The parabola x = y² +1 is y² x = −1. It is symmetrical about x-axis and has the vertex at (1, 0) and focus at \(\left( \frac { 5 }{ 4 } ,0 \right) \). The region for revolution is shaded. Hence, the required volume is given by
\(V=\pi \int _{ -1 }^{ 1 }{ { x }^{ 2 }dy } \)
\(=\pi \int _{ -1 }^{ 1 }{ { ({ y }^{ 2 }+1) }^{ 2 }dy } \)
\(=2\pi \int _{ 0 }^{ 1 }{ \left( { y }^{ 4 }+{ 1y }^{ 2 }+1 \right) dy } \), since the integrand is an even function
\(=2\pi { \left( \frac { { y }^{ 5 } }{ 5 } +2\frac { { y }^{ 3 } }{ 3 } +y \right) }_{ 0 }^{ 1 }=2\pi \left( \frac { 1 }{ 5 } +\frac { 2 }{ 3 } +1 \right) \pi \)

The possible events of selection are
(i) both balls may be black, or
(ii) one white and one black or
(iii) both are white.
Therefore X is a random variable that take the values,
X (both are black balls) = Rs. 2(30) = Rs. 60
X (one black and one white ball) = Rs. 30 − Rs. 20 = Rs. 10
X (both are white balls) = Rs. 2( − 20) = - Rs. 40
Therefore X takes on the values 60,10, and − 40.

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

f(x) = x³ - 5x + 12, x₀ = 2
f(x_o) = 2³ - 5(2) + 12
= 8 - 10 + 12 = 10
f'(x) = 3x² - 5
⇒ f'(x_o) = 3 (2²) - 5 = 7
∴ L(x) = f(x_o) +f'(x_o) (x - x_o)
= 10 + 7(x - 2)
= 10 + 7x - 14
L(x) = 7x- 4 

Let \(\sqrt{x}\) = u
Then x = u², and so dx = 2u du
When x = 0, u = 0 
When x = 9, u = 3
\(\therefore \int _{ 0 }^{ 9 }{ \frac { 1 }{ x+\sqrt { x } } dx=\int _{ 0 }^{ 3 }{ \frac { 1 }{ { u }^{ 2 }+u } (2u)du=2\int _{ 0 }^{ 3 }{ \frac { 1 }{ 1+u } du=2{ \left[ log|1+u \right] }_{ 0 }^{ 3 }=2[log4-0]=log16 } } } \)

Let X be the random variable denotes the total C score is two throws of a die.
Sample space S

n (S) = 36
X = {4,5,6,7,8}
From the sample space

This is an indeterminate of the form \((\frac{\infty}{\infty})\).
To evaluate this limit, we apply l ’Hôpital Rule.
\(\underset{x\rightarrow \infty }{lim}(\frac{x^{2}+17x+29}{x^{4}})=\underset{x\rightarrow \infty}{lim}(\frac{2x+17}{4x^{3}})\)
= \(\underset{x\rightarrow \infty}{lim}(\frac{2}{12x^{2}})=0\)

f(x) = x² -12x + 10; [1, 2]
Given f(x) = x² -12x + 10 ; [1, 2]
f'(x) = 2x - 12
f'(x) = 0
\(\Rightarrow\) 2x-12 = 0
\(\Rightarrow\) 2x = 12
\(\Rightarrow\) x = 6
\(\therefore\) The critical number is 6
Evaluating f (x) at the end points x = 1,
x = 2 and at the critical number x = 6 we get
f(1) = 1²-12(1)+10 = -1
f(2) = 2²-12(2)+10 = -10
Absolute maximum f(1) = -1
Absolute minimum f(2) = -10

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.

Im[(1−i)z + 1] = 0
(1-i)z + 1 = (1-i)( x + iy) +1
= x + iy-ix-i²y+1
= x+iy-ix+y+1
= (x + y + 1) + i(y - x)
∴ Im[(1-i)z + 1] = y-x = 0
⇒ x - y = 0
Hence, the Cartesian equation is x - y = 0

