The entries in the columns corresponding to p → q and ㄱp ν q are identical and hence they are equivalent.

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

Since α and β are the roots of the quadratic equation, we have α + β =\(\frac { 7 }{ 2 } \) and αβ = \(\frac { 13 }{ 2 } \).
Thus, to construct a new quadratic equation,
Sum of the roots = α²+β² = (α+β)²-2αβ =\(\frac { -3 }{ 4 } \)
Product of the roots = α²β² = (αβ)² = \(\frac { 169 }{ 4 }\)
Thus a required quadratic equation is  x²+\(\frac { 3 }{ 4 } x+\frac { 169 }{ 4 } \)= 0. 
From this we see that 4x²+3x+169 = 0 is a quadratic equation with roots α² and β².

tan^-1y+tan^-1(ex)=c

prove

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

\(-\frac { 15 }{ 2 } \)

0.677

(i) 1
(ii) 2
(iii) 1

2:π

Clearly 1 is the identity element of (G₁)
Inverse of 1 is 1 [∴ (1)(1) = 1]
Inverse of i is -i [∴ (i) (-i) = -i² = 1]
Inverse of -1 is -1 [∴ (-1)(-1) = 1]
Inverse of is i [∴ (-i)(i) = -i² = 1]

Given f(x, y) = \(\frac { x }{ { x }^{ 2 }+{ y }^{ 2 } } \)
f(tx, ty) = \(\frac { tx }{ { t }^{ 2 }{ x }^{ 2 }+{ t }^{ 2 }{ y }^{ 2 } } =\frac { tx }{ { t }^{ 2 }(x^{ 2 }+{ y }^{ 2 }) } \)
= \(\frac { x }{ t({ x }^{ 2 }+{ y }^{ 2 }) } ={ t }^{ -1 }.f(x,y)\)
∴ f(x,y) is a homogeneous function of degree - 1
∴ By Euler's theorem,
\(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } \) = nf = -1.f
∴ \(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } \) = -f
Hence proved

Given w (x, y) = x³ - 3xy+ 2.0, x, y ∊ R
w_x = \(\frac { \partial w }{ \partial x } \) = 3x² - 3y + 0 = 3x² - 3y 
w_x = at (1, -1)= 3 (1p - 3- (-1) = 3 + 3 = 6
wy = \(\frac { \partial w }{ \partial y } \) = 0 -3x+4y = -3x+4y
w_y (1, -1) = -3 (1)+4(-1)= -3 - 4 = -7.
w (x_o, y_o) = w (1, -1)
= 1³- 3(1) (-1) + 2 (-1)²
= 1+3+2 = 6
Linear approximation is given by
L(x,y) = w(x_o,y_o) + \((\frac { \partial w }{ \partial x })_{(x_o,y_o) }\)(x-x₀) + \((\frac { \partial w }{ \partial xy})_{(x_o,y_o) }\) (y-y_o)
= 6 + 6 (x - 1) + -7 (y + 1)
= 6 + 6x - 6 - 7y - 7
= 6x - 7y - 7

Taking u = 1 + x − 2x³ and v = e^-2x, and applying the Bernoulli’s formula, we get
I = \(\\ \int _{ 0 }^{ 1 }{ { e }^{ -2x }(1+x-{ 2x }^{ 3 })dx } \)
\(={ \left[ (1+x-{ 2x }^{ 3 })\left( \frac { { e }^{ -2x } }{ -2 } \right) -(1-6{ x }^{ 2 })\left( \frac { { e }^{ -2x } }{ -4 } \right) +(-12x)\left( \frac { { e }^{ -2x } }{ -8 } \right) -(-12)\left( \frac { { e }^{ -2x } }{ 16 } \right) \right] }_{ 0 }^{ 1 }\)
\(={ \left[ \frac { { e }^{ -2x } }{ 16 } (16{ x }^{ 3 }+24{ x }^{ 2 }+16x) \right] }_{ 0 }^{ 1 }\)
\(\\ =\frac { 7 }{ 2{ e }^{ 2 } } \)

Let X be the random variable denotes the Winning amount.
X (Both are black balls) = Rs. 2 (-10) = Rs. -20
X (one red and oneblack ball) = Rs.15-Rs. 10 = Rs. 5
X (both are red ball) = Rs. 2 (15) = Rs. 30
= {-20, 5, 30}
The sample space consists of ¹⁴C₂ = 91
X = -20, Both are black balls= ⁸C₁ = 28
X = 5, One black, one redball = ⁸C₁ x ⁶C₁ = 8 x 6 = 48
X = 30, Both are white balls = ⁶C₁ = 15

