Let p: 19 is a prime number.
q: All the angles of a triangle are equal be two simple statements.
(i) 19 is not a prime number and all the angles of a triangle are equal.
~p ∧ q
(ii) 19 is a prime number or all the angles of a triangle are not equal.
p ∧ ~q
(iii) 19 is a prime number and all the angles of a triangle are equal.
p ∧ q
(iv) 19 is not a prime number.
~p.

Take 2a = 2\(\pi\) and f(x) = g(cosx)
Then, f (2a−x) = f(2\(\pi\)-x) = g(cos(2\(\pi\)-x)) = g(cos x) = f(x)
\(\therefore \int _{ 0 }^{ 2a }{ f(x)dx=2 } \int _{ 0 }^{ a }{ f(x)dx } \)
\(\therefore \int _{ 0 }^{ 2\pi }{ g(cosx)dx=2\int _{ 0 }^{ \pi }{ g(cosx)dx } } \)

The normal vectors of the two given planes \(\vec { r } .(2\hat { i } +2\hat { j } +2\hat { k } )\)= 11 and 4x+2y+2z = 15 are \(\vec { { n }_{ 1 } } =2\hat { i } +2\hat { j } +2\hat { k } \) and \(\vec { { n }_{ 2 } } =4\hat { i } -2\hat { j } +2\hat { k } \) respectively.
If θ is the acute angle between the planes, then we have
\(\theta =cos^{ -1 }\left( \frac { |\vec { { n }_{ 1 } } .\vec { { n }_{ 2 } } | }{ |\vec { { n }_{ 1 } } .\vec { { n }_{ 2 } } | } \right) =cos^{ -1 }\left( \frac { |((2\hat { i } +2\hat { j } +2\hat { k } ).(4\hat { i } -2\hat { j } +2\hat { k } ))| }{ |(2\hat { i } +2\hat { j } +2\hat { k } )||4\hat { i } -2\hat { j } +2\hat { k } | } \right) =cos^{ -1 }\left( \frac { \sqrt { 2 } }{ 3 } \right) \).

Since α and β are the roots of 17x²+ 43x −73 = 0 , we have α + β =\(\frac { -43 }{ 17 } \) and  αβ =\(\frac { -73 }{ 17 } \).
We wish to construct a quadratic equation with roots α + 2 and  β + 2. Thus, to construct such a quadratic equation, calculate
the sum of the roots =  α + β + 4 = \(\frac { -4 }{ 17 } +4=\frac { 25 }{ 17 } \) and
the product of the roots = αβ + 2(α+β)+4 = \(\frac { -73 }{ 17 } +2\left( \frac { -43 }{ 17 } \right) +4=\frac { -91 }{ 17 } \)
Hence a quadratic equation with required roots is x²-\(\frac { 25 }{ 17 } x-\frac { 91 }{ 17 } \) = 0
Multiplying this equation by 17, gives 17x²−25x−91 = 0
which is also a quadratic equation having roots α + 2 and β + 2

Given p = 0.4, n = 3
q = 1-0.4-0.6
\( \mathrm{P}(\mathrm{X} \geq 1)=\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2) \)
\(=3 \mathrm{C}_{1}(0.4)^{1}(0.6)^{2}+3 \mathrm{C}_{2}(0.4)^{2}(0.6)^{1}+\mathrm{P}(\mathrm{X}=3) \)
\(=3 \times\left(\frac{4}{10}\right)\left(\frac{36}{100}\right)+3\left(\frac{16}{100}\right)\left(\frac{6}{10}\right)+\frac{64}{100} \)
\(=\frac{1}{1000}(432+288+64)=\frac{784}{1000}=0.784\)

