y = log |e^x+e^-x|+c

\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +y=0\)

order 2;degree not defined

order – 2 : degree 2

\(\frac { 3 }{ 2 } \)

\(\frac { 1 }{ 2 } \)

\(\frac { 13 }{ 3 } \)

\(\frac { \pi }{ 60 } \)

1.1

\(\frac { { \vartheta }^{ 2 }u }{ \vartheta x\vartheta y } \)

concave upward everywhere

f(x) is strictly increasing \(\left[ -\frac { 1 }{ 2 } ,\frac { 1 }{ 2 } \right] \)

Absolute minimum value = 1,
Absolute maximum value = 6

(i) π
(ii) πX2^n-1
(iii) 2
(iv) 0

\({ e }^{ 2x }=1+\frac { 2x }{ 1! } +\frac { \left( 2x \right) ^{ 2 } }{ 2! } +\frac { \left( 3x \right) ^{ 3 } }{ 3! } +...\)

Rolle’s theorem is not valid.

2x – y + 3 = 0

Let a be any element and e be the identity element.
The a * e = e * a = a
a * e = a ⇒ a + e -1 = a ⇒ e-1 = 0 ⇒ e = 1
∴ The identity element is 1

Limits are from 0 to 4

3x - 4y = 0
4y = 3x
\(\Rightarrow y=\frac { 3 }{ 4 } x\)
∴ Required volume = \(\pi \int _{ 0 }^{ 4 }{ { y }^{ 2 }dx } \)
= \(\pi \int _{ 0 }^{ 4 }{ \frac { 9 }{ 16 } { x }^{ 2 }dx } \)
= \(\frac { 9\pi }{ 16 } { \left[ \frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 4 }=\frac { 9\pi }{ 16 } \times \frac { 64 }{ 3 } \)
∴ V = \(12\pi \) Cubic units

Given curve is y = \(\sqrt x\) + 1

Required area
\(\int _{ 0 }^{ 4 }{ ydx } =\int _{ 0 }^{ 4 }{ (\sqrt { x } +1)dx } \)
\({ \left[ \frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 } }+x \right] }_{ 0 }^{ 4 }=\frac { 2 }{ 3 } { (4) }^{ \frac { 3 }{ 2 } }+4\)
\(\frac { 2 }{ 3 } (4)\sqrt { 4 } +4=\frac { 16 }{ 3 } +4\)
\(\frac { 16+12 }{ 3 } =\frac { 28 }{ 3 } \) sq.units

Given u(x, y) = x² + 3xy + y²
u(x_o, y_o) = u(2,1)
= 2² + 3(2)(1) + 1²
= 4 + 6 + 1 = 11
\(\frac { \partial u }{ \partial x } \) = 2x+ 3y
\({ \left( \frac { \partial u }{ \partial x } \right) }_{ (2,1) }\)= 2 + 3 = 5
\(\frac { \partial u }{ \partial y } \) = 3x+ 2y
\({ \left( \frac { \partial u }{ \partial y } \right) }_{ (2,1) }\) = 6 + 2 = 8
Linear approximation 
L(x,y) = U(x_o, y_o) + \({ \left( \frac { \partial u }{ \partial x } \right) }_{ ({ x }_{ 0 },{ y }_{ 0 }) }\) (x - x_o) + \({ \left( \frac { \partial u }{ \partial y} \right) }_{ ({ x }_{ 0 }{ ,y }_{ 0 }) }\)(y - y_o)
L (x,y) = 11 + 5 (x - 2) + 8 (y - 1)
= 11 + 5x - 10 + 8y - 8
L(x,y) = 5x + 8y - 7

Given A = {a, b, c, d} and * is defined as follows.

1) Commulative Property:
\(a * b=c\)  but  \(b \times a=d \Rightarrow a * b \neq b * a\)
\(a * c=b\) but \(c * a=c \Rightarrow a * c \neq c * a\) 
\(\therefore\) \(\text { * }\) is not commutative on A..
2) Associative Property: 
\((a+b) * c=c * c=a\) 
\(a *(b * c)=a * b=c\) 
\(\therefore(a * b) * c \neq a *(b * c)\) 
\(\text { * }\) is not associative on A.