The given equation is rewritten as \(\frac { dy }{ dx } =\frac { { y }^{ 2 } }{ xy-{ x }^{ 2 } } \)
This is a homogeneous differential equation 
Put y = vx . Then, we have \(x\frac { dv }{ dx } =\frac { v }{ v-1 } \)
By separating the variables, \(\frac { v-1 }{ v } dv=\frac { dx }{ x } .\)
Integrating, we obtain v − log |v| = log |x| + log |C| or v = log |vxC|.
Replacing v by \(\frac{y}{x}\), we get, \(\frac{y}{x}\) = log |Cy| = e^y/x or y = ke^y/x (how!) which is the required solution.

We have, \(\frac{dy}{dx}=2x+3\). Hence at (1, 2), \((\frac{dy}{dx})=5\)

Therefore, the required equation of tangent is.
\((y-2)=5(x-1)\Rightarrow 5x-y-3=0\) 
The slope of the normal at the point (1, 2) is -\(\frac{1}{5}\).
therefore, the required equation of normal is
\((y-2)=-\frac{1}{5}(x-1)\Rightarrow x+5y-11=0\) 

Given y = ae^-3x+ b   ........(1)
Differentiating cquation (1) w.r.t 'x', we get
\(\frac { d y }{ d x } =ae^{ -3x }(-3)+0\)
\(\frac { d y }{ d{ x } } =ae^{ -3x }(-3) \)
Again differentiating, we get
\(\Rightarrow \frac { d^{ 2 }y }{ d{ x }^{ 2 } } ={ ae^{-3 x } }(+9)\)
\(\Rightarrow \frac { d^{ 2 }y }{ d{ x }^{ 2 } } =-\frac{1}{3}\frac{dy}{dx}\times 9\)
\(\Rightarrow \frac { d^{ 2 }y }{ d{ x }^{ 2 } } = {-3}\frac{dy}{dx} \)
\(\Rightarrow \frac { d^{ 2 }y }{ d{ x }^{ 2 } } +3\frac{dy}{dx} \) = 0 
Therefore, y = ae^-3x + b is a solution of the given differential equation.

Let z = x+iy ⇒ \(\bar { z } \) = x+iy
∴ \(\\ \frac { \bar { z } +1 }{ z-1 } =\frac { z-iy+1 }{ x-iy-i } =\frac { (x+1)iy }{ x-i(y+1) }\)
= \(\frac { (x+1)-iy }{ x-i(y+1) } \times \frac { x+i(y+1) }{ x+i(y+1) } \)
Choosing the real part alone we get,
\(\frac { x(x+1)+y(y+1) }{ { x }^{ 2 }+(y+1)^{ 2 } } \) = 0
⇒ x(x+1) + y(y+1) = 0
⇒ x²+x+y²+y = 0 which is the locus of z.

-1 = i² = i \(\times\) i =\(\sqrt { -1 } \times \sqrt { -1 } =\sqrt { (-1) } \times \sqrt { (-1) } \)
= \(\sqrt { 1 } \)
⇒ -1 = i
In the above proof we have used \(\sqrt { -1 } \times \sqrt { -1 } \)
= \(\sqrt { (-1)(-1) } \) which is wrong
Since \(\sqrt { ab } =\sqrt { a } .\sqrt { b } \) is true only at least one of a and b is non-negative.

Given equation of the hyperbola is
3x² - 6y² = -18; \(\div \)by (-18) we get \(\frac{y^2}{3}- \frac{x^2}{6}\) = 1
The transverse axis is long y-axis.
Here a² = 3, b² = 6
Length of transverse axis is 2a = 2\(\sqrt { 3 } \)
Length of conjugate axis is 2b = 2\(\sqrt { 3 } \)
\(e=\sqrt { 1+\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1+\frac { 6 }{ 3 } } =\sqrt { 1+2 } =\sqrt { 3 } \)

Given equation is x² + y² - 2x + 5y + 7 = 0
Here 2g = -2 ⇒ g = -1 ⇒ 2f = 5 ⇒ f = \(\frac { 5 }{ 2 } \)
c = 7
Centre is (-g, -f) = \(\left( 1,\frac { -5 }{ 2 } \right) \)
r = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } =\sqrt { { 1 }^{ 2 }+{ \left( \frac { -5 }{ 2 } \right) }^{ 2 }-7 } \)
= \(\sqrt { 1+\frac { 25 }{ 4 } -7 } =\sqrt { \frac { 25 }{ 4 } -6 } =\sqrt { \frac { 1 }{ 4 } } =\frac { 1 }{ 2 } \)
∴ Cireumferenee of elrele = 2πr = 2π\(\left( \frac { 1 }{ 2 } \right) \) = π units
Area of the circle = πr² = π\({ \left( \frac { 1 }{ 2 } \right) }^{ 2 }\) = \(\frac { \pi }{ 4 } \) sq.units

