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

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.

By (14), Linear approximation is given by
L (x, y, z) =  U(x₀, y₀, z₀) + \({ \frac { { \partial }U }{ { \partial x } } | }_{ { x }_{ 0 },{ y }_{ 0 },{ z }_{ 0 } }\) (x-x₀)+\({ \frac { { \partial }U }{ { \partial y } } | }_{ { x }_{ 0 },{ y }_{ 0 },{ z }_{ 0 } }\) (y-y₀)+\({ \frac { { \partial }U }{ { \partial z } } | }_{ { x }_{ 0 },{ y }_{ 0 },{ z }_{ 0 } }\) (z-z₀)
Now U_x = 2x -y, Uy = -xand U_z = 3cos z.
Here (x₀, y₀, z₀) = (2,−1,0 )
hence U_x (2, −1,0) = 5, U_y (2, −1,0) = −2 and U_z (2,-1,0) = 3.
Thus L(x, y, z) = 6 + 5(x − 2) − 2( y +1) + 3(z − 0) = 5x − 2y + 3z − 6 is the required linear approximation for U at (2,−1,0).

\(\int _{ 0 }^{ 1 }{ { x }^{ m } } { (1-x) }^{ n }dx=\frac { m!\times n! }{ (m+n+1)! } \)
\(\therefore \int _{ 0 }^{ 1 }{ { x }^{ 3 }{ (1-x) }^{ 4 }dx } =\frac { 3!\times 4! }{ (3+4+1)! } =\frac { 3!\times 4! }{ 8! } =\frac { 3\times 2\times 1\times 4\times 3\times 2\times 1 }{ 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 } =\frac { 1 }{ 280 } \)

\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { cos }^{ \frac { 1 }{ 2 } }\theta (1-{ cos }^{ 2 }) } sin\ \theta\ d\theta \)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( { cos }^{ \frac { 1 }{ 2 } }\theta -{ cos }^{ \frac { 5 }{ 2 } }\theta \right) } sin\theta\ d\theta \)
\(Put\ cos\theta =t\)
\(\\ \Rightarrow -sin\theta\ d\theta =dt\)
\(\Rightarrow sin\theta\ d\theta =-dt\)

\(=-\int _{ 1 }^{ 0 }{ ({ t }^{ \frac { 1 }{ 2 } }-{ t }^{ \frac { 5 }{ 2 } }) } dt=\int _{ 0 }^{ 1 }{ ({ t }^{ \frac { 1 }{ 2 } }-{ t }^{ \frac { 5 }{ 2 } }) } dt\)
\(={ \left[ \frac { { t }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } -\frac { { t }^{ \frac { 7 }{ 2 } } }{ \frac { 7 }{ 2 } } \right] }_{ 0 }^{ 1 }=\left[ \frac { 2 }{ 3 } (1)-\frac { 2 }{ 7 } (1)-0 \right] \)
\(=\frac { 2 }{ 3 } -\frac { 2 }{ 7 } =\frac { 14-6 }{ 21 } =\frac { 8 }{ 21 } \)

Given \(f(x)=\sqrt [ 3 ]{ x } \)
Let x₀ = 27 and \(\triangle x=0.2\)
We know L(x) = \(f({ x }_{ 0 })+{ f }^{ ' }({ x }_{ o })\) (x - x₀) ∀ x ∈ (a, b)
∴ \(\sqrt [ 3 ]{ 27.2 } =f(27)+{ f }^{ ' }(27)(0.2)\) .. (1)
Now f(27) = \(\sqrt [ 3 ]{ 27 } =3\)
\({ f }^{ ' }(27)=\frac { 1 }{ 3 } x^{ \frac { 1 }{ 3 } -1 }=\frac { 1 }{ 3 } { x }^{ \frac { 2 }{ 3 } }=\frac { 1 }{ { 3x }^{ \frac { 2 }{ 3 } } } \)
\(\therefore \) becomes,
\(\sqrt [ 3 ]{ 27.2 } =3+\frac { 1 }{ 27 } (0.2)\)
= 3 + .0074 = 3.0074
\(\therefore \) \(\sqrt [ 3 ]{ 27.2 } =3.0074\)

