Let f(x) = \(f(x)={ x }^{ \frac { 2 }{ 3 } },{ x }_{ 0 }=125,\triangle x=-2\)
∴ (123)^\(\frac23\) = f(125) +1'(125) (-2) ... (1)
\(f(125)={ (125) }^{ \frac { 2 }{ 3 } }={ { (5 }^{ 3 }) }^{ \frac { 2 }{ 3 } }\) = 5² = 25
\({ f }^{ ' }(x)={ \frac { 2 }{ 3 } x }^{ \frac { 2 }{ 3 } -1 }={ \frac { 2 }{ 3 } x }^{ \frac { 1 }{ 3 } }=\frac { 2 }{ { 3x }^{ \frac { 1 }{ 3 } } } \)
\({ f }^{ ' }(125)=\frac { 2 }{ { 3(125)x }^{ \frac { 1 }{ 3 } } } =\frac { 2 }{ { 3{ (5 }^{ 3 } })^{ \frac { 1 }{ 3 } } } =\frac { 2 }{ 3(5) } =\frac { 2 }{ 15 } \)
∴ \({ (123) }^{ \frac { 2 }{ 3 } }=25+\frac { 2 }{ 15 } (-2)\)
\(=25-\frac { 4 }{ 15 } =25-0.27\)
\({ (123) }^{ \frac { 2 }{ 3 } }=24.73\)

We know from (4), that L(x) = f(x₀) +f'(x₀)(x-x₀) We have x₀ = 3, \(\Delta \)x = 0.2 and hence f(3) = \(\sqrt { 1+3 } \) = 2. Also,
f′(x) = \(\frac { 1 }{ 2\sqrt { 1+x } } \) and hence f'(3) = \(\frac { 1 }{ 2\sqrt { 1+3 } } \) = \(\frac { 1 }{ 4 } \)
Thus, L(x) = 2 +\(\frac { 1 }{ 4 } \)(x-3) = \(\frac { x }{ 4 } \)+\(\frac { 5 }{ 4 } \) gives the required linear approximation.
Now, f(3.2) = \(\sqrt { 4.2 } \) ≈ L(3.2) = \(\frac { 3.2 }{ 4 } \)+\(\frac { 5 }{ 4 } \) = 2.050
Actually, if we use a calculator to calculate we get \(\sqrt { 4.2 } \) = 2.04939

Let X be the random variable of number of black cards occur.
X = {0,1,2}
Sample space ⁵²C₂ = 1326
Let X denote the number of black cards drawn.
X = 0, X (both are red cards) = ²⁶C₂ = 325  
X = 1 (1 black card and 1 red card) = ²⁶C₁ \(\times\) ²⁶C₁ = 676  
X = 2 (both are black cards) = ²⁶C₂ = 325  
∴ X takes the values 0, 1, 2

A circle passing through the points (a, 0) and (−a, 0) has its centre on y - axis.
Let (0, b) be the centre of the circle. S o, the radius of the circle is \(\sqrt { { a }^{ 2 }+{ b }^{ 2 } } \) .
Therefore the equation of the family of circles passing through the points (a, 0) and (−a, 0) is x² + ( y − b)² = a² + b², b is an arbitrary constant. ...(1)
Differentiating both sides of (1) with respect to x, we get
\(2x+2(y-b)\frac { dy }{ dx } =0\Rightarrow y-b=\frac { x }{ \frac { dy }{ dx } } \Rightarrow b=\frac { x }{ \frac { dy }{ dx } } +y\)
Substituting the value of b in equation (1), we get
\(2x+2(y-b)\frac { dy }{ dx } =0\Rightarrow y-b=\frac { x }{ \frac { dy }{ dx } } \Rightarrow b=\frac { x }{ \frac { dy }{ dx } } +y\)
\({ x }^{ 2 }+\frac { { x }^{ 2 } }{ { \left( \frac { dy }{ dx } \right) }^{ 2 } } ={ a }^{ 2 }+{ \left[ \frac { x }{ \frac { dy }{ dx } } +y \right] }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }{ \left( \frac { dy }{ dx } \right) }^{ 2 }+{ x }^{ 2 }={ a }^{ 2 }{ \left( \frac { dy }{ dx } \right) }^{ 2 }+{ \left[ x+y{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ 2 }\)
\(\Rightarrow ({ x }^{ 2 }-{ y }^{ 2 }-{ a }^{ 2 })\frac { dy }{ dx } -2xy=0\)
which is the required differential equation