\({ tan }^{ -1 }\left( \frac { 2x }{ 1-{ x }^{ 2 } } \right) +{ tan }^{ -1 }\left( \frac { 2x }{ 1-{ x }^{ 2 } } \right) =\frac { \pi }{ 3 } \)
\(\left[ \because { co }^{ -1 }\left( \frac { 1 }{ x } \right) ={ tan }^{ -1 }\left( x \right) \right] \)
\(\Rightarrow 2{ tan }^{ -1 }\left( \frac { 2x }{ 1-{ x }^{ 2 } } \right) =\frac { \pi }{ 3 } \)
\(\Rightarrow { tan }^{ -1 }\left( \frac { 2x }{ 1-{ x }^{ 2 } } \right) =\frac { \pi }{ 6 } \)
\(\Rightarrow \cfrac { 2x }{ 1-{ x }^{ 2 } } =tan\left( \cfrac { \pi }{ 6 } \right) =\cfrac { 1 }{ \sqrt { 3 } } \)
\(\Rightarrow 2\sqrt { 3x } =1-{ x }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+2\sqrt { 3x } -1=0\)
\(\Rightarrow x=\frac { -2\sqrt { 3 } \pm \sqrt { 12-4(1)(-1) } }{ 2 } \) \(\left[\because x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\right]\)
\(\Rightarrow x=\cfrac { -2\sqrt { 3 } \pm \sqrt { 16 } }{ 2 } \Rightarrow x=\cfrac { -2\sqrt { 3 } \pm 4 }{ 2 } \)
\(\Rightarrow x=2\left( \frac { -\sqrt { 3 } \pm 2 }{ 2 } \right) \)
\(\Rightarrow x=-\sqrt { 3 } \pm 2\)
\(\Rightarrow x=-2-\sqrt { 3 } \text { or} -2-\sqrt { 3 } \)
Since \(x>0,x=-2-\sqrt { 3 } \) is not possible
\(\therefore x=2-\sqrt { 3 } \)

Let \({ sin }^{ -1 }\left( \frac { 3 }{ 5 } \right) =A\Rightarrow \frac { 3 }{ 5 } =sinA\)

\(cosA=\frac { adj }{ hyp } =\frac { 4 }{ 5 } \)
Let \({ sin }^{ -1 }\left( \frac { 5 }{ 13 } \right) =B\Rightarrow sinB=\frac { 5 }{ 13 } \)

\(\Rightarrow cosB=\frac { 12 }{ 13 } \)
\(\therefore cos\left[ { sin }^{ -1 }\frac { 3 }{ 5 } +{ sin }^{ -1 }\frac { 5 }{ 13 } \right] =cos(A+B)\)
= cos A cos B-sin A sin B
= \(\frac { 4 }{ 5 } .\frac { 12 }{ 13 } -\frac { 3 }{ 5 } .\frac { 5 }{ 13 } =\frac { 48 }{ 65 } -\frac { 15 }{ 65 } \)
= \(\frac { 33 }{ 65 } \)

Given \(\overset { \rightarrow }{ OA } =6\overset { \wedge }{ i } -7\overset { \wedge }{ j } ,\overset { \rightarrow }{ OB } =16\overset { \wedge }{ i } -29\overset { \wedge }{ j } -4\overset { \wedge }{ k } ,\overset { \rightarrow }{ OC } =3\overset { \wedge }{ i } -6\overset { \wedge }{ j } \) and \(\overset { \rightarrow }{ OD } =2\overset { \wedge }{ i } +5\overset { \wedge }{ j } +10\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ AB } =\overset { \rightarrow }{ OB } -\overset { \rightarrow }{ OA } =10\overset { \wedge }{ i } -22\overset { \wedge }{ j } -4\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ AC } =\overset { \rightarrow }{ OC } -\overset { \rightarrow }{ OA } =-6\overset { \wedge }{ i } +10\overset { \wedge }{ j } -6\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ AD } =\overset { \rightarrow }{ OD } -\overset { \rightarrow }{ OA } =-4\overset { \wedge }{ i } +12\overset { \wedge }{ j } +10\overset { \wedge }{ k } \)
\(\therefore \left[ \overset { \rightarrow }{ AB } \overset { \rightarrow }{ AC } \overset { \rightarrow }{ AD } \right] =\left| \begin{matrix} 10 \\ -6 \\ -4 \end{matrix}\begin{matrix} -22 \\ -10 \\ 12 \end{matrix}\begin{matrix} -4 \\ -6 \\ 10 \end{matrix} \right| \)
= 10 (100 + 75) + 22(-60 -24) -4(-72 +40)
= 1720 - 1848 + 128 = 0
Hence , the given points are coplanar.