\(\underset { x\rightarrow { 1 }^{ + } }{ lim } \left( \frac { 2 }{ { x }^{ 2 }-1 } -\frac { x }{ x-1 } \right) =\underset { x\rightarrow { 1 }^{ + } }{ lim } \left( \frac { 2-x(x+1) }{ { x }^{ 2 }-1 } \right) \)
\(\underset { x\rightarrow { 1 }^{ + } }{ lim } \left( \frac { 2-{ x }^{ 2 }-x }{ { x }^{ 2 }-1 } \right) =\frac { 0 }{ 0 } \)
Form which is indeterminate,
Applying L' Hopital rule we get,
\(\underset { x\rightarrow { 1 }^{ + } }{ lim } \left( \frac { 2-{ x }^{ 2 }-x }{ { x }^{ 2 }-1 } \right) =\frac { 0 }{ 0 } =\frac { -2-1 }{ 2(1) } =\frac { -3 }{ 2 } \)

f'(x) = -2 sin x + 2 cos 2x
f'(x) = 0
\(\Rightarrow\) 2 sin x + 2 cos 2x = 0
\(\Rightarrow\) 2 sin x + 2(1 - 2 sin²x) = 0
\(\Rightarrow\) 4 sin²x + 2 sin x - 2 = 0
\(\Rightarrow\) 4 sin² x + 2 sin x - 2 = 0
\(\Rightarrow\) 2 sin² x + sin x-1 = 0
\(\Rightarrow\) (sin x+1) (2sin x-1) = 0
\(\Rightarrow\) \(sinx=-1\ or\ sinx=\frac { 1 }{ 2 } \) 
\(\Rightarrow\) \(sinx=-sin\frac { \pi }{ 2 } \) or 
\(sinx=sin\frac { \pi }{ 6 } \) 
 \(\Rightarrow sinx=sin\left( -\frac { \pi }{ 2 } \right) \) 
\(sinx=sin\frac { \pi }{ 6 } \)  
\(\Rightarrow x=-\frac { \pi }{ 2 } or \ x=\frac { \pi }{ 6 } \)
\(\Rightarrow x=\frac { \pi }{ 6 } \) 
\(\\ \\ \\ \\ \\ \\ \left[ \because x=-\frac { \pi }{ 2 }∉ \left[ 0.\frac { \pi }{ 2 } \right] \right] \)
\(\therefore\) The critical number is \(x=\frac { \pi }{ 6 } \) 
Evaluating f(x) at the end points x = 0, \(x=\frac { \pi }{ 2 } \) and at the critical number \(x=\frac { \pi }{ 6 } \) we get.
f(0) = 2 cos0 + sin0 = 2
\(f(\frac { \pi }{ 2 } )=2cos\frac { \pi }{ 2 } +sin\pi =0\) 
\(f\left( \frac { \pi }{ 6 } \right) =2cos\frac { \pi }{ 2 } +sin\frac { \pi }{ 3 } \) 
\(2\left( \frac { \sqrt { 3 } }{ 2 } \right) +\frac { \sqrt { 3 } }{ 2 } =\frac { 3\sqrt { 3 } }{ 2 } \) 
From these values, the absolute maximum is \(\frac { 3\sqrt { 3 } }{ 2 } \) which occurs at \(x=\frac { \pi }{ 6 } \) and the absolute minimum is 0 which occurs at \(x=\frac { \pi }{ 2 } \)

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 equation of family of curves is
y = Ae^8x + Be^-8x ..(1)
where A & B are arbitrary constants. Differentiating cquation (1) twice successively (because we have two arbitrary constant), we get
Differentiating with respect to 'x' we get,
\(\frac { dy }{ dx } \\ \\ \) = 8Ae^8x + 8Be^-8x
Differentiating again with respect to 'x' we get,
\(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = 64Ae^8x + 64Be^-8x
= 64(Ae^8x + Be^-8x)
\(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = 64 y [using (1)]
\(\frac { d^{ 2 }y }{ dx^{ 2 } } \) - 64 y = 0
Which is the required differential equation.

