Then A∨ B =\(\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\vee \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 0\vee 1 & 1\vee 1 \\ 1\vee 0 & 1\vee 1 \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\)
\(A\wedge B=\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\wedge \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 0\wedge 1 & 1\wedge 1 \\ 1\wedge 0 & 1\wedge 1 \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}\)

\(S=\left|\begin{array}{l} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{array}\right|\)
Let X be a random variable whose values x are the number of fours.
The sample space S is given in the table.
It can also be written as
S = {(i, j)} , where i = 1, 2, 3, 6 and j = 1, 2, 3, 6
Therefore X takes on the values of 0, 1 and 2.
We observe that
(i) X = 0, if (i,  j) for i ≠ 4, j≠ 4,
(ii) X = 1, if (1, 4), (2, 4), (3, 4), (5, 4), (6, 4), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6)
(iii) X = 2, if (4, 4) ,
Therefore,

Values of the Random Variable X	0	1	2	Toatal
Number of elements in inverse images	25	10	1	36

The probabilities are
\(f(0)=P(X=0)\cfrac { 25 }{ 36 } \) 
\(f(1)=P(X=1)=\cfrac { 10 }{ 36 } \) 
and \(f(20=P(X=2)=\cfrac { 1 }{ 36 } \)
Clearly the function f(x) satisfies the conditions
(i) f (x) ≥ 0, for x = 0, 1, 2 and
(ii) \(\underset { x }{ \Sigma } f(x)=\sum _{ x=0 }^{ x=-2 }{ f(x) } =f(0)+f(1)+f(2)=1\) 
\(=\frac{25}{36}+\frac{10}{36}+\frac{1}{36}=1\)
The probability mass function is presented as

x	0	1	2
f(x)	\(\frac { 25 }{ 36 } \)	\(\frac { 10 }{ 36 } \)	\(\frac { 1 }{ 36 } \)

(or) 
\(f(x)=\begin{cases} \begin{matrix} \frac { 25 }{ 36 } & for \ x=0 \end{matrix} \\ \begin{matrix} \frac { 10 }{ 36 } & for \ x=1 \end{matrix} \\ \begin{matrix} \frac { 1 }{ 36 } & for \ x=2 \end{matrix} \end{cases}\)

\( \because \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -ax }dx}=\frac { n! }{ { a }^{ n+1 } } \)
\(n=5,\quad a=3 \)
\(=\frac { 5! }{ { 3 }^{ 6 } } \)

Given r = 2 mm
dr = (2.1 - 2) = 0.1 mm
Area = πr²
Approximate area dA = 2πr dr
= 2π (2) (0.1)
= 4 π (0.1) = 0.4 π mm²

If we directly substitute x = 0 we get an indeterminate form \(\frac{0}{0}\) and hence we apply the l’Hôpital’s rule to evaluate the limit as,
\(\underset{x\rightarrow 0^{+}}{lim}(\frac{sin \ x}{x^{2}})=\underset{x\rightarrow 0^{+}}{lim}(\frac{cos \ x}{2x})=\infty\)
\(\underset{x\rightarrow 0^{-}}{lim}(\frac{sin \ x}{x^{2}})=\underset{x\rightarrow 0^{-}}{lim}(\frac{cos \ x}{2x})=\infty\)
As the left limit and the right limit are not the same we conclude that the limit does not exist.
Remark
One may be tempted to use the l’Hôpital’s rule once again in \(\underset{x\rightarrow 0^{+}}{lim} (\frac{cos \ x}{2x})\) to conclude
\(\underset{x\rightarrow 0^{+}}{lim} (\frac{cos \ x}{2x})\)\(\underset{x\rightarrow 0^{+}}{lim} (\frac{-sin \ x}{2})\)=0
which is not true because it was not an indeterminate form.

Equation of the circle is
2x² + 2y² - 6x + 4y + 2 = 0
Dividing by 2, we get
x² + y² - 3x + 2y + 1 = 0
Here 2g = -3 ⇒ g = \(\frac { -3 }{ 2 } \)
2f = 2 ⇒ f = 1
and c = 1
∴ Centre is (-g, -f) = \(\left( \frac { 3 }{ 2 } ,-1 \right) \)
and r = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \) = \(\sqrt { { \left( \frac { 3 }{ 2 } \right) }^{ 2 }+{ 1 }^{ 2 }-1 } \)
= \(\sqrt { \frac { 9 }{ 4 } } =\frac { 3 }{ 2 } \) units.

