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
\(\frac { \partial U }{ \partial z } =\frac { { 3z }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \)
\(\therefore \frac { \partial U }{ \partial x } +\frac { \partial U }{ \partial y } +\frac { \partial U }{ \partial z } =\frac { { 3x }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } +\frac { { 3y }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } +\frac { { 3z }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \)
\(=\frac { { 3({ x }^{ 2 }+y }^{ 2 }+{ z }^{ 2 }) }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \)

\({ tan }^{ -1 }\sqrt { x(x+1) } +{ sin }^{ -1 }\sqrt { { x }^{ 2 }+x+1 } =\frac { \pi }{ 2 } \)
This equation holds if
\({ x }^{ 2 }+x\ge 0\) and \({ x }^{ 2 }+x+1\le 1\)
Now, \({ x }^{ 2 }+x\le 0\) and \(0\le { x }^{ 2 }+x+1\le 1\)
\(\Rightarrow { x }^{ 2 }+x\ge 0\quad { x }^{ 2 }+x+1\le 1\) [\( \because { x }^{ 2 }+x+1\ge 0 \) for all x]
\( \Rightarrow x^{2}+x \geq 0 \text { and } x^{2}+x \leq 0 \)
\( \Rightarrow x^{2}+x=0 \Rightarrow x=0,-1 \)
Hence, x = 0, -1 are the solutions of the given equation.

\(tan\left( { sin }^{ -1 }\frac { 3 }{ 5 } +{ cot }^{ -1 }\frac { 3 }{ 2 } \right) \)
Let \(sin^{ -1 }\left( \frac { 3 }{ 5 } \right) =x\)

\(\Rightarrow \frac { 3 }{ 5 } =sinx\)
\(\therefore tanx=\frac { opp }{ adj } =\frac { 3 }{ 4 } \)
\({ cot }^{ -1 }\left( \frac { 3 }{ 2 } \right) =y\)
\(\Rightarrow \frac { 3 }{ 2 } =coty\Rightarrow tany=\frac { 2 }{ 3 } \)
\(\therefore tan\left( { sin }^{ -1 }\frac { 3 }{ 5 } +{ cot }^{ -1 }\frac { 3 }{ 2 } \right) =tan(x+y)\)
\(\frac { tanx+tany }{ 1-tanxtany } \)
\(\frac { \frac { 3 }{ 2 } +\frac { 2 }{ 3 } }{ 1-\left( \frac { 3 }{ 4 } \right) \left( \frac { 2 }{ 3 } \right) } =\frac { \frac { 9+6 }{ 12 } }{ 1-\frac { 6 }{ 12 } } \)
\(\frac { \frac { 3 }{ 2 } +\frac { 2 }{ 3 } }{ 1-\left( \frac { 3 }{ 4 } \right) \left( \frac { 2 }{ 3 } \right) } =\frac { \frac { 9+6 }{ 12 } }{ 1-\frac { 6 }{ 12 } } \)
\(\therefore tan\left( { sin }^{ -1 }\frac { 3 }{ 5 } +{ cot }^{ -1 }\frac { 3 }{ 2 } \right) =\frac { 17 }{ 6 } \)

We have |z| - |z - i|
\(\Rightarrow\)|x + iy| = |x + iy - i|
\(\Rightarrow\) \(\sqrt { { x }^{ 2 }+{ y }^{ 2 } } =\sqrt { { x }^{ 2 }+\left( y-1 \right) ^{ 2 } } \)
\(\Rightarrow x^{2}+y^{2}=x^{2}+y^{2}-2 y+1\)
\(\Rightarrow\) 2y -1 = 0

Since axis is parallel to y-axis the required equation of the parabola is (x +1)² = 4a(y+2).
Since this passes through (3, 6) (3 +1)² = 4a(6+ 2) 
a = \(\frac { { 1 } }{ { 2 } } \)
Then the equation of parabola is (x+1)² = 2(y+2) which on simplifying yields,
x²+2x−2y − 3 = 0.