The slope of the line y = x is 1. The tangent to the given curve will be parallel to the line, if the slope of the tangent to the curve at a point is also 1. Hence,
\(\frac{dy}{dx}=3x^{2}-6x+1=1\)
which gives \( 3x^{2}-6x=0\)
Hence, x = 0 and x = 2.
Therefore, at (0, –2) and (2, –4) the tangent is parallel to the line y = x.

Consider \(\frac { 19-7i }{ 9+i } =\frac { 19-7i }{ 9+i } \times \frac { 9-i }{ 9-i } \)
= \(\frac { 171-19i-63i+7i^{ 2 } }{ (9)^{ 2 }-{ i }^{ 2 } } =\frac { 171-82i-7 }{ 81+1 } \)
= \(\frac { 164-82i }{ 82 } =\frac { 82(2-i) }{ 82 } \) = 2- i
Also \(\frac { 20-5i }{ 7-6i } =\frac { 20-5i }{ 7-6i } \times \frac { 7+6i }{ 7+6i } \)
= \(\frac { 140+120i-35i-30^{ 2 } }{ { 7 }^{ 2 }-(6i)^{ 2 } } \)
\(=\frac{140+85 i+30}{49+36}=\frac{170+85 i}{85}=\frac{\not 85(2+i)}{\not 85}=2+i\)
∴ \(\left( \frac { 19-7i }{ 9+i } \right) ^{ 12 }+\left( \frac { 20-5i }{ 7-6i } \right) ^{ 12 }\) = (2-i)12+(2+i)12
Let z = (2-i)¹²+(2+i)¹²
∴ \(\overline { z } \) = \(\overline { (2-i)^{ 12 }+(2+i)^{ 12 } } =\overline { (2-i)^{ 12 } } +\overline { (2+i)^{ 12 } } \)
[∵ \(\overline { { z }_{ 1 }+{ z }_{ 2 } } =\overline { { z }_{ 1 } } +\overline { { z }_{ 2 } } \)]
= (2 + i)¹²+(2 - i)¹²= z
∴ \(\overline { z } \) = z ⇒ z is purely real
∴ \(\left( \frac { 19-7i }{ 9+i } \right) ^{ 12 }+\left( \frac { 20-5i }{ 7-6i } \right) ^{ 12 }\) is real.

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.

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.

Given equation of the circle is
3x² + (3 - p)xy + qy² - 2px = 8pq
For the circle, co-efficient of xy = 0
⇒ 3-p = 0 ⇒ p = 3
Also, co-efficient of x² = co-efficient of y²
⇒ 3 = q
∴ Equation of the circle is
3x² + 3y² - 6x = 8(3)(3)
3x² + 3y² - 6x - 72 = 0
Dividing by 3, we get
x² + y² - 2x - 24 = 0
Here 2g = - 2⇒ g = - 1
f = 0 and c = - 24
Centre is (-g, -f) (1, 0)
and r = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \)
= \(\sqrt { { (-1 })^{ 2 }+0+24 } \)
= \(\sqrt { 25 } \) = 5 units.

Since \(\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \) is a root, x-\(\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \) is a factor. To remove the outermost square root, we take x +\(\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \) as another factor and find their product.
\(\left( x+\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \right) \left( x-\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \right) ={ x }^{ 2 }-\frac { \sqrt { 2 } }{ \sqrt { 3 } } \)
Still we didn’t achieve our goal. So we include another factor x²+\(\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \) and get the product.
\(\left( { x }^{ 2 }-\frac { \sqrt { 2 } }{ \sqrt { 3 } } \right) \left( { x }^{ 2 }+\frac { \sqrt { 2 } }{ \sqrt { 3 } } \right) ={ x }^{ 4 }-\frac { 2 }{ 3 } \)
So, 3x⁴- 2 = 0 is a required polynomial equation with the integer coefficients.
Now we identify the nature of roots of the given equation without solving the equation. The idea comes from the negativity, equality to 0, positivity of Δ = b²- 4ac.