Let z = -2
z = 2(-1) = 2(cos π + i sin π)
∴ arg(z) = π

Z = \(\left( \frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \right) ^{ 107 }+\left( \frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } \right) ^{ 107 }\)
\(\bar { z } =\left( \overline { \frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } } \right) ^{ 107 }+\left( \overline { \frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } } \right) ^{ 107 }\)
= \(\left( \frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } \right) ^{ 107 }+\left( \frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \right) ^{ 107 }\) = z
Since z = \(\bar { z } \), Im(z) = 0

Given z = 5i¹¹ + 7i³
= 5i⁴ . i⁴ . i² . i¹ + 7. i² . i¹
= 5(1)(1)(-1)(i) + 7(-1)(i)
= -5i - 7i = -12i
∴ Re(z) = 0 and In(z) = -12

Given equation is x² + 3y² = a²
\(\div \) a² we get, \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ \frac { { a }^{ 2 } }{ 3 } } =1\)
Here a² and b2 = \(\frac{a^2}{3}\) ⇒ b = \(\frac{a}{\sqrt3}\)
Length of major axis is 2a and
Length of minor axis is 2b = \(\frac { 2a }{ \sqrt { 3 } } \times \frac { \sqrt { 3 } }{ \sqrt { 3 } } =\frac { 2a\sqrt { 3 } }{ 3 } \)

Given equation of tangent is y = 3x + 1
The condition for any line y = mx + c to be a tangent to y² = 4ax is c = \(\frac{a}{m}\)
1= \(\frac{a}{m}\) ⇒ 1 = \(\frac{a}{m}\) ⇒ a = 3
Length of the latus rectum is 4a = 4(3) = 12 units.

Given circle is  x² +y² + 2x - 3y - 8 = 0
Equation of tangent is xx₁ + yy₁ + 1(x + x₁) -\(\frac{3}{2}\)
(y + y₁) - 8 = 0
AE(2, 3), the tangent is
x(2) + y(3) + x + 2 - \(\frac{3}{2}\) (y + 3) -8 = 0
⇒ 3x + 3y + 2 - \(\frac{3y}{2}\) - \(\frac92\) - 8 = 0
Multiply by 2 we get,
⇒ 6x + 6y + 4 - 3y - 9 - 16 = 0
⇒ 6x + 3y - 21 = 0

Let \({ cos }^{ -1 }\left( \frac { 3 }{ 5 } \right) =\theta \Rightarrow \frac { 3 }{ 5 } =cos\theta \)
\(\therefore sin\theta =\sqrt { 1-{ cos }^{ 2 }\theta } =\sqrt { 1-\frac { 9 }{ 25 } } =\sqrt { \frac { 25-9 }{ 25 } } \)
= \(\sqrt { \frac { 16 }{ 25 } } =\frac { 4 }{ 5 } \)
\(\therefore sin\left( { cos }^{ -1 }\left( \frac { 3 }{ 5 } \right) \right) =\frac { 4 }{ 5 } \)

Given
\({ cot }^{ -1 }\left( \frac { 1 }{ 7 } \right) =0\Rightarrow \theta =\frac { 1 }{ 7 } \)
\(\Rightarrow tan\theta =7\)
\(\Rightarrow sec\theta =\sqrt { 1+{ tan }^{ 2 }\theta } =\sqrt { 1+{ 7 }^{ 2 } } =\sqrt { 50 } =5\sqrt { 2 } \)
\(\Rightarrow cos\theta ={ \frac { 1 }{ 5\sqrt { 2 } } }\)

Let p(x) = x⁷-6x⁶+7x⁵+5x²+2x+2
It has only one change of sign. Now,
p(-x) = (-x)⁷ -6(-x)⁶ +7(-x)⁵ +5(-x)² +2(-x)+2
= -x⁷ -6x⁶ -7x⁵ + 5x² - 2x + 2
It has two, change of sign.
Hence, p(x) has one positive root and has at least two negative roots.

Matrix form of the given system of equations is
\(\left[ \begin{matrix} 1 & 2 & 2 \\ 1 & -3 & -3 \\ 2 & 1 & k \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
AX = B where A =\(\left[ \begin{matrix} 1 & 2 & 2 \\ 1 & -3 & -3 \\ 2 & 1 & k \end{matrix} \right] \)
Homogeneous system of equations has trivial solution only if |A| ≠ 0.
∴ \(\left[ \begin{matrix} 1 & 2 & 2 \\ 1 & -3 & -3 \\ 2 & 1 & k \end{matrix} \right] \) ≠ 0
Expanding along R₁,
\(1\left| \begin{matrix} -3 & -3 \\ 1 & k \end{matrix} \right| -2\left| \begin{matrix} 1 & -3 \\ 2 & k \end{matrix} \right| +2\left| \begin{matrix} 1 & -3 \\ 2 & 1 \end{matrix} \right| \neq 0\)
⇒ 1 (-3k + 3) - 2 (k + 6) + 2 (1 + 6) ≠ 0
⇒ -3k + 3 -2k - 12 + 14 ≠ 0
⇒ -5k + 5 ≠ 0
⇒ -5k ≠ -5 ⇒ k ≠ \(\frac { -5 }{ -5 } \) =1
⇒ k ≠ 1