By definition, the domain of yx = cos^-1 x is -1. This leads to \(-1\le\frac{2+sinx}{3}\le1\) which is same as -3\(\le\)2+sinx\(\le\)3
so, -5\(\le sin\ x\le1 \) reduces to -1\(\le sin\ x\le1 \), which gives
-sin^-1(1)\(\le x\le sin^-1(1) or -\frac{\pi}{2}\le x\le \frac{\pi}{2}\)
Thus, the domain of cos-1\((\frac{2+sin\ x}{3}) is [-\frac{\pi}{2},\frac{\pi}{2}].\)

Given polynomial equation is 
2x⁴+ 5x³−7x² + 8 = 0
Here a = 2, b = 5, c = -7, d = 0, e = 8
By vieta's formula, 
\(\alpha +\beta +\gamma +\delta =\frac { -b }{ a } =\frac { -5 }{ 2 } \)
\(\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta =\frac { c }{ a } =\frac { -7 }{ 2 } \)
\(\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta =\frac { -d }{ a } =0\)
\(\alpha \beta \gamma \delta =\frac { e }{ a } =\frac { 8 }{ 2 } =4\)
Given roots of quadratic equation are 
∝ + β + ૪ + \(\delta \) and ∝β૪\(\delta \)
∴ sum of the roots = (∝+β+૪+\(\delta \)) (∝β૪\(\delta \))
\(=\left( \frac { -5 }{ 2 } +4 \right) =\frac { -5+8 }{ 2 } =\frac { 3 }{ 2 } \)
\(=\left( \alpha +\beta +\gamma +\delta \right) (\alpha \beta \gamma \delta )\)
\(=\left( \frac { -5 }{ 2 } \right) (4)=\frac { -20 }{ 2 } =-10\)
∴ The sum required quadrate equation is x²-x 
(sum of the roots) + product of the roots = 0
\(\Rightarrow { x }^{ 2 }-x\left( \frac { 3 }{ 2 } \right) -10=0\)
\(\Rightarrow { 2x }^{ 2 }-3x-20=0\)

\({ f }_{ y }=\frac { 1 }{ 1+\frac { { x }^{ 2 } }{ { y }^{ 2 } } } \left( \frac { -x }{ { y }^{ 2 } } \right) =\frac { -\frac { x }{ { y }^{ 2 } } }{ \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { { x }{ x }^{ 2 } } } } \)
= \(\frac { -x }{ { x }^{ 2 }+{ y }^{ 2 } } \)
\({ f }_{ x }=\frac { 1 }{ 1+\frac { { x }^{ 2 } }{ { { y }^{ 2 } } } } \left( \frac { 1 }{ y } \right) =\frac { \frac { 1 }{ y } }{ \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { { y }^{ 2 } } } } \)
\(=\frac { y }{ { { x }^{ 2 }+{ y }^{ 2 } } } \)
\({ f }_{ xy }=\frac { \partial }{ \partial x } ({ f }_{ y })\)
\({ f }_{ xy }=-\left[ \frac { { (x }^{ 2 }+{ y }^{ 2 })(1)-x{ (2x) } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=-\left[ \frac { { x }^{ 2 }+{ y }^{ 2 }-2{ x }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=-\left[ \frac { { y }^{ 2 }-{ x }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=\frac { x^{ 2 }-{ y }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \) ...(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

1.1

(x-2)+(x-2)^z

Equation of family of straight lines in my plane is y = mx - c where m and c are arbitrary constraints.
Differentiating, y' = m
Differentiating again, y" = 0, is the required differential equation.

Let ∝, β are the roots of the equation
Sum of the roots \(\alpha +\beta =\frac { -b }{ a } \)
 \(=\frac { [-(a-2)] }{ 1 } =a-2\)
Product of the roots \(=\alpha \beta =\frac { c }{ a } \)
\(=\frac { -(a+1) }{ 1 } =-(a+1)\)
we have \({ \alpha }^{ 2 }{ \beta }^{ 2 }=({ \alpha +\beta ) }^{ 2 }-2\alpha \beta \)
\(={ (a-2) }^{ 2 }+2(a+1)\)
\(={ a }^{ 2 }-4a+4+2a+2\)
\(=(a-1{ ) }^{ 2 }+5\)
Thus \(\\ { \alpha }^{ 2 }+{ \beta }^{ 2 }\) is least if a = 1

