The condition for the line y = mx+c to be a tangent to
\(x^{2}+y^{2}=a^{2} \text { is } \mathrm{c}=\pm a \sqrt{1+m^{2}}\)
Here a = 3, m = 3
\(\therefore c=\pm 3 \sqrt{10}\)

a *b = min (a, b) on A = {1,2,3,4, 5}  Let a,b ∈A
A = {1,2, 3, 4, 5}
a*b = min {(a, b)}
Now, 1,2 ∈ A \(\Rightarrow\)1 * 2 = 1 ∈ A
3, 5 ∈ A \(\Rightarrow\) 3 * 5 = 3 ∈ A
Hence * is a binary operation on A

Given n = 6, \(p=\frac { 1 }{ 3 } \), k = 3
\(P(X=k)=\left( \begin{matrix} n \\ k \end{matrix} \right) { p }^{ k }\left( 1-p \right) ^{ n-k },\) 
n = 0,1,2, ... n
\(\therefore P(X=k)=\left( \begin{matrix} n \\ k \end{matrix} \right) { p }^{ k }(1-p)^{ n-k }\) 
n = 0,1,2, ... n
\(P(X=3)=\left( \begin{matrix} 6 \\ 3 \end{matrix} \right) \left( \cfrac { 1 }{ 3 } \right) ^{ 3 }\left( 1-p \right) ^{ 6-3 }\) 
= \(\left( \begin{matrix} 6 \\ 3 \end{matrix} \right) \left( \cfrac { 1 }{ 3 } \right) ^{ 3 }\left( \cfrac { 2 }{ 3 } \right) ^{ 2 }\) 

\(P(X=3)=\frac { 160 }{ 729 } \) 

\({ I }_{ n }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { n-1 }{ n } { I }_{ n-2 },n\ge 2 } \)
\(\therefore { I }_{ 7 }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { 6 }{ 7 } \times \frac { 4 }{ 5 } \times \frac { 2 }{ 3 } \times 1 } =\frac { 16 }{ 35 } \)

Given \(f(x)=|\frac{1}{x}|, x\in [-1,1]\)
Rolle's theorem is not applicable since \(f(x)=(\frac{1}{x})\) is not continuous at x = 0 in [-1, 1] and not differentiable in (-1, 1)

y = Acos x + Bsin x ... (1)
Differentiating (1) twice successively, we get
\(\frac{dy}{dx}\)= −Asin x + Bcos x. ...(2)
\(\frac{d^2y}{dx^2}\) = -Acos x − Bsin x = −(A cos x + B sin x). ...(3)
Substituting (1) in (3), we get \(\frac{d^2y}{dx^2}\) + = 0 as the required differential equation

Equation of the circle is
x² + y² + 6x - 4y + 4 = 0.
Here 2g = 6 ⇒ g = 3
2f  = -4 ⇒ f = -2 and c = 4
Centre is (-g, -f) ⇒ (-3, 2)
r = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \) = \(\sqrt { { 3 }^{ 2 }+(-2){ }^{ 2 }-{ 4 } } \)
= \(\sqrt { 9+4-4 } \)
= \(\sqrt { 9 } \)
= 3 unit

Let \(\vec { a } =7\hat { i } +\lambda \hat { j } -3\hat { k } ,\vec { b } =\hat { i } +2\hat { j } -\hat { k } \) and \(\vec { c } =-3\hat { i } +7\hat { j } -5\hat { k } \)
∴ volume of the parallelepiped
= \(\vec { a } .(\vec { b } \times \vec { c } )\)
Given \(\vec { a } .(\vec { b } \times \vec { c } )\) = 90
⇒ \(\left| \begin{matrix} 7 & \lambda & -3 \\ 1 & 2 & -1 \\ -3 & 7 & 5 \end{matrix} \right| \) = 90
⇒ \(-6\left| \begin{matrix} 2 & -1 \\ 7 & 5 \end{matrix} \right| -\lambda \left| \begin{matrix} 1 & -1 \\ -3 & 5 \end{matrix} \right| -3\left| \begin{matrix} 1 & 2 \\ -3 & 7 \end{matrix} \right| \) = 90
⇒ 7(10+7)-λ(5-3)-3(7+6) = 90
⇒ 7(17)-λ(2)-3(13) = 90
⇒ 119-2λ-39 = 90
⇒ 119-39-90 = 2λ
⇒ -10 = 2λ
⇒ λ = -5