|z-4|²-|z-1|² = 16
|x+iy-4|² - |x+iy-1|² = 16
⇒ |(x-4)+iy|² - |(x-1)+iy2|² = 16
⇒ [(x-4)²+y²] - [(x-1)²+y²] = 16
⇒ x²-8x+16+y²-[x²-2x+1+y²] = 16
\(\Rightarrow \not x^{2}-8 x+16+\not y^{2}-\not x^{2}+2 x-1-\not y^{2}=16\)
⇒ -6x+15-16 = 0
⇒ -6x-1 = 0
⇒ 6x+1 = 0 Which is the required Cartesian equation.
The locus of the point is a straight line.

Given g(x) = sin^-1 x + cos^-1x
From the definition of sin^-1x.
\(-1\le x\le 1\) ...(1)
Also from the definition of cos^-1x
\(-1\le x\le 1\)   .........(2)
\(\therefore \) From (1) & (2),
Domain ofg(x) = [-1, 1] U [-1, 1]
= [-1, 1]
Hence the domain of g(x) is [-1, 1].

2x-y = 8, 3x+2y+2 = -2
The matrix form of the system is
\(\left[ \begin{matrix} 2 & -1 \\ 3 & 2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 8 \\ -2 \end{matrix} \right] \)
⇒ AX = B where A =\(\\ \left[ \begin{matrix} 2 & -1 \\ 3 & 2 \end{matrix} \right] \)
B =\(\left[ \begin{matrix} 8 \\ -2 \end{matrix} \right] \)
⇒ X = A^-1N
Now, |A| =\(\left[ \begin{matrix} 2 & -1 \\ 3 & 2 \end{matrix} \right] \)= 4 + 3 = 7
∴ A^-1= \(\frac { 1 }{ |A| } \)adj A
= \(\frac { 1 }{ 7 } \left[ \begin{matrix} 2 & 1 \\ -3 & 2 \end{matrix} \right] \)
∴ X = A^-1B = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 2 & 1 \\ -3 & 2 \end{matrix} \right] \left[ \begin{matrix} 8 \\ -2 \end{matrix} \right] \)
= \(\frac { 1 }{ 7 } \left[ \begin{matrix} 16-2 \\ -24-4 \end{matrix} \right] \)
= \(\frac { 1 }{ 7 } \left[ \begin{matrix} 14 \\ -28 \end{matrix} \right] =\left[ \begin{matrix} \frac { 14 }{ 7 } \\ \frac { -28 }{ 7 } \end{matrix} \right] =\left[ \begin{matrix} 2 \\ -4 \end{matrix} \right] \)
∴ x = 2, y = -4
Hence, the solution set is {2, -4}

Here, \(\vec { a } =(2\hat { i } -\hat { j } +2\hat { k } ),\vec { b } =(3\hat { i } +4\hat { j } +2\hat { k } )\).
The vector form of the given plane is \(\vec { r } .(\hat { i } -\hat { j } +\hat { k } )=5\). Then \(\vec { r } .(\hat { i } -\hat { j } +\hat { k } )=5\) and p = 5
We know that the position vector of the point of intersection of the line  \(\vec { r } =\vec { a } +t\vec { b } \) and the plane
\(\vec { r } .\vec { d } =p\vec { u } =\vec { a } +\left( \frac { p-(\vec { a } .\vec { n } ) }{ \vec { b } .\vec { n } } \right) \vec { b } \), where \(\vec { b } .\vec { n } \neq \vec { 0 } \)
Clearly, we observe that \(\vec { b } .\vec { n } \neq \vec { 0 } \)
Now, \(\frac { p-(\vec { a } .\vec { n } ) }{ \vec { b } .\vec { n } } =\frac { 5-(2\hat { i } -\hat { j } +2\hat { k } ).(\hat { i } -\hat { j } +\hat { k } ) }{ (3\hat { i } +4\hat { j } +2\hat { k } ).(\hat { i } -\hat { j } +\hat { k } ) } =0\). Therefore, the position vector of the point of intersection of the given line and the given plane is
\(\hat { r } =(2\hat { i } -\hat { j } +2\hat { k } )+(0)(3\hat { i } +4\hat { j } +2\hat { k } )=2\hat { i } -\hat { j } +2\hat { k } \)
That is, the given straight line intersects the plane at the point (2, −1, 2)
Aliter:
The Cartesian equation of the given straight line is \(\frac { x-2 }{ 3 } =\frac { y+1 }{ 4 } =\frac { z-2 }{ 2 } =t\)(say)
We know that any point on the given straight line is of the form (3t+2, 4 t−1, 2 t+2). If the given line and the plane intersects, then this point lies on the given pane x−y+z−5 = 0.
So, (3t + 2)−(4t − 1) + (2t + 2) − 5 = 0 ⇒ t = 0.
Therefore, the given line intersects the given plane at the point (2, -1, 2)