Let z =\(\left( \frac { 19+9i }{ 5-3i } \right) ^{ 15 }-\left( \frac { 8+i }{ I+2i } \right) ^{ 15 }\)
Here, \(\frac { 19+9i }{ 5-3i } =\frac { (19+9i)(5+3i) }{ (5-3i)(5+3i) } \)
= \(\frac { (95-27)+i(45+57) }{ { 5 }^{ 2 }+{ 3 }^{ 2 } } =\frac { 68+102i }{ 34 } \)
= 2 + 3i ................(1)
and \(\frac { 8+i }{ 1+2i } =\frac { (8+i)(1-2i) }{ (1+2i)(1-2i) } \)
= \(\frac { (8+2)+i(1-16) }{ { 1 }^{ 2 }+{ 2 }^{ 2 } } =\frac { 10-15i }{ 5 } \)
= 2 - 3i .............. (2)
Now z =\(\left( \frac { 19+9i }{ 5-3i } \right) ^{ 15 }-\left( \frac { 8+i }{ 1+2i } \right) ^{ 15 }\)
⇒ z = (2 + 3i)¹⁵ - (2 - 3i)¹⁵   (by (1) and (2))
Then by definition, \(\bar { z } =\left( \overline { (2+3i)^{ 15 }-(2-3i)^{ 15 } } \right) \)
= \(\left( \overline { 2+3i } \right) ^{ 15 }-\left( \overline { 2-3i } \right) ^{ 15 }\) (using properties of conjugates)
= (2 - 3i)¹⁵ - (2 + 3i)¹⁵ = -((2 + 3i)¹⁵ - (2-3i)¹⁵)
⇒ \(\\ \overline { z } \) = -z
Therefore, \(\left( \frac { 19+9i }{ 5-3i } \right) ^{ 15 }-\left( \frac { 8+i }{ 1+2i } \right) ^{ 15 }\) is purely imaginary.

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.

C(2, 1) is the centre and 3x + 4y + 10 = 0 cuts a chord AB on the circle. Let M be the midpoint of AB,
then AM = BM = 3. Now BMC is a right triangle.
So, we have CM = \(\frac { \left| 3\left( 2 \right) +4\left( 1 \right) +10 \right| }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } } =4\)
By Pythagoras theorem BC² = BM² + MC² = 3² + 4² = 25
BC = 5 = radius
Equation of the required circle is
(x−2)²+(y−1) = 5²
x²+y²−4x−2y−20 = 0 .

We know that the domain of sin⁻¹(x) is [-1, 1].
This leads to −1\(\le\)2 - 3x²\(\le\)1, Which implies -3\(\le\) -3x²\(\le\)-1
Now, -3\(\le\) -3x², gives x²\(\le\)1 and ............(1)
-3\(\le\)-3x²\(\le\)-1, gives x²\(\ge\)\(\frac{1}{2}\) .........(2)
Combining the equations (1) and (2), we get \(\frac{1}{3}\le x^2\le 1\). That is \(\frac{1}{\sqrt3}\le |x|\le1\), Which gives \(x\in[-1,-1\frac{1}{\sqrt3}]\cup[\frac{1}{\sqrt3},1]\)
since  a\(\le|x|\le b\) implies x \(\in[-b,-a]\cup[a,b]\).

Let z₁ = (2+i)x + (1−i)y + 2i−3 = (2x+y−3) + i(x−y+ 2)and
z₂ = x+(−1+2i)y+1+i = (x−y+1) + i(2y+1)
Given that z₁ = z₂ 
Therefore (2x+y−3) + i(x−y+2) = (x−y+1) + i(2y+1).
Equating real and imaginary parts separately, gives
2x+y−3 = x−y+1 \(\Rightarrow\) x+2y = 4
x−y+2 = 2y +1 \(\Rightarrow\) x−3y = −1
Solving the above equations, gives
x = 2 and y = 1.