Let tan⁻¹(\(\sqrt3\)) = y
Then, tan y =\(\sqrt3\)
Thus, y = \(\frac{\pi}{3}\)
Since \(\in(-\frac{\pi}{2},\frac{\pi}{2})\)
Thus the principal value of tan⁻¹(\(\sqrt3\)) = \(\frac{\pi}{3}\)

Since α, β, and γ are the roots of the equation x³+ px²+ qx + r = 0, we have
Σ₁ α + β + γ = -p and Σ₃ αβγ = -r
\(\Sigma \frac { 1 }{ \beta \gamma } =\frac { 1 }{ \beta \gamma } +\frac { 1 }{ \gamma \alpha } +\frac { 1 }{ \alpha \beta } =\frac { \alpha +\beta +\gamma }{ \alpha \beta \gamma } =\frac { -p }{ -r } =\frac { p }{ r } \).

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.

Given 2a = 20 cm \(\Rightarrow\) a = 10 cm;
2b = 10 cm \(\Rightarrow\)a = 5 cm
\(\therefore\) Equation of the ellipse is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
\(\Rightarrow \frac { { x }^{ 2 } }{ 100 } +\frac { { y }^{ 2 } }{ 25 } =1\Rightarrow \frac { { y }^{ 2 } }{ 25 } =1-\frac { { x }^{ 2 } }{ 100 } =\frac { 100-{ x }^{ 2 } }{ 100 } \)
\(\Rightarrow { y }^{ 2 }=\frac { 25 }{ 100 } (100-{ x }^{ 2 })\)

\(\therefore\) Required volume \(=2\pi \int _{ 0 }^{ 10 }{ { y }^{ 2 }dx } \)
\(=2\pi \int _{ 0 }^{ 10 }{ \frac { 25 }{ 100 } (100-{ x }^{ 2 })dx=\frac { 50\pi }{ 100 } \int _{ 0 }^{ 10 }{ (100-{ x }^{ 2 })dx } } \)
\(=\frac { \pi }{ 2 } { \left[ 100x-\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 10 }=\frac { \pi }{ 2 } \left[ 100-\frac { 1000 }{ 3 } \right] \)
\(V=\frac { \pi }{ 2 } \left( \frac { 3000-1000 }{ 3 } \right) =\frac { \pi }{ 2 } \left( \frac { 2000 }{ 3 } \right) \)
\(=\frac { 1000\pi }{ 3 } \)

Since the given function is a probability mass function, the total probability is one. That is \(\underset { x }{ \Sigma } f(x)=1\)
 From the given data k + 2k + 6k + 5k + 6k +10k+1
\(30k=1\Rightarrow k=\frac { 1 }{ 30 } \) 
Therefore the probability mass function is

x	1	2	3	4	5	6
f(x)	\(\cfrac { 1 }{ 30 } \)	\(\cfrac { 2 }{ 30 } \)	\(\cfrac { 6 }{ 30 } \)	\(\cfrac { 5 }{ 30 } \)	\(\cfrac { 6 }{ 30 } \)	\(\cfrac { 10 }{ 30 } \)

(i) P(2 < X < 6) = f(3)+ f(4)+ f(5) = \(\frac { 6 }{ 30 } +\frac { 5 }{ 30 } +\frac { 6 }{ 30 } =\frac { 17 }{ 30 } \) 
(ii) P(2≤X≤5) = f(2)+f(3)+f(4) = \(\frac { 2 }{ 30 } +\frac { 6 }{ 30 } +\frac { 5 }{ 30 } =\frac { 13 }{ 30 } \) 
(iii) P(2≤4) = f(1)+f(2)+f(3)+f(4) = \(\frac { 1 }{ 30 } +\frac { 2 }{ 30 } +\frac { 6 }{ 30 } +\frac { 5 }{ 30 } =\frac { 14 }{ 30 } \) 
(iv) P(3>X) = f(4)+f(5)+f(6) = \(\frac { 5 }{ 30 } +\frac { 6 }{ 30 } +\frac { 10 }{ 30 } =\frac { 21 }{ 30 } \)   