\(\left[ \begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix} \right] \)
Let A =\(\left( \begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix} \right) \)
adj A =\(\left( \begin{matrix} +\left| \begin{matrix} 4 & 1 \\ 7 & 2 \end{matrix} \right| & -\left| \begin{matrix} 3 & 1 \\ 3 & 2 \end{matrix} \right| & +\left| \begin{matrix} 3 & 4 \\ 3 & 7 \end{matrix} \right| \\ -\left| \begin{matrix} 3 & 1 \\ 7 & 2 \end{matrix} \right| & +\left| \begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix} \right| & -\left| \begin{matrix} 2 & 3 \\ 3 & 7 \end{matrix} \right| \\ +\left| \begin{matrix} 3 & 1 \\ 4 & 1 \end{matrix} \right| & -\left| \begin{matrix} 2 & 1 \\ 3 & 1 \end{matrix} \right| & +\left| \begin{matrix} 2 & 3 \\ 3 & 4 \end{matrix} \right| \end{matrix} \right) \)
=\(\left[ \begin{matrix} +(8-7)-(6-3)+(21-12) \\ -(6-7)+(4-3)-(14-9) \\ +(3-4)-(2-3)+(8-9) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} 1 & -3 & 9 \\ 1 & 1 & -5 \\ -1 & 1 & -1 \end{matrix} \right] ^{ T }\)
adj A =\(\left[ \begin{matrix} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1 \end{matrix} \right] \)

Rewriting the given equations as\(​​\frac { x+2 }{ 4 } =\frac { y+3 }{ -2 } =\frac { 2z-6 }{ 3/2 } \) and comparing with \(\frac { x-{ x }_{ 1 } }{ { b }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { b }_{ 2 } } =\frac { z-{ z }_{ 1 } }{ { b }_{ 3 } } \) we have \(\vec { b } ={ b }_{ 1 }\hat { i } +{ b }_{ 2 }\hat { j } +{ b }_{ 3 }\hat { k } \) = \(-4\hat { i } -2\hat { j } +\frac { 3 }{ 2 } \hat { k } =-\frac { 1 }{ 2 } (8\hat { i } +4\hat { j } -3\hat { k } )\). Clearly, \(\vec { b } \) is parallel to the vector \(8\hat { i } +4\hat { j } -3\hat { k } \). Therefore, a vector equation of the required straight line passing through the given point (-4, 2, -3) and parallel to the vector \(8\hat { i } +4\hat { j } -3\hat { k } \) in parametric form is
\(\vec { r } =(-4\hat { i } +2\hat { j } -3\hat { k } )+t(8\hat { i } +4\hat { j } -3\hat { k } )\), t ∈ R
Therefore, Cartesian equations of the required straight line are given by
\(\frac { x-4 }{ 8 } =\frac { y-2 }{ 4 } =\frac { z+3 }{ -3 } \)

\(\frac { 2\left( cos\frac { 9\pi }{ 4 } +isin\frac { 9\pi }{ 4 } \right) }{ 4\left( cos\left( \frac { -3\pi }{ 2 } + \right) sin\left( \frac { -3\pi }{ 2 } \right) \right) } \)
= \(\frac { 1 }{ 2 } \left( cos\left( \frac { 9\pi }{ 4 } -\left( \frac { -3\pi }{ 2 } \right) \right) +isin\left( \frac { 9\pi }{ 4 } -\left( \frac { -3\pi }{ 2 } \right) \right) \right) \)
= \(\frac { 1 }{ 2 } \left( cos\left( \frac { 9\pi }{ 4 } +\frac { 3\pi }{ 2 } \right) +isin\left( \frac { 9\pi }{ 4 } +\frac { 3\pi }{ 2 } \right) \right) \)
= \(\frac { 1 }{ 2 } \left( cos\left( \frac { 15\pi }{ 4 } \right) +isin\left( \frac { 15\pi }{ 4 } \right) \right) =\frac { 1 }{ 2 } \left( cos\left( 4\pi -\frac { \pi }{ 4 } \right) +isin\left( 4\pi -\frac { \pi }{ 4 } \right) \right) \)
= \(\frac { 1 }{ 2 } \left( cos\left( \frac { \pi }{ 4 } \right) -isin\left( \frac { \pi }{ 4 } \right) \right) =\frac { 1 }{ 2 } \left( \frac { 1 }{ \sqrt { 2 } } +i\frac { 1 }{ \sqrt { 2 } } \right) \)
\(\frac { 2\left( cos\frac { 9\pi }{ 4 } +isin\frac { 9\pi }{ 4 } \right) }{ 4\left( cos\left( \frac { -3\pi }{ 2 } + \right) sin\left( \frac { -3\pi }{ 2 } \right) \right) } \) = \(\frac { 1 }{ 2\sqrt { 2 } } -i\frac { 1 }{ 2\sqrt { 2 } } =\frac { \sqrt { 2 } }{ 4 } +i\frac { \sqrt { 2 } }{ 4 } \) Which is in rectangular form.