\(\left[ \begin{matrix} -1 & 3 \\ 4 & -7 \\ 3 & -4 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} -1 & 3 \\ 4 & -7 \\ 3 & -4 \end{matrix} \right] \)
A is a matrix of order 3 \(\times\) 2
∴ \(\rho \)(A) ≤ min (3, 2) = 2
The highest order of minor of A is 2
It is \(\left| \begin{matrix} -1 & 3 \\ 4 & -7 \end{matrix} \right| \)= 7-12 = 5 ≠ 0
∴ \(\rho \)(A) = 2

ang \(z=\frac { -2 }{ 1+i\sqrt { 3 } } \)
= arg(-2)-arg \(\left( 1+i\sqrt { 3 } \right) \) \(\left( \because arg\left( \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right) =arg{ z }_{ 1 }=g_{ 2 } \right) \)
= \(\left( \pi -{ tan }^{ -1 }\left( \frac { 0 }{ 2 } \right) \right) -tan^{ -1 }\left( \frac { \sqrt { 3 } }{ 1 } \right) \)
= \(\pi -\frac { \pi }{ 3 } =\frac { 2\pi }{ 3 } \)
This implies that one of the values of arg z is \(\frac { 2\pi }{ 3 } \)
Since \(\frac { 2\pi }{ 3 } \) lies between \(-\pi \), the principal argument Argz is \(\frac { 2\pi }{ 3 } \)

Since the given three vectors are coplanar, we have \(\left| \begin{matrix} 2 & -1 & 3 \\ 3 & 2 & 1 \\ 1 & m & 4 \end{matrix} \right| \) = 0 ⇒ m = -3

cos^-1(-x) = \(\pi\)-cos⁻¹(x)
Let cos^-1(-x) = \(\theta \)   ..(1)
\(\Rightarrow -x=cos\theta \)
\(\Rightarrow x=-cos\theta =cos\theta =cos\left( \pi -\theta \right) \)
\(\Rightarrow \pi -\theta ={ cos }^{ -1 }\left( x \right) \)
\(\Rightarrow \theta =\pi -{ cos }^{ -1 }x\) ...(2)
From (1) & (2) \({ cos }^{ -1 }\left( -x \right) =\pi -{ cos }^{ -1 }\left( x \right) \)
\({ cos }^{ -1 }\left( -x \right) =\pi -{ cos }^{ -1 }\left( x \right) \) is true.

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

Let \(f(x)=xcos\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) \)
\(f(-x)=-x\quad cos\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) \)
\(=-x\quad cos\left( \frac { { e }^{ \frac { 1 }{ x } }-1 }{ { e }^{ \frac { 1 }{ x } }+1 } \right) \)
\(=-x\quad cos\left( \frac { 1-{ e }^{ x } }{ 1+{ e }^{ x } } \right) \)
\(=-x\quad cos\left( -\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) \right) \)
\(=-xcos\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) \)
\([\because cos(-\theta )=cos\theta ]\)
= -f(x)
\(\therefore\) f(x) is an odd function
\(\therefore \int _{ -5 }^{ 5 }{ xcos\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) } dx=0\)

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

Values of random variable	30	5	-20	Total
Number of points in inverse image	15	48	28	91

Observe that, f(0) = 0 = f (1), is continuous in the interval [0,1] and is differentiable in (0,1). Now,
 \(f'{x}=2x(1-x)(1-2x)\).
Therefore,  \(f'(c)=0 \) gives c = 0, 1 and \(\frac{1}{2}\)
which \(\Rightarrow c= \frac{1}{2}\in (0,1)\).

Let the parabola be y² = 4ax 
Since focus is 2m from the vertex a = 2
Equation of the parabola is y² = 8x
Let P be a point on the parabola whose x -coordinate is 3m from the
vertex P (3, y) 
y² = 8 × 3
y =\(\sqrt { 8\times 3 } \)
= \(2\sqrt { 6 } \)
The width of the antenna 3m from the vertex is 4\(\sqrt { 6 } \) m.