Let V be the velocity and the retardation (negative acceleration) be -\(\frac{dv}{dt}\)
Given \(\frac{dv}{dt}\) = -V
Separating the variables,
\(\frac{dv}{v}=-dt\)
\(\Rightarrow \int { \frac { dv }{ v } } =-\int { dt } \)
\(\Rightarrow log\quad v=-t+logC\)
\(\Rightarrow logv-logC=-t\)
\(\Rightarrow log\left( \frac { v }{ { C }_{ v } } \right) =-t\)
\(\Rightarrow ={ e }^{ -t }\)
\(\Rightarrow \frac { v }{ { C }_{ v } } ={ Ce }^{ -t }...(1)\)
Given when t = 0, v = m/sec
\(\therefore\) (1) become 10 = C_e0 \(\Rightarrow\) C = 10
\(\therefore\) (1)  v = 10e^-t
When t = 2, v = 10e^-2
\(\Rightarrow v=\frac { 10 }{ { e }^{ 2 } } \)

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 AB be the position of the ladder at any time t such that OA = x and OB = y
Then OA² + OB² = AB²
⇒ x² + y² = 17²
Given \(\frac { dx }{ dt } \) = 5 and x = 8
When x = 8, 8² + y² = 17²
 ⇒ y² = 289 - 64 = 225
 ⇒ y = 15
Differentiating (1) with respect to 't' we get,
\(2x\frac { dx }{ dt } +2y\frac { dy }{ dt } =0\)
⇒ 8(5) + 15 \(\frac { dy }{ dt } \) = 0 [∵ x = 8, \(\frac { dx }{ dt } \) = 15, y = 15]
⇒ 40 + 15\(\frac { dy }{ dt } \) = 0
⇒ \(\frac { dy }{ dt } =\frac { -40 }{ 15 } =\frac { -8 }{ 3 } \) m/sec
∴ The rate of top of the ladder moving down the wall is \(\frac{-8}{3}\) m/sec
(ii) The ladder, the wall and the floor forms a right angled triangle.
 Area = \(\frac12\)xy
Differentiating with respect to 't' we get,
\(\frac { dA }{ dt } =\frac { 1 }{ 2 } \left[ x\frac { dy }{ dx } +y\frac { dx }{ dt } \right] \)
\(=\frac { 1 }{ 2 } \left[ 8\left( -\frac { 8 }{ 3 } \right) +15(5) \right] \)
= \(\frac { 1 }{ 2 } \left[ \frac { -64 }{ 3 } +75 \right] =\frac { 1 }{ 2 } \)
= \(\frac { 1 }{ 2 } \left[ \frac { -64+225 }{ 3 } \right] =\frac { 1 }{ 2 } \left( \frac { 161 }{ 3 } \right) \)
= \(\frac { dA }{ dt } \) = 26.83 sq.m/sec