\( \mathrm{E}(\mathrm{X}) =\Sigma x p(x) \)
\(=-2\left(\frac{1}{4}\right)+2\left(\frac{1}{4}\right)+5\left(\frac{1}{2}\right) \)
\( =\frac{5}{2} \)
\(4 \cdot \mathrm{E}\left(\mathrm{X}^{2}\right)-\operatorname{Var}(2 \mathrm{X}) =4 \mathrm{E}\left(\mathrm{X}^{2}\right)-4\left[\mathrm{E}\left(\mathrm{X}^{2}\right)-\left(\mathrm{E}\left(\mathrm{X}_{1}\right)^{2}\right)\right] \)
\( =4 \mathrm{E}\left(\mathrm{X}^{2}\right)-4 \mathrm{E}\left(\mathrm{X}^{2}\right)+4[\mathrm{E}(\mathrm{X})]^{2} \)
\( =4[\mathrm{E}(\mathrm{X})]^{2} \)
\( =4\left(\frac{5}{2}\right)^{2}=4\left(\frac{25}{4}\right)=25 \)

Given 2cos α = x + \(\frac { 1 }{ x } \)
⇒ 2cos α = \(\frac { { x }^{ 2 }+1 }{ x } \)
⇒ x²+1 = 2x cos α
⇒ x²-2x cos α+1 = 0
⇒ \(\frac { 2cos\alpha \pm \sqrt { (-2cos\alpha )^{ 2 }-4(1)(1) } }{ 2 } \)
= \(\frac { 2cos\alpha \pm \sqrt { 4cos^{ 2 }\alpha -4 } }{ 2 } \)  \(\left[ \because \frac { b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \right] \)
= \(\frac { 2cos\alpha \pm \sqrt { -sin^{ 2 }\alpha } }{ 2 } \)
= \(\frac { 2cos\alpha \pm isin\alpha }{ 2 } \) [∵ sin²α + cos²α = 1]
⇒ x² = cos α ± sin α
Also, 2cos β = y + \(\frac { 1 }{ y } \)
⇒ 2cos β = \(\frac { { y }^{ 2 }+1 }{ y } \)
⇒ y²-2y cos β + 1 = 0
⇒ \(\frac { 2cos\beta \pm \sqrt { (-2cos^{ 2 }\beta ^{ 2 }-4(1)(1) } }{ 2 } \)
= \(\frac { 2cos\beta \pm \sqrt { 4cos^{ 2 }\beta } -4 }{ 2 } =\frac { 2cos\beta \pm 2isin\beta }{ 2 } \)
⇒ y = cos β ±i sin β
\(\frac { { x }^{ m } }{ { y }^{ n } } -\frac { { y }^{ n } }{ { x }^{ m } } =2isin\left( m\alpha -n\beta \right) \)
x^m = (cos α+sin α)^m = cos mα + i sin mα  [By De moivre's theorem]
yⁿ = (cos β + i sin β)ⁿ = cos nβ+i sin nβ
∴ \(\frac { { x }^{ m } }{ { y }^{ n } } =\frac { cosm\alpha +isinm\alpha }{ cosn\beta +isinn\beta } \)
= cos(mα-nβ) +i sin(mα+nβ)
and \(\frac { { y }^{ n } }{ { x }^{ m } } =\frac { \frac { 1 }{ { x }^{ m } } }{ { y }^{ n } } \)
= cos(mα-nβ)-i sin(mα-nβ)

\( \mathrm{f}(x)=|x-1|=\left\{\begin{array}{cc} -(x-1) \text { if } x<1 \\ x-1 & \text { if } x \geq 1 \end{array}\right. \)
\( I=\int_{0}^{1}|x-1| d x \)
\( =\int_{0}^{1}-(x-1) d x=-\left(\frac{x^{2}}{2}-x\right)_{0}^{1} \)
\( =-\left(\frac{1}{2}-1-0\right)=-\left(\frac{-1}{2}\right)=\frac{1}{2} \)