The parametric form of vector equations of the given straight lines are
\(\vec { r } =(2\hat { i } +3\hat { j } +4\hat { k } )+t(-2\hat { i } +\hat { j } -2\hat { k } )\)
and \(\vec { r } =(3\hat { i } -2\hat { k } )+t(2\hat { i } -\hat { j } +2\hat { k } )\)
Comparing the given two equations with \(\vec { r } =\vec { a } +t\vec { b } ,\vec { r } =\vec { c } +s\vec { d } \)
we have \(\vec { a } =2\hat { i } +3\hat { j } +4\hat { k } ,\vec { b } =-2\hat { i } +\hat { j } -2\hat { k } ,\vec { c } =3\hat { i } -2\hat { k } ,\vec { d } =2\hat { i } -\hat { j } +2\hat { k } \)
Clearly, \(\vec { b } \) is a scalar multiple of \(\vec { d } \), and hence the two straight lines are parallel. We know that the shortest distance between two parallel straight lines is given by \(d=\frac { \left| (\vec { c } -\vec { a } )\times \vec { b } \right| }{ \left| \vec { b } \right| } \)
\((\vec { c } -\vec { a } )\times \vec { b } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & -3 & -6 \\ -2 & 1 & -2 \end{matrix} \right| =12\vec { i } +14\vec { j } -5\vec { k } \)
\(d=\frac { \left| 12\hat { i } +14\hat { j } -5\hat { k } \right| }{ \left| -2\hat { i } +\hat { j } -2\hat { k } \right| } =\frac { \sqrt { 365 } }{ 3 } \)

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.

Using the definition of the scalar triple product, we get 
\([\vec { a } \times \vec { b } ,\vec { b } \times \vec { c } ,\vec { c } \times \vec { a } ]\) = \((\vec { a } \times \vec { b } ).[(\vec { b } \times \vec { c } )\times (\vec { c } \times \vec { a } )]\) ....(1)
By treating \((\vec { b } \times \vec { c } )\) as the first vector in the vector triple product, we find
\((\vec { b } \times \vec { c } )\times (\vec { c } \times \vec { a } )\) = \(((\vec { b } \times \vec { c } ).\vec { a } )\vec { c } \) -  \(((\vec { b } \times \vec { c } ).\vec { c } )\vec { a } )\) = \([{ \vec { a } ,\vec { b } ,\vec { c } }]\vec { c } \)
Using this value in (1), we get
\([\vec { a } \times \vec { b } ,\vec { b } \times \vec { c } ,\vec { c } \times \vec { a } ]\) = \((\vec { a } \times \vec { b } ).([\vec { a } ,\vec { b } ,\vec { c } ]\vec { c } )=[\vec { a } ,\vec { b } ,\vec { c } ](\vec { a } \times \vec { b } ).\vec { c } ={ [\vec { a } ,\vec { b } ,\vec { c } ] }^{ 2 }\)

We have |z| - |z - i|
\(\Rightarrow\)|x + iy| = |x + iy - i|
\(\Rightarrow\) \(\sqrt { { x }^{ 2 }+{ y }^{ 2 } } =\sqrt { { x }^{ 2 }+\left( y-1 \right) ^{ 2 } } \)
\(\Rightarrow x^{2}+y^{2}=x^{2}+y^{2}-2 y+1\)
\(\Rightarrow\) 2y -1 = 0