Resultant of the given forces is \(\hat{F}\) = ( \(\hat { 2j } +\hat { 5j } +\hat { 6k } \) )+ (\(-\hat { i } -\hat { 2j } -\hat { k } \) ) = \(\hat { i } +\hat { 3j } +\hat {5 k } \)
Let A and B be the points (4, −3, −2) and (6, 1,  −3) respectively.
Then the displacement vector of the particle is 
\(\vec { d } =\vec { AB } =\vec { OB } -\vec { OA } =(\hat { 6i } +\hat { j } -\hat { 3k } )-(\hat { 4i } -\hat { 3j } -\hat { 2k } )=\hat { 2i } +\hat { 4j } -\hat { k } \)
Therefore the work done
w = \(\vec { f } .\vec { d } =(\hat { i } +\hat { 3j } +\hat { 5k } ).(\hat { 2j } +\hat { 4j } -\hat { k } )\) = 9 units.

If P(x) denotes the polynomial in the equation, then P(2) = 0.
Hence 2 is a root of the polynomial.
To find other roots, we divide the given polynomial x³−5x²−4x + 20 by x − 2 and get Q(x) = x² −3x−10 as the quotient.
Solving Q(x) = 0 we get −2 and 5 as roots.
Thus 2, −2, 5 are the solutions of the given equation.

\(\left| z+3+4i \right| \le \left| z \right| +\left| 3+4i \right| =2+5=7\)
\(\left| z+3+4i \right| \le 7\) .............. (1)
\(\left| z+3+4i \right| \ge \left| \left| z \right| -\left| 3+4i \right| \right| =\left| 2-5 \right| =3\)
\(\left| z+3+4i \right| \ge 3\) ............ (2)
From (1) and (2) we get, \(3\le \left| z+3+4i \right| \le 7\) 

 \(tan\left( { cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) -{ sin }^{ -1 }\left( -\frac { 1 }{ 2 } \right) \right) \)
Let \({ cos }^{ -1 }\left( \frac { -1 }{ 2 } \right) =x\)
\(\Rightarrow \frac { 1 }{ 2 } =cosx\)
\(\Rightarrow cosc=cos\frac { \pi }{ 3 } \)
\(\Rightarrow x=\frac { \pi }{ 3 } \)
Let \({ sin }^{ -1 }\left( \frac { -1 }{ 2 } \right) =y\)
\(\Rightarrow \left( \frac { -1 }{ 2 } \right) =siny\)
\(\Rightarrow siny=\frac { -1 }{ 2 } =-sin\frac { \pi }{ 6 } =\left( \frac { -\pi }{ 6 } \right) \)
\(\Rightarrow y=\frac { -\pi }{ 6 } \)
\(\therefore { tan }^{ -1 }\left( cos^{ -1 }\left( \frac { 1 }{ 2 } \right) -{ sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) \right) \)
= \({ tan }^{ -1 }\left( \frac { \pi }{ 3 } -\left( \frac { -\pi }{ 6 } \right) \right) ={ tan }^{ -1 }\left( \frac { \pi }{ 3 } +\frac { \pi }{ 6 } \right) \)
= \(tan\left( \frac { 2\pi +\pi }{ 0 } \right) =tan\left( \frac { 3\pi }{ 6 } \right) =tan\left( \frac { \pi }{ 2 } \right) \)
= \(\infty \)

Given A =\(\left[ \begin{matrix} 8 & -4 \\ -5 & 3 \end{matrix} \right] \)
adj A =\(\left[ \begin{matrix} 3 & 4 \\ 5 & 8 \end{matrix} \right] \)
[Interchange the elements in the leading diagonal and change the sign of the elements in the off diagonal]
|A| = 24 - 20 = 4
∴ A(adj A) =\(\\ \left[ \begin{matrix} 8 & -4 \\ -5 & 3 \end{matrix} \right] \left[ \begin{matrix} 3 & 4 \\ 5 & 8 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 24-20 & 32-32 \\ -15+15 & -20+24 \end{matrix} \right] =\left[ \begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix} \right] \) ....(1)
(adj A)(A) =\(\left[ \begin{matrix} 3 & 4 \\ 5 & 8 \end{matrix} \right] =\left[ \begin{matrix} 8 & -4 \\ -5 & 3 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 24-20 & -12+12 \\ 40-40 & -20+24 \end{matrix} \right] =\left[ \begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix} \right] \)...(2)
|A|I₂ = 4\(\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] =\left[ \begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix} \right] \) .....(3)
From (1), (2) and (3), it is proved that
A (adj A) = (adj A) A = |A|I₂

p	q	r	q ⟶ r	p⟶ q	q ⟶ r	(p⟶ q) ∧ (q ⟶ r)	((p⟶ q) ∧ (q ⟶ r)) ⟶ (p ⟶ r)
T	T	T	T	T	T	T	T
T	T	F	F	T	F	F	T
T	F	T	T	F	T	F	T
T	F	F	F	F	T	F	T
F	T	T	T	T	T	F	T
F	T	F	T	T	F	F	T
F	F	T	T	T	T	T	T
F	F	F	T	T	T	T	T

∴ ((p⟶ q) ∧ (q ⟶ r)) ⟶ (p ⟶ r)
The given statement is a tautology.