Mean:
\( \mathrm{E}(\mathrm{X}) =\int_{-\infty}^{\infty} x f(x) d x \)
\( =\int_{0}^{2} x \frac{3}{4} x(2-x) d x \)
\( =\frac{3}{4} \int_{0}^{2} x^{2}(2-x) d x \)
\( =\frac{3}{4} \int_{0}^{2}\left(2 x^{2}-x^{3}\right) d x \)
\( =\frac{3}{4}\left[2 \frac{x^{3}}{3}-\frac{x^{4}}{4}\right]_{0}^{2}=\frac{3}{4}\left[\frac{2}{3}(8)-\frac{16}{4}\right]\)
Mean = 1
Variance:
\( \mathrm{E}\left(\mathrm{X}^{2}\right) =\int_{-x}^{x} x^{2} f(x) d x=\int_{0}^{2} x^{2} \frac{3}{4} x(2-x) d x \)
\( =\frac{3}{4} \int_{0}^{2}\left(2 x^{3}-x^{4}\right) d x \)
\( =\frac{3}{4}\left[2 \cdot \frac{x^{4}}{4}-\frac{x^{5}}{5}\right]_{0}^{2} \)
\( =\frac{3}{4}\left[\frac{16}{2}-\frac{32}{5}\right]=\frac{6}{5} \)
Variance \(=E\left(X^{2}\right)-[E(X)]^{2} \)
\(=\frac{6}{5}-1=\frac{1}{5} \)

If θ is the acute angle between two planes \(\vec{r} \cdot \vec{n}_{1}=p_{1} \text { and } \vec{r} \cdot \vec{n}_{2}=p_{2}\) then \(\theta\) is the acute angle between their normal vectors \(\vec{n}_{1} \text { and } \vec{n}_{2}\) .
Therefore, \(\cos \theta=\left(\frac{\left|\vec{n}_{1} \cdot \vec{n}_{2}\right|}{\left|\vec{n}_{1}\right|\left|\vec{n}_{2}\right|}\right) \Rightarrow \theta=\cos ^{-1}\left(\frac{\left|\vec{n}_{1} \cdot \vec{n}_{2}\right|}{\left|\vec{n}_{1}\right|\left|\vec{n}_{2}\right|}\right)\)

Equation of the circle is x² + y² + 2gx + 2 fy + c = 0 with centre C (-g, - f ) and radius \(r=\sqrt{g^{2}+f^{2}-c}\) 
Let P(x₁, y₁) be a point in the plane. Join CP and let it meet the circle at Q. Then the point P is outside, on or within the circle according as
\(|C P| \text { is } \begin{cases}>|C Q| & \text { or, } \\ =|C Q| & \text { or, } \\ <|C Q| . & \end{cases}\)
\(\Rightarrow \quad C P^{2} \text { is } \begin{cases}>r^{2} & \text { or, } \\ =r^{2} & \text { or } \quad\{C Q=r\}, \\ <r^{2} .\end{cases}\)
\(\Rightarrow\left(x_{1}+g\right)^{2}+\left(y_{1}+f\right)^{2} \text { is } \begin{cases}>g^{2}+f^{2}-c & \text { or, } \\ =g^{2}+f^{2}-c & \text { or, } \\ <g^{2}+f^{2}-c\end{cases}\)
\(\Rightarrow \quad x_{1}^{2}+y_{1}^{2}+2 g x_{1}+2 f y_{1}+c \text { is } \begin{cases}>0 & \text { or, } \\ =0 & \text { or, } \\ <0 & \end{cases}\)

We prove the theorem by assuming that the quadratic equation is a monic polynomial equation. The result for non-monic polynomial equation can be proved in a similar way.
Let p and q be rational numbers such that \(\sqrt{q}\) is irrational. Let p + \(\sqrt{q}\) be a root of the equation x² + bx + c = 0 where b and c are rational numbers.
Let α be the other root. Computing the sum of the roots, we get
α + p + \(\sqrt{q}\) = −b
and hence \(\alpha+\sqrt{q}=-b-p \in \mathbb{Q}\). Taking − b − p as s, we have α + \(\sqrt{q}\)  = s
This implies that
α = s −\(\sqrt{q}\)
Computing the product of the roots, we get
(s − \(\sqrt{q}\))( p + \(\sqrt{q}\)) = c
and hence \((s p-q)+(s-p) \sqrt{q}=c \in \mathbb{Q}\) . Thus s − p = 0 . This implies that s = p and hence we get α = p − \(\sqrt{q}\) . So, the other root is p − \(\sqrt{q}\) .

