of the associative property. So it is to be verified as usual
For instance, (2+₅3)+₅ 4 = 0+₅ 4 = 4(mod 5)
and 2+₅(3+₅4) = 2 +₅ 2 = 4(mod5)
Hence (2+₅3)+₅4 = 2+₅(3+₅4)
Proceeding like this one can verify this for all possible triples and ultimately it can be shown that +₅ is associative
(iv) The row headed by 0 and the column headed by 0 are identical. Hence the identity element is 0.
(v) The existence of inverse is guaranteed provided the identity 0 exists in each row and each column. From Table, it is clear that this property is true in this case. The method of finding the inverse of any one of the elements of Z₅, say 2 is outlined below.
First find the position of the identity element 0 in the III row headed by 2. Move horizontally along the III row and after reaching 0, move vertically above 0 in the IV column, because 0 is in the III row and IV column. The element reached at the topmost position of IV column is 3. This element 3 is nothing but the inverse of 2, because, 2+₅ 5+ = 0 (mod5). In this way, the inverse of each and every element of Z₅ can be obtained. Note that the inverse of 0 is 0, that of 1 is 4, that of 2 is 3, that of 3 is 2, and, that of 4 is 1.

W(x, y, z) = xy + yz + zx, x =u -v, y = uv, z = u + v; y = uv; z = u
\(\frac { \partial W }{ \partial x } \) = y + z; \(\frac { \partial W }{ \partial y } \) = x + z
∴ \(\frac { \partial W }{ \partial x } \) = uv + u +  v;
\(\frac { \partial W }{ \partial y} \) = u - v + u + v;
\(\frac { \partial W }{ \partial z} \) = uv +  u - v
\(\frac { dx }{ du } =1;\frac { dy }{ du } =v;\frac { dz }{ du } =1\)
\(\frac { dx }{ dv } =1;\frac { dy }{ dv } =v;\frac { dz }{ dv} =1\)
By chain rule
\(\frac { \partial W }{ \partial u } =\frac { \partial w }{ \partial x } .\frac { dx }{ du } +\frac { \partial w }{ \partial y } .\frac { dy }{ du } +\frac { \partial w }{ \partial z } .\frac { dz }{ du } \)
= (uv +u +v) (1) +2u (v) + (uv +u - v)(1)

\(\frac { \partial W }{ \partial u } \) = 4uv + 2u = 12u (2v + 1)
\({ \left( \frac { \partial W }{ \partial u } \right) }_{ \left( \frac { 1 }{ 2 } ,1 \right) }\) = 2 x \(\frac12\) (2+ 1) = 1(2+ 1) = 3
= (uv + u + v) (-1) + (2u) (u) + (uv + u - v)(1)

= 2u² - 2v = 2 (u² - v)
∴  \({ \left( \frac { \partial W }{ \partial v } \right) }_{ \left( \frac { 1 }{ 2 } ,1 \right) }\) = \(2\left( \frac { 1 }{ 4 } -1 \right) =2\left( -\frac { 3 }{ 4 } \right) =-\frac { 3 }{ 2 } \)

\({ f }_{ y }=\frac { 1 }{ 1+\frac { { x }^{ 2 } }{ { y }^{ 2 } } } \left( \frac { -x }{ { y }^{ 2 } } \right) =\frac { -\frac { x }{ { y }^{ 2 } } }{ \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { { x }{ x }^{ 2 } } } } \)
= \(\frac { -x }{ { x }^{ 2 }+{ y }^{ 2 } } \)
\({ f }_{ x }=\frac { 1 }{ 1+\frac { { x }^{ 2 } }{ { { y }^{ 2 } } } } \left( \frac { 1 }{ y } \right) =\frac { \frac { 1 }{ y } }{ \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { { y }^{ 2 } } } } \)
\(=\frac { y }{ { { x }^{ 2 }+{ y }^{ 2 } } } \)
\({ f }_{ xy }=\frac { \partial }{ \partial x } ({ f }_{ y })\)
\({ f }_{ xy }=-\left[ \frac { { (x }^{ 2 }+{ y }^{ 2 })(1)-x{ (2x) } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=-\left[ \frac { { x }^{ 2 }+{ y }^{ 2 }-2{ x }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=-\left[ \frac { { y }^{ 2 }-{ x }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=\frac { x^{ 2 }-{ y }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \) ...(1)
\({ f }_{ xy }=\frac { \partial }{ \partial y } ({ f }_{ x })\)
\({ f }_{ xy }=\frac { ({ x }^{ 2 }+{ y }^{ 2 })(1)-y(2y) }{ ({ x }^{ 2 }+{ { y }^{ 2 }) }^{ 2 } } \)
\(=\frac { { x }^{ 2 }+{ y }^{ 2 }-2{ y }^{ 2 } }{ ({ x }^{ 2 }+{ { y }^{ 2 }) }^{ 2 } } \)
\(=\frac { { x }^{ 2 }-{ y }^{ 2 } }{ ({ x }^{ 2 }+{ { y }^{ 2 }) }^{ 2 } } \)  ...(2)
∴ From (1) and (2), f_xy = f_yz

