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.

Note that the function u is not homogeneous. So we cannot apply Euler’s Theorem for u.
However, note that f(x,y) = \(\frac { x+y }{ \sqrt { x } +\sqrt { y } }\) = sin u is homogeneous; because
f(tx,ty) = \(\frac { tx+ty }{ \sqrt { tx } +\sqrt { ty } } \) = t^1/2 f(x, y), \(\forall \) x, y, t\(\ge \)0
Thus f is homogeneous with degree \(\frac { 1 }{ 2 } \) and so by Euler’s Theorem we have
\(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } =\frac { 1 }{ 2 } f(x,y)\).
Now substituting f = sin u in the above equation, we obtain
\(x\frac { \partial (sinu) }{ \partial x } +y\frac { \partial (sinu) }{ \partial y } =\frac { 1 }{ 2 } sin \ u\)
\(x\quad cosu\frac { \partial u }{ \partial x } +y\quad cosu\frac { \partial u }{ \partial x } =\frac { 1 }{ 2 } sin \ u\)  ...(19)
Dividing both sides by cosu we obtain
\(x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } =\frac { 1 }{ 2 } tan \ u\)
Note:
Solving this problem by direct calculation will be possible; but will involve lengthy calculations.

The equation of the parabola is ( y + 2)² = −(x−9). The parabola crosses the y- axis at (0, −5) and (0, 1). The vertex is at (9, −2) and the axis of the parabola is y = −2. The required area is sketched
Viewing in the positive direction of x-axis, and making horizontal strips, the required area A is given by

\(A=\int _{ -5 }^{ 1 }{ xdy } =\int _{ -5 }^{ 1 }{ (5-4y-{ y }^{ 2 })dy={ \left[ 5y-2{ y }^{ 2 }-\frac { { y }^{ 3 } }{ 3 } \right] }_{ -5 }^{ 1 }=\frac { 8 }{ 3 } -\left( -\frac { 100 }{ 3 } \right) =36 } \)

Let us put I = \(\int ^\frac{\pi}{4}_{0}\) log(1 + tan x) dx
Applying the property \(\int ^{a}_{0}\) f(x) dx =\(\int ^{a}_{0}\)f(a-x) dx in equation (1), we get
I = \(\int ^\frac{\pi}{4}_{0}\) log \([1+tan (\frac{\pi}{4}-x)]\)dx = \(\int ^\frac{\pi}{4}_{0}\) log\([1 + \frac{tan \frac {\pi}{4}- tan x}{1 + tan {\frac {\pi}{4} tan x}}]\) dx
=\(\int ^\frac{\pi}{4}_{0}\) log \([1 + \frac {1 - tan x}{1 + tan x}]\)dx = \(\int ^\frac{\pi}{4}_{0}\) log\([\frac {1+tan x +1 - tan x}{1 + tan x}]\) dx
=\(\int ^\frac{\pi}{4}_{0}\) log \([\frac{2}{1+tanx}]\) dx = \(\int ^\frac{\pi}{4}_{0}\) [log 2 - log (1+tan x)] dx
= log 2  \(\int ^\frac{\pi}{4}_{0}\) dx - \(\int ^\frac{\pi}{4}_{0}\)log (1+tan x)] dx
= \(\frac{\pi}{4}\)log 2 - I
So, we get 2I = \(\frac{\pi}{4}\)log 2.
Hence, we get I = \(\frac{\pi}{8}\)log 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

Let A be the mass of the isotope remaining after t years, and let −k be the constant of proportionality, where k > 0. Then the rate of decomposition is modeled by \(\frac{da}{dt}=-kA,\) where the minus sign indicates that the mass is decreasing. It is a separable equation. Separating the variables,we get\(\frac{da}{dt}=-kdt\).
Integrating on both sides, we get log |A| = −kt + log |C| or A = Ce−^kt.
Given that the initial mass is 200mg. That is, A = 200 when t = 0 and thus, C = 200.
Thus, we get A = − 200e^-kt.
Also, A =150when t = 2 and therefore, k = \(\frac{1}{2}log(\frac{4}{3})\)
Hence, A(t) = 200e^{\(\frac{1}{2}log(\frac{4}{3})\)} is the mass of isotope remaining after t years.
The half-life t_h is the time corresponding to A = 100 mg
Thus, \({ t }_{ k }=\frac { 2log\left( \frac { 1 }{ 2 } \right) }{ log\left( \frac { 3 }{ 4 } \right) } \).