Given z = x + iy
Consider \(\left| \frac { z-4i }{ z+4i } \right| =1\Rightarrow \left| \frac { x+iy-4i }{ x+iy+4i } \right| \)=1
⇒ \(\left| \frac { x+i(y-4) }{ x+i(y+4) } \right| \)
⇒ \(\frac { \sqrt { { x }^{ 2 }+(y-4)^{ 2 } } }{ \sqrt { { x }^{ 2 }+((y+4)^{ 2 } } } \) = 1
⇒ \(\sqrt { { x }^{ 2 }+(y-4)^{ 2 } } =\sqrt { { x }^{ 2 }+(y+4)^{ 2 } } \)
Squaring both sides we get,
x²+(y-4)² = x²+(y+4)²
\(\Rightarrow \not x^{2}+\not y^{2}-8 y+\not 16=\not x^{2}+\not y^{2}+8 y+\not 16\)
⇒ 8y+8y = 0
⇒ 16y = 0
⇒ y = 0    [∵ 16 ≠ 0]
y = 0 is the equation of real axis Locus of z is the real axis.

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.

SS′ = 2c and 2c = 4; A'A = 2a = 6
c = 2 and a = 3,
b² = a²−c² = 9−4 = 5.
Major axis is along x-axis, since a > b.
Centre (0, 0) and Foci are (±2, 0) 
Therefore, equation of the ellipse is \(\frac { { x }^{ 2 } }{ 9 } \frac { { y }^{ 2 } }{ 5 } =1\)
 

Let tan⁻¹(−1) = y. Then, tan y = -1 = -tan\(\frac{\pi}{4}=tan(-\frac{\pi}{4})\)
As -\(\frac{\pi}{4}\in (-\frac{\pi}{2},\frac{\pi}{2}), tan^-1(-1)=-\frac{\pi}{ 3}\)
Now, cos^-1\((\frac{1}{2})\) = y implies cos y = \(\frac{1}{2}\) = cos\(\frac{\pi}{3}\)
As \(\frac{\pi}{3}\)\(\in\)[0, \(\pi\)], cos^-1 \((\frac{1}{2})=\frac{\pi}{3}\)
Now, sin^-1\((-\frac{1}{2})\) = y implies sin y = -\(\frac{1}{2}\) = sin(-\(\frac{\pi}{3}\)).
As -\(\frac{\pi}{6}\in[-\frac{\pi}{2},\frac{\pi}{2}], sin^-1(-\frac{1}{2})=-\frac{\pi}{6}\)
Therefore, tan⁻¹(−1)+cos^-1\((\frac{1}{2})+sin^-1(-\frac{1}{2})=-\frac{\pi}{4}+\frac{\pi}{3}-\frac{\pi}{6}=-\frac{\pi}{12}\)

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 \(\frac { 7 }{ 2 } \) - \(\frac { i }{ 2 } \)

We get AB = \(\left[ \begin{matrix} 0 & -3 \\ 1 & 4 \end{matrix} \right] \left[ \begin{matrix} -2 & -3 \\ 0 & -1 \end{matrix} \right] =\left[ \begin{matrix} 0+0 & 0+3 \\ -2+0 & -3-4 \end{matrix} \right] =\left[ \begin{matrix} 0 & 3 \\ -2 & -7 \end{matrix} \right] \)
(AB)^-1 = \(\frac { 1 }{ \left( 0+6 \right) } \left[ \begin{matrix} -7 & -3 \\ 2 & 0 \end{matrix} \right] =\frac { 1 }{ 6 } \left[ \begin{matrix} -7 & -3 \\ 2 & 0 \end{matrix} \right] \)...(1)
A^-1 = \(\frac { 1 }{ \left( 0+3 \right) } \left[ \begin{matrix} 4 & 3 \\ -1 & 0 \end{matrix} \right] =\frac { 1 }{ 3 } \left[ \begin{matrix} 4 & 3 \\ -1 & 0 \end{matrix} \right] \)
B^-1 = \(\frac { 1 }{ \left( 2-0 \right) } \left[ \begin{matrix} -1 & 3 \\ 0 & -2 \end{matrix} \right] =\frac { 1 }{ 2 } \left[ \begin{matrix} -1 & 3 \\ 0 & -2 \end{matrix} \right] \)
B^-1A^-1 = \(\frac { 1 }{ 2 } \left[ \begin{matrix} -1 & 3 \\ 0 & -2 \end{matrix} \right] \frac { 1 }{ 3 } \left[ \begin{matrix} 4 & 3 \\ -1 & 0 \end{matrix} \right] =\frac { 1 }{ 6 } \left[ \begin{matrix} -7 & -3 \\ 2 & 0 \end{matrix} \right] \) ...(2).
As the matrices in (1) and (2) are same, (AB)⁻¹ = B⁻¹A⁻¹ is verified.