(i) m + n∈Z, ∀m, n∈Z. Hence + is a binary operation on Z.
(ii) Also m + n = n + m,∀m, n∈Z. So the commutative property is satisfied
(iii) ∀m, n, p∈Z, m+ (n + p) = (m+ n) + p. Hence the associative property is satisfied.
(iv) m + e = e + m = m ⇒ e = 0. Thus ヨ 0∈Z⋺(m+ 0) = (0 + m) = m. Hence the existence of identity is assured.
(v) m + m' = m'+ m = 0 ⇒ m' = −m. Thus ∀∈Z,ョ−m∈Z ⋺ m+ (−m) = (−m) + m = 0. Hence, the existence of inverse property is also assured. Thus we see that the usual addition + on Z satisfies all the above five properties.

I =\(\int ^{1}_{0} (tan ^{-1} x + tan ^{-1}(1-x))\) dx
=\(\int ^{1}_{0} (tan ^{-1} x dx + \int ^{1}_{0} tan ^{-1}(1-x)) dx\)
= \(\int ^{1}_{0} (tan ^{-1} x dx + \int ^{1}_{0} tan ^{-1}(1-(1-x)) dx\),  Since\(\int ^{a}_{0} f (x) dx = \int ^{a}_{0} f(a-x) dx \)
=\(\int ^{1}_{0} tan ^{-1} x dx + \int ^{1}_{0} tan ^{-1} x dx\)
= 2\(\int ^{1}_{0} tan^{-1} x dx\)
= \([2 \int udv]^1_0\) ,  Where u = tan^-1 x and dν = dx
= 2\([uν - \int udv]^1_0,\) applying integration by parts
= 2\((x tan^{-1} x - \int x \frac{dx}{1+x^2})^{1}_{0}\)
= 2\((x tan^{-1} x - \frac {1}{2} log (1+x^2))^{1}_{0}\)
= \(\frac{\pi}{2} \)- log 2

Recall that volume of a right circular cylinder is given by V = \(\pi \)r²h where r is the radius and h is the height. So we have V (r) = \(\pi \)r²h = 20\(\pi \)r²
V (10.1) −V (10)≈ \(\frac { dV }{ dr } { { | }_{ r=10 } }\) (10.1 10) = 20\(\pi \)²(10(0.1))
Thus the estimate for the change in the volume is 40 \(\pi \) cm³
Exact calculation of the volume change gives
V (10.1) −V (10) = 2040.2\(\pi \) -2000\(\pi \) = 40.2\(\pi \) cm³.
So relative error = \(\frac { 40.2\pi -40\pi }{ 40.2\pi } \) = \(\frac { 1 }{ 201 } \) = 0.00497 and hence
the percentage error = relative error x 100 = \(\frac { 1 }{ 201 } \)x100 = 0.497% 

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

To find \(\int _{ 1 }^{ 1.5 }{ xdx } \) in  {1.1, 1.2, 1.3, 1.4, 1.5}.
Here a = 1,
b = 1.5, n = 5, f(x) = x
\(\therefore h=\Delta x=\frac { b-a }{ n } =\frac { 1.5-1 }{ 5 } =\frac { 0.5 }{ 5 } =0.1\)
The partition of the interval is given by 
x₀ = 1
x₁ = x₀ + h = 1 + 0.1 = 1.1
x₂ = x₁ + h = 1.1 + 0.1 = 1.2
x₃ = x₂ + h = 1.2 + 0.1 = 1.3
x₄ = x₃ + h = 1.3 + 0.1 = 1.4
x₅ = x₄ + h = 1.4 + 0.1 = 1.5
The left end rule for Riemann sum with equal width \(\Delta\)x is
\(\int _{ a }^{ b }{ xdx } =[f({ x }_{ 0 })+f({ x }_{ 1 })+...+f({ x }_{ (n-1 })]\Delta x\)
S = [f(1) + f(1.1) + f(1.2) + f(1.3) + f(1.4)](0.1)
= (1 + 1.1 + 1.2 + 1.3 + 1.4) (0.1)
= 6 \(\times\) 0.1
S = 0.6