\(\frac { 3 }{ x } -\frac { 4 }{ y } -\frac { 2 }{ z } \) - 0, \(\frac { 1 }{ x } +\frac { 2 }{ y } +\frac { 1 }{ z } \) - 2 = 0, \(\frac { 2 }{ x } -\frac { 5 }{ y } -\frac { 4 }{ z } \) + 1 = 0
Put \(\frac { 1 }{ x } =u,\frac { 1 }{ y } =v,\frac { 1 }{ z } =w\)
We get 3u - 4v - 2w = 1, u + 2v + w = 2, 2u - 5v - 4w = -1
∴ \(\left| \begin{matrix} 3 & -4 & -2 \\ 1 & 2 & 1 \\ 2 & -5 & -4 \end{matrix} \right| =3\left| \begin{matrix} 2 & 1 \\ -5 & -4 \end{matrix} \right| +4\left| \begin{matrix} 1 & 1 \\ 2 & -4 \end{matrix} \right| -2\left| \begin{matrix} 1 & 2 \\ 2 & -5 \end{matrix} \right| \)
= 3(- 8 + 5) + 4'(- 4 - 2) - 2(- 5 - 4)
= 3(- 3) + 4(- 6) - 2(- 9)
= - 9 - 24 + 18 = -15
Δ₁ = \(\left| \begin{matrix} 1 & -4 & -2 \\ 2 & 2 & 1 \\ -1 & -5 & -4 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 2 & 1 \\ -5 & -4 \end{matrix} \right| +4\left| \begin{matrix} 2 & 1 \\ -1 & -4 \end{matrix} \right| -2\left| \begin{matrix} 2 & 2 \\ -1 & -5 \end{matrix} \right| \)
= 1(- 8 + 5) + 4(- 8 + 11 -2(-10 + 2)
= 1(- 3) + 4(-7) - 2(- 8)
= - 3 - 28 + 16= -15
Δ₂ = \(\left| \begin{matrix} 3 & 1 & -2 \\ 1 & 2 & 1 \\ 2 & -1 & -4 \end{matrix} \right| \)
\(3\left| \begin{matrix} 2 & 1 \\ -1 & -4 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 2 & -4 \end{matrix} \right| -2\left| \begin{matrix} 1 & 2 \\ 2 & -1 \end{matrix} \right| \)
= 3(- 8 + 1) - 1(- 4 - 2) - 2(- 1 - 4)
= 3(-7) - 1(- 6) - 2(- 5)
= -21 + 6 + 10 = -5
Δ₃ = \(\left| \begin{matrix} 3 & -4 & 1 \\ 1 & 2 & 2 \\ 2 & -5 & -1 \end{matrix} \right| \)
\(3\left| \begin{matrix} 2 & 2 \\ -5 & -1 \end{matrix} \right| +4\left| \begin{matrix} 1 & 2 \\ 2 & -1 \end{matrix} \right| -2\left| \begin{matrix} 1 & 2 \\ 2 & -5 \end{matrix} \right| \)
= 3(-2 + 10) + 4(-1 - 4)+ 1(-5 - 4)
= 3(8) + 4(- 5) + 1(- 9)
= 24 - 20 - 9 = - 5
∴ \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { -15 }{ -15 } =1\Rightarrow \frac { 1 }{ x } =1\Rightarrow \)x = 1
v = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { -5 }{ -15 } =\frac { 1 }{ 3 } \Rightarrow \frac { 1 }{ y } =\frac { 1 }{ 3 } \Rightarrow \)y = 3
w = \(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { 5 }{ -15 } =\frac { 1 }{ 3 } \Rightarrow \frac { 1 }{ z } =\frac { 1 }{ 3 } \Rightarrow \)z = 3
∴ Solution set is {1, 3, 3}

The plane passing through three points namely
\(\vec { a } =3\hat { i } +6\hat { k } -2\hat { k } \),
\(\vec { b } =-\hat { i } -2\hat { j } +6\hat { k } \) and
\(\vec { c } =6\hat { i } +4\hat { j } -2\hat { k } \)
Now, \(\vec { b } -\vec { a } =(-1-3)\hat { i } +(-2-6)\hat { j } +(6+2)\hat { k } \)
\(=-4\hat { i } -8\hat { j } +8\hat { k } \)
\(\vec { c } -\vec { a } =(6-3)\hat { i } +(-4-6)\hat { j } +(-2+2)\hat { k } \)
\(=3\hat { i } -10\hat { j } \)
The parametric form of vector equation of the plane passing through three points is
\(\vec { r } =\vec { a } +s(\vec { b } -\vec { a } )+t(\vec { c } -\vec { a } ),s,t\in R\)
\(\Rightarrow \vec { r } =(3i+6i-2k)+s(4\hat { i } -8\hat { j } +8\hat { k } )+(3\hat { i } -10\hat { j } )s,t\in R\)
The parametric form of vector equation of the plane passing through three points is
\([\vec { r } -\vec { a } ,\vec { b } -\vec { a } ,\vec { c } -\vec { a } ]\) = 0
\(\Rightarrow \vec { r } -(3i+6i-2k).[(4\hat { i } -8\hat { j } 8\hat { k } )+(3\hat { i } -10\hat { j } )]=0\)
Cartesian equation is 
\( \Rightarrow \vec{r}(2 \hat{i}+3 \hat{j}+4 \hat{k})=16 \)
\( \Rightarrow 2 x+3 y+4 z-16=0 \)