The length and breadth of the cuboid are x + 1.
x + 2 and x + 3
[∵ they are increased by 1, 2, 3 units]
Also volume = V + 52 .
[since V is increased by 52]
∴ V + 52 = (x +1)(x + 2)(x + 3) .........(1)
⇒ V = (x + 1) (x + 2) (x + 3) - 52
Here a = -1, β = -2, ૪ = -3
⇒ V = x³-x²(α+β+૪)+x(αβ+β૪+૪α)-αβ૪ = 52
⇒ V = x³-x²(-1-2-3)+x(2+6+3)-(-1)(-2)(-3) = 52
⇒ V = x³-x²(-6)+x(11)+6-52
⇒ x³ = x²+6x²+11x+6-52
⇒ 6x²+11x-46 = 0
⇒ (6x+23)(x-2) = 0
⇒ (6x+23)(x-2) = 0
⇒ x = 2

∴ Volume of the cube = x³ = 2³ = 8.
Volume of a cuboid = 52 + 8 = 60
[∵ x = \(\frac{-23}{6}\) is not possible as x represents the side of the cube]

Also the above result can be proved without using truth tables. This proof will be provided after studying the logical equivalence

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?).

The given curve is \(y=\begin{cases} cosx,0\le x\le \frac { \pi }{ 2 } \\ -cosx,\frac { \pi }{ 2 } \le x\le \pi \end{cases}\)
It lies above the x − axis. The required area is sketched. So, the required area is given by

\(A=\int _{ 0 }^{ \pi }{ ydx=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ cosxdx } +\int _{ \frac { \pi }{ 2 } }^{ \pi }{ (-cosx)dx } ={ [sin\quad x] }_{ 0 }^{ \frac { \pi }{ 2 } }-{ [sin\quad x] }_{ \frac { \pi }{ 2 } }^{ \pi } } \)
= [1-0]-[0-1] = 2

Here f (x) = x, a = 0 and b = 1. Hence, we get
\(\int _{ a }^{ b }{ f(x)dx } =\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ f } \left( \frac { r }{ n } \right) \Rightarrow \int _{ 0 }^{ 1 }{ xdx } =\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ \frac { r }{ n } } \)
\(=\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ { n }^{ 2 } } [1+2+...+n]\)
\(=\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ { n }^{ 2 } } \frac { n(n+1) }{ 2 } =\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ 2 } \left( 1+\frac { 1 }{ n } \right) =\frac { 1 }{ 2 } \)