Let \(\overset { \rightarrow }{ r } =x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } \)
\(\therefore x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } =\left( s-2t \right) \overset { \wedge }{ i } +\left( 3-t \right) \overset { \wedge }{ j } +\left( 2s+t \right) \overset { \wedge }{ k } \)
Equating the co-efficients of like components both sides,
We get, x = s - 2t
y = 3 - t
z = 2s + t
Eliminating x and t using determinates we get
\(\left| \begin{matrix} x \\ y-3 \\ z \end{matrix}\begin{matrix} 1 \\ 0 \\ 2 \end{matrix}\begin{matrix} -2 \\ -1 \\ 1 \end{matrix} \right| =0\)
⇒ x (0+2) -1(y - 3 + z) -2 (2y - 6 - 0) = 0
⇒ 2x - y + 3 - z- 4y + 12 = 0
⇒ 2x - 5y - z + 15 = 0

|A| =\(\left[ \begin{matrix} -2 & -3 \\ 5 & -6 \end{matrix} \right] \) = 12+15 = 27
∴ A^-1 = \(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 27 } \left[ \begin{matrix} -6 & 3 \\ -5 & 2 \end{matrix} \right] \)
(A^-1)^T = \(\frac { 1 }{ 27 } \left[ \begin{matrix} -6 & -5 \\ 3 & 2 \end{matrix} \right] \)...(1)
A^T =\(\left[ \begin{matrix} -2 & 5 \\ -3 & -6 \end{matrix} \right] \)
|A^T| =\(\left[ \begin{matrix} -2 & 5 \\ -3 & -6 \end{matrix} \right] \) = 12+15 = 27
∴ (A^T)^-1 = \(\frac { 1 }{ |A^{ T }| } adj(A^{ T })=\frac { 1 }{ 27 } \left[ \begin{matrix} -6 & -5 \\ 3 & -2 \end{matrix} \right] \)...(2)
From (1) and (2), (A^-1)^T = (A^T)^-1

Δ = \(\left| \begin{matrix} 2 & 3 \\ 1 & 6 \end{matrix} \right| \) = 12 - 3 = 9 ≠ 0
Δ₁ = \(\left| \begin{matrix} 10 & 3 \\ 4 & 6 \end{matrix} \right| \) = 60 - 12 = 48
Δ₂ = \(\left| \begin{matrix} 2 & 10 \\ 1 & 4 \end{matrix} \right| \) = 8 - 10 = -2
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 48 }{ 9 } =\frac { 16 }{ 3 } \)
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { -2 }{ 9 } \)
∴ Solution set is \(\left\{ \frac { 16 }{ 3 } ,\frac { -2 }{ 9 } \right\} \).

Put y = \(\frac { x-1+x-5 }{ 2 } =-3\)
⇒ x = y + 3
∴ (x-1)⁴+(x-5)⁴ = 82
⇒ (y+3-1)⁴+(y+3-5)⁴ = 82
⇒ (y+2)⁴+(y-2)⁴ = 82
⇒ 2(y⁴+24y²+16) = 82
⇒ y⁴+24y²+16 = 41
⇒ y⁴+24y²-25 = 0
⇒ (y²+25)(y²-1) = 0
⇒ y = 土5i, y = 士1
∴ x = 3土5i, 4, 2.

Given sin (e^x) -5^x + 5^-x
We have \({ 5 }^{ x }+{ 5 }^{ -x }={ ({ 5 }^{ \frac { x }{ 2 } }-{ 5 }^{ \frac { -x }{ 2 } }) }^{ 2 }+2\ge 2\)
If sin (e^x) -5^x + 5^-x has a solution
We get sin (e^x) ≥ 2 which i not possible for any
real x as |sin 0|≤ 1 for all θ∈ R.
∴ Sin (e^x) = 5^X + 5^-x has no solution.

