Given a = edge of the circle
= 10 cm and da
= 0.2 cm
Volume of cube = a³
Approximate amount of cubic centimeters of paint is used to paint this cube = 3a² da
= 3(10²)(0.2)
= 300 \(\left( \frac { 2 }{ 10 } \right) \) = 60 cm³
Exact amount of paint used = f(x + ∆x) - f(x)
= f(10.2) - f(10)
= 10.2³ -10³
= 1061.208 - 1000
= 61.208 cm²

I \(=\int _{ 0 }^{ 1 }{ \sqrt { \frac { (1-x)(1-x) }{ (1+x)(1-x) } } } dx\)
\(=\int _{ 0 }^{ 1 }{ \sqrt { \frac { (1-x) 2}{ (1-x)^2 } } } dx\)
\(=\int _{ 0 }^{ 1 }{ \frac { 1-x }{ \sqrt { 1-{ x }^{ 2 } } } } dx\)
Put x = sin u, dx = cosu.du, 0u = sin^-1(x)
For x = 0, u = 0 and x = 1, u = \(\frac{\pi}{2}\)
\(=\int _{ 0 }^{ 1 }{ \frac { 1-sin\ u }{ \sqrt { 1-{ sin }^{ 2 }u } } } \times cos u.du\)
\(=\int _{ 0 }^{ 1 } \frac { 1-sin\ u }{cos u} \times cos u.du\)
= \( [u+cos u]^\frac{\pi }{2}_0\)= (\(\frac{\pi}{2}\)+0)-(0+1) = \(\frac{\pi}{2}\) -1
\(=\frac { \pi }{ 2 } -0-1=\frac { \pi }{ 2 } -1\)

We write \(\int _{ 0 }^{ 2a }{ f(x)dx\int _{ 0 }^{ a }{ f(x)dx } } +\int _{ 0 }^{ 2a }{ f(x)dx } \) ....(1)
Consider \(\int _{ 0 }^{ 2a }{ f(x)dx } \)
Substituting x = a + u, we have dx = du ; when x = a, u = 0 and when x = 2a,u = a.
\(\therefore \int _{ 0 }^{ 2a }{ f(x)dx=\int _{ 0 }^{ a }{ f(a+u)du } =\int _{ 0 }^{ a }{ d(u)du } } \), since f(x) = f(a+x)
\(\\ \\ \\ =\int _{ 0 }^{ a }{ f(x)dx } \) ..(2)
Substituting (2) in (1), we get
\(\int _{ 0 }^{ 2a }{ f(x)dx } =2\int _{ 0 }^{ a }{ f(x)dx } \)

We need to find an approximate value of \(\sqrt { 9.2 } \) using linear approximation. Now by (3), we have f(x₀+Δx) ≈ f(x₀)+f'(x₀)Δx. To do this, we have to identify an appropriate function f, a point x₀ and Δx. Our choice should be such that the right side of the above approximate equality, should be computable without the help of a calculator. So, we choose  
f(x) = \(\sqrt { x,{ x }_{ 0 } } \) = 9  and Δx = 0.2. Then f'(x₀) = \(\frac { 1 }{ 2\sqrt { 9 } } \) and hence.
 \(\sqrt { 9.2 } \) ≈ f(9) + f'(9)(0.2) = 3+\(\frac { 0.2 }{ 6 } \) = 3.03333
Now if we use a calculator, just to compare, we find \(\sqrt { 9.2 } \) = 3.03315. We see that our approximation is accurate to three decimal places and the error is 3.03315 - 3.03333 = 0.00018. [Also note that one could choose f (x) = \(\sqrt { 1+x,{ x }_{ 0 } } =8\) and Δx = 0.2. So the choice of f and x₀ - are not necessarily unique].
 So in the above example, the absolute error is 3.03315-3.03333 = -0.00018. Note that the absolute error says how much the error; but it does not say how good the approximation is. For instance, let us consider two simple cases
Case 1 : Suppose that the actual value of something is 5 and its approximated value is 4, then the absolute error is 5 − 4 = 1.
Case 2 : Suppose that the actual value of something is 100 and its approximated value is 95. In this case, the absolute error is 100 − 95 = 5. So the absolute error in the first case is smaller when compared to the second case.
Among these two approximations, which is a better approximation; and why? The absolute error does not give a clear picture about whether an approximation is a good one or not. On the other hand, if we calculate relative error or percentage of error (defined below), it will be easy to see how good an approximation is. If the actual value is zero, then we do know how close our approximate answer is to the actual value. So if the actual value is not zero.