Length of latus rectum = 4,
distance between foci = 4\(\sqrt { 2 } \) major axis is y-axis
Given \(\frac { { 2b }^{ 2 } }{ a } =4\) and distance between foci = 2ae = 4\(\sqrt { 2 } \)
⇒ ae = \(2\sqrt { 2 } \)
⇒a²e² = 8 ....(1)
\(\frac { { 2b }^{ 2 } }{ a } =4\Rightarrow { b }^{ 2 }=2a\) ...(2)
We know b² = a²(1 - e²)
b² = a² - a²e²
2a = a2 - 8
[using (1) and (2)]
a² - 2a - 8 = 0
On factorising we get
(a - 4)(a + 2) = 0
a = 4 or-2
a = 4
⇒ [∴ a = -2 is not possible]
⇒ ae = 16
∴ From (2), b² = 2(4) = 8
Hence, the equation of the ellipse is
\(\frac { { x }^{ 2 } }{ 8 } +\frac { { y }^{ 2 } }{ 16 } =1\) [∵ Major axis is y -axis]

Let the position vectors of the parts A and B on the circle lie \(\vec { a } \) and \(\vec { b } \) respectively.
Since O is the centre of the circle
\(|\vec { OA } |=|\vec { OB } |\Rightarrow |\vec { a } |=|\vec { b } |\) ....(1)
Also D is the mid-point of AB,
⇒ \(\vec { OD } =\frac { \vec { a } +\vec { b } }{ 2 } \) (mid-point formula)
\(\left( \frac { \vec { a } +\vec { b } }{ 2 } \right) .(\vec { OB } -\vec { OA } )\)
=\(\left( \frac { \vec { a } +\vec { b } }{ 2 } \right) .(\vec { Ob } -\vec { Oa } )\)
= \(\frac { 1 }{ 2 } \left[ |\vec { b } |^{ 2 }-|\vec { a}| ^{ 2 } \right] \)
=\(\left[ \because (\vec { a } +\vec { b } ).(\vec { b } -\vec { a } )=|\vec { b } |^{ 2 }-|\vec { a } |^{ 2 } \right] \)
= \(\frac { 1 }{ 2 } \left[ |\vec { b } |^{ 2 }-|\vec { b| } ^{ 2 } \right] \) (using (1))
= \(\frac{1}{2}\)(0) = 0
⇒ \(\vec { OD } .\vec { AB } =0\Rightarrow \vec { OD } \bot \vec { AB } \)
Hence, if a line is drawn from the centre of a to the mid-point of a chord, then that line is perpendicular to the chord.

|z²-3| ≤ |z²|+|-3| [Triangle law of inequality]
≤ |z|²+3≤1+3  [∴ |z| = 1]
|z²-3| ≤ 4 ..............(1)
Also, |z²- 3| ≥ ||z²|-|-3||
≥ ||z|²-3|   [∵ |-3| = 3]
≥ |1²-3|  [∵ |z| = 1]
≥ |-2| .
|z²-3| ≥ 2.............(2)
From (1) and (2) we get 2 ≤ |z²-3| ≤ 4
Hence proved.