The equation of the ellipse is \(\frac { { \left( x-11 \right) }^{ 2 } }{ 484 } +\frac { { y }^{ 2 } }{ 64 } =1\). The origin of the sound wave and the kidney stone of patient should be at the foci in order to crush the stones.
a² = 484 and b² = 64
c² = a² -b²
= 484-64
= 420
c \(\simeq \) 20.5
Therefore the patient’s kidney stone should be placed 20.5 cm from the centre of the ellipse.

Given \(\vec { a } =-2\hat { i } +5\hat { j } +3\hat { k } \), \(\hat { b } =\hat { i } +3\hat { j } -2\hat { k } \) and \(\vec { c } =-3\vec { i } +\vec { j } +4\vec { k } \)
Volume of the parallelepiped = \(\vec { a } .(\vec { b } \times \vec { c } )\)
= \(\left| \begin{matrix} -2 & 5 & 3 \\ 1 & 3 & -2 \\ -3 & 1 & 4 \end{matrix} \right| \)
= \(-2\left| \begin{matrix} 3 & -2 \\ 1 & 4 \end{matrix} \right| -5\left| \begin{matrix} 1 & -2 \\ - & 4 \end{matrix} \right| +3\left| \begin{matrix} 1 & 3 \\ -3 & 1 \end{matrix} \right| \)
= -2(12+2)-5(4-6)+3(1+9)
= -2(14)-5(-2)+3(10) = -28+10+30
= 12
Vector Area of the base parallelogram = \(\vec { b } \times \vec { c } \)
=\(\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 3 & -2 \\ -3 & 1 & 4 \end{matrix} \right| =\hat { i } \left| \begin{matrix} 3 & -2 \\ 1 & 4 \end{matrix} \right| \hat { j } \left| \begin{matrix} 1 & -2 \\ -3 & 4 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 1 & 3 \\ -3 & 1 \end{matrix} \right| \)
= \(\hat { i } \)(12+2)-\(\hat { j } \)(4-6)+\(\hat { k } \)(1+9)
= \(\hat { i } \)(14)-\(\hat { j } \)(-2)+\(\hat { k } \)(10)
= 14\(\hat { k } \)+2\(\hat { j } \)+10\(\hat { k } \)
Area of the parallelogram =\(\sqrt { { 14 }^{ 2 }+{ 2 }^{ 2 }+{ 10 }^{ 2 } } \)
= \(\sqrt { 196+4+100 } \)
= \(\\ \sqrt { 300 } =10\sqrt { 3 } \)..............(2)
Volume of the parallelepiped = Base area x attitude
∴ Altitude = \(\frac { Volume }{ Base\quad area } =\frac { 12 }{ 10\sqrt { 3 } } =\frac { 6 }{ 5\sqrt { 3 } } \times \frac { \sqrt { 3 } }{ \sqrt { 3 } } \)

= \(\frac { 2\sqrt { 3 } }{ 5 } \) units.

2x+5y = -2, x+2y = -3
The matrix form of the system is
\(\left( \begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} -2 \\ -3 \end{matrix} \right) \)
⇒ AX = B where
A =\(\left( \begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix} \right) ,B=\left( \begin{matrix} -2 \\ -3 \end{matrix} \right) \)
X =\(\left( \begin{matrix} x \\ y \end{matrix} \right) \)
⇒ = A^-1B
|A| = \(\left| \begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix} \right| \)= 4 - 5 = -1 ≠ 0
∴ A^-1 =\(\frac { 1 }{ |A| } adjA=\frac { 1 }{ -1 } \left[ \begin{matrix} 2 & -5 \\ -1 & 2 \end{matrix} \right] \)
=\(\left[ \begin{matrix} -2 & 5 \\ 1 & -2 \end{matrix} \right] \)
∴ X = A^-1B =\(\left[ \begin{matrix} -2 & 5 \\ 1 & -2 \end{matrix} \right] \left[ \begin{matrix} -2 \\ -3 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 4-15 \\ -2+6 \end{matrix} \right] =\left[ \begin{matrix} -11 \\ 4 \end{matrix} \right] \)
∴ Solution set is x = -11, y = 4