Let f(x) = log x
f^l(x) = \(\frac{1}{x}\) = x^-1
f^lI(x) = -1x^-2
f^III(x) = 2x^-3
f^IV(x) = -6 x^-4
⇒ f(1) = log 1 = 0
f^l(1) = \(\frac11\) = 1
f^ll(1) = -1
f^lll(1) = 2
f^IV(1) = -6
Taylor series for f(x) at x = 1 is
\(f(x)=f(1)+\frac { { f }^{ 1 }(1) }{ 1! } (x-1)+\frac { { f }^{ II }(1) }{ 2! } ({ x-1) }^{ 2 }+\frac { { f }^{ III }(1) }{ 3! } { (x-1) }^{ 3 }+\).....
log x = \(\frac { 1 }{ 1! } (x-1)+\frac { -1 }{ 2! } { (x-1) }^{ 2 }+\frac { 2 }{ 3! } ({ x-1) }^{ 3 }-\frac { 6 }{ 4! } ({ x-1) }^{ 4 }+..\)
log x = \((x-1)-\frac { 1 }{ 2 } { (x-1) }^{ 2 }+\frac { 1 }{ 3 } { (x-1) }^{ 3 }-\frac { 1 }{ 4 } ({ x-1) }^{ 4 }\)+ ....

We have, f (x) is continuous in \([\frac{1}{2},2 ]\) and differentiable in \((\frac{1}{2},2 )\) with  \(f(\frac{1}{2})=\frac{5}{2}=f(2) \). 
By  the Rolle’s theorem there must exist a \(c \in (\frac{1}{2},2 )\) such that, \(f'(c)=1-\frac{1}{c^{2}}=0 \Rightarrow c^{2}=1 \) gives \(\Rightarrow c=\pm1, \) As \(1\in(\frac{1}{2},2)\) we choose c = 1.

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
\((y-2)=-\frac{1}{5}(x-1)\Rightarrow x+5y-11=0\) 

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.

\(\left[ \begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix} \right] \)
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 =\(\left[ \begin{matrix} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1 \end{matrix} \right] \)

Let \(\vec { b } =5\hat { i } +6\hat { j } +7\hat { k } \) and \(\vec { a } =7\hat { i } +9\hat { j } +13\hat { k } \) 
The parametric form of vector equation of a straight line passing through two points \(\vec { a } \) and \(\vec { b } \) is
\(\vec { r } =\vec { a } +t(\vec { b } -\vec { a } )\)
∴ \(\vec { r } =(7\hat { i } +9\hat { j } +13\hat { k } )+t(7-5)\hat { i } +(9-6)\hat { j } +(13-7)\hat { k } \)
\(\vec { r } =(7\hat { i } +9\hat { j } +13\hat { k } )+t(2\hat { i } +3\hat { j } +6\hat { k } ),t\in R\)
The Cartesian equation of a straight line passing through two points as
\(\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { z-{ z }_{ 1 } }{ { z }_{ 2 }-{ z }_{ 1 } } \) [(x₁,y₁,z₁) is (7,9,13) (x₂,y₂,z₂) is (5,6,7))
\(\frac { x-7 }{ 5-7 } =\frac { y-9 }{ 6-9 } =\frac { z-13 }{ 7-13 } \)
⇒ \(\frac { x-7 }{ -2 } =\frac { y-9 }{ -3 } =\frac { z-13 }{ -6 } \)
= \(\frac { x-7 }{ 2 } =\frac { y-9 }{ 3 } =\frac { z-13 }{ 6 } \)