Let A be non-singular. Then \(|A| \neq 0\) and A⁻¹ exists. By definition
\(A A^{-1}=A^{-1} A=I_{n}\)
(i) By (1), we get \(\left|A A^{-1}\right|=\left|A^{-1} A\right|=\left|I_{n}\right|\)
Using the product rule for determinants, we get \(|A|\left|A^{-1}\right|=\left|I_{n}\right|=1\)
Hence, \(\left|A^{-1}\right|=\frac{1}{|A|}\)
(ii) From (1), we get \(\left(A A^{-1}\right)^{T}=\left(A^{-1} A\right)^{T}=\left(I_{n}\right)^{T} .\)
Using the reversal law of transpose, we get \(\left(A^{-1}\right)^{T} A^{T}=A^{T}\left(A^{-1}\right)^{T}=I_{n}\). Hence \(\left(A^{T}\right)^{-1}=\left(A^{-1}\right)^{T}\)
(iii) Since λ is a non-zero scalar, from (1), we get \((\lambda A)\left(\frac{1}{\lambda} A^{-1}\right)=\left(\frac{1}{\lambda} A^{-1}\right)(\lambda A)=I_{n}\)
\(\text { So, }(\lambda A)^{-1}=\frac{1}{\lambda} A^{-1}\)

\(y=2tan\frac { x }{ 2 } -x+c\)

\(\int _{ 0 }^{ 50 }{ \left[ x-\left| x \right| \right] dx } \)

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).

\(put\ t={ sin }^{ -1 }x\Rightarrow dt=\frac { dx }{ \sqrt { 1-{ x }^{ 2 } } } \)
\(\therefore \int _{ 0 }^{ \frac { \pi }{ 4 } }{ { e }^{ at }t\quad dt=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ t } { e }^{ at }dt } \)
\( u=t\ \ v={ e }^{ at }dt\)
\(u'=1\quad { v }_{ 1 }= e ^t\)
\(u'' = 0 \ \ { v }_{ 2 }=e^t\)
\(\int { uvdx } ={ uv }_{ 1 }-u'{ v }_{ 2 }\)
\(\int _{ 0 }^{ \frac { \pi }{ 4 } }{ t{ e }^{ at }dt } ={ \left[ t\frac { { e }^{ at } }{ a } -1\frac { { e }^{ at } }{ { a }^{ 2 } } \right] }_{ 0 }^{ \frac { \pi }{ 4 } }\)
\(dt = \frac{1}{\sqrt 1-x^2}dx\)
I = \(\int ^\frac{\pi}{4}_0 e^t t \ dt\)
\([t e^t-t]^{\frac{\pi}{4}}_0\)
\(=\frac { { e }^{ \frac { \pi }{ 4 } } }{ { a }^{ 2 } } \left( \frac { a\pi }{ 4 } -1 \right) -\frac { { e }^{ 0 } }{ { a }^{ 2 } } (-1)\)
\(=1+ e ^ \frac { \pi }{ 4 } (\frac { \pi }{ 4 }-1) \)

Let (a, b)  ∈ R² an arbitrary point we shall investigate continuity of f at (a,b).
That is, we shall check if all the three conditions for continuity hold for f at (a, b).
(i) f(a, b) = \(\frac { { a }^{ 2 }-{ b }^{ 2 } }{ b^{ 2 }+1 } \) is defined
(ii) \(\begin{matrix} lim \\ (x,y)\rightarrow (a,b) \end{matrix}=\frac { \begin{matrix} lim \\ (x,y)\rightarrow (a,b) \end{matrix}{ x }^{ 2 }-{ y }^{ 2 } }{ \begin{matrix} lim \\ (x,y)\rightarrow (a,b) \end{matrix}{ y }^{ 2 }+1 } \) = L exists
= \(\frac { { a }^{ 2 }-{ b }^{ 2 } }{ b^{ 2 }+1 } \) = L exists
(iii) Also, f(a, b) = \(\frac { { a }^{ 2 }-{ b }^{ 2 } }{ b^{ 2 }+1 } \)
∴ \(\begin{matrix} lim \\ (x,y)\rightarrow (a,b) \end{matrix}\)= L = f(a, b)
Hence f satisfies all the three conditions since (a, b) is an arbitrary point on R², we conclude that f is continuous at every point of R².

