\(\frac { { e }^{ 2x } }{ 2 } +c\)

x =\(\sqrt { t } \)
V = \(\frac { dx }{ dt } =\frac { 1 }{ 2 } t^{ -\frac { 1 }{ 2 } }\) ..(1)
Acceleration = \(\frac { { d }^{ 2 }x }{ { dt }^{ 2 } } =\frac { 1 }{ 2 } \left( -\frac { 1 }{ 2 } t^{ -\frac { 3 }{ 2 } } \right) =\frac { -t^{ -\frac { 3 }{ 2 } } }{ 4 } \)
∴ Acceleration is negative
Acceleration = \(-\frac { 1 }{ 4 } \left( { t }^{ -\frac { 1 }{ 2 } } \right) ^{ 3 }\)
= \(-2\left( \frac { 1 }{ 2 } t^{ -\frac { 1 }{ 2 } } \right) ^{ 3 }\) = 2V³    [using (1)]
Hence, acceleration is negative proportional to the cube of the velocity.

Given \(f(x)=\begin{cases} \begin{matrix} { 3e }^{ -3x } & x>0 \end{matrix} \\ \begin{matrix} 0 & elsewhere \end{matrix} \end{cases}\)
\(E(X)=\int _{ 0 }^{ \infty }{ x.f(x)dx } =\int _{ 0 }^{ \infty }{ x.3.{ e }^{ -3x }dx } \) 
= \(3\int _{ 0 }^{ \infty }{ x.{ e }^{ -3x }dx } \left[ \therefore \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -ax }dx=\cfrac { n! }{ { a }^{ n+1 } } } \right] \) 
= \(3\times \frac { 1! }{ { 3 }^{ 2 } } =\frac { 3 }{ 9 } =\frac { 1 }{ 3 } \)
ஃ Expected life of the electronic equipment is = \(\frac { 1 }{ 3 } \) 

Given g(x, y) = \(\frac { { 3x }^{ 2 }-xy }{ { x }^{ 2 }+{ y }^{ 2 }+3 } \)
\(\begin{matrix} lim \\ (x,y)\rightarrow (1,2) \end{matrix}g(x,y)=\begin{matrix} lim \\ (x,y)\rightarrow (1,2) \end{matrix}\frac { { 3x }^{ 2 }-xy }{ { x }^{ 2 }+{ y }^{ 2 }+3 } \)
\(=\frac { { 3(1) }^{ 2 }-1(2) }{ { 1 }^{ 2 }+{ 2 }^{ 2 }+3 } =\frac { 3-2 }{ 8 } =\frac { 1 }{ 8 } \)

In the given differential equation, the highest order derivative is \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) whose power is 1.
Therefore, the given differential equation is of order 2.
The given differential equation is not a polynomial equation in its derivatives and so its degree is not defined.

Since the parabola is symmetric about x-axis, it is either open upward or downward.
Let the equation be x² = 4ay ...(1)
Since (2, -3) lies on the parabola,
2² = 4a(-3) ⇒ a = \(\frac { -1 }{ 3 } \)
Substituting a = \(\frac { -1 }{ 3 } \) in (1) we get,
x² = 4 \(\left( \frac { -1 }{ 3 } \right) \) y ⇒ 3x² = -4y. Which is the required equation of the parabola.

Let p(x) = x⁷-6x⁶+7x⁵+5x²+2x+2
It has only one change of sign. Now,
p(-x) = (-x)⁷ -6(-x)⁶ +7(-x)⁵ +5(-x)² +2(-x)+2
= -x⁷ -6x⁶ -7x⁵ + 5x² - 2x + 2
It has two, change of sign.
Hence, p(x) has one positive root and has at least two negative roots.

A =\(\left[ \begin{matrix} 4 \\ 7 \end{matrix}\begin{matrix} 5 \\ -3 \end{matrix}\begin{matrix} -6 \\ 0 \end{matrix}\begin{matrix} 1 \\ 8 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 4 \\ -1 \end{matrix}\begin{matrix} 5 \\ -3 \end{matrix}\begin{matrix} -6 \\ 12 \end{matrix}\begin{matrix} 1 \\ 6 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} -1 \\ 4 \end{matrix}\begin{matrix} -3 \\ 5 \end{matrix}\begin{matrix} 12 \\ -6 \end{matrix}\begin{matrix} 6 \\ 1 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }+(-1){ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 4 \end{matrix}\begin{matrix} 13 \\ 5 \end{matrix}\begin{matrix} -12 \\ -6 \end{matrix}\begin{matrix} 6 \\ 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-4{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \end{matrix}\begin{matrix} 13 \\ -47 \end{matrix}\begin{matrix} -12 \\ 42 \end{matrix}\begin{matrix} -6 \\ 25 \end{matrix} \right] \)
The equivalent row-echelon matrix hats two non zero rows.
∴ \(\rho\) (A) = 2