Let A be the matrix. Performing elementary row operations, we get
A = \(\left[ \begin{matrix} 2 \\ \begin{matrix} -3 \\ 6 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 4 \\ 2 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} -2 \\ -1 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} -1 \\ 7 \end{matrix} \end{matrix} \right] \)\(\overset { { R }_{ 2 }\longrightarrow 2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} -6 \\ 6 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 8 \\ 2 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} -4 \\ -1 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} -2 \\ 7 \end{matrix} \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }+3{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-3{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 2 \\ 8 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} 8 \\ -13 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 7 \\ -2 \end{matrix} \end{matrix} \right] \).
\(\overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-4{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 2 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} 8 \\ -45 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 7 \\ -30 \end{matrix} \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }\div \left( -15 \right) }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 2 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} 8 \\ 3 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 7 \\ 2 \end{matrix} \end{matrix} \right] \).
The last equivalent matrix is in row-echelon form. It has three non-zero rows. So, ρ(A) = 3.

Given \(\frac { \pi }{ 2 } <{ cos }^{ -1 }(3x-1)<\pi \)
\((cos2\frac { \pi }{ 2 } <3x-1\))
\(\Rightarrow 0<3x-1<-1\)
\(0+1<3x<-1+1\)
\(1<3x<0\)
This inequality is true, only when 0 < x < \(\frac { 1 }{ 3 }\)

With usual notations in triangle ABC, we have \(\vec { BC } =\vec { a } ,\vec { CA } =\vec { b } \) \(\vec { AB } =\vec { c } \).
Then \(\left| \vec { BC } \right| =a\) , \(\left| \vec { CA } \right| =b\), \(\left| \vec { AB} \right| =c\), and \(\vec { BC } +\vec { CA } +\vec { AB } =\vec { 0 } \)
So, \(\vec { BC } =-\vec { CA } -\vec { AB } \)
Then applying dot product, we get
\(\vec { BC } .\vec { BC } =(-\vec { CA } -\vec { AB } ).(-\vec { CA } -\vec { AB } )\)
⇒ \({ \left| \vec { BC } \right| }^{ 2 }={ \left| \vec { CA } \right| }^{ 2 }+{ \left| \vec { AB } \right| }^{ 2 }+\vec { 2CA } .\vec { AB } \)
⇒ a² = b²+c²+2bc cos (\(\pi\) - A)
⇒ a² = b²+c²−2bc cos A.
The results (ii) and (iii) are proved in a similar way.

Given \(\int _{ 0 }^{ \infty }{ { e }^{ -a{ x }^{ 2 } }{ x }^{ 3 }dx=32,\alpha >0} \)

\(\Rightarrow \int _{ 0 }^{ \infty }{ { e }^{ { -\alpha x }^{ 2 } } } .{ x }^{ 2 }.dx=32\)
\(put\ t= { x }^{ 2 }\)
\(\Rightarrow dt=2x\ dx\)
\(\\ \Rightarrow \frac { dt }{ 2 } = xdx\)
\( \int _{ 0 }^{ \infty }{ { e }^{ -\alpha t }\frac{dt}{2}} \frac { 1 }{ 2 } \int _{ 0 }^{ \infty }{ { e }^{ -\alpha t }.tdt } \)
\( \frac { 1 }{ 2 } \times\frac{1} { \alpha}^{ 2 } \) ...... (1)
Given \(\int^x_0e^{-ax^2}x^3 dx = 32\)
By (1),
\(\Rightarrow \frac { 1! }{ { \alpha }^{ 2 } } =32\times 2\left[ \because \int _{ 0 }^{ \infty }{ { e }^{ -ax }{ x }^{ n }dx=\frac { n! }{ { a }^{ n+1 } } } \right] \)
\(\frac{1}{2}\times \frac{1}{\alpha^2}= 32 \Rightarrow \frac {1}{\alpha ^2}= 64\\ \alpha ^2 = \frac{1}{64}\)
\(\Rightarrow \alpha =\frac { 1 }{ 8 } \)