Given v = 3-4i, w = 4+3i and \(\frac { 1 }{ u } =\frac { 1 }{ v } +\frac { 1 }{ w } \)
∴ \(\frac { 1 }{ u } =\frac { 1 }{ 3-4i } +\frac { 1 }{ 4+3i } \)
= \(\frac { 3+4i }{ (3-4i)(3+4i) } +\frac { 4-3i }{ (4+3i)(4-3i) } \)
= \(\\ \frac { 3+4i }{ 9-(4i)^{ 2 } } +\frac { 4-3i }{ 16-(3i)^{ 2 } } =\frac { 3+4i }{ 9+16 } +\frac { 4-3i }{ 16+9 } \)
= \(\frac { 3+4i }{ 25 } +\frac { 4-3i }{ 25 } =\frac { 3+4i+4-3i }{ 25 } \)
\(\frac { 1 }{ u } =\frac { 7+i }{ 25 } \)
∴ u = \(\frac { 25 }{ 7+i } \times \frac { 7-i }{ 7-i } =\frac { 25(7-i }{ 7^{ 2 }-({ i }^{ 2 }) } \) 
= \(\frac { 25(7-i) }{ 49+1 } =\frac { 25(7-i) }{ 50 } =\frac { 1 }{ 2 } \)(7-i)
∴ u = \(\frac { 1 }{ 2 } \)(7-i) or \(\frac { 7 }{ 2 } \) - \(\frac { i }{ 2 } \)

Given \(f(x)=sin^{ -1 }\left( \frac { x^{ 2 }+1 }{ 2x } \right) \le 1\)
We know that the domain of sin^-1(x) is [-1, 1]
\(\Rightarrow -1\le \frac { { x }^{ 2 }+1 }{ 2x } \le 1\)
Consider \(\Rightarrow -1\le \frac { { x }^{ 2 }+1 }{ 2x } \)
\(\Rightarrow 0\le \frac { { x }^{ 2 }+1 }{ 2x } +1\)
\(\Rightarrow \frac { { x }^{ 2 }+1+2x }{ 2x } \ge 0\)
\(\Rightarrow \frac { \left( x+1 \right) ^{ 2 } }{ 2x } \ge 0\)
\(\Rightarrow \) x = -1 and x < 0
Consider \(\cfrac { { x }^{ 2 }+1 }{ 2x } \le 1\)
\(\Rightarrow \frac { { x }^{ 2 }+1 }{ 2x } -1\le 0\)
\(\Rightarrow \frac { { x }^{ 2 }-2x+1 }{ 2x } \le 0\)
\(\Rightarrow \frac { \left( x-1 \right) ^{ 2 } }{ 2x } \le 0\)
From (1) and (2) Domain {-1, 1}

z₁(z₂ + z₃) = z₁z₂ + z₁z₃
Given z₁= 3, z₂ = -7i, z₃ = 5+4i
LHS = z₁(z₂ + z₃)
= 3 [-7i + 5 + 4i]
= 3[5-3i]
= 15-9i
RHS = z₁z₂ + z₁z₃
= 3(-7i) + 3(5 + 4i)
= -21i +15 +12i
= -9i +15
= 15-9i
LHS = RHS
∴ z₁(z₁ + z₃) = z₁z₂ + z₁z₃
Hence proved

The statement (p ∧ q) ∧ ~(p ∨ q)

We shall follow the tree diagram to calculate
So first we need to find \(\frac { \partial g }{ \partial x } ,\frac { \partial g }{ \partial y } ,\frac { dx }{ dt } \) and \(\frac { dx }{ dt } \).
Now, \(\frac { \partial g }{ \partial x } \) = 2x-y + cos(x + y), \(\frac { \partial g }{ \partial x } \) = -x+cos(x + y), \(\frac { dx }{ dt } \) = 3e^3t and \(\frac { dx }{ dt } \) = 2t.
Thus, \(\frac { dg }{ dt } =\frac { \partial g }{ \partial x } \frac { dx }{ dt } +\frac { \partial g }{ \partial y } \frac { dy }{ dt } \)
= (2x − y + cos(x + y)) 3e^3t + (−x + cos(x + y))( 2t)
= (2e^3t - t² + cos(e^3t - t²))3e^3t +( -e^3t + cos(e^3t - t²))(2t ) 
= 6e^6t - 3t² e^3t +3e^3t cos(e^3t - t²) -2te^3t +2t cos(e^3t - t²) 
Also, some times our W(x, y) will be such that x = x(s, t) , and y = y(s, t) where s, t ∈ R. Then W can be considered as a function that depends on s and t. If x, y both have partial derivatives with respect to s, t and W has partial derivatives with respect to x and y, then we can calculate the partial derivatives of W with respect to s and t using the following theorem.