 \(tan\left( { cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) -{ sin }^{ -1 }\left( -\frac { 1 }{ 2 } \right) \right) \)
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 =\(\left[ \begin{matrix} 3 & 2 \\ 7 & 5 \end{matrix} \right] \) and B =\(\left[ \begin{matrix} -1 & -3 \\ 5 & 2 \end{matrix} \right] \)
AB =\(\left[ \begin{matrix} 3 & 2 \\ 7 & 5 \end{matrix} \right] \left[ \begin{matrix} -1 & -3 \\ 5 & 2 \end{matrix} \right] =\left[ \begin{matrix} -3+10 & -9+4 \\ -7+25 & -21+10 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 7 & -5 \\ 18 & -11 \end{matrix} \right] \)
|AB| = -77+90 = 13 ≠ 0 ⇒ (AB)^-1 exists
|A| = 15-14 = 1 ≠ 0 ⇒ A^-1 exists
|B| = -2+15 = 13 ≠ 0 ⇒ B^-1 exists
(AB)^-1 = \(\frac { 1 }{ |AB| } adj(AB)=\frac { 1 }{ 13 } \left( \begin{matrix} -11 & 5 \\ -18 & 7 \end{matrix} \right) \) ...............(1)
B^-1 = \(\frac { 1 }{ |B| } adj(B)=\frac { 1 }{ 13 } \left( \begin{matrix} 2 & 3 \\ -5 & -1 \end{matrix} \right) \)
A^-1 = \(\frac { 1 }{ |A| } \)(adj A)
= \(\frac { 1 }{ 1 } \left( \begin{matrix} 5 & -2 \\ -7 & 3 \end{matrix} \right) =\left( \begin{matrix} 5 & -2 \\ -7 & 3 \end{matrix} \right) \)
∴ B^-1A^-1 = \(\frac { 1 }{ 13 } \left( \begin{matrix} 2 & 3 \\ -5 & -1 \end{matrix} \right) \left( \begin{matrix} 5 & -2 \\ -7 & 3 \end{matrix} \right) \)
= \(\frac { 1 }{ 13 } \left( \begin{matrix} 10-21 & -4+9 \\ -25+7 & 10-3 \end{matrix} \right) \)
= \(\frac { 1 }{ 13 } \left( \begin{matrix} -11 & 5 \\ -18 & 7 \end{matrix} \right) \) ..............(2)
 From (1) or (2) it is proved that
(AB)^-1 = B^-1 A^-1

Given (3 -i) x - (2 - i) y + 2i + 5
= 2x + (-1 + 2i) y + 3 + 2i
⇒ 3x - ix - 2y + iy + 2i + 5 = 2x - y + 2iy + 3 + 2i
choosing the real and imaginary parts
(3x-2y + 5) + i (-x + y + 2) = 2x - y + 3 + i (2y+ 2)
Equating the real and imaginary parts both sides, we get
3x- 2y+ 5 = 2x-y+3
⇒ 3x - 2y + 5 - 2x +y - 3 = 0
⇒ x-y = -2... (1)
-x+y+2 = 2y+2
⇒ -x+y+2-2y-2 = 0
⇒ -x-y = 0 ⇒ x+y = 0.. (2)
(1)-(2) we get,

y = 1
Substituting y = 1 in (2) we get.
x+1 = 0 ⇒ x = -1
∴ x = -1 and y = 1

Equation of the circle passing through the points of intersection of the chord and circle by
Theorem  is x² + y²−16+\(\lambda \)(3x + y + 5) = 0 .
The chord 3x + y + 5 = 0 is a diameter of this circle if the centre\(\left( \frac { -3\lambda }{ 2 } \frac { -\lambda }{ 2 } \right) \) lies on the chord.
So we have 3\(\left( \frac { -3\lambda }{ 2 } \right) \)-\(\frac { -\lambda }{ 2 } \)+5 = 0,
\(\frac { -9\lambda }{ 2 } \)-\(\frac { \lambda }{ 2 } \)+5 = 0,
−5λ + 5 = 0 ,
λ = 1.
Therefore, the equation of the required circle is x² + y²+3x + y −11 = 0.

i) Though - is not binary on N; it is binary on Z. To check the validity of any more properties satisfied by – on Z, it is better to check them for some particular simple values.
ii) Take m = 4 , n = 5 and (m− n) = (4 − 5) = −1and (n −m) = (5 − 4) = 1.
Hence (m− n) ≠ (n −m). So the operation - is not commutative on Z.
iii) In order to check the associative property, let us put m = 4, n = 5 and p = 7 in both (m- n) - p and m- (n - p).
(m−n)− p = (4−5)−7 = (−1−7) = −8 …(1)
m−(n− p) = 4−(5−7) = (4+2) = 6  …(2)
From (1) and (2), it follows that (m - n) - p m - (n - p).
Hence – is not associative on Z.
iv) Identity does not exist (why?).
v) Inverse does not exist (why?).