Given g (x, y) = x² + 3xy - 7y + cos(5x)
g_x = 2x + 3y - 0 - 5 sin 5x
= 2x + 3y - 5 sin 5x
g_y = 0 + 3x (1) - 7 + 0
= 3x - 7
\({ g }_{ xy }=\frac { \partial }{ \partial x } ({ g }_{ y })\) = 3
\({ g }_{ xx }=\frac { \partial }{ \partial x } ({ g }_{ x })\)
= 2(1) + 0 - 5(5) cos (5x)
= 2 - 25 cos (5x)
\({ g }_{ yy }=\frac { \partial }{ \partial y } ({ g }_{ y })=\) 0
\({ g }_{ yx }=\frac { \partial }{ \partial y } ({ g }_{ x })\)
= 0 + 3(1) - 0 = 3

Note that the function f takes value 1 on the x, y-axes and 0 everywhere else on R². So let us calculate
\(\frac { \partial f }{ \partial x } (0,0)\) = \(\underset { h\longrightarrow 0 }{ lim } \) \(\frac { f\left( 0+h,0 \right) -f(0,0) }{ h } =\underset { h\longrightarrow 0 }{ lim } \frac { 1-1 }{ h } =0;\)
\(\frac { \partial f }{ \partial y } (0,0)\) = \(\underset { k\longrightarrow 0 }{ lim } \) \(\frac { f\left( 0,0+k \right) -f(0,0) }{ k } =\underset { k\longrightarrow 0 }{ lim } \frac { 1-1 }{ k } =0\)
This completes (i).

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.

Given cos α + cos β + cos \(\gamma\) = sin α + sin β + sin \(\gamma\)
∴ (cos α + cos β + cos \(\gamma\)) + i(sin α + sin β + sin \(\gamma\)) = 0
⇒ (cos α + i sin α) + (cos β + i sin β) + (cos \(\gamma\)+i sin \(\gamma\)) = 0
⇒ a + b + c = 0 where a = cos α + i sin α, b = cos β + i sin β, c = cos \(\gamma\) + i sin\(\gamma\)
If a + b + c = 0, then a³+b³+c³ = 3abc
∴ (cos α + i sin α)³ + (cos β + i sin β)³ + (cos \(\gamma\) + i sin \(\gamma\))³ = 3[ (cos α + i sin α) + (cos β + i sin β) + (cos \(\gamma\) + i sin \(\gamma\))
= 3[(cos(α + β + \(\gamma\)) + i sin(α + β + \(\gamma\))]
⇒ (cos 3α + cos β + cos \(\gamma\)) + i[sin 3α + sin 3β + sin 3\(\gamma\))]
= 3(cos(α + β + \(\gamma\)) + i sin(α + β + \(\gamma\))
Equating the real and imaginary parts, we get 
\( \cos 3 \alpha+\cos 3 \beta+\cos 3 \gamma=3 \cos (\alpha+\beta+\gamma) \)
\( \sin 3 \alpha+\sin 3 \beta+\sin 3 \gamma=3 \sin (\alpha+\beta+\gamma) \)

Equation of the circle is x² + y² − 6x + 6y − 8 = 0
∴ Equation of the tangent at (x₁, y₁) is
xx₁ + yy₁ - \(\frac 62\) (x + x₁) + \(\frac 62\) (y + y₁) - 8 = 0
Given (x₁, y₁) is (2, 2)
Equation of the tangent at (2, 2) is
x(2) + y(2) - 3(x + 2) + 3(y + 2) - 8 = 0
\(\Rightarrow 2 x-2 y-3 x+\not 6+3 y+\not 6-8=0\)
⇒ -x + 5y - 8 = 0
⇒ x − 5y + 8 = 0
Equation of the normal is
yx₁ - xy₁ + g(y - y₁) - f(x - x₁) = 0
⇒ y(2) -x(2) - 3(y - 2) -3(x - 2) = 0
∵ 2g = -6
⇒ g = -3
2f = 6
⇒ f = 3
⇒ 2y - 2x - 3y + 6 - 3x + 6 = 0
⇒ -5x - y + 12 = 0
Another method for Normal:
Equation of tangent is perpendicular to normal
x - 5y + 8 = 0
Perpendicular equation be 5x + y + k = 0
At (2, 2)
10 + 2 + k = 0
k = -12
Therefore 5x + y - 12 = 0 be equation of normal.