Let \(\vec { a } ={ a }_{ 1 }\hat { i } +{ a }_{ 2 }\hat { j } +{ a }_{ 3 }\hat { k } \)
∴ LHS = \(\hat { i } \times (\vec { a } \times \hat { i } )+\hat { j } \times (\vec { a } \times \hat { j } )+\hat { k } \times \vec { a } \times \hat { k }\)
\((\hat { i }. \hat { i } )\vec { a } -(\hat { i } .\hat { a } )\hat { i } +(\hat { j } .\hat { j } )\vec { a } -(\hat { j } .\vec { a } )\hat { j } +(\hat { k } .\hat { k } )\vec { a } -(\hat { k } .\vec { a } )\hat { k } \)
\([\because \vec { a } \times (\vec { b } \times \vec { c } )=(\vec { a } .\vec { c } )\vec { b } -(\vec { a } .\vec { b } )\vec { c } ]\)
\(1.\vec { a } -{ a }_{ 1 }\hat { i } +1.\vec { a } -{ a }_{ 2 }\hat { j } +1.\vec { a } -{ a }_{ 3 }\hat { k } ]\)
\([\because \hat { i } .\hat { i } =\hat { j } .\hat { j } =\hat { k } .\hat { k } =1\)and
\(\hat { i } \vec { a } =\hat { i } ({ a }_{ 1 }\hat { i } +{ a }_{ 2 }\hat { j } +{ a }_{ 3 }\hat { k } )={ a }_{ 1 }\hat { j } .\vec { a } ={ a }_{ 2 }\quad \hat { k } .\vec { a } ={ a }_{ 3 }\)
\(3\vec { a } -({ a }_{ 1 }\hat { i } +{ a }_{ 2 }\hat { j } +{ a }_{ 3 }\hat { k } )\)
= \(3\vec { a } -\vec { a } =2\vec { a } \)
= RHS .
∴ LHS = RHS. Hence proved

\(\frac { 9\sqrt { 3 } { a }^{ 2 } }{ 45 } \)

100 m / s, t = 4 sec, 200 m / s, −100 m / s

Let P_o be the initial population and the population after t year be P.
Given \(\frac { dp }{ dt } =\frac { 2p }{ 100 } \Rightarrow \frac { dp }{ dt } =\frac { p }{ 50 } \)
⇒ \(\frac { dp }{ p } =\frac { dt }{ 50 } \Rightarrow \int { \frac { dp }{ p } } =\int { \frac { dt }{ 50 } } \)
⇒ log p =\(\frac { t }{ 50 } \) + c ...(1)
when t = 0, p = 0
⇒ log p₀ = 0+c ⇒ log p₀  ....(1)
∴ (1) becomes, log p =\(\frac { t }{ 50 } \)+log P₀.
⇒ log\(\left( \frac { P }{ { p }_{ 0 } } \right) =\frac { t }{ 50 } \)
⇒ t = 50 log\(\left( \frac { P }{ { p }_{ 0 } } \right) \)
when p = 2p₀, t = 50 log\(\left( \frac { 2P_{ 0 } }{ { p }_{ 0 } } \right) \) = 50 log 2
= 50(0.3) = 15 years.
Hence the population doubles in 15 years.

Since the entries in the last column are F, (p ∧ ~p) ∧ (~q ∧ P) is a contradiction

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]}.

(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.

Given (x,y) = x³ - 2x²y + 3xy² + y³ ...(1)
f(tx, ty) = (tx)³ - 2(tx)² (ty) + 3 (tx) (ty)² + (ty)³
= t³ x³ - 2t² x²ty + 3txt²y²+ t³y³
= t³ (x³ - 2x²y + 3xy² +y³)
f(tx, ty) = t³.f(x, y)
∴ f is a homogeneous function and its degree is 3.
Differentiate (1) partially with respect to 'x' and 'y' we get
\(\frac { \partial f }{ \partial x } { =3 }^{ 2 }-4xy+3{ y }^{ 2 }\)
\(\Rightarrow x\frac { \partial f }{ \partial x } ={ 3x }^{ 2 }-4xy+{ 3xy }^{ 2 }\)  ...(2)
\(\frac { \partial f }{ \partial y } =-{ 2x }^{ 2 }+6xy+3{ xy }^{ 2 }\)
\(\Rightarrow y\frac { \partial f }{ \partial y } =-2{ x }^{ 2 }+{ 6xy }^{ 2 }+{ 3y }^{ 3 }\) ...(3)
Adding (2) and (3) we get,
\(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } \) = 3x² - 4x²y + 3x²y- 2x²y + 6xy² + 3y³
= 3x² - 6x²y + 9xy² + 3y³
= 3 (x³ - 2x²y - 3xy² +y³)
= 3f [using (1)]
∴ \(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } \) = 3f = nf where 3 is the degree of (x, y)
Hence Euler's theorem is verified.