By taking the vertex; at the origin, the parabola is open downward.

Its equation is x² = -4ay
It passes through (6, -4)
∴ 36 = -4a(-4) ⇒ 4a = - \(\frac { 36 }{ 4 } \) = 9
∴ (1) becomes, x² = -9y
To find the slope at (-6, -4)
Differentiating (1) with respect to 'x' we get,
2x = -9\(\frac { dy }{ dx } \)
⇒ \(\frac { dy }{ dx } =\frac { -2x }{ 9 } \)
At (-6, -4), \(\frac { dy }{ dx } =-2\frac { (-6) }{ 9 } =\frac { 12 }{ 9 } =\frac { 4 }{ 3 } \)
∴ \(tan\theta =\frac { 4 }{ 3 } \Rightarrow \theta ={ tan }^{ -1 }\left( \frac { 4 }{ 3 } \right) \)
∴ The angle of projection is tan^-1 \(\left( \frac { 4 }{ 3 } \right) \)

The parametric equation of tangent at 't₁' to the parabola y² = 4ax is yt₁ = x + at₁²  ...(1)
Also, the parametric equation of tangent at 't₂' to the parabola y = 4ax is yt₂ = x+ at₂² ...(2)
(1) ➝ yt₁ = x + at₁²
(2) ➝ yt₂ = x + at₂²
(1) - (2) y(t₁ -t₂) = a(t₁² - t₂²)

⇒ y = a(t₁ + t₂)
Substitutingy = a(t₁ + t₂) in (1) we get,
a(t₁ + t₂)t₁ = x + at₁²

⇒ x = at₁t₂
Hence, the point of intersection of two lengths is
[at₁t₂, a(t₁ + t₂)]

Let cos⁻¹x = \(\alpha\) and cos-1 y = \(\beta\).
Then, x = cos\(\alpha\) and y cos =\(\beta\)
cos^-1x + x + cos^-1 y + cos^-1x = \(\pi\) gives \(\alpha\)+\(\beta\) = \(\pi\) -cos^-1z.
Now, cos(\(\alpha\)+\(\beta\)) = cos\(\alpha\)cos\(\beta\)-sin\(\alpha\)sin\(\beta\) = xy-\(\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-{ y }^{ 2 } } \)
-cos(cos^-1 z) = xy\(\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-{ y }^{ 2 } } \)
so, \(-z=xy-\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-{ y }^{ 2 } } \), Which gives -xy - z = -\(\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-{ y }^{ 2 } } \)
Squaring on both sides and simplifying, we get x² + y² + z² + 2xyz = 1.

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

Let f(x) x⁶-3x⁵-5x⁴+22x³-39x²-39x+135
Given (1+2i) is a root \(\Rightarrow\)(-2i) is also a root
Also \(\sqrt3\) is a root \(\Rightarrow\)-\(\sqrt3\) is also a root.
Hence, the factors of f(x) are [x - (1 + 2i)]
[x-(1-2i)] [x\(\sqrt3\)] [x+\(\sqrt3\)]
[(x-1)-2i] [(x-1)+2i] [x-\(\sqrt3\)][x+\(\sqrt3\)]
((x-1)²+2²)(x²-3) = (x²-2x+1+4)(x²-3)
\(\Rightarrow\) factor of f(x) is (x²-2x+5)(x²-3)
\(\Rightarrow\)x⁴-3x²-2x³+6x+5x²-15
\(\Rightarrow\)(x⁴-3x²-2x³+6x-15) is a factor of f(x) 
To find the other factor, let us divide f(x) by
x⁴ - 2x³ + 2x² + 6x - 15