This is an indeterminate of the form \(1^{\infty}\).
Let \(g(x)=(1+x)^{\frac{1}{x}}\). Taking the logarithm, we get
\(log \ g(x)=\frac{log(1+x)}{x}\)
\(\underset{x\rightarrow 0^{+}}{lim} log (g(x))=\underset{x\rightarrow 0^{+}}{lim}(\frac{log(1+x)}{x})\) \((\frac{0}{0})\) 
=\(\underset{x\rightarrow0^{+}}{lim}(\frac{\frac{1}{1+x}}{1})\)  (by 1’Hôpital Rule)
= 1.
But, \(\underset{x\rightarrow0^{+}}{lim}log g(x)=log(\underset{x\rightarrow 0^{+}}{lim} g(x))\)
Therefore, log\((\underset{x\rightarrow 0^{+}}{lim} g(x))=1\).
Hence by exponentiating, we get, \(\underset{x\rightarrow 0^{+}}{lim}g(x)=e.\)

Given AC = b and CD = a
Let hand r be the height and radius of the cylinder
ஃ AB = AC - BC = b - h
In the diagram

ΔABE - ΔACD
\(\therefore \frac { AB }{ BE } =\frac { AC }{ CD } \) 
\(\Rightarrow \frac { b-h }{ r } =\frac { b }{ a } \)
 \(\Rightarrow r=\frac { a\left( b-h \right) }{ b } \)
Volume of the cylinder = \({ \pi r }^{ 2 }h\)
= \(\pi \left( \frac { a }{ b } \left( b-b \right) ^{ 2 }h \right) \) 
= \(\pi \frac { { a }^{ 2 } }{ { b }^{ 2 } } \left( b-h \right) ^{ 2 }h\)
= \(\pi \frac { { a }^{ 2 } }{ { b }^{ 2 } } \left( { b }^{ 2 }+{ h }^{ 2 }-2bh \right) h\)
 \(V=\frac { { \pi a }^{ 2 } }{ { b }^{ 2 } } \left( { b }^{ 2 }{ h }^{ 2 }+{ h }^{ 3 } \right) \)
Differentiating with respect to 'h',
\(\frac { dV }{ dh } =\frac { { a }^{ 2 } }{ { b }^{ 2 } } \pi \left( { b }^{ 2 }+{ 3h }^{ 2 }-4bh \right) \)
\(\frac { dV }{ dh } =0\Rightarrow { b }^{ 2 }-4bh+{ 3j }^{ 2 }=0\)
\(\Rightarrow (b-3h)+(b-h)=0\Rightarrow b-3h=0\quad \)or b = h
\(\Rightarrow h=\frac { b }{ 3 } asb\neq h\) 
\(\frac { dV }{ { dh }^{ 2 } } =\frac { { a }^{ 2 } }{ { b }^{ 2 } } \pi \left( 6\left( \frac { b }{ 3 } \right) -4b \right) \)
\(\left( \frac { { d }^{ 2 }V }{ { dh }^{ 2 } } \right) _{ at\ h=\frac { b }{ 3 } }=\frac { { a }^{ 2 } }{ { b }^{ 2 } } \pi \left( 6\left( \frac { b }{ 3 } \right) -4b \right) \)
= \(\frac { { a }^{ 2 } }{ { b }^{ 2 } } \pi (-2b)<0\) 
ஃ V is maximum at \(h=\frac { b }{ 3 } \)
ஃ Maximum volume of the cylinder
\(V=\frac { { \pi a }^{ 2 } }{ { b }^{ 2 } } \left( b-h \right) ^{ 2 }h\) 
= \(\frac { { \pi a }^{ 2 } }{ { b }^{ 2 } } \left( b-\frac { b }{ 3 } \right) ^{ 2 }\left( \frac { b }{ 3 } \right) \left[ \therefore h=\frac { b }{ 3 } \right] \)
= \(\frac { { \pi a }^{ 2 } }{ { b }^{ 2 } } \left( \frac { 2b }{ 3 } \right) ^{ 2 }\left( \frac { b }{ 3 } \right) \)
\(=\frac{\pi a^{2}}{\not b^{2}}\left(\frac{4\not b^{2}}{9}\right)\left(\frac{b}{3}\right)=\frac{4}{9}\left(\frac{1}{3} \pi a^{2} b\right)\)
= \(\frac { 4 }{ 9 } \)(Volume of the cone)
Hence, volume of the largest cylinder that can be hidden underneath is \(\frac { 4 }{ 9 } \) times volume of the cone.