Given w (x, y, z) = x² +y² +y², 
x = e^t, y = e^t sin t, z = e^t cos t
\(\frac { \partial u }{ \partial x } \) = 2x; \(\frac { \partial u }{ \partial y} \) = 2y; \(\frac { \partial u }{ \partial z} \) = 2z
\(\frac { \partial u }{ \partial x } \) = 2e^t
\(\frac { \partial u }{ \partial y} \) = 2e^t sin t
\(\frac { \partial u }{ \partial z} \) = 2e^t cos t
\(\frac{dx}{dt}\) = e^t
\(\frac{dy}{dt}\) = e^t cas t + sin t e^t
⇒ \(\frac{dz}{dt}\) = et (- sin t ) + cos t et
By chain rule
\(\frac { dw }{ dt } =\frac { \partial u }{ \partial x } .\frac { dx }{ dt } +\frac { dw }{ \partial y } .\frac { dy }{ dt } +\frac { \partial w }{ \partial z } .\frac { dz }{ dt } \)
∴ \(\frac { dw }{ dt } \) = 2e^t(e^t) + 2e^t sin t (e^t cos t +sin t e^t) - e^t sin t + 2e^t cas t (e^t cos t - e^t sin t )

= e^2t [2 + 2] = 4e^2t

Given \(f(x)=\begin{cases} \begin{matrix} 16{ xe }^{ -4x } & foex>0 \end{matrix} \\ \begin{matrix} 0 & forx\le 0 \end{matrix} \end{cases}\) 
Mean :
= \(E(x)=\int _{ 0 }^{ \infty }{ x.f(x)dx } \) 
= \(\int _{ 0 }^{ \infty }{ x.16.xe^{ -4x }dx } \) 
= \(16\int _{ 0 }^{ \infty }{ { x }^{ 2 } } .{ e }^{ -4x }dx=16\times \frac { 2! }{ { 4 }^{ 3 } } \)  \(\left[ \because \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -ax }dx=\frac { n! }{ { a }^{ n+1 } } } \right] \) 
= \(16\times \frac { 2 }{ 64 } \)
\(=\frac { 1 }{ 2 } \) 
Variance :
\(E({ x }^{ 2 })=\int _{ 0 }^{ \infty }{ { x }^{ 2 }.f(x)dx } \) 
= \(\int _{ 0 }^{ \infty }{ { x }^{ 2 }.{ e }^{ -4x } } dx\) 
= \(16\int _{ 0 }^{ \infty }{ { x }^{ 3 }{ e }^{ -4x }dx } \) 

ஃ Var(X) = E(X²) - [E(x)]²
\( =\frac{3}{8}-\frac{1}{4} \)
\(=\frac{1}{8} \)
∴ Var(X) \(=\frac{1}{8} \)

Using the substitution X = \(\sqrt u\), we get dx = \(\frac { 1 }{ 2\sqrt { u } } du\)
When x = 0, we get u = 0
When x = \(\infty\), we get u = \(\infty\)
\(\therefore \int _{ 0 }^{ \infty }{ { e }^{ -{ x }^{ 2 } }{ x }^{ 2n-1 }dx } =2\int _{ 0 }^{ \infty }{ { e }^{ -u }{ \left( \sqrt { u } \right) }^{ 2n-1 } } \frac { 1 }{ 2\sqrt { u } } du=\int _{ 0 }^{ \infty }{ { e }^{ -u }{ u }^{ n-1 }du } \) = Γ(n)

Given v(x, y, z) = x³ + y³ + z³ + xyz³
\(\frac { \partial v }{ \partial z } \) = 0 + 0 +3z² + 3xy = 3z² + 3xy
\(\frac { \partial v }{ \partial y } \) = 0 + 3y² + 0 + 3xz = 3y² + 3xz
Now, \(\frac { { \partial }^{ 2 }v }{ \partial y\partial z } =\frac { { \partial } }{ \partial { y } } \left( \frac { \partial v }{ \partial z } \right) \) = 0 + 3x = 3x ...(1)
\(\frac { { \partial }^{ 2 }v }{ \partial z\partial y } =\frac { { \partial } }{ \partial { z } } \left( \frac { \partial v }{ \partial y } \right) \) = 0 + 3x = 3x ..... (2)
From (1) and (2),
\(\frac { { \partial }^{ 2 }v }{ \partial y\partial z } =\frac { { \partial }^{ 2 }v }{ \partial z\partial y } \)

Diameter = 30 cm
Radius = 15 cm
Circumference (c) = 2πr
\(\frac{dc}{dr}\) = 2π(3) = 6πcm
dc = 2πdr
\(\frac{6}{2π}\) cm = dr
\(\frac{3}{π}\) cm = dr
Approximate growth of the diameter
= 2dr = 2 \(\times\) \(\frac{3}{π}\) cm = \(\frac{6}{2π}\)cm
(ii) A = πr²
dA = π ²r dr
\(d \mathrm{~A}=\not \pi 2(15) \frac{3}{\not \pi} \mathrm{cm}^{2}\)
dA = 90 cm²
Area = πr² = π \(\times\)15 \(\times\) 15 cm²
 

Given f(0) = -1, f(2) = 4
∴ f(x) is a continuous function in [0,2]
By Lagrange's mean value theorem,
f'(x) = \(\frac { f(b)-f(a) }{ b-a } \) = \(\frac{f(2)-f(0)}{2-0}\)
= \(\frac{4-(-1)}{2}\) = \(\frac{5}{2}\) = 2.5
= 2.5 ∉ [0, 2]
Since f'(x) cannot be 2.5 at any point in [0, 2], there does not exist a differentiable function f(x).