\(x\frac { dy }{ dx } +2y=x^2logx\)
Dividing by x we get,
\(\frac { dy }{ dx } +\frac { 2 }{ x } y=xlogx\)
This is a linear differential equation 
\(\therefore P=\frac { 2 }{ x } ;Q=logx\)
\(\int { pdx } =\int { \frac { 2 }{ x } } dx=2logx=logx^2\)
\(\therefore I.F={ e }^{ \int { pdx } }={ e }^{ log\ x ^2}=x ^2\)
\(\therefore\) The solution is \({ e }^{ \int { u\ dv} }=uv-\int { vdu}\)
\(u=log\ x;dv=x^3\)
\(du=\frac { 1 }{ x }dx;v=\frac { { x }^{ 4} }{ 4} \)
\({ ye }^{ \int { pdx } }=\int { Q{ e }^{ \int { pdx } }dx+c } \)
\(\Rightarrow { yx }^{ 2 }=\int { xlogx.({ x }^{ 2 })dx } \)
\(\Rightarrow { x }^{ 2 }y=\int { { x }^{ 3 } } log\quad xdx\)
\(\Rightarrow { x }^{ 2 }y=\frac { { x }^{ 4 } }{ 4 } logx-\int { \frac { { x }^{ 4 } }{ 4 } .\frac { 1 }{ x } } dx\)
\(\Rightarrow { x }^{ 2 }y=\frac { { x }^{ 4 } }{ 4 } logx-\frac { 1 }{ 4 } \int { { x }^{ 3 }dx } \)
\(\Rightarrow { x }^{ 2 }y=\frac { { x }^{ 4 } }{ 4 } logx-\frac { { x }^{ 4 } }{ 16 } +c\)

Let \(f(x)=\frac{1}{x}\). then the Taylor series of f (x) is
\(f(x)=\sum^{n=\infty}_{n=0}a_{n}(x-2)^{n},\) where \(a_{n}=\frac{f^{(n)}(2)}{n!}\)
Various derivatives of the function f (x) evaluated at x = 2 are given below.

Substituting these values, we get the required expansion of the function as:
\(\frac{1}{x}=\frac{1}{2}-\frac{1}{4}\frac{(x-2)}{1!}+\frac{1}{4}\frac{(x-2)^{2}}{2!}-\frac{3}{8}\frac{3(x-2)^{3}}{3!}+...\)
which is, \(\frac{1}{x}=\frac{1}{2}-\frac{(x-2)}{4}+\frac{(x-2)^{2}}{8}-\frac{(x-2)^{3}}{16}+...\)

ydx + (1 + x²) tan^-1 xdy = 0
\(\mathrm{yd} x=-\left(1+x^2\right) \tan ^{-1} x \mathrm{dy} \)
\(\frac{d x}{\left(1+x^2\right) \tan ^{-1} x}=-\frac{d y}{y} \)
Take \(\mathrm{t}=\tan ^{-1} x \)
\(\mathrm{dt}=\frac{1}{1+x^2} d x\)
The equation can be written as
\(\frac{d t}{t}=-\frac{d y}{y}\)
Taking Integration on both sides, we get
\(\int \frac{d t}{t}=-\int \frac{d y}{y}\)
log t = - log y + log C
log (tan^-1 x) = -log y+ log C
log (tan^-1 x) + log y = log C
log y(tan^-1 x) = log c
y tan^-1 x = c

12x + 8x = 29x²- 4
This equation can be re-written as
12x³ - 29x² + 8x + 4 = 0

∴ x = 2 is a root and the remaining factor is 
12x²- 5x - 2
⇒ (3x-2)(4x+1) = 0
⇒ 3x-2 = 0 or 4x+1 = 0

⇒ x = \(\frac{2}{3}\)
x = \(\frac{-1}{4}\)
∴ The roots are 2, \(\frac{2}{3}\), \(\frac{-1}{4}\)

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

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

Here the number of unknowns is 3.
The matrix form of the system is AX = B, where
A = \(\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & -3 & 3 \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \), B = \(\left[ \begin{matrix} -9 \\ -18 \\ -27 \end{matrix} \right] \).
Applying elementary row operations on the augmented matrix[A | B], we get
[A | B] = \(\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & -3 & 3 \end{matrix}|\begin{matrix} -9 \\ -18 \\ -27 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-2{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-3{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} -9 \\ 0 \\ 0 \end{matrix} \right] \).
So, ρ(A) = ρ ([A | B]) = 1 < 3.
From the echelon form, we get the equivalent equations x - y + z = -9, 0 = 0, 0 = 0.
The equivalent system has one non-trivial equation and three unknowns.
Taking y = s, z = t arbitrarily, we get x - s + t = -9; x = -9 + s - t.
So, the solution is (x = -9 + s - t, y = s, z = t), where s and t are parameters.
The above solution set is a two-parameter family of solutions.
Here, the given system of equations is consistent and has infinitely many solutions which form a two parameter family of solutions.