The volume of the baloon of radius r is \(V =\frac{4}{3}\pi r ^{3} \)
We are given \(\frac{dV }{dt }=1000 \) and we need to find \(\frac{dr }{dt} \) when r = 7. Now,
\(\frac{dV}{dt}=3\times\frac{4}{3}\pi r^{2}\times \frac{dr}{dt}\)
Substituting r = 7 an \(\frac{dV}{dt}\) = 1000, we get 1000 \(= 4\pi \times 49 \times \frac{dr}{dt}\)
Hence, \(\frac{dr}{dt}=\frac{1000}{4\times49\times \pi}=\frac{250}{49\pi}\)

The surface area S of the baloon is S = 4ㅠr². Therefore, \(\frac{dS}{dt}=8\pi \times r \times \frac{dr}{dt}\)
Substituting\(\frac{dr}{dt}=\frac{250}{49\pi}\) and r = 7, we get \(\frac{dS}{dt}=8\pi\times7\times \frac{250}{49\pi}=\frac{2000}{7}\)
Therefore, the rate of change of radius is \(\frac{250}{49 \pi}\) cm/sec and the change of surface area is \(\frac{2000}{7}\) cm² / sec

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

Given A =\(\left[ \begin{matrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{matrix} \right] \), B=\(\left[ \begin{matrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{matrix} \right] \)
AB =\(\left[ \begin{matrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{matrix} \right] \left[ \begin{matrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{matrix} \right] \)
=\(\left[ \begin{matrix} -5+3+6 & -5+2+3 & -10+1+9 \\ 7+3-10 & 7+2-3 & 14+1-15 \\ 1-3+2 & 1-2+1 & 2-1+3 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{matrix} \right] \)= 4. I₃
BA =\(\left[ \begin{matrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{matrix} \right] \left[ \begin{matrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{matrix} \right] \)
=\(\left[ \begin{matrix} -5+7+2 & 1+1-2 & 3-5+2 \\ -15+14+1 & 3+2-1 & 9-10+1 \\ -10+7+3 & 2+1-3 & 6-5+3 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{matrix} \right] \)= 4. I₃.
So, we get AB = BA = 4. I₃
⇒ \(\left( \frac { 1 }{ 4 } A \right) B=B\left( \frac { 1 }{ 4 } A \right) =1\)
⇒ B^-1 = \(\frac { 1 }{ 4 } \) = 1
Writing the given set of equations in matrix form we get,
\(\left[ \begin{matrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 7 \\ 2 \end{matrix} \right] \)
⇒ \(B=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 7 \\ 2 \end{matrix} \right] \)
⇒ \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ={ B }^{ -1 }\left[ \begin{matrix} 1 \\ 7 \\ 2 \end{matrix} \right] =\left[ \frac { 1 }{ 4 } A \right] \left[ \begin{matrix} 1 \\ 7 \\ 2 \end{matrix} \right] \)
= \(\frac { 1 }{ 4 } \left[ \begin{matrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{matrix} \right] \left[ \begin{matrix} 1 \\ 7 \\ 2 \end{matrix} \right] \)
= \(\frac { 1 }{ 4 } \left[ \begin{matrix} -5+7+6 \\ 7+7-10 \\ 1-7+2 \end{matrix} \right] =\frac { 1 }{ 4 } \left[ \begin{matrix} 8 \\ 4 \\ -4 \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 1 \\ -1 \end{matrix} \right] \)
∴ x = 2, y = 1, z = -1
Hence, the solution set is {2, 1, - 1}.

Let A be the origin, \(\vec { b } \) be the position vector of B and \(\vec {c } \) be the position vector of C .
Now D is the midpoint of BC , and so the position vector of D \(\frac{\vec{b}+\vec{c}}{2}\). There, we get