Given w(x, y, z) = \(\left( \frac { { 5x }^{ 3 }{ y }^{ 4 }+7{ y }^{ 2 }{ xz }^{ 4 }-{ 75y }^{ 3 }{ z }^{ 4 } }{ { x }^{ 2 }+{ y }^{ 2 } } \right) \)
Let (x, y, z) = \(\frac { { 5x }^{ 3 }{ y }^{ 4 }+7{ y }^{ 2 }{ xz }^{ 4 }-{ 75y }^{ 3 }{ z }^{ 4 } }{ { x }^{ 2 }+{ y }^{ 2 } } \)
⇒ w = log f  ...(1)
⇒ e^w = f
f(λx, λy, λz) = \(\frac { { 5\lambda }^{ 3 }{ x }^{ 3 }{ \lambda }^{ 4 }{ y }^{ 4 }+7{ \lambda }^{ 2 }{ y }^{ 2 }\lambda x{ \lambda }^{ 4 }{ z }^{ 4 }-75{ \lambda }^{ 3 }{ y }^{ 3 }{ \lambda }^{ 4 }{ z }^{ 4 }{ }^{ } }{ { \lambda }^{ 2 }{ x }^{ 2 }+{ \lambda }^{ 2 }{ y }^{ 2 } } \)
= \(\frac { { \lambda }^{ 7 }(5{ x }^{ 3 }{ y }^{ 4 }+7{ y }^{ 2 }{ xz }^{ 4 }-{ 75 }y^{ 3 }{ z }^{ 4 } }{ { \lambda }^{ 2 }({ x }^{ 2 }+{ y }^{ 2 }) } ={ \lambda }^{ 5 }f(x,y,z)\)
∴ f(x, y, z) is a homogeneous function of degree 5.
∴ By Euler's theorem,
\(x.\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } +z\frac { \partial f }{ \partial z } =5.f\)
⇒ \(x.\frac { \partial }{ \partial x } ({ e }^{ w })+y.\frac { \partial }{ \partial y } ({ e }^{ w })+z.\frac { \partial }{ \partial z } ({ e }^{ w })=5.{ e }^{ w }\) [using (1)]
⇒ \(x.{ e }^{ w }\frac { \partial w }{ \partial x } +y.{ e }^{ w }\frac { \partial w }{ \partial y } +z.{ e }^{ w }\frac { \partial w }{ \partial z } ({ e }^{ w })=5{ e }^{ w }\)
⇒ \(x\frac { \partial w }{ \partial x } +y\frac { \partial w }{ \partial y } +z\frac { \partial w }{ \partial z } \)  [Divided by e^w]

Let p be the probability of the useful life hours of a fluorescent light.
n = 12
P = 0.9
q = 1-p = 0.1
P(X= x)= ⁿC_x p^x q^n-x, x = 0, 1,2, .., n
(i) Exactly 10
P(X = 10) = ¹²C₁₀(0.9)¹⁰(1 - 0.9)²
= ¹²C₁₀(0.9)¹⁰ (0.1)²
(ii) Atleast 11
P(X≥11) = R(X = 11) + P(X = 12)
= ¹²C₁₁(0.9)¹¹ (0.1)¹ + ¹²C₁₂(0.9)¹²(0.1)⁶
= ¹²C₁ (0.9)¹¹ (0.1) + (0.9)¹²
= ¹²(0.9)¹¹ (0.1) + (0.9)¹²
= (0.9)¹¹ ((12)(0.1) + 0.9)
= (0.9)¹¹ (1.2 + 0.9)
= (0.9)¹¹ (2.1)
(ii) Atleast 2 will not have a useful
P(X,10) = 1 -P(X > 10)
= 1 - [P(X = 11) + P(X = 12)]
= 1-(2.1) (0.9)¹¹