Given equation is \(\frac { { (x-1) }^{ 2 } }{ 100 } +\frac { ({ y-2) }^{ 2 } }{ 64 } =1\)
This is an equation of the ellipse
∴ a² = 100,b² = 64
⇒ c² = a² - b²
⇒ c² = 100 - 64 = 36
∴ c = 6
\(\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 64 }{ 100 } } =\sqrt { \frac { 100-64 }{ 100 } } \)
\(=\sqrt{\frac{36}{100}}=\frac{\not 6} {\not {10}}\)
∴ c = \(\frac { 3 }{ 5 } \)
(a) Center is (-1, 2) ⇒ h = -1, k = 2
(b) Foci are (h - c, k), (h + c, k) ⇒ (-1 - 6, 2), (-1 + 6, -2) ⇒ (-7, 2), (5, 2)
(c) Vertices are (h - a, k) and (h + a, k) ⇒ (-1 -10,2), (-1 + 10,2) ⇒ (-11, 2), (9, 2)
(d) Equation of directrices are x + 1 = \(\pm \frac { a }{ e } \)
\(\Rightarrow x+1=\pm \frac { 10 }{ \frac { 3 }{ 5 } } \Rightarrow x+1=\pm \frac { 50 }{ 3 } \)
\(\therefore x+1=\frac { 50 }{ 3 } and x+1=-\frac { 50 }{ 3 } \)
\(\Rightarrow x=\frac { 50 }{ 3 } -1\) and \(x=-\frac { 50 }{ 3 } -1\)
\(\Rightarrow x=\frac { 47 }{ 3 } \) and \(x=\frac { -53 }{ 3 } \)

We know that the vector equation of a plane passing through the line of intersection of the planes
\(\vec { r } .\vec { { n }_{ 1 } } ={ d }_{ 1 }\) and \(\vec { r } .\vec { { n }_{ 2 } } ={ d }_{ 2 }\) is given by \((\vec { r } .\vec { { n }_{ 1 } } -{ d }_{ 1 })+\lambda (\vec { r } .\vec { { n }_{ 2 } } -{ d }_{ 2 })=0\)
Substituting \(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } ,\vec { { n }_{ 1 } } =\hat { i } +\hat { j } +\hat { k } ,\vec { { n }_{ 2 } } =2\hat { i } -3\hat { j } +5\hat { k } \), \({ d }_{ 1 }=1,{ d }_{ 2 }=-2\) in the above equation, we get
(x + y + z + 1) + \(\lambda \) (2x - 3y + 5z - 2) = 0
Since this plane passes through the point (−1, 2,1) , we get λ = \(\frac{3}{5}\), and hence the required equation
of the plane is 11x−4y+20z=1 .