Let the point of intersection of the two curves be (a,b) . Hence,
 \(a^{2}+4b^{2}=8\) and  \(a^{2}-2b^{2}=4\) ...(4)
It is enough if we show that the product of the slopes of the two curves evaluated at (a, b) is −1.
Differentiation of \(x^{2}+4y^{2}=8\)  with respect x, gives
\(2x+8y=\frac{dy}{dx}=0\).
Therefore \(\frac{dy}{dx}= -\frac{-x}{4y}\),
\((\frac{dy}{dx})_{(a,b)}=m_{1}= -\frac{a}{4b}\)
Differentiation of x²-2y² = 4 with respect to x, gives
\(2x-4y\frac{dy}{dx}=0\)
Therefore, \(\frac{dy}{dx}=\frac{x}{2y}\), 
at \((a,b)(\frac{dy}{dx})=m_{2}= \frac{a}{4b}\).
Therefore, \(m_{1}\times m_{2}=(-\frac{a}{4b})\times (\frac{a}{2b})= -\frac{a^{2}}{8b^{2}}\)  ...(5)
Applying the ratio of proportions in (4), we get
\(\frac{a^{2}}{-16-16}=\frac{b^{2}}{-8+4}=\frac{1}{-2-4}\)
Therefore, \(\frac{a^{2}}{b^{2}}=\frac{32}{4}=8\) Substituting in (5), we get \(m_{1}\times m_{2}=-1\) Hence, the curves cut orthogonally.

Let h and r be the height and the base radius. Therefore h = 2r. Let V be the volume of the salt cone.

\(V=\frac{1}{3}\pi r^{2}h=\frac{1}{12}\pi h^{3}; \frac{dV}{dt}=30\) mtr³ / min.
Hence,  \(\frac{dV}{dt}=\frac{1}{4}\pi h^{2}\frac{dh}{dt}\)
Therefore,  \(\frac{dh}{dt}=4 \frac{dV}{dt}.\frac{1}{\pi h^{2}}\)
That is,  \(\frac{dh}{dt}=4\times30\times \frac{1}{100 \pi}\)
=\(\frac{6}{5\pi}\) mtr / min. 