Let y = sec^-1 (-2). Then, sec y = -2
By the definition, the range of the principal value branch of y = sec⁻¹x is [0, \(\pi\)]\{\({{\frac{\pi}{2}}}\)}
Let us find y in [0, \(\pi\)] - {\({{\frac{\pi}{2}}}\)} such that sec y = -2
But, sec y = −2 \(\Rightarrow\) cos y = -\(\frac{1}{2}\)
Now, cos y = -\(\frac { 1 }{ 2 } =-cos\frac { \pi }{ 3 } =cos\left( \pi -\frac { \pi }{ 3 } \right) =cos\frac { 2\pi }{ 3 } \). Therefore, y = \(\frac{2\pi}{3}\)
since \(\frac{2\pi}{3}\in[0,\pi]\)\{\({{\frac{\pi}{2}}}\)}, the principal value of sec^-1(-2) is \(\frac{2\pi}{3}\)

Let \(\vec { a } =7\hat { i } +\lambda \hat { j } -3\hat { k } ,\vec { b } =\hat { i } +2\hat { j } -\hat { k } \) and \(\vec { c } =-3\hat { i } +7\hat { j } -5\hat { k } \)
∴ volume of the parallelepiped
= \(\vec { a } .(\vec { b } \times \vec { c } )\)
Given \(\vec { a } .(\vec { b } \times \vec { c } )\) = 90
⇒ \(\left| \begin{matrix} 7 & \lambda & -3 \\ 1 & 2 & -1 \\ -3 & 7 & 5 \end{matrix} \right| \) = 90
⇒ \(-6\left| \begin{matrix} 2 & -1 \\ 7 & 5 \end{matrix} \right| -\lambda \left| \begin{matrix} 1 & -1 \\ -3 & 5 \end{matrix} \right| -3\left| \begin{matrix} 1 & 2 \\ -3 & 7 \end{matrix} \right| \) = 90
⇒ 7(10+7)-λ(5-3)-3(7+6) = 90
⇒ 7(17)-λ(2)-3(13) = 90
⇒ 119-2λ-39 = 90
⇒ 119-39-90 = 2λ
⇒ -10 = 2λ
⇒ λ = -5

We have = \(\frac { z+3 }{ z-5i } =\frac { 1+4i }{ 2 } \)
\(\Rightarrow\) 2(z + 3) = (1 + 4i) (z− 5i)
\(\Rightarrow\) 2z + 6 = (1 + 4i)z + 20−5i
\(\Rightarrow\) (2−1−4i)z = 20− 5i− 6 
\(\Rightarrow\) \(z=\frac { 14-5i }{ 1-4i } =\frac { \left( 14-5i \right) \left( 1+4i \right) }{ \left( 1-4i \right) \left( 1+4i \right) } =\frac { 34+51i }{ 17 } =2+3i\)

Since α and β are the roots of the quadratic equation, we have α + β =\(\frac { 7 }{ 2 } \) and αβ = \(\frac { 13 }{ 2 } \).
Thus, to construct a new quadratic equation,
Sum of the roots = α²+β² = (α+β)²-2αβ =\(\frac { -3 }{ 4 } \)
Product of the roots = α²β² = (αβ)² = \(\frac { 169 }{ 4 }\)
Thus a required quadratic equation is  x²+\(\frac { 3 }{ 4 } x+\frac { 169 }{ 4 } \)= 0. 
From this we see that 4x²+3x+169 = 0 is a quadratic equation with roots α² and β².

y=tan^-1(x+y+1)