Given f(x) = x³ - 2x² ; x = 2, ∆ x = dx = 0.5
(1) df = f'(x).∆x = (3.x² - 4x) ∆x
= [3(2)2 - 4(2)] (0.5)
= 4(0.5) = 2.0
∆f = f(x + ∆x) - f(x)
= f(2.5) - f(2)
= [(2.5)³ - 2 (2.5)²] - [2³ - 2(2²)]
= 15.625 -12.5 -0 = 3.125
(2) df = f'(x) ∆x (2x + 2) (∆x)
x = -0.5, ∆x = dx = 0.1
= (2(-0.05) + 2) = 0.1
∆f = f(x + ∆x) - f(x)
= f(-0.5 + 0.1) - f(-0.5)
= f(0.4) - f(-0.5)
= [(-0.4)² + 2(-0.4) + 3] - [(-0.5)² + 2(-0.5) + 3]
= (16 - 0.8 + 3) - (0.25 - 1 + 3)
= 2.36 - 2.25 = 0.11 

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
\(\therefore \) \(\sqrt [ 3 ]{ 27.2 } =3.0074\)

a) f(x) is continuous in [0, 9]
b) f(x) is differentiable in (0, 9)
c) f(0) = 0
\(f(9)=\sqrt { 9 } -\frac { 9 }{ 3 } =3-3=0\)
∴ f(0) = f(9)
∴ By Rolle's theorem, there exists C ∈ [0, 9] such that f'(c) = 0
⇒ \({ \frac { 1 }{ 2 } c }^{ \frac { 1 }{ 2 } -1 }-\frac { 1 }{ 3 } =0\)
⇒ \({ \frac { 1 }{ 2 } c }^{ \frac { 1 }{ 2 } -1 }=\frac { 1 }{ 3 } \)
⇒ \(\frac { 1 }{ 2\sqrt { c } } =\frac { 1 }{ 3 } \)
⇒ \(\frac { 1 }{ \sqrt { c } } =\frac { 2 }{ 3 } \)
⇒ \(\sqrt { c } =\frac { 2 }{ 3 } \)
Squaring both sides, c = \(\frac94\) ∈ [0, 9]

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

Δ = \(\left| \begin{matrix} 3 & a \\ 2 & a \end{matrix} \right| \) = 3a - 2a = a
Δ₁ =\(\left| \begin{matrix} 4 & a \\ 2 & a \end{matrix} \right| \) = 4a - 2a = 2a
Δ₂ = \(\left| \begin{matrix} 3 & 4 \\ 2 & 2 \end{matrix} \right| \)= 6 - 8 = -2

\(\therefore x=\frac{\Delta_{1}}{\Delta}=\frac{2 \not a}{\not a}=2=y y=\frac{\Delta_{2}}{\Delta}=\frac{-2}{a}\)
∴ Solution set is {2,\(\frac { -2 }{ a } \)}

\(cos\left( { sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) -{ tan }^{ -1 }\left( \frac { 3 }{ 4 } \right) \right) \)
\(\Rightarrow \frac { 4 }{ 5 } =sinx\)

\(\therefore cosx=\frac { adj }{ hyp } =\frac { 3 }{ 5 } \)
Let \({ tan }^{ -1 }\left( \frac { 3 }{ 4 } \right) =y\)
\(\Rightarrow tany=\frac { 3 }{ 4 } \)

\(\Rightarrow siny=\frac { opp }{ hyp } =\frac { 3 }{ 5 } \ cosy=\frac { adj }{ hup } =\frac { 4 }{ 5 } \)
\(\therefore cos\left( { sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) \right) -{ tan }^{ -1 }\left( \frac { 3 }{ 4 } \right) \)
= cos (x - y) = cos x cos y + sin x sin y
= \(\frac { 3 }{ 5 } .\frac { 4 }{ 5 } +\frac { 4 }{ 5 } .\frac { 3 }{ 5 } \)
= \(\frac { 12 }{ 25 } +\frac { 12 }{ 25 } =\frac { 24 }{ 25 } \)
= \(\frac { 24 }{ 25 } \)