Let x and y be the length, breadth of the printed rectangular page

Given xy = 24
\(\Rightarrow y=\frac { 24 }{ x } \)    ..(1)
Length of the page with margin
= x+1+1 = x+2
breadth of the page with margin
= y + 1.5 + 1.5 = y + 3
Area of the page = (x + 2) (y + 3)
Let f(x) = (x + 2) (y + 3)
= \((x+2)\left( \frac { 24 }{ x } +3 \right) \) 
= \(24+3x+\frac { 48 }{ x } +30\) 
= \(3x+\frac { 48 }{ x } +30\) 
\(f'\left( x \right) =3-\frac { 48 }{ { x }^{ 2 } } \)
\(\Rightarrow { x }^{ 2 }=16\Rightarrow x=\pm 4\)
ஃ The critical number are 4,-4
\(f''\left( x \right) =-48\left( \frac { -2 }{ { x }^{ 3 } } \right) =\frac { 96 }{ { x }^{ 3 } } \) 
\(f''\left( 4 \right) =\frac { 96 }{ 64 } >0\) 
ஃ f(x) is minimum when x = 4
When \(x=4,y=\frac { 24 }{ 4 } =6\)  [From (1)]
ஃ Length of the page = x + 2 = 4 + 2 = 9 cm
Breadth of the page = y + 3 = 6 + 3 = 6 cm

\(\\ \\ \\ \frac { dy }{ dx } ={ tan }^{ 2 }(x+y)...(1)\)
Take x + y = t
\(\Rightarrow 1+\frac { dy }{ dx } =\frac { dt }{ dx } \)
\(\Rightarrow \frac { dy }{ dx } =\frac { dt }{ dx } -1\)
∴ (1) becomes,
\(\frac { dt }{ dx } -1={ tan }^{ 2 }t\)
\(\Rightarrow \frac { dt }{ dx } ={ tan }^{ 2 }t1\)
\(\Rightarrow \frac { dt }{ dx } ={ sec }^{ 2 }(t)\)
\(\Rightarrow \frac { dt }{ { sec }^{ 2 }t } =dx\)
\(\Rightarrow { cos }^{ 2 }t\quad dt=dx\)
\(\left(\frac{1+\cos 2 t^{\circ}}{2}\right) d t=\mathrm{d} x \quad\left(\because \cos ^2 \theta=\frac{1+\cos 2 \theta}{2}\right)\)
\(\left[ cos\quad 2x=2{ cos }^{ 2 }x-1{ cos }^{ 2 }x=\frac { 1+cos2x }{ 2 } \right] \)
Taking integration on both sides, we get
\(\Rightarrow \left( \frac { 1+cos2\quad t }{ 2 } \right) dt=dx\)
\(\Rightarrow \frac { 1 }{ 2 } \int { (1+cos2t)dt=\int { dx } } \)
\(\Rightarrow \frac { 1 }{ 2 } \left[ t+\frac { sin2t }{ 2 } \right] =x+c\)
\(\Rightarrow \frac { 1 }{ 2 } \left[ t+\frac { 2sintcost }{ 2 } \right] =x+c\)
\(\Rightarrow \frac { 1 }{ 2 } [t+sin\ t\ cost]=x+c\ [\because t=x+y]\)
\(\Rightarrow \frac { 1 }{ 2 } [x+y+sin(x+y)cos(x+y)=x+c\)

Given s (t) = 16t², height = 400 ft.
⇒ t² = \(\frac { 400 }{ 16 } =\frac { 100 }{ 4 } \)
t² = 25
t = 5 sec
(ii) Average velocity = \(\frac { ds }{ dt } \) = 32 t
When t = 2 sec
Average in the last
2 sec = \(\frac { V \ at \ t=3+V \ at \ t=5 }{ 2 } \)
= \(\frac { 32(3)+32(5) }{ 2 } \)
= \(\frac { 96+160 }{ 2 } =\frac { 256 }{ 2 } \)
= 128 f/sec
(iii) Instantaneous Velocity
=\(\frac { ds }{ dt } \) = 32t
When t = 5 sec
Velocity = \(\frac { ds }{ dt } \) = 32(5)
= 160 ft/sec