Given lines are x+1 = 2y = -12z
\(\Rightarrow \frac { x+1 }{ 1 } =\frac { y-0 }{ 1 } =\frac { z-0 }{ \frac { -1 }{ 12 } } \)
and x = y + 2 = 6z - 6
\(\Rightarrow \frac { x-0 }{ 1 } =\frac { y+2 }{ 1 } =\frac { z-1 }{ \frac { 1 }{ 6 } } \)
\(\therefore \vec { a } =-\hat { i } ,\vec { b } =\hat { i } +\frac { 1 }{ 2 } \vec { j } -\frac { 1 }{ 12 } \hat { k } \)
\(\vec { c } =-2\hat { j } +\hat { k } \ and\ \vec { d } =\hat { i } +\hat { j } +\frac { 1 }{ 6 } \hat { k } \)
\(\vec { c } -\vec { a } =2\hat { j } +\hat { k } -(-\hat { i } )=\hat { i } -2\hat { j } +\hat { k } \)
\(\vec { b } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & \frac { 1 }{ 2 } & -\frac { 1 }{ 12 } \\ 1 & 1 & \frac { 1 }{ 6 } \end{matrix} \right| \)
\(=\hat { i } \left( \frac { 1 }{ 12 } +\frac { 1 }{ 12 } \right) -\hat { j } \left( \frac { 1 }{ 6 } +\frac { 1 }{ 12 } \right) +\hat { k } \left( 1-\frac { 1 }{ 12 } \right) \)
\(=\frac { 1 }{ 6 } \hat { i } -\frac { 1 }{ 4 } \hat { j } +\frac { 1 }{ 2 } \hat { k } \)
Now \((\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )\)
\(=\left( \hat { i } -2\hat { j } +\hat { k } \right) .\left( \frac { 1 }{ 6 } \hat { i } -\frac { 1 }{ 4 } \hat { j } +\frac { 1 }{ 2 } \hat { k } \right) \)
\(=\frac { 1 }{ 6 } +\frac { 2 }{ 4 } +\frac { 1 }{ 2 } =\frac { 1 }{ 6 } +\frac { 1 }{ 2 } +\frac { 1 }{ 2 } \)
\(=\frac { 1 }{ 6 } +1=\frac { 7 }{ 6 } \)
Since \((\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )\)0, the given lines are skew lines. 
\(\therefore |\vec { b } \times \vec { d } |=\sqrt { \frac { 1 }{ 36 } +\frac { 1 }{ 16 } +\frac { 1 }{ 4 } } \)
\(\sqrt { \frac { 4+9+36 }{ 144 } } =\sqrt { \frac { 49 }{ 144 } } =\frac { 7 }{ 12 } \)
Shortest distances between the skew lines

Equation of the hyperbola is 12x²- 9y² = 108
\(\div 108\) we get, \(\frac { { 12x }^{ 2 } }{ 108 } -\frac { 9{ y }^{ 2 } }{ 108 } =1\)
⇒ \(\frac { { x }^{ 2 } }{ 9 } -\frac { { y }^{ 2 } }{ 12 } =1\)
∴ a² = 9, b² = 12
Parametric equation of tangent to the hyperbola is \(\frac { x \ sec\ \theta }{ a } -\frac { y \ tan \ \theta }{ b } =1\)
When \(\theta =\frac { \pi }{ 3 } \), the equation is
\(\frac { xsec\frac { \pi }{ 3 } }{ 3 } -\frac { ytan\frac { \pi }{ 3 } }{ 2\sqrt { 3 } } =1\)

⇒ \(\frac { 4x-3y }{ 6 } =1\) ⇒ 4x - 3y - 6 = 0 is the required equation of tangent.
Parametric equation of normal to the hyperbola is
\(\frac { ax }{ sec\theta } +\frac { by }{ tan\theta } ={ a }^{ 2 }+{ b }^{ 2 }\)
At \(\theta =\frac { \pi }{ 3 } ,\frac { 3x }{ sec\frac { \pi }{ 3 } } +\frac { 2\sqrt { 3 } }{ tan\frac { \pi }{ 3 } } =9+12\)
\(\Rightarrow \frac{3 x}{2}+\frac{2 \sqrt{\not 3} y}{\sqrt{\not 3}}=21\)
⇒ \(\frac { 3x }{ 2 } \) + 2y = 21 ⇒ 3x + 4y = 42
⇒ 3x + 4y - 42 = 0 is the required equation of normal.

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 = \(\left[ \begin{matrix} \left( 6/5 \right) & -1 \\ \left( 1/5 \right) & 0 \end{matrix} \right] =\frac { 1 }{ 5 } \left[ \begin{matrix} 6 & -5 \\ 1 & 0 \end{matrix} \right] \).

Let the roots be a−d, a, a+d.
Then the sum of the roots is 3a which is equal to 6 from the given equation.
Thus 3a = 6 and hence a = 2.
The product of the roots is a³− ad² which is equal to −24 from the given equation.
Substituting the value of a, we get 8−2d² = −24 and hence d = ±4.
If we take d = 4 we get −2, 2, 6 as roots and if we take d = −4, we get 6, 2, −2 as roots (same roots given in reverse order) of the equation.