Given equation are \({ x }^{ 2 }px+q=0\)  ............(1)
and \({ x }^{ 2 }+p'x+q'=0\)    ..........(2)
Let ∝ be the common root for (1) and (2)
∴ ∝² + p∝ + q = 0   .........(3)
and ∝²+ p'∝ + q' = 0    .............(4)
Solving (3) and (4) by cross multiplication method we get
p  q 1 p
p' q' 1 p'
\(\Rightarrow \frac { { \alpha }^{ 2 } }{ pq'-p'q } =\frac { \alpha }{ q-q' } =\frac { 1 }{ p'-p } \)
consider \(\frac { { \alpha }^{ 2 } }{ pq'-p'q } =\frac { \alpha }{ q-q' } \)
\(\Rightarrow \frac { { \alpha }^{ 2 } }{ \alpha } =\frac { pq'-p'q }{ q-q' } \)
\(\alpha =\frac { pq'-p'q }{ q-q } \)
Consider \(\frac { \alpha }{ q-q' } =\frac { 1 }{ p'-p } \Rightarrow \alpha =\frac { q-q' }{ p'-p } \)
Hence its roots are \(\frac { pq'-p'q }{ q-q' } or\quad \frac { q-q' }{ p'-p } \)

Given \(\vec { a } =\hat { i } -2\hat { j } +3\hat { k } ,\vec { b } =2\hat { i } +\hat { j } +\hat { k } \) and \(\vec { c } =3\hat { i } +2\hat { j } +\hat { k } \)
(i) \((\vec { a } \times \vec { b } )\) = \(\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & -2 & 3 \\ 2 & 1 & -2 \end{matrix} \right| \)
= \(\hat { i } \left| \begin{matrix} -2 & 3 \\ 1 & -2 \end{matrix} \right| -j\left| \begin{matrix} 1 & 3 \\ 2 & -2 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 1 & -2 \\ 2 & 1 \end{matrix} \right| \)
= \(\\ \hat { i } \)(4-3)-\(\\ \hat { j } \)(-2-6)+\(\\ \hat { k } \)(1+4)
= \(\hat { i } +8\hat { j } +5\hat { k } \)
\((\vec { a } \times \vec { b } )\times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 8 & 5 \\ 3 & 2 & 1 \end{matrix} \right| \)
=\(\hat { i } \left| \begin{matrix} 8 & 5 \\ 2 & 1 \end{matrix} \right| \hat { -j } \left| \begin{matrix} 1 & 5 \\ 3 & 1 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 1 & 8 \\ 3 & 2 \end{matrix} \right| \)
 = \(\\ \hat { i } \)(8-10)-\(\\ \hat { j } \)(1-15)+\(\\ \hat { k } \)(2-24)
= \(-2\hat { i } +14\hat { j } -22\hat { k } \)
(ii) \(\vec { a } \times (\vec { b } \times \vec { c } )\)
\(\vec { b } \times \vec { c } \) = \(\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 1 & -2 \\ 3 & 2 & 1 \end{matrix} \right| \)
= \(\hat { i } \left| \begin{matrix} 1 & -2 \\ 2 & 1 \end{matrix} \right| \hat { -j } \left| \begin{matrix} 2 & -2 \\ 3 & 1 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix} \right| \)
= \(\\ \hat { i } \)(1+4)-\(\\ \hat { j } \)(2+6)+\(\\ \hat { k } \)(4-3) = \(5\hat { i } -8\hat { j } +\hat { k } \)
∴ \(\vec { a } \times (\vec { b } \times \vec { c } )=\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & -2 & 3 \\ 5 & -8 & 1 \end{matrix} \right| \)
= \(\hat { i } \left| \begin{matrix} -2 & -3 \\ -8 & 1 \end{matrix} \right| -\hat { j } \left| \begin{matrix} 1 & 3 \\ 5 & 1 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 1 & -2 \\ 5 & -8 \end{matrix} \right| \)
= \(\\ \hat { i } \)(-2-24)-\(\\ \hat { j } \)(1-15)+\(\\ \hat { k } \)(-8-10)
= \(22\hat { i } +14\hat { j } +2\hat { k } \).