Let T represent the temperature of the boiling water and Tm represents the temperature of the kitchen.
By Newton's law of cooling
\(\Rightarrow \int { \frac { dT }{ T-{ T }_{ m } } =K\int { dt } } \)
\(\Rightarrow log(T-{ T }_{ m })=Kt+logC\)
\(\Rightarrow log(T-{ T }_{ m })-logC=Kt\)
\(\Rightarrow log\left( \frac { T-{ T }_{ m } }{ C } \right) =Kt\)
\(\Rightarrow T-{ T }_{ m }={ Ce }^{ Kt } ...(1)\)
when t=0,T=100
\(\therefore 100-{ T }_{ m }={ Ce }^{ 0 }\)
\(\Rightarrow C=100-{ T }_{ m }\)
\(\Rightarrow becomes,\ T-{ T }_{ m }=(100-{ T }_{ m }){ e }^{ Kt }\)
Also when t = 5, T = 80
\(\therefore 80-{ T }_{ m }=(100-{ T }_{ m }){ e }^{ 5K }\)
\(\Rightarrow { e }^{ 5K }=\frac { 80-{ T }_{ m } }{ 100-{ T }_{ m } } ..(2)\)
When t = 10, T = 65
(2) \(\Rightarrow\) 65 - T = (100-T_m)e^10K
= (100-T_m)(e^5K)²
\(=(100-{ T }_{ m }){ \left( \frac { 80-{ T }_{ m } }{ 100-{ T }_{ m } } \right) }^{ 2 }\)
[using(2)]
\(\Rightarrow 65-{ T }_{ m }=\frac { { (80-{ T }_{ m } })^{ 2 } }{ 100-{ T }_{ m } } \)
\(\Rightarrow\) 6500-65T_m-100T_m+T_m² = 6400+T_m²-160T_m
\(\Rightarrow\) 6500-6400 = 165Tm-160Tm
\(\Rightarrow\) 100 = 5Tm
\(\\ \Rightarrow { T }_{ m }=\frac { 100 }{ 5 } ={ 20 }^{ o }C\)
Hence the temperature of the kitchen is 20^oC

\(\frac { dy }{ dx } +\frac { { 3x }^{ 2 }y }{ 1+{ x }^{ 3 } } =\frac { { sin }^{ 2 }x }{ 1+{ x }^{ 3 } } \)
This is a linear differential equation 
\(\therefore P=\frac { { 3x }^{ 2 } }{ 1+{ x }^{ 3 } } ;Q=\frac { { sin }^{ 2 }x }{ 1+{ x }^{ 3 } } \)
\(\therefore \int { pdx } =\int { \frac { { 3x }^{ 2 } }{ 1+{ x }^{ 3 } } dx } =log(1+{ x }^{ 3 })\)
\(\therefore I.F.={ e }^{ \int { pdx } }={ e }^{ log(1+{ x }^{ 3 }) }=(1+{ x }^{ 3 })\)
\(\therefore\)The solution is \({ ye }^{ \int { pdx } }=\int { Q{ e }^{ \int { pdx } }dx+c } \)
\(\Rightarrow y(1+{ x }^{ 3 })=\int { \frac { { sin }^{ 2 }x }{ 1+{ x }^{ 3 } } (1+{ x }^{ 3 })dx+c } \)
\(cos2x=1-2{ sin }^{ 2 }x\)
\(sin2x=\frac { 1-cos2x }{ 2 } =\int { { sin }^{ 2 }xdx+c } \)
\(\Rightarrow y(1+{ x }^{ 3 })=\int { \frac { 1-cos2x }{ 2 } } dx+c\)
\(\Rightarrow y(1+{ x }^{ 3 })=\frac { x }{ 2 } -\frac { sin2x }{ 4 } +c\)

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 h and r be the height and the base radius. Therefore h = 2r. Let V be the volume of the salt cone.

\(V=\frac{1}{3}\pi r^{2}h=\frac{1}{12}\pi h^{3}; \frac{dV}{dt}=30\) mtr³ / min.
Hence,  \(\frac{dV}{dt}=\frac{1}{4}\pi h^{2}\frac{dh}{dt}\)
Therefore,  \(\frac{dh}{dt}=4 \frac{dV}{dt}.\frac{1}{\pi h^{2}}\)
That is,  \(\frac{dh}{dt}=4\times30\times \frac{1}{100 \pi}\)
=\(\frac{6}{5\pi}\) mtr / min. 