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
\(I=\int _{ 0 }^{ 2a }{ { x }^{ 2 }\sqrt { 2ax-{ x }^{ 2 } } dx } \)
\(\int _{ \frac { \pi }{ 2 } }^{ 0 }{ { 4a }^{ 2 }{ cos }^{ 2 }\theta \sqrt { { 4a }^{ 2 }{ cos }^{ 2 }\theta -4{ a }^{ 2 }{ cos }^{ 4 }\theta } } (-4a\ cos\theta sin\ \theta )d\theta \)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { 4a }^{ 2 }{ cos }^{ 2 }\theta\ 2a\ cos\ \theta sin\ \theta (4a\ cos\ \theta sin\theta )d\theta } \)
\(=32{ a }^{ 4 }\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { cos }^{ 4 }\theta { sin }^{ 2 }\theta d\theta } \)
\(=32{ a }^{ 4 }\times \frac { 1 }{ 6 } \times \frac { 3 }{ 4 } \times \frac { 1 }{ 2 } \times \frac { \pi }{ 2 } =\pi { a }^{ 4 }\) 

Let f(x) = e^-x, x ∈ [a, b]
a) e-x is continuous in [a, b]
b) e-x is differentiable in (a, b)
c) f(b) = e^-b, f(a) = e^-a
By Lagrange's mean value I theorem, there exerise c ∈ [a, b] such that f'(c) = \(\frac { f(b)-f(a) }{ b-a } \)
⇒ -e^-c =\(\frac { { e }^{ -b }-{ e }^{ -a } }{ b-a } \)
⇒ -e^-c = \(\frac { { e }^{ -a }-{ e }^{ -b} }{ b-a } \)
⇒ |e^-c| = \(\left| \frac { { e }^{ -b }-{ e }^{ -a } }{ b-a } \right| \)
⇒ \(\frac { \left| { e }^{ -a }-{ e }^{ -b } \right| }{ \left| a-b \right| } \) < 1
[∵ |-a^-c| < 1 for c ∊ [a, b] a > 0, b > 0]
⇒ |e^-a - e^-b| < |a -b|
Hence proved.

\({ cot }^{ -1 }x-{ xot }^{ -1 }\left( x+2 \right) =\frac { \pi }{ 12 } ,x>0\)
\({ tan }^{ -1 }\left( \frac { 1 }{ x } \right) -{ tan }^{ -1 }\left( \frac { 1 }{ x+2 } \right) =\frac { \pi }{ 2 } \)
\(\left[ \because { cot }^{ 1 }\left( x \right) ={ tan }^{ -1 }\left( \frac { 1 }{ x } \right) ifx>0 \right] \)
\(\Rightarrow { tan }^{ -1 }\left( \frac { \frac { 1 }{ x } +\frac { 1 }{ x+2 } }{ 1+\frac { 1 }{ x } .\frac { 1 }{ x+2 } } \right) =\frac { \pi }{ 2 } \)
\(\Rightarrow \left( \frac { \frac { x+2-x }{ x(x+2) } }{ \frac { x(x+2)+1 }{ x(x+2) } } \right) ={ tan15 }^{ 0 }\)
\(\left[ \because { tan15 }^{ 0 }=tan\left( { 45 }^{ 0 }-30^{ 0 } \right) \right] \)
\(\frac { { tan45 }^{ 0 }-{ tan30 }^{ 0 } }{ 1+tan{ 45 }^{ 0 }tan{ 30 }^{ 0 } } \)
\(\frac { 1-\frac { 1 }{ \sqrt { 3 } } }{ 1+\frac { 1 }{ \sqrt { 3 } } } =\frac { \sqrt { 3 } -1 }{ \sqrt { 3 } +1 } \times \frac { \sqrt { 3 } -1 }{ \sqrt { 3 } -1 } \)
\(\frac { \left( \sqrt { 3 } -1 \right) ^{ 2 } }{ 3-1 } =\frac { \left( \sqrt { 3 } -1 \right) ^{ 2 } }{ 2 } \)

\(\Rightarrow \frac { 2 }{ { x }^{ 2 }+2x+1 } =\frac { \left( \sqrt { 3 } -1 \right) ^{ 2 } }{ 2 } \)
\(\Rightarrow \frac { 7 }{ (x+1)^{ 2 } } =\frac { \left( \sqrt { 3 } -1 \right) ^{ 2 } }{ 2 } \)
\(\Rightarrow 4=(x+1)^{ 2 }\left( \sqrt { 3 } -1 \right) ^{ 2 }\)
Taking square root both sides
\(2=(x+1)(\sqrt { 3 } -1)\)
\(\Rightarrow x+1=\frac { 2 }{ \sqrt { 3- } 1 } \times \frac { \sqrt { 3 } +1 }{ \sqrt { 3 } +1 } =\frac { 2\left( \sqrt { 3 } +1 \right) }{ 3-1 } =\sqrt { 3 } +1\)
\(x+1=\sqrt { 3 }+ 1\)
\(x=\sqrt { 3 } \)