Given equation is 2x⁴- 8x + 6x²- 3 = 0
Here a = 2, b = -8, c = 6, d = 0, e = -3
Let ∝, β, ૪ and \(\delta \) be the roots of equation (1)
Then by Vieta's formula,
\(\sum { _{ 1 }= } \alpha +\beta +\gamma +\delta =\frac { -b }{ a } =\frac { -(-8) }{ 2 } =4\)
\(\sum { _{ 2 } } =\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta =\frac { c }{ a } =\frac { 6 }{ 2 } =3\)
\(\sum { _{ 3 } } =\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta =\frac { -d }{ a } =\frac { 0 }{ a } \)
\(\sum { _{ 4 } } =\alpha \beta \gamma \delta =\frac { e }{ a } =\frac { -3 }{ 2 } \)
Now, (a+b+c+d)² = a²+b²+c²+d²+2(ab+ac+ad+bc+cd)
⇒ n∝²+β²+૪²+\(\delta\)² = (∝ + β + ૪ + \(\delta\))²-2(\(\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta \))
∝² + β² + ૪² = 4²-2(3) = 16 - 6 = 10

Given \(A=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) ,B=\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) and\quad C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
\(\left( \begin{matrix} 1 & 1 \\ 1 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 1 \\ 1 & 1 \\ 0 & 1 \end{matrix} \right) \wedge \left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) =\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) \)

Given \(A=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) ,B=\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) and\quad C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) \vee \left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 1\vee 0 & 0\vee 1 \\ 0\vee 1 & 1\vee 0 \\ 1\vee 1 & 0\vee 0 \end{matrix}\begin{matrix} 1\vee 0 & 0\vee 1 \\ 0\vee 1 & 1\vee 0 \\ 0\vee 0 & 1\vee 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 1 & 1 \\ 1 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 1 \\ 1 & 1 \\ 0 & 1 \end{matrix} \right) \) [∵ a∨b=max(a,b)]

\(\begin{matrix} lim \\ (x,y)\rightarrow (0,0) \end{matrix}cos=\left( \frac { { e }^{ x }siny }{ y } \right) \) = \(cos\left( { e }^{ 0 }\frac { siny }{ y } \right) \)
= cos[(1)(1)] = cos (1) \(\left[ \because \begin{matrix} lim \\ y\rightarrow 0 \end{matrix}\frac { siny }{ y } =1 \right] \)

We have,
\(\underset { x\rightarrow { 0 }^{ - } }{ lim } =-\infty \ and\ \underset { x\rightarrow { 0 }^{ x } }{ lim } =\frac { 1 }{ x } =\infty \). Hence, the required vertical asymptote is x = 0 or the y -axis.
As the curve is symmetric with respect to both the axes, y = 0 or the x -axis is also an asymptote.
Hence this (rectangular hyperbola) curve has both the vertical and horizontal asymptotes.

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

Let \(g(x)=\frac { 1 }{ 2 } { tan }^{ -1 }\left( 1-{ x }^{ 2 } \right) -\frac { \pi }{ 4 } \)
By the definition of tan^-1 x, it is a function with
the entire real line \(\left( -\infty ,\infty \right) \) as its domain.
\(\therefore \) Domain of \(g(x)=\frac { 1 }{ 2 } { tan }^{ -1 }\left( 1-{ x }^{ 2 } \right) -\frac { \pi }{ 4 } \)is R
\(\therefore \) Domain of g(x) is R.

Given \(\vec { a } =\hat { i } +\hat { j } +\hat { k } \), \(\vec { b } =\hat { i } \), \(\vec { c } ={ c }_{ 1 }\hat { i } +{ c }_{ 2 }\hat { j } +{ c }_{ 3 }\hat { k } \)
∴ \(\vec { c } =\hat { i } +2\hat { j } +{ c }_{ 3 }\hat { k } \)
Also, it given that \(\vec { a } ,\vec { b } \) and \(\vec { c } \) are co-planar.
∴ \(\vec { a } .(\vec { b } \times \vec { c } )\)
⇒ \(\left| \begin{matrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 2 & { c }_{ 3 } \end{matrix} \right| \)= 0
⇒ \(1\left| \begin{matrix} 0 & 0 \\ 2 & { c }_{ 3 } \end{matrix} \right| -1\left| \begin{matrix} 1 & 0 \\ 1 & { c }_{ 3 } \end{matrix} \right| +1\left| \begin{matrix} 1 & 0 \\ 1 & 2 \end{matrix} \right| \) = 0
⇒ 1(0-0)-1(c₃-0)+1(2-0) = 0
⇒ 0 - c₃+2 = 0 ⇒ c₃ = 2
∴ c₃ = 2