The plane containing the line
\(\vec { r } =(\hat { i } -\hat { j } +3\hat { k } )+t(2\hat { i } -\hat { j } +4\hat { k } )\)
∴ The required plane is passing through the point \(\vec { a } =\hat { i } -\hat { j } +3\hat { k } \) and parallel to a vector \(\vec { b } =2\hat { i } -\hat { j } +4\hat { k } \) Also, the plane is perpendicular to the plane 
\(\vec { c } =\hat { i } +2\hat { j } +\hat { k } \) 
∴ The parametric form of vector equation of the plane passing through one point (\(\vec { a } \)) and parallel to two vectors \(\vec { b } \) and \(\vec { c } \) 
\(\vec { r } =\vec { a } +s\vec { b } +t\vec { c } \) where s, t ∈ R
⇒ \(\vec { r } .(\hat { i } -\hat { j } +3\hat { k } )+s(2\hat { i } -\hat { j } +4\hat { k } )+t(\hat { i } +2\hat { j } +\hat { k } )\) s, t  ∈ R 
Cartesian equation is
\(\left| \begin{matrix} x-{ x }_{ 1 } & y-{ y }_{ 1 } & z-{ z }_{ 1 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \\ { c }_{ 1 } & { c }_{ 2 } & { c }_{ 3 } \end{matrix} \right| =0\)
\(\Rightarrow \left| \begin{matrix} x-1 & y+1 & z-3 \\ 2 & -1 & 4 \\ 1 & 2 & 1 \end{matrix} \right| =0\)
⇒ (x - 1)(-1 - 8) - (y + 1)(2 - 4) + (z - 3)(4 + 1) = 0
⇒ (x - 1)(-9) - (y + 1)(-2) + (z - 3)5 = 0
⇒ -9x + 9 +2y + 2 + 5z - 15 = 0
⇒ -9x + 2y + 5z - 4 = 0
⇒ 9x - 2y - 5z + 4 = 0

Let AB be the rod and P(x₁, y₁) be a point on the rod such that AP = 0.3 m.
Draw PD ⊥ x-axis and PC ⊥ y - axis.

Δ ADP ≅ Δ PCB
∴ \(\frac { PC }{ DA } =\frac { PB }{ AP } =\frac { BC }{ PD } \)
⇒ \(\frac { x_{ 1 } }{ DA } =\frac { 0.9 }{ 0.3 } =\frac { BC }{ { y }_{ 1 } } \)
⇒ \(DA=\frac { 0.3{ x }_{ 1 } }{ 0.9 } =\frac { { x }_{ 1 } }{ 3 } \)
and BC = \(\frac { 0.9{ y }_{ 1 } }{ 0.3 } =\frac { 9 }{ 3 } { y }_{ 1 }=3{ y }_{ 1 }\)
Now OA = OD + DA
= \({ x }_{ 1 }+\frac { { x }_{ 1 } }{ 3 } =\frac { 4{ x }_{ 1 } }{ 3 } \)
OB = OC + BC = y₁ + 3y₁ = 4y₁
But OA² + OB² = AB²
⇒ \({ \left( \frac { 4{ x }_{ 1 } }{ 3 } \right) }^{ 2 }+{ \left( 4{ y }_{ 1 } \right) }^{ 2 }={ \left( 1.2 \right) }^{ 2 }\)
⇒ \(\frac { { { x }_{ 1 } }^{ 2 } }{ 9 } +\frac { { { y }_{ 1 } }^{ 2 } }{ 9 } =\frac { 1.44 }{ 16 } =0.09\) ≅ 1
∴ Locus of (x₁, y₁) is \(\frac { { x }^{ 2 } }{ 9 } +\frac { { y }^{ 2 } }{ 1 } =1\)
Here a² = 9, b² = 1
∴ \(e=\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 1 }{ 9 } } =\sqrt { \frac { 9-1 }{ 9 } } \)
= \(\sqrt { \frac { 8 }{ 9 } } \)
e = \(\frac { 2\sqrt { 2 } }{ 3 } \) 