The other factor is x² - x - 9
\(\Rightarrow x=\frac { 1\pm \sqrt { { (-1) }^{ 2 }-4(1)(-9) } }{ 2 } \left[ \because x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \right] \)
\(\Rightarrow x=\frac { 1\pm \sqrt { 37 } }{ 2 } \)
Hence the roots are
1 - 2i, 1 + 2i, \(\sqrt { 3 }, -\sqrt { 3 }, \frac { 1+\sqrt { 37 } }{ 2 } ,\frac { 1-\sqrt { 37 } }{ 2 } \).

Let \(\hat { a } =\vec { OA } \) and \(\hat { b } =\vec { OB } \) be the unit vectors and which make angles α and β, respectively, with positive x-axis, where A and B are as in the diagram.
Draw AL and BM perpendicular to the x-axis. Then \(\left| \vec { OL } \right| =\left| \vec { OA } \right| \) cos α = cos α, \(\left| \vec { LA } \right| =\left| \vec { OA } \right| \) sin α =  sin α
So, \(\vec { OL } =\left| \vec { OL } \right| \)\(\hat { i } \) =  cos,α \(\hat { i } \), \(\overrightarrow { LA } \) = sin α (-\(\hat { j } \))
Therefore, \(\hat { a } =\overrightarrow { OA} = \overrightarrow { OL } +\overrightarrow { LA } \) = cos α \(\hat { i } \) - sin α \(\hat { j } \) ..(1)
Similarly \(\hat { b } \) = cos β \(\hat { i } \)+ sin β \(\hat { j } \) ....(2)
The angle between \(\hat { a } \) and \(\hat{b}\) is α + β and so,
\(\hat { a } .\hat { b } =\left| \hat { a } \right| \left| \hat { b } \right| \) cos (α + β) = cos (α + β) ... (3)

On the other hand, from (1) and (2)
\(\hat { a } .\hat { b } =(cos\alpha \hat { i } -sina\hat { j } )(cos\beta \hat { i } -sin\beta \hat { j } )\) = cos α cos β - sin α sin β....(4)
From (3) and (4), we get cos(α + β) = cos α cos β - sin α sin β

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

p	q	q ➝ p	~p	~q	~q ➝ ~p
T	T	T	F	F	T
T	F	T	F	T	T
F	T	F	T	F	F
F	F	T	T	T	T

The entries in the columns corresponding q ➝ p and ~p ➝ ~q are identical and hence they are equivalent.
q ➝ p ≡ ~p ➝ ~q
Hence proved

Given absolute error = 2%
⇒ \(\frac { dl }{ l } =2 \% =\frac { 2 }{ 100 } =0.02\)
Given T = 2ㅠ\(\sqrt { \frac { 1 }{ g } } \)
Taking logarithm on both sides,
log T = log 2ㅠ + \(\frac12\) log l - \(\frac12\) log g
Taking differential on both sides we get,
\(\frac{1}{T}dT=0+\frac{1}{2}.\frac{1}{l}.dl\)
\(\frac { \Delta T }{ T } =\frac { 1 }{ 2 } (.02)\)
\(\frac { \Delta T }{ T } =0\)
ஃ Percentage error = \(\frac { \Delta T }{ T } \times100=.01\times100=1 \%\)