9x²- y²- 36x - 6y + 18 = 0
Given equation is 9x²- y²- 36x - 6y + 18 = 0
⇒ 9x² - 36x - (y² + 6y) = -18
⇒ 9(x²-4x)-(y²+6y) =-18
⇒ 9(x² - 4x + 4 - 4) - (y² + 6y + 9 - 9) = -18
⇒ 9(x-2)²-36-(y+3)²+9 =-18
⇒ 9(x-2)² - (y+3)² = -18+36-9
⇒ 9(x - 2)² - (y + 3)² = 9
Dividing by 9 we get, \(\frac { { (x-2) }^{ 2 } }{ 1 } -\frac { ({ y+3) }^{ 2 } }{ 9 } =1\)
This is an equation of the hyperbola whose transverse axis is parallel to x-axis.
a² = 1, b² = 9
∴ c² = a² + b² = 1 + 9 = 10 ⇒ c = \(\sqrt { 10 } \)
\(e=\sqrt {1-\frac { { b }^{ 2 } }{ { a }^{ 2 } }} =\sqrt { 1-\frac { 9 }{ 1 } } =\sqrt { 10 } \)
a) Center is (2, -3)
⇒ h = 2, k = -3
(b) Foci are (h + c, k), (17 - c, k)
⇒ (2 +\(\sqrt { 10 } \), -3), (2 - \(\sqrt { 10 } \), -3)
(c) Vertic ar (h + a, k) (h - a, k)
⇒ (2+ 1,-3), (2-1,-3)
⇒ (3, -3) (1, -3)
(d) Equation of directrices are x - 2 = \(\pm \frac { a }{ e } \)
⇒ \(x-2=\pm \frac { 1 }{ \sqrt { 10 } } \)
\(x=2\pm \frac { 1 }{ \sqrt { 10 } } \)
⇒ \(x=2+\frac { 1 }{ \sqrt { 10 } } \) and \(x=2-\frac { 1 }{ \sqrt { 10 } } \)

Given plane is passing through the points
\(\vec { a } =2\hat { i } +2\hat { j } +2\hat { k }, \vec { b } =9\hat { i } +3\hat { j } +6\hat { k } \)
Equation of the given plane is 2x + 6y + 6z = 9. It can be written as \(\vec { r } .(2\hat { i } +6\hat { j } +6\hat { k } )=9\)
Since the given plane is perpendicular to \(2\hat { i } +6\hat { j } +6\hat { k } \), the required plane is parallel to \(\vec { c } =2\hat { i } +6\hat { j } +6\hat { k } \). Hence, parametric form of vector equation of plane passing through two points and parallel to a vector is
\(\vec { r } =\vec { a } +s(\vec { b } -\vec { a } )+t\vec { c } ,s,t\in R\)
\(\vec { r } =2\hat { i } +2\hat { j } +\hat { k } +s(7\hat { i } +\hat { j } +5\hat { k } )+t(2\hat { i } +6\hat { j } +6\hat { k } ),s,t\in R\)
Cartesian equation of the plane is
\(\left| \begin{matrix} x-{ x }_{ 1 } & y-{ y }_{ 1 } & z-{ z }_{ 1 } \\ { x }_{ 2 }-{ x }_{ 1 } & { y }_{ 2 }-{ y }_{ 1 } & { z }_{ 2 }-{ z }_{ 1 } \\ { c }_{ 1 } & { c }_{ 2 } & { c }_{ 3 } \end{matrix} \right| =0\)
\(\Rightarrow \left| \begin{matrix} x-2 & y-2 & z-1 \\ 7 & 1 & 5 \\ 2 & 6 & 6 \end{matrix} \right| =0\)
⇒ (x-2)(6-30) - (y-2)(42-10) + (z-1)(42-2) = 0
⇒ (x - 2)(-24) - (y - 2)(32) + (z - 1)(40) = 0
⇒ 24x + 48 - 32y + 64 + 40z - 40 = 0
⇒ -24x - 32y + 40z + 72 = 0
\(\div\) - 8 we get
3x+ 4y - 5z - 9 = 0 is the Cartesian form.
∴ The parametric form of vector equation is
\(\vec { r } =\vec { r } (3\vec { i } +4\vec { j } -5\vec { k } )=9\)