Here the number of unknowns is 3.
The matrix form of the system is AX = B, where A = \(\left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 1 & \lambda \\ 1 & 3 & -5 \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \), B = \(\left[ \begin{matrix} 7 \\ \mu \\ 5 \end{matrix} \right] \).
Applying elementary row operations on the augmented matrix [A | B], we get
[A | B] = \(\left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 1 & \lambda \\ 1 & 3 & -5 \end{matrix}|\begin{matrix} 7 \\ \mu \\ 5 \end{matrix} \right] \overset { { R }_{ 2 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 3 & -5 \\ 1 & 1 & \lambda \end{matrix}|\begin{matrix} 7 \\ 5 \\ \mu \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 1 \\ 0 & 1 & -6 \\ 0 & -1 & \lambda -1 \end{matrix}|\begin{matrix} 7 \\ -2 \\ \mu -7 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }+{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 1 \\ 0 & 1 & -6 \\ 0 & 0 & \lambda -7 \end{matrix}|\begin{matrix} 7 \\ -2 \\ \mu -9 \end{matrix} \right] \).
(i) If λ =7 and μ \(\neq\) 9, then ρ(A) = 2 and ρ([A | B]) = 3. So ρ(A) ≠ ρ([A | B]) Hence the given system is inconsistent and has no solution.
(ii) If λ ≠ 7 and μ is any real number, then ρ(A) = 3 and ρ([A | B]) = 3.
So, ρ(A) = ρ([A | B]) = 3 = Number of unknown. Hence the given system is consistent and has a unique solution.
(iii) If λ = 7 and μ = 9, then ρ(A) = 2 and ρ([A | B]) = 2.
So, ρ(A) = ρ([A | B]) = 2 < Number of unknown. Hence the given system is consistent and has infinite number of solutions.

We know that \({ tan }^{ -1 }x+{ tan }^{ -1 }y={ tan }^{ -1 }\left( \frac { x+y }{ 1-xy } \right) \)
\(\therefore LHS={ tan }^{ -1 }(x)+{ tan }^{ -1 }(y)+{ tan }^{ -1 }(z)\)
= \({ tan }^{ -1 }\left( \frac { x+y }{ 1-xy } \right) +{ tan }^{ -1 }\left( z \right) \)
= \({ tan }^{ -1 }\left( \frac { \frac { x+y }{ 1-xy } +z }{ 1-z\left( \frac { x+y }{ 1-xy } \right) } \right) \) by(1)
= \({ tan }^{ -1 }\left( \frac { \frac { x+y+z(1-xy) }{ 1-xy } }{ \frac { (1-xy)-z(x+y) }{ 1-xy } } \right) \)
= \({ tan }^{ -1 }\left( \frac { \frac { x+y+z-xyz }{ 1-xy } }{ \frac { 1-xy-zx-zy }{ 1-xy } } \right) \)

= \({ tan }^{ -1 }\left( \frac { x+y+z-xyz }{ 1-xy-yz-zx } \right) =RHS\)
Hence proved.

Given z = x + iy
Im \(\left( \frac { 2z+1 }{ iz+1 } \right) \)= 0
⇒ Im\(\left( \frac { 2(x+iy)+1 }{ i(x+iy)+1 } \right) \)= 0
⇒ Im\(\left( \frac { (2x+1)+2iy }{ ix+i^{ 2 }y+1 } \right) \)
⇒ Im\(\left( \frac { (2x+1)+2iy }{ ix-y+1 } \right) \)
\(\left( \frac { (2x+1)+iy }{ (1-y)+ix } \right) \)
Multiply and divide by the conjugate of the denominator
We get Im\(\left( \frac { (2x+1)+2iy }{ (1-y)+ix } \times \frac { (1-y)-ix }{ (1-y)-ix } \right) \)=0
⇒ Im\(\left( \frac { (2x+1)+2iy\times (1-y)-ix }{ (1-y)^{ 2 }+{ x }^{ 2 } } \right) \)
Choosing the imaginably part we get,
\(\frac { (2x+1)(-x)+2y(1-y) }{ (1-y)^{ 2 }+{ x }^{ 2 } } \)
⇒ (2x+1)-x+2y(1-y) = 0
⇒ -2x²-x+2y-2y² = 0
⇒ 2x²+2y²+x-2y = 0
Hence, locus of z is 2x²+2y²+x-2y = 0