\(\left[ \begin{matrix} 5 & 1 & 1 \\ 1 & 5 & 1 \\ 1 & 1 & 5 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 5 & 1 & 1 \\ 1 & 5 & 1 \\ 1 & 1 & 5 \end{matrix} \right] \)
Expending along R₁,
|A| = \(5\left| \begin{matrix} 5 & 1 \\ 1 & 5 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & 5 \end{matrix} \right| +1\left| \begin{matrix} 1 & 5 \\ 1 & 1 \end{matrix} \right| \)
= 5 (25 - 1)-1 (5 - 1)+ 1 (1 - 5)
= 5 (24) - 1(4) + 1(- 4)
= 120 - 4 - 4 = 120 - 8 = 112 ≠ 0
Since A is non singular, A^-1 exit
adj A =\(\left[ \begin{matrix} +\left| \begin{matrix} 5 & 1 \\ 1 & 5 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 1 & 5 \end{matrix} \right| & +\left| \begin{matrix} 1 & 5 \\ 1 & 1 \end{matrix} \right| \\ -\left| \begin{matrix} 1 & 1 \\ 1 & 5 \end{matrix} \right| & +\left| \begin{matrix} 5 & 1 \\ 1 & 5 \end{matrix} \right| & -\left| \begin{matrix} 5 & 1 \\ 1 & 1 \end{matrix} \right| \\ +\left| \begin{matrix} 1 & 1 \\ 5 & 1 \end{matrix} \right| & -\left| \begin{matrix} 5 & 1 \\ 1 & 1 \end{matrix} \right| & +\left| \begin{matrix} 5 & 1 \\ 1 & 5 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} +(25-1)-(5-1)+(1-5) \\ -(5-1)+(25-1)-(5-1) \\ +(1-5)+(5-1)+(25-1) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} 24 & -4 & -4 \\ -4 & 24 & -4 \\ -4 & -4 & 24 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 24 & -4 & -4 \\ -4 & 24 & -24 \\ -4 & -4 & 24 \end{matrix} \right] \)
Taking 4 common from every entry we get,
adj A = \(4\left[ \begin{matrix} 6 & -1 & -1 \\ -1 & 6 & -1 \\ -1 & -1 & 6 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 112 } .4\left[ \begin{matrix} 6 & -1 & -1 \\ -1 & 6 & -1 \\ -1 & -1 & 6 \end{matrix} \right] \)
= \(\frac { 1 }{ 28 } \left[ \begin{matrix} 6 & -1 & -1 \\ -1 & 6 & -1 \\ -1 & -1 & 6 \end{matrix} \right] \).

\(\left( cos\frac { \pi }{ 6 } +isin\frac { \pi }{ 6 } \right) \left( cos\frac { \pi }{ 12 } +isin\frac { \pi }{ 12 } \right) \)
[∵ (cosθ₁+isinθ₁)(cosθ₂+isonθ₂)
= cos(θ₁+θ₂)+isin(θ₁+θ₂)
= \(cos\left( \frac { 2\pi +\pi }{ 12 } \right) +isin\left( \frac { 2\pi +\pi }{ 12 } \right) \)
\(cos\left( \frac { 3\pi }{ 12 } \right) +isin\left( \frac { 3\pi }{ 12 } \right) \)
\(cos\left( \frac { \pi }{ 4 } \right) +isin\left( \frac { \pi }{ 4 } \right) \)
\(\frac { 1 }{ \sqrt { 2 } } +i\frac { 1 }{ \sqrt { 2 } } =\frac { 1 }{ \sqrt { 2 } } \)(1+i)
Aliter:
\( \left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)\left(\cos \frac{\pi}{12}+i \sin \frac{\pi}{12}\right) \)
\( =\cos \left(\frac{\pi}{6}+\frac{\pi}{12}\right)+i \sin \left(\frac{\pi}{6}+\frac{\pi}{12}\right) \)
\( =\left(\cos \frac{3 \pi}{12}+i \sin \frac{3 \pi}{12}\right) \)
\( =\cos \frac{\pi}{4}+i \sin \frac{\pi}{4} \)
\( =\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}=\frac{1+i}{\sqrt{2}} \)