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 in H.P. Then, their reciprocals are in A.P. and roots of the equation
\(\left( \frac { 1 }{ x } \right) ^{ 3 }+p\left( \frac { 1 }{ x } \right) ^{ 2 }+q\left( \frac { 1 }{ x } \right) \)+ r = 0 ⇔ rx³ + qx² + px + 1 = 0.....(1)
Since the roots of (1) are in A.P., we can assume them as α-d, α, α+d
Applying the Vieta’s formula, we get
Σ₁ = (α-d)+α+(α+d) = -\(\frac { q }{ r } \) ⇒ 3α = -\(\frac { q }{ r } \) ⇒ α = -\(\frac { q }{ 3r } \)
But, we note that α is a root of (1). Therefore, we get
\(r\left( -\frac { q }{ 3r } \right) ^{ 2 }+q\left( -\frac { q }{ 3r } \right) ^{ 2 }+p\left( -\frac { q }{ 3r } \right) \) + 1 = 0 ⇒ q³ + 3q³ - 9pqr + 27r² = 0 ⇒ 2q³ + 27r².

Given \(p=\frac { 3 }{ 4 } \) 
n = 5
P(X = x) = nC_xp^x (1-p)^n-x
\(P(X=3)={ 5C }_{ 3 }\left( \frac { 3 }{ 4 } \right) ^{ 3 }\left( 1-\frac { 3 }{ 4 } \right) ^{ 2 }\) 
\(P(X=3)={ 5C }_{ 2 }\left( \frac { 3 }{ 4 } \right) ^{ 3 }\left( \frac { 1 }{ 4 } \right) ^{ 2 }\) [∵nCr = nC_n-r]

= \(\frac { 135 }{ 512 } \)

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

\(f(x)= \begin{cases}2(x-1) & 1
\(Mean=E(X)=\int _{ 1 }^{ 2 }{ f(x)dx=\int _{ 1 }^{ 2 }{ 2((x-1)dx } } \) 
\( =2\left[\frac{8}{3}-\frac{4}{2}-\frac{1}{3}+\frac{1}{2}\right] \)
\( =2\left(\frac{7}{3}-\frac{3}{2}\right) \)
\( =2 \times \frac{5}{6} \)
\( =\frac{5}{3} \)
\(E({ x }^{ 2 })=\int _{ 1 }^{ 2 }{ { x }^{ 2 }f(x)dx } \) 
= \(\int _{ 1 }^{ 2 }{ { x }^{ 2 }.2\left( x-1 \right) } dx\) 
= \(2\int _{ 1 }^{ 2 }{ ({ x }^{ 3 }-{ x }^{ 2 })dx } \) 
= \(2\left[ \frac { { x }^{ 4 } }{ 4 } -\frac { { x }^{ 3 } }{ 3 } \right] _{ 1 }^{ 2 }\) 
= \(2\left[ \left( 4-\frac { 8 }{ 3 } \right) -\left( \frac { 1 }{ 4 } -\frac { 1 }{ 3 } \right) \right] \) 
= \(2\left[ \frac { 4 }{ 3 } +\frac { 1 }{ 12 } \right] =2\left[ \frac { 16+1 }{ 12 } \right] \) 
= \(\frac { 17 }{ 6 } \) 
ஃ Var(X) = E(X²) - [E(X)]²
= \(\frac { 17 }{ 6 } -(\frac{5}{ 3 }^{ 2 })=\frac { 17 }{ 6 } -\frac { 25 }{ 9 } \)
= \(\frac{51-50}{18}\)
= \(\frac{1}{18}\)

Volume of sphere = \(\frac43\) πr³
Given r = 5 mm
⇒ dr = (5.3 - 5) = 0.3 mm
\(\text { Approximate volume }=\frac{4}{\not 3} \pi \cdot \not 3 r^{2} d r\)
= 4π (5²) (0.3)
= 100 π (0.3)
= 30π mm³