A =\(\left[ \begin{matrix} 4 \\ 7 \end{matrix}\begin{matrix} 5 \\ -3 \end{matrix}\begin{matrix} -6 \\ 0 \end{matrix}\begin{matrix} 1 \\ 8 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 4 \\ -1 \end{matrix}\begin{matrix} 5 \\ -3 \end{matrix}\begin{matrix} -6 \\ 12 \end{matrix}\begin{matrix} 1 \\ 6 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} -1 \\ 4 \end{matrix}\begin{matrix} -3 \\ 5 \end{matrix}\begin{matrix} 12 \\ -6 \end{matrix}\begin{matrix} 6 \\ 1 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }+(-1){ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 4 \end{matrix}\begin{matrix} 13 \\ 5 \end{matrix}\begin{matrix} -12 \\ -6 \end{matrix}\begin{matrix} 6 \\ 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-4{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \end{matrix}\begin{matrix} 13 \\ -47 \end{matrix}\begin{matrix} -12 \\ 42 \end{matrix}\begin{matrix} -6 \\ 25 \end{matrix} \right] \)
The equivalent row-echelon matrix hats two non zero rows.
∴ \(\rho\) (A) = 2

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

Agumented matrix
[A|B] \(\left[ \begin{matrix} 2 & 5 \\ 6 & 15 \end{matrix}|\begin{matrix} 7 \\ 13 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 2 & 5 \\ 0 & 0 \end{matrix}|\begin{matrix} 7 \\ -8 \end{matrix} \right] \)
Here \(\rho\) (A) = 2 and \(\rho\)([A|B]) = 3
∴ \(\rho\) (a) ≠ \(\rho\) ([AIB])
Hence the system is inconsistent.

Given A³ = 1
Pre multiply by A^-1 we get,
A^-1. A³ = A^-1. I
⇒ (A^-1. A) A² = A^-1 [∵ A^-1 I = A^-1]
⇒ I. A² = A^-1  I∵  A^-1. A = I]
⇒ A² = A^-1  [∵ I. A² = A²]
∴ A^-1 = A²

The equation of the required plane through the intersection of the given planes is
( x + y + z - 6 ) + λ (2x + 3y + 4z + 5) = 0 ......(1)
This passes through (1, 1, 1)
∴ ( 1+ 1 + z - 6) + λ (2 + 3+ 4 + 5) = 0
⇒ -3 +14λ = 0 \(\Rightarrow \lambda =\frac { 3 }{ 14 } \)
Substituting \(\lambda =\frac { 3 }{ 14 } \) in (1) we get
( x + y + z - 6 )+\(\frac { 3 }{ 14 } \) (2x + 3y + 4z + 5) = 0
⇒ 14( x + y + z - 6 ) +3 (2 + 3+ 4 + 5) = 0
⇒ 20x + 23y + 26z - 69 = 0

Since the plane passing through the point \(\overset { \rightarrow }{ a } =\overset { \wedge }{ i } -\overset { \wedge }{ j } +2\overset { \wedge }{ k } \) and is normal to the vector \(\overset { \rightarrow }{ n } =2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \)
the vector equation of the plane is \(\overset { \rightarrow }{ r } .\overset { \rightarrow }{ n } =\overset { \rightarrow }{ a } .\overset { \rightarrow }{ n } \)
\(\Rightarrow { \overset { \rightarrow }{ r } }.\left( 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) =\left( \overset { \wedge }{ i } -\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) .\left( 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) \)
\(\Rightarrow { \overset { \rightarrow }{ r } }.\left( 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) \)= 2 - 3 + 4 = 3
\(\therefore { \overset { \rightarrow }{ r } }.\left( 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) \)= 3

Given (x₁, y₁, z₁) is (2, -1, 3) (x₂, y₂, z₂) is (4, 2, 1)
Cartesian equation of a line passing through two points is \(\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { z-{ z }_{ 1 } }{ { z }_{ 2 }-{ z }_{ 1 } } \)
\(\Rightarrow \frac { x-2 }{ 2 } =\frac { y+1 }{ 3 } =\frac { z-3 }{ -2 } \)