The given differential equation can be written as
\(\frac{\cos x}{\cos x} \frac{d y}{d x}+y \frac{\sin x}{\cos x}   =\frac{1}{\cos x}  \)
\(\frac{d y}{d x}+\left(\frac{\sin x}{\cos x}\right) y   =\sec x  \)
\(\frac{d y}{d x}+(\tan x) y   =\sec x\)
This is of the form \(\frac{d y}{d x}+P y=Q \)
where
\(\mathrm{P}  =\tan x  \)
\(\mathrm{Q}  =\sec x\)
Thus, the given differential equation is linear.
\(I.F=e^{\int \operatorname{Pdx}}=e^{\int \operatorname{Lin} x d x}=e^{\log (\sec x)}=\sec x\)
So, the required solution is given by 
\({[\mathrm{y} \times \mathrm{I} . \mathrm{F}] }  =\int[Q \times I F] d x+c  \)
\(\mathrm{y} \times \sec x  =\int \sec x \times \sec x d x+c  \)
\(\mathrm{y} \sec x  =\int \sec ^2 x d x+c \\ \mathrm{y} \sec x  =\tan x+\mathrm{c}  \)
\(\div \sec x, \frac{y \sec x}{\sec x}  =\frac{\tan x}{\sec x}+\frac{c}{\sec x} \)
\(\mathrm{y}  =\frac{\sin x}{\cos x} \times \frac{1}{\sec x}+\frac{c}{\sec x}  \)
\(\mathrm{y}  =\frac{\sin x}{\cos x} \times \cos x+c \cos x \)
\(=\sin x+c \cos x\)

f(x) = x³ − 3x + 2, x ∈ [-2, 2]
a) f(x) is continuous in [-2, 2]
b) f(x) is differentiable in (-2, 2)
f(-2) = (-2)³ - 3 (-2) + 2
= -8 + 6 + 2 = 0
f(2) = 2³ - 3(2) +2
= 8 - 6 + 2 = 4
By Lagrange's mean value theorem, there exists c ∈ [-2,2] such that f'(c) = \(\frac { f(b)-f(a) }{ b-a } \)
⇒ 3c³ - 3 = \(\frac { 4-0 }{ 2-(-2) } =\frac { 4 }{ 4 } =1\)
⇒ 3²-3 = 1
⇒ 3² =4 ⇒ c² = \(\frac43\)
⇒ c = 土 \(\frac { 2 }{ \sqrt { 3 } } \) ∈ [-2, 2] 

\({ cos }^{ -1 }\left( cos\frac { \pi }{ 7 } cos\frac { \pi }{ 17 } -sin\frac { \pi }{ 17 } sin\frac { \pi }{ 17 } \right) .\)
\({ cos }^{ -1 }\left( cos\left( \frac { \pi }{ 7 } +\frac { \pi }{ 17 } \right) \right) \)
\(\left[ \therefore cosA\ cosB-sinA\ sinB=cos(A+B) \right] \)
= \({ cos }^{ -1 }\left( cos\left( \frac { 24\pi }{ 119 } \right) \right) \)    \(\left[ \therefore \frac { 24\pi }{ 119 } \varepsilon \left[ 0,\pi \right] \right] \)
= \(\frac { 24\pi }{ 119 } \) 

Given planes are \(\vec { r } .(\hat { i } +\hat { j } -2\hat { k } )\) and 
\(2x-2y+z=2\Rightarrow \vec { r } .\left( 2\hat { i } -2\hat { j } +\hat { k } \right) =3\)
\(\therefore { \vec { n } }_{ 1 }=\hat { i } +\hat { j } -2\hat { k } \) and \({ \vec { n } }_{ 2 }=2\hat { i } -2\hat { j } +\hat { k } \)
Angle between the plane is'
\(cos\theta =\frac { { \vec { n } }_{ 1 }.{ \vec { n } }_{ 2 } }{ \left| { \vec { n } }_{ 1 } \right| \left| { { \vec { n } }_{ 2 } } \right| } =\frac { \left| 1(2)+1(-2)-2(1) \right| }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 }+\left( -2 \right) ^{ 2 }.\sqrt { { 2 }^{ 2 }+\left( -2 \right) ^{ 2 }+{ 1 }^{ 2 } } } } \)
= \(\frac { \left| -2 \right| }{ \sqrt { 6 } .\sqrt { 9 } } =\frac { 2 }{ \sqrt { 6 } (3) } =\frac { 2 }{ 3\sqrt { 6 } } \)
\(\theta ={ { cos }^{ -1 }\left( \frac { 2 }{ 3\sqrt { 6 } } \right) }\)