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

Given V(x, y) = e^x(x cos y - y sin y)
\(\frac { \partial V }{ \partial x } \) = e^x (cos y) +(x cos y - y sin y)e^x
= e^x (cos y + x cos y - y sin y)
\(\frac { { \partial }^{ 2 }V }{ \partial { x }^{ 2 } } \) = e^x(0 + cos y - 0) + (cos y +x cos y.- y sin y)e^x
= e^x(2 cos y + x cos y - y sin y) ... (1)
\(\frac { \partial V }{ \partial y } \) = e^x(-x sin y- y cos y- sin y)

\(\frac { { \partial }^{ 2 }V }{ \partial { y }^{ 2 } } \) = e^x (-x cos y - (-y sin y + cos y) - cos y)
= e^x(- x cos y + y sin y - cos y - cos y)
= e^x (- x cos y + y sin y - 2 cos y) ... (2)
(1)+(2)➝ 
\(\frac { { \partial }^{ 2 }V }{ \partial { x }^{ 2 } } \) + \(\frac { { \partial }^{ 2 }V }{ \partial { y }^{ 2 } } \) = e^x(2 cos y + x cos y - y sin y - x cos y + y sin y - 2 cos y]
= e^x (0) = 0
Hence proved

  \(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1\)
Given equation is \(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1\)
This is an equation of the hyperbola where the transverse axis is parallel to the y-axis.
∴ a² = 16, b² = 9, c² = a² + b²
⇒ c² = 16 + 9 = 25 ⇒ c = 5
\(e=\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1+\frac { 9 }{ 16 } } =\sqrt { \frac { 16+9 }{ 16 } } =\sqrt { \frac { 25 }{ 16 } } =\frac { 5 }{ 4 } \)
(a) Center is (0, 0)
⇒ h = 0, k = 0

(b) Vertices are (h, k + a), (h, k-a)
⇒ (0, 0 + 4), (0, 0 - 4) ⇒ (0, 4) (0,-4)
(c) Foci are (h, k + c), (h, k- c)
⇒ (0, 0 + 5), (0,  0 - 5) ⇒ (0, 5) (0, -5)
(d) Equations of Directrices are y = \(x=\pm \frac { a }{ e } \)
\(\Rightarrow y=\pm \frac { 4 }{ \frac { 5 }{ 4 } } \Rightarrow y=\pm \frac { 16 }{ 5 } \)

Comparing the given equation with equation of a straight line \(\vec { r } =\vec { a } +t\vec { b } \), we have \(\vec { a } =3\hat { i } -2\hat { j } +6\hat { k } \) and \(\vec { b } =3\hat { i } -2\hat { j } +6\hat { k } \). Therefore,
(i) If \(\vec { b } ={ b }_{ 1 }\hat { i } +{ b }_{ 2 }\hat { j } +{ b }_{ 3 }\hat { k } \), then direction ratios of the straight line are \({ b }_{ 1 },{ b }_{ 2 },{ b }_{ 3 }\). Therefore, direction ratios of the given straight line are proportional to 2, −1,3, and hence the direction cosines of the given straight line are \(\frac { 2 }{ \sqrt { 14 } } ,\frac { -1 }{ \sqrt { 14 } } ,\frac { 3 }{ \sqrt { 14 } } \)
(ii) vector equation of the straight line in non-parametric form is given by \((\vec { r } -\vec { a } )\times \vec { b } =\vec { 0 } \), Therefore, \((\vec{r}-(3\hat { i } -2\hat { j } +6\hat { k } )\times(2\hat { i } -\hat { j } +3\hat { k } )=\vec{0}\)
(iii) Here (x₁, y₁ ,z₁ ) = (3, −2,6) and the direction ratios are proportional to 2, −1,3. Therefore, Cartesian equations of the straight line are \(\frac { x-3 }{ 2 } =\frac { y+2 }{ -1 } =\frac { z-6 }{ 3 } \)