For the parabola,
x²- 4x - 5y -1 = 0
x²- 4x = 5y +1
x²−4x +4 = 5y +1+ 4
(x − 2)² = 5(y +1) which is in standard form.
Therefore 4a = 5 and the vertex is (2, -1) , and focus is \(\left( 2,\frac { 1 }{ 4 } \right) \)
Equation of directrix is 
y-k+ a = 0
y+1+\(\frac { 5 }{ 4 } \)
4y +9 = 0 
Length of latus rectum is 5 units.

Let \(\hat { a } =\vec { OA } \) and \(\hat { b } =\vec { OB } \) be the unit vectors making angles α and β respectively, with positive x -axis, where A and B are as shown in the diagram.
Then, we get
\(\hat { a } \) = cosα\(​​\hat{i}\)+ sinα\(\hat {j } \) and \(\hat{b}\)ˆ = cosβ\(​​\hat{i}\)+ sin β \(\hat {j } \),

The angle between \(\hat { a } \) and \(\hat{b}\) is α −β and, the vectors \(\hat{b}\), \(\hat { a } \), \(\hat{k}\) form a right-handed system.
Hence, we get \(\hat { b } \times \hat { a } =\left| \hat { b } \right| \left| \hat { a } \right| \)sin(α −β )\(\hat {k}\) = sin(α −β)\(\hat {k}\)
On the other hand, \(\hat { b } \times \hat { a } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ cos\beta & sin\beta & 0 \\ sin\alpha & sin\alpha & 0 \end{matrix} \right| \) = (sin α cos β -cos α sin β)\(\hat { k } \)
Hence, by equations (1) and (2), we get
sin(α −β ) = sin α cos β − cos α sin β

Since the coefficients of the equation are all rational numbers, 2+i and 3-\(\sqrt{2}\) are roots, we get 2-i and 3+\(\sqrt{2}\) are also roots of the given equation. Thus (x-(2+i)), (x-(2-i)), (x-(3-\(\sqrt{2}\))) and (x-(3+\(\sqrt{2}\))) are factors. Thus their product.
((x-(2+i))(x-(2-i))(x-(3-\(\sqrt{2}\)))(x-(3+\(\sqrt{2}\))) is a factor of the given polynomial equation.
That is, (x²-4x+5)(x²-6x+7) is a factor. Dividing the given polynomial equation by this factor, we get the other factor as (x²-3x-4) which implies that 4 and −1 are the other two roots. Thus
2+i, 2-i, 3+\(\sqrt{2}\), 3-\(\sqrt{2}\), -1, and 4 are the roots of the given polynomial equation.

Let the general equation of the circle be
x² +y² +2gx + 2fy + c = 0 ........ (1)
It passes through points (1, 1), (2, -1) and(3, 2) .
Therefore,
2g+2f+c = -2 ........ (2)
4g−2f+c = -5 ........(3)
6g+4f+c = -13 ...... (4)
(2) – (3) gives −2g + 4f = 3 ... (5)
(4) – (3) gives 2g + 6f = -8  ....(6)
(5) + (6) gives f = \(-\frac { 1 }{ 2 } \) 
Substituting f = \(-\frac { 1 }{ 2 } \) in (6), g = -\(\frac { 5 }{ 2 } \)
Substituting f = \(-\frac { 1 }{ 2 } \) and g = - \(\frac { 5 }{ 2 } \) in (2), c = 4 .
Therefore the required equation of the circle is
x²+ y²+ 2\(\left( -\frac { 5 }{ 2 } \right) x+2\left( -\frac { 1 }{ 2 } \right) \)y+4 = 0 and x² + y² − 5x − y + 4 = 0.