Let x be the length of the edge of a cube and V be its volume at any time 't'
\( \mathrm{V}=\mathrm{x}^{3} \text { and } \frac{d x}{d t} =10 \mathrm{~cm} / \mathrm{sec} \)
\(\frac{d V}{d t} =3 \mathbf{x}^{2} \frac{d x}{d t} \)
\( =3 \mathrm{x}^{2}(10) \)
\(\frac{d V}{d t} =30 \mathrm{x}^{2} \)
\(\left(\frac{d V}{d t}\right)_{x=5} =30(5)^{2}=750 \mathrm{~cm}^{3} / \mathrm{sec}\)
Volume of the cube is increasing at the rate of 750 cm³ / sec

Volume of sphere = \(\frac43\)πr²
Given r = 10 cm 
\(\frac{dr}{dt}\) = - 0.2
V = \(\frac43\)πr³
Change in Volume
= \(\frac{4}{\not 3} \pi . \not 3 r^{2} \frac{d r}{d t}\)
= 4π(10)² (-0.2)
= 400 π (-0.2) = -80 πcm³
∴ Volume decreases by 80 π cm³
Surface area of sphere = 4πr²
Change 10 surrace area = 4 π2r\(\frac{dr}{dt}\)
= 8π(10) (-0.2)
= -\(\frac{80π\times2}{10}\) = -16 π cm²
∴ Surface area decreases by 16 π cm²

Let f(x) = ^{\(x^\frac13\)}, xo = 27, ∆x = -1
∴ \(\sqrt [ 3 ]{ 26 } =f(27)+{ f }^{ ' }(27)\)
\(f(27)={ (27) }^{ \frac { 1 }{ 3 } }={ (2^{ 3 }) }^{ \frac { 1 }{ 3 } }={ 3 }^{ 1 }=3\)
\({ f }^{ ' }(x)=\frac { 1 }{ 3 } x^{ \frac { 1 }{ 3 } -1 }=\frac { 1 }{ 3 } { x }^{ \frac { 2 }{ 3 } }=\frac { 1 }{ { 3x }^{ \frac { 2 }{ 3 } } } \)
∴ \({ f }^{ ' }(27)=\frac { 1 }{ { 3(27)x }^{ \frac { 2 }{ 3 } } } =\frac { 1 }{ { { { 3(3 }^{ 3 } }) }^{ \frac { 2 }{ 3 } } } =\frac { 1 }{ 3({ 3 }^{ 2 }) } =\frac { 1 }{ 27 } \)
∴ (1) becomes
\(\sqrt [ 3 ]{ 26 } =3+\frac { 1 }{ 27 } (-1)\)
= 3 - \(\frac{1}{27}\)(-1)
= 3 - \(\frac{1}{27}\) = 3 - 0.037
\(\sqrt [ 3 ]{ 26 } \) = 2.963

Given curve is y = x² − 5x + 4 and the line is 3x + y = 7
Slope of the tangent to the curve
\({ m }_{ 1 }=\frac { dx }{ dt } \) = 2x - 5
Slope of the line = \({ m }_{ 2}=\frac { dx }{ dt } \) = -3
\(\left[ \because m=\frac { co-efficient \ of \ x }{ co-efficient \ of \ y } \right] \)
Since the tangent of the curve and the lines are parallel, their slopes are equal.
∴ m₁ = m₂
⇒ 2x - 5 = -3
⇒ 2x = 2
⇒ x = 1
Substituting x = 1 in y = x² - 5x + 4 we get
y = 12-5(1)+4 = 0
∴ The required point is (1, 0).

Then A∨ B =\(\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\vee \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 0\vee 1 & 1\vee 1 \\ 1\vee 0 & 1\vee 1 \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\)
\(A\wedge B=\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\wedge \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 0\wedge 1 & 1\wedge 1 \\ 1\wedge 0 & 1\wedge 1 \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}\)

Since the polynomial in the numerator is a higher degree (2^nd) than the denominator (1^st), we know we have a slant asymptote. To find it, we must divide the numerator by the denominator. We can use long division to do that:
\(\begin{array}{r} x-11 \\ x + 5 \sqrt { x ^ { 2 } - 6 x + 7 } \\ \frac{x^{2}+5 x}{-11 x+7} \\ \frac{-11 x-55}{62} \end{array}\)
Notice that we don't need to finish the long division problem to find the remainder. We only need the terms that will make up the equation of the line. The slant asymptote is y = x −11.

As you can see in this graph of the function, the curve approaches the slant asymptote y = x −11 but never crosses it:

f(x) = |3x + 1|, x ∈ |-1, 3|

Since \(LHL \neq RHL, f'(-\frac{1}{3})\) does not exist.
Hence, Lagrange's mean value theorem is not applicable.