Given \(\vec { a } =\hat { i } -\hat { k } ,\vec { b } =x\hat { i } +\hat { j } +(1-x)\hat { k } ,\vec { c } =y\hat { i } +x\hat { j } +(1+x+y)\hat { k } \)
\([\vec { a } ,\vec { b } \vec { c } ]=\vec { a } .(\vec { b } \times \vec { c } )=\left| \begin{matrix} 1 & 0 & -1 \\ x & 1 & 1-x \\ y & x & 1+x-y \end{matrix} \right| \)
= \(1\left| \begin{matrix} 1 & 1-x \\ x & 1+x-y \end{matrix} \right| +0-y\left| \begin{matrix} x & y \\ y & x \end{matrix} \right| \)
= [(1+x-y)-x(1-x)]-[x²-y]
\(=1+\not x-\not y-\not x+\not x^{2}-\not x^{x}+\not y\)
= 1
∴ \([\vec { a } \vec { b } \vec { c } ]\) = 1 for all values of x and y
∴ \([\vec { a } \vec { b } \vec { c } ]\) depends on neither x nor y.

Since tan(tan^-1 x) = x, x ∈ R,
We have tan(tan^-1(2019)) = 2019

Given z₁= 2 - i and z₂= -4+3i
z₁z₂ = (2-i)(-4+3i)
= -8 + 6i + 4i - 3i²
= -8 +10i - 3(-1)
= -8 +10i + 3 = -5 +10i
Inverse of z₁z₁ is \(\frac { 1 }{ { z }_{ 1 }{ z }_{ 2 } } \)
= \(\frac { 1 }{ -5+10i } \times \frac { -5-10i }{ -5-10i } \)
= \(\frac { -5-10i }{ (-5)^{ 2 }-(10i)^{ 2 } } \)
= \(\frac { -5-10i }{ 25-100i^{ 2 } } \)
= \(\frac { -5-10i }{ 25+100 } \)  [∵ i² = -1]
\(=\frac{\not{5}(-1-2 i)}{\not{5}\langle(2 5)}=\frac{-1-2 i}{25}\)
∴ Inverse of z₁z₂ is \(\frac { 1 }{ 25 } \) (-1-2i)
Inverse of \(\frac { { z }_{ 1 } }{ { z }_{ 2 } } \) is \(\frac { 1 }{ \frac { { z }_{ 1 } }{ { z }_{ 2 } } } =\frac { { z }_{ 2 } }{ { z }_{ 2 } } \)
∴ Inverse of \(\frac { { z }_{ 1 } }{ { z }_{ 2 } } =\frac { { z }_{ 2 } }{ { z }_{ 1 } } =\frac { -4+3i }{ 2-i } \times \frac { 2+i }{ 2+i } \)
= \(\frac { -4+3i }{ 2-i } \times \frac { 2+i }{ 2+i } \)
= \(\frac { -8-4i+6i+3i^{ 2 } }{ 2^{ 2 }-({ i }^{ 2 }) } \)
= \(\frac { -8+2i-3 }{ 4+1 } =\frac { -11+2i }{ 5 }\)
\( =\frac { 1 }{ 5 } \)(-11 + 2i)
∴ Inverse of \(\frac { { z }_{ 1 } }{ { z }_{ 2 } } \) is \(\frac { -11+2i }{ 5 }\)or \(\frac { 1 }{ 5 } \)(-11 + 2i)