Now, \(\frac { z-1 }{ z+1 } =\frac { x+iy-1 }{ x+iy+1 } =\frac { \left( x-1 \right) +iy }{ \left( x+1 \right) +iy } =\frac { \left[ \left( x-1 \right) +iy \right] \left[ \left( x+1 \right) -iy \right] }{ \left[ \left( x+1 \right) +iy \right] \left[ \left( x+1 \right) -iy \right] } \)
\(\Rightarrow \frac { z-1 }{ z+1 } =\frac { \left( { x }^{ 2 }+{ y }^{ 2 }-1 \right) +i\left( 2y \right) }{ \left( x+1 \right) ^{ 2 }+{ y }^{ 2 } } \)
Since, arg \(\left( \frac { z-1 }{ z+2 } \right) =\frac { \pi }{ 2 } \Rightarrow { tan }^{ -1 }\left( \frac { 2y }{ { x }^{ 2 }+{ y }^{ 2 }-1 } \right) \)= \(\frac { \pi }{ 2 } \)
\(\Rightarrow \frac { 2y }{ { x }^{ 2 }+{ y }^{ 2 }-1 } =tan\frac { \pi }{ 2 } \) ⇒ x²+ y² − 1 = 0
\(\Rightarrow { x }^{ 2 }+{ y }^{ 2 }=1\)

\({ tan }^{ -1 }x+{ tan }^{ -1 }y+{ tan }^{ -1 }z={ tan }^{ -1 }\left( \frac { x+y+z-xyz }{ 1-xy-yz-zx } \right) \)
Given \({ tan }^{ -1 }x+{ tan }^{ -1 }x+{ tan }^{ -o }y+{ tan }^{ -1 }z=\pi \)
\(\Rightarrow \pi ={ tan }^{ -1 }\left( \frac { x+y+zxyz }{ 1-xy-yz-zx } \right) \)
\(\Rightarrow tan\pi =\frac { x+y+z-xyz }{ 1-xy-yz-zx } \)
\(\Rightarrow 0=\frac { x+y+z-xyz }{ 1-xy-yz-zx } \quad [\therefore tan\pi =0]\)
\(\Rightarrow x+y+z-xyz=0\)
\(\Rightarrow x+y+z=xyz\)

Let P(x) = ax²+ bx + c
Given P(-3) = 21
[∵ P(x) ÷ x + 3, the remainder is 21]
⇒ a(-3)2 + b(-3) + c = 21
⇒ 9a - 3b + c = 21
Also, P(5) = 61
⇒ a(5)2 + b(5) + c = 61
[using remainder theorem]
⇒ 25a +5b + c = 61..........(2)
and P(1) = 9
⇒ a(1)2 + b(1) + c = 9
⇒ a + b + c = 9 ............(3)
Reducing the augment matrix to an equivalent row-echelon form using elementary row operations, we get
\(\left[ \begin{matrix} 9 & - & 1 \\ 25 & 5 & 1 \\ -1 & 1 & 1 \end{matrix}|\begin{matrix} 21 \\ 61 \\ 9 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 25 & 5 & 1 \\ 9 & -3 & 1 \end{matrix}|\begin{matrix} 9 \\ 61 \\ 21 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-9{ R }_{ 1 }\\ { R }_{ 2 }\rightarrow { R }_{ 2 }-25{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & -20 & -24 \\ 0 & -12 & -8 \end{matrix}|\begin{matrix} 9 \\ -164 \\ -60 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }\div \\ { R }_{ 3 }\rightarrow { R }_{ 3 }\div 4 }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & -5 & -6 \\ 0 & -3 & -2 \end{matrix}|\begin{matrix} 9 \\ -41 \\ -15 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-\frac { 3 }{ 5 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & -5 & -6 \\ 0 & 0 & \frac { 8 }{ 5 } \end{matrix}|\begin{matrix} 9 \\ -41 \\ \frac { 48 }{ 5 } \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow 5{ R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & -5 & -6 \\ 0 & 0 & 8 \end{matrix}|\begin{matrix} 9 \\ -41 \\ 48 \end{matrix} \right] \)
Writing the equivalent equations from the row-echelon matrix we get,
a + b + c = 9 ............(1)
5b + 6c = 41 ................(2)
-8c = -48
⇒ c = 6
Substituting c = 6 
⇒  5b + 36 = 41
⇒ 5b = 5  
b = 1
Substituting b = 1, c = 6
a + 1 + 6 = 9
⇒ a + 7 = 9
⇒ a = 9 - 7
⇒ a = 2
∴ a = 2, b = 1, and c = 6

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.