\({ \left| \vec { AD } \right| }^{ 2 }=\vec { AD } .\vec { AD } \)= \(\left( \frac { \vec { b } +\vec { c } }{ 2 } \right) .\left( \frac { \vec { b } +\vec { c } }{ 2 } \right) \)= \(\frac { 1 }{ 4 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 }+2\overrightarrow { b } .\vec { c } )\)....(1)
Now, \(\vec { BD } =\vec { AD } -\vec { AB } \) = \(\frac { \vec { b } +\vec { c } }{ 2 } -\vec { b }=\frac { \vec {c } -\vec { b} }{ 2 }\)
Then, we get, =\({ \left| \vec { BD } \right| }^{ 2 }=\vec { BD } .\vec {BD } \) = \(\left( \frac { \vec { c } -\vec { b} }{ 2 } \right) .\left( \frac { \vec { c } -\vec { b} }{ 2 } \right) \) = \(\frac { 1 }{ 4 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 }-2\overrightarrow { b } .\vec { c } )\)....(2)
Now, adding (1) and (2), we get
Therefore, \({ \left| \vec { AD } \right| }^{ 2 }+{ \left| \vec { BD } \right| }^{ 2 }=\frac { 1 }{ 4 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 }+2\overrightarrow { b } .\vec { c } )+\frac { 1 }{ 4 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 }-2\overrightarrow { b } .\vec { c } )=\frac { 1 }{ 2 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 })\)
⇒ \({ \left| \vec { AD } \right| }^{ 2 }+{ \left| \vec { BD } \right| }^{ 2 }=\frac { 1 }{ 2 } ({ \left| \vec { AB } \right| }^{ 2 }+{ \left| \vec { AC } \right| }^{ 2 })\)
Hence, \({ \left| \vec { AB } \right| }^{ 2 }+{ \left| \vec { AC } \right| }^{ 2 }=2({ \left| \vec { AD } \right| }^{ 2 }+{ \left| \vec { BD } \right| }^{ 2 })\)

Given p, q are the roots of lx² + nx + n = 0 
\(p+q=\frac { -b }{ a } =\frac { -n }{ l } ...(1)\)
\(pq=\frac { c }{ a } =\frac { n }{ l } ...(2)\)
L.H.S 
\( \sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}} =\frac{\sqrt{p}}{\sqrt{q}}+\frac{\sqrt{q}}{\sqrt{p}}+\frac{\sqrt{n}}{\sqrt{l}}=\frac{p+q}{\sqrt{p} \sqrt{q}}+\frac{\sqrt{n}}{\sqrt{l}} \)
\( =\frac{-\frac{n}{l}}{\sqrt{p q}}+\frac{\sqrt{n}}{\sqrt{l}}=\frac{-\frac{n}{l}}{\frac{\sqrt{n}}{\sqrt{l}}}+\frac{\sqrt{n}}{\sqrt{l}} \)
\(=\frac{-\frac{n}{l}+\frac{n}{l}}{\sqrt{\frac{n}{l}}}=0\) R.H.S

We find that |A| = \(\left| \begin{matrix} 8 & -6 & 2 \\ -6 & 7 & 4 \\ 2 & -4 & 3 \end{matrix} \right| \) = 8(21 - 16) + 6(-18 + 8) + 2(24 - 14) = 40 - 60 + 20 = 0
By the definition of adjoint, we get
adj A = \({ \left[ \begin{matrix} \left( 21-16 \right) & -\left( -18+8 \right) & \left( 24-14 \right) \\ -\left( -18+8 \right) & \left( 24-4 \right) & -\left( 32+12 \right) \\ \left( 24-14 \right) & -\left( -32+12 \right) & \left( 56-36 \right) \end{matrix} \right] }^{ T }=\left[ \begin{matrix} 5 & 10 & 10 \\ 10 & 20 & 20 \\ 10 & 20 & 20 \end{matrix} \right] \)
So, we get
A(adj A) = \(\left[ \begin{matrix} 8 & -6 & 2 \\ -6 & 7 & -4 \\ 2 & -4 & 3 \end{matrix} \right] \left[ \begin{matrix} 5 & 10 & 10 \\ 10 & 20 & 20 \\ 10 & 20 & 20 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 40-60+20 & 80-120+40 & 80-120+40 \\ -30+70-40 & -60+140-80 & -60+140-80 \\ 10-40+30 & 20-80+60 & 20-80+60 \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \) = 0I₃ = |A|I₃,
Similarly, we get
(adj A)A = \(\left[ \begin{matrix} 5 & 10 & 10 \\ 10 & 20 & 20 \\ 10 & 20 & 20 \end{matrix} \right] \left[ \begin{matrix} 8 & -6 & 2 \\ -6 & 7 & -4 \\ 2 & -4 & 3 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 40-60+20 & -30+70-40 & 10-40+30 \\ 80-120+40 & -60+140-80 & 20-80+60 \\ 80-120+40 & -60+140-80 & 20-80+60 \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \) = 0I₃ = |A|I₃.
Hence, A(adj A) = (adj A)A = |A|I₃.