The equation of the plane passing through the intersection of the planes 2x + 3y−z + 7 = 0 and x + y− 2z + 5 = 0 is (2x + 3y −z + 7) +λ (x + y −2z + 5) = 0 or (2 + λ )x+ (3 + λ )y+ (−1 − 2λ ) z+ (7 + 5λ) = 0
since this plane is perpendicular to the given plane x+y−3z−5 = 0, the normals of these two planes are perpendicular to each other.
Therefore, we have (1)(2 + λ) + (1)(3 + λ) + (−3)(−1 − 2λ)z = 0
which implies that λ = −1.
Thus the required equation of the plane is
(2x + 3y − z + 7)−(x + y −2z + 5) = 0
⇒ x + 2y + z + 2 = 0

\(\frac { { (x-3) }^{ 2 } }{ 225 } +\frac { ({ y-4) }^{ 2 } }{ 289 } =1\)
Given equation is \(\frac { { (x-3) }^{ 2 } }{ 225 } +\frac { ({ y-4) }^{ 2 } }{ 289 } =1\)
This is an equation of the ellipse a² = 289
b² = 225 and
c² = a² - b² ⇒ 289 - 225 = 64 ⇒ c = 8.
\(e=\sqrt { \frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 225 }{ 289 } } =\sqrt { \frac { 289-225 }{ 289 } } \)
 \(=\sqrt { \frac { 64 }{ 289 } } =\frac { 8 }{ 17 } \)
(a) Center is (3, 4) ⇒ h = 3, k = 4
(b) foci are (h, k+c), (h, k-c)
⇒ (3, 4 + 8), (3, 4 - 8) ⇒ (3, 12), (3,-4)
(c) Vertices are (h, k - a), (h, k + a)
⇒ (3, 4 -17), (3, 4 + 17) ⇒ (3, -13), (3, 21)
(d) Equations of directrices are y - 4 = \(\pm \frac { a }{ e } \)
\(\Rightarrow y-4=\pm \frac { 17 }{ \frac { 8 }{ 17 } } +4\Rightarrow y-4=\pm \frac { -289 }{ 8 } +4\)
\(\Rightarrow y=\frac { 289 }{ 8 } +4\) and \(y=\frac { -289 }{ 8 } +4\)
\(\Rightarrow y=\frac { 289+32 }{ 2 } \) and \(y=\frac { -289+32 }{ 8 } \)
\(\Rightarrow y=\frac { 321 }{ 8 } \) and \(y=\frac { -257 }{ 8 } \)

Applying Gauss-Jordan method, we get
[A | I₂] = \(\left[ \begin{matrix} 0 & 5 \\ -1 & 6 \end{matrix}|\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} -1 & 6 \\ 0 & 5 \end{matrix}|\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right] \overset { { R }_{ 1 }\longrightarrow \left( -1 \right) { R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -6 \\ 0 & 5 \end{matrix}|\begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right] \)\(\overset { { R }_{ 1 }\longrightarrow \left( -1 \right) { R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -6 \\ 0 & 5 \end{matrix}|\begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right] \overset { { R }_{ 2 }\longrightarrow \frac { 1 }{ 5 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -6 \\ 0 & 1 \end{matrix}|\begin{matrix} 0 & -1 \\ \left( 1/5 \right) & 0 \end{matrix} \right] \)\(\overset { { R }_{ 1 }\longrightarrow { R }_{ 1 }+6{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}|\begin{matrix} \left( 6/5 \right) & -1 \\ \left( 1/5 \right) & 0 \end{matrix} \right] \).
So, we get A^-1 = \(\left[ \begin{matrix} \left( 6/5 \right) & -1 \\ \left( 1/5 \right) & 0 \end{matrix} \right] =\frac { 1 }{ 5 } \left[ \begin{matrix} 6 & -5 \\ 1 & 0 \end{matrix} \right] \).

The sum of the coefficients of the polynomial is 0. Hence 1 is a root of the polynomial. To find other roots, we divide x³- 3x²- 33x + 35 by x-1 and get x²-2x-35 as the quotient. Solving this we get 7 and −5 as roots. Thus 1, 7, −5 form the solution set of the given equation.