Given 2cos α = x+\(\frac { 1 }{ x } \)
⇒ 2cos α = \(\frac { { x }^{ 2 }+1 }{ x } \)
⇒ x²+1 = 2xcos α
⇒ x²-2x cos α+1 = 0
⇒ \(\frac { 2cos\alpha \pm \sqrt { (-2cos\alpha )^{ 2 }-4(1)(1) } }{ 2 } \)
= \(\frac { 2cos\alpha \pm \sqrt { 4cos^{ 2 }\alpha -4 } }{ 2 } \)  \(\left[ \because \frac { b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \right] \)
= \(\frac { 2cos\alpha \pm \sqrt { -sin^{ 2 }\alpha } }{ 2 } \)
= \(\frac { 2cos\alpha \pm isin\alpha }{ 2 } \) [∵ sin²α+cos²α = 1]
⇒ x² = cos α ± sin α
Also, 2cos β = y+\(\frac { 1 }{ y } \)
⇒ 2cos β = \(\frac { { y }^{ 2 }+1 }{ y } \)
⇒ y²-2y cos β+1 = 0
⇒ \(\frac { 2cos\beta \pm \sqrt { (-2cos^{ 2 }\beta ^{ 2 }-4(1)(1) } }{ 2 } \)
=\(\frac { 2cos\beta \pm \sqrt { 4cos^{ 2 }\beta } -4 }{ 2 } =\frac { 2cos\beta \pm 2isin\beta }{ 2 } \)
⇒ y = cosβ ± i sinβ
\({ x }^{ m }{ y }^{ n }+\cfrac { 1 }{ { x }^{ m }{ y }^{ n } } =2cos(m\alpha +n\beta )\)
x^myⁿ = (cos α + i sin mα) (cos nβ + i sin nβ)
cos(mα+nβ)+i sin(mα+nβ)
\(\frac { 1 }{ { x }^{ m }{ y }^{ n } } \) = cos(mα+nβ)-i sin(mα+nβ)

= 2cos(mα+nβ)

Let P(x, y) be any point on the hyperbola
\(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)

Then by definition, SP = ePM and S'P = ePM'
SP = ePM ⇒ SP = e(NK)
= e(CN - CK)
= \(e\left( x-\frac { a }{ e } \right) \)  = ex - a
= and S'P = ePM' ⇒ S'P = e(NK')
⇒ e(CN + CK') = \(e\left( x+\frac { a }{ e } \right) \) = ex + a
∴ S'P -SP = (ex + a) - (ex - a)
ex + a - ex + a = 2a (constant)
= length of the transverse axis.

The matrix form of the system is AX = B , where A = \(\left[ \begin{matrix} 5 & 2 \\ 3 & 2 \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \end{matrix} \right] \), B = \(\left[ \begin{matrix} 3 \\ 5 \end{matrix} \right] \)
We find |A| = \(\left| \begin{matrix} 5 & 2 \\ 3 & 2 \end{matrix} \right| \) = 10 - 6 = 4 ≠ 0. So, A⁻¹ exists and A⁻¹ = \(\frac { 1 }{ 4 } \left[ \begin{matrix} 2 & -2 \\ -3 & 5 \end{matrix} \right] \)
Then, applying the formula X = A⁻¹B, we get
\(\left[ \begin{matrix} x \\ y \end{matrix} \right] =\frac { 1 }{ 4 } \left[ \begin{matrix} 2 & -2 \\ -3 & 5 \end{matrix} \right] \left[ \begin{matrix} 3 \\ 5 \end{matrix} \right] =\frac { 1 }{ 4 } \left[ \begin{matrix} -4 \\ 16 \end{matrix} \right] =\left[ \begin{matrix} \frac { -4 }{ 4 } \\ \frac { 16 }{ 4 } \end{matrix} \right] =\left[ \begin{matrix} -1 \\ 4 \end{matrix} \right] \).
So the solution is (x = −1, y = 4).