The cubic equation is
x³ – x² (α + β + γ) + x (αβ + βγ + γα) – αβγ = 0
\( x^{3}-x^{2}\left(2+\frac{1}{2}+1\right)+x\left(1+\frac{1}{2}+2\right)-(2)\left(\frac{1}{2}\right)(1)=0\)
\(x^{3}-x^{2}\left(\frac{4+1+2}{2}\right)+x\left(\frac{2+1+4}{2}\right)-1=0 \)
\(x^{3}-x^{2}\left(\frac{7}{2}\right)+x\left(\frac{7}{2}\right)-1=0\)
2x³ – 7x² + 7x – 2 = 0

We are given that, T = 10x − x²
Hence, the rate of change at any distance from one end is given by \(\frac{dT}{dx}=10-2x \) 
The mid point of the rod is at x = 5
Substituting x = 5, we get \(\frac{dT}{dx}=0\)

Given y = e^−x + mx + n .... (1) 
Derentiating cquation (1) wr.t 'x', we get
\(\frac { dx }{ dx } =-e^{ -x }(-1)+m\)
\(\frac { d^{ 2 }y }{ d{ x }^{ 2 } } =-e^{ -x }+m\)
Again differentiating, we get
\(\Rightarrow \frac { d^{ 2 }y }{ d{ x }^{ 2 } } ={ e^{ x } }+0\)
\(\Rightarrow \left( \frac { d^{ 2 }y }{ d{ x }^{ 2 } } \right) =e^x\)
Substituting the value of \(\frac{d^2y}{dx^2}\) in the given differential equation, we get
\(\Rightarrow e^{ x }\left( \frac { d^{ 2 }y }{ d{ x }^{ 2 } } \right) -1=e^x(e^x)-1\\= e^{x-x}-1\\ e^0-1= 1-1=0\)
Thus the solution of the given differential equation e^x \(\left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) \) -1 = 0

\(\frac { dy }{ dx } +xy=cotx\)
Given differential equation is
\(\frac { dy }{ dx } +xy=cotx\)
The highest derivative is 1 and its power is 1 order 1, degree 1.

Prove the following properties
Z is real if and only if z =\(\bar { z } \)
Let z = x+iy
Then \(\bar { z } \) = x+iy
z = \(\bar { z } \)
⇔ x + iy = x-iy
⇔ x + iy - x + iy = 0
⇔ 2iy = 0
⇔ y = 0
[∴ 2 and i are constants]
when y = 0, z = x which is real
∴ z is purely ⇔ z = \(\bar { z } \)

Let y = sin^-1x
When y = 0, 0 = sin^-1Ix
\(\Rightarrow\) sin(0) = sin (sin^-1)(x))
\(\Rightarrow\)sin 0 = x
\(\Rightarrow\)x = 0
Hence, solution to (1) is x = 0. Also, graph of sin x and sin^-1x intersect at origin (0, 0).

Centre of the circle is (3,-2) which lies on x + y − 1 = 0. So the line x + y − 1 = 0 passes through the centre and therefore the line x + y −1 = 0 is a diameter of the circle for all possible values of c .

We first find adj A. By definition, we get adj A = \({ \left[ \begin{matrix} +{ M }_{ 11 } & -{ M }_{ 12 } \\ -{ M }_{ 21 } & +{ M }_{ 22 } \end{matrix} \right] }^{ T }={ \left[ \begin{matrix} d & -c \\ -b & a \end{matrix} \right] }^{ T }=\left[ \begin{matrix} d & -c \\ -c & a \end{matrix} \right] \).
Since A is non-singular, |A| = ad - bc ≠ 0.
As \({ A }^{ -1 }=\frac { 1 }{ \left| A \right| } \) adj A, we get A^-1 = \(\frac { 1 }{ ad-bc } \left[ \begin{matrix} d & -b \\ -c & a \end{matrix} \right] \).