Let A = \(\left[ \begin{matrix} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{matrix} \right] \). Then |A| = \(\left| \begin{matrix} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{matrix} \right| \) = 2(7) + (-12) + 3(-1) = -1 ≠ 0.
Therefore, A⁻¹ exists. Now, we get
adj A = \({ \left[ \begin{matrix} +\left| \begin{matrix} 3 & 1 \\ 2 & 3 \end{matrix} \right| & -\left| \begin{matrix} -5 & 1 \\ -3 & 3 \end{matrix} \right| & +\left| \begin{matrix} -5 & 3 \\ -3 & 2 \end{matrix} \right| \\ -\left| \begin{matrix} -1 & 3 \\ 2 & 3 \end{matrix} \right| & +\left| \begin{matrix} 2 & 3 \\ -3 & 3 \end{matrix} \right| & -\left| \begin{matrix} 2 & -1 \\ -3 & 2 \end{matrix} \right| \\ +\left| \begin{matrix} -1 & 3 \\ 3 & 1 \end{matrix} \right| & -\left| \begin{matrix} 2 & 3 \\ -5 & 1 \end{matrix} \right| & +\left| \begin{matrix} 2 & -1 \\ -5 & 3 \end{matrix} \right| \end{matrix} \right] }^{ T }={ \left[ \begin{matrix} 7 & 12 & -1 \\ 9 & 15 & -1 \\ -10 & -17 & 1 \end{matrix} \right] }^{ T }=\left[ \begin{matrix} 7 & 9 & -10 \\ 12 & 15 & -17 \\ -1 & -1 & 1 \end{matrix} \right] \).
Hence, A^-1 = \(\frac { 1 }{ \left| A \right| } \)(adj A) = \(\frac { 1 }{ \left( -1 \right) } \left[ \begin{matrix} 7 & 9 & -10 \\ 12 & 15 & -17 \\ -1 & -1 & 1 \end{matrix} \right] =\left[ \begin{matrix} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{matrix} \right] \).

This an indeterminate of the form 1∞
Let \(g(x)={ \left( 1+\frac { 1 }{ x } \right) }^{ x }\)
Taking logarithm, we get,
\(log(g(x))=log{ \left( 1+\frac { 1 }{ x } \right) }^{ x }\)
\(=xlog{ \left( 1+\frac { 1 }{ x } \right) }\)
\(=\frac { { \left( 1+\frac { 1 }{ x } \right) } }{ \frac { 1 }{ x } } \)
\(\therefore ​​​​​​\underset { x\rightarrow \infty }{ lim } log(g(x))=​​​​​​\underset { x\rightarrow \infty }{ lim } \frac { { \left( 1+\frac { 1 }{ x } \right) } }{ \frac { 1 }{ x } } =\left( \frac { 0 }{ 0 } form \right) \)

\(=​​​​​​\underset { x\rightarrow \infty }{ lim } \frac { { 1 } }{ \left( 1+\frac { 1 }{ x } \right) } =1\)
But \(​​​​​​\underset { x\rightarrow \infty }{ lim } log(g(x))=log​​​​​​\underset { x\rightarrow \infty }{ lim } g(x)\)
\(\therefore log(\underset { x\rightarrow \infty }{ lim } g(x))=1\)
\({ e }^{ log }(\underset { x\rightarrow \infty }{ lim } g(x))={ e }^{ 1 }=e\Rightarrow (\underset { x\rightarrow \infty }{ lim } g(x))=e\)
\(\Rightarrow \underset { x\rightarrow \infty }{ lim } { \left( 1+\frac { 1 }{ x } \right) }^{ x }=e\)

Since the sum of the co-efficients is
2 - 9 + 10 - 3 = 12 - 12 = 0
⇒ x = 1 is a root of (x)
∴ (x - 1) is a factor of (x)
To find the other factor, let us divide f(x) by x-1 

[Using synthetic division]
f(x) = (x-1)(x - 3)(2x - 1) = 0
⇒x - 1 = 0, x - 3 = 0 or 2x - 1 = 0
⇒ x = 1, x = 3x, x = \(\frac{1}{2}\)
Hence the roots are 1, 3, \(\frac{1}{2}\).