Area under the curve y = sin x between x = 0 and x = \(\frac { \pi }{ 3 } \) is
\({ A }_{ 1 }=\int _{ 0 }^{ \frac { \pi }{ 3 } }{ ydx } =\int _{ 0 }^{ \frac { \pi }{ 3 } }{ sinxdx } =-{ \left[ cosx \right] }_{ 0 }^{ \frac { \pi }{ 3 } }\)
\(=-(cos\frac { \pi }{ 3 } -cos0)=-\left( \frac { 1 }{ 2 } -1 \right) \)
\(=-\left( -\frac { 1 }{ 2 } \right) =\frac { 1 }{ 2 } \)
Area under the curve y = sin 2x between x = 0 and \(\frac { \pi }{ 3 } \) is
\({ A }_{ 2 }=\int _{ 0 }^{ \frac { \pi }{ 3 } }{ { sin2 \ x \ dx=-\left[ \frac { cos2 \ x }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 3 } } } \)
\(=-\frac { 1 }{ 2 } [cos2\frac { \pi }{ 3 } -cos0]-\frac { 1 }{ 2 } \left[ -\frac { 1 }{ 2 } -1 \right] =-\frac { 1 }{ 2 } \left( -\frac { 3 }{ 2 } \right) =\frac { 3 }{ 4 } \)
\(\therefore \frac { { A }_{ 1 } }{ { A }_{ 2 } } =\frac { \frac { 1 }{ 2 } }{ \frac { 3 }{ 4 } } =\frac { 1 }{ 2 } \times \frac { 4 }{ 3 } =\frac { 2 }{ 3 } \)
∴ A₁:A₂ =2 : 3 

Let p be the probability of the useful life hours of a fluorescent light.
n = 12
P = 0.9
q = 1-p = 0.1
P(X= x)= ⁿC_x p^x q^n-x, x = 0, 1,2, .., n
(i) Exactly 10
P(X = 10) = ¹²C₁₀(0.9)¹⁰(1 - 0.9)²
= ¹²C₁₀(0.9)¹⁰ (0.1)²
(ii) Atleast 11
P(X≥11) = R(X = 11) + P(X = 12)
= ¹²C₁₁(0.9)¹¹ (0.1)¹ + ¹²C₁₂(0.9)¹²(0.1)⁶
= ¹²C₁ (0.9)¹¹ (0.1) + (0.9)¹²
= ¹²(0.9)¹¹ (0.1) + (0.9)¹²
= (0.9)¹¹ ((12)(0.1) + 0.9)
= (0.9)¹¹ (1.2 + 0.9)
= (0.9)¹¹ (2.1)
(ii) Atleast 2 will not have a useful
P(X,10) = 1 -P(X > 10)
= 1 - [P(X = 11) + P(X = 12)]
= 1-(2.1) (0.9)¹¹

First we shall calculate \(\frac { \partial f }{ \partial x } \) (x, y). Note that f is a sum of two functions and so
\(\frac { \partial f }{ \partial x } =\frac { \partial }{ \partial x } sin({ xy }^{ 2 })+\frac { \partial }{ \partial x } \left( { e }^{ { x }^{ 3 }+5y } \right) \)
\(=cos(x{ y }^{ 2 })\frac { \partial }{ \partial x } (x{ y }^{ 2 })+{ e }^{ { x }^{ 3 }+5y }\frac { \partial }{ \partial x } \left( { x }^{ 3 }+5y \right) \)
= cos(xy² ) y² + \({ e }^{ { x }^{ 3 }+5y }\) 3x²
Similarly,
\(\frac { { \partial }^{ }f }{ { \partial y\ } } =\frac { \partial }{ \partial y } sin(x{ y }^{ 2 })+\frac { \partial }{ \partial y } \left( { e }^{ { x }^{ 3 }+5y } \right) \)
\(=cos(x{ y }^{ 2 })\frac { \partial }{ \partial y } (x{ y }^{ 2 })+{ e }^{ { x }^{ 3 }+5y }\frac { \partial }{ \partial y } \left( { x }^{ 3 }+5y \right) \)
= cos(xy²)2xy + 5\({ e }^{ { x }^{ 3 }+5y }\)
Next we consider,
\(\frac { { \partial }^{ 2 }f }{ { \partial y\partial x } } =\frac { \partial }{ \partial y } \left( \frac { \partial f }{ \partial x } \right) =\frac { \partial }{ \partial y } ({ y }^{ 2 }cos(x{ y }^{ 2 })+3{ x }^{ 2 }{ e }^{ { x }^{ 3 }+5y })\)
\(=\frac { \partial }{ \partial y } ({ y }^{ 2 }cos(x{ y }^{ 2 }))+\frac { \partial }{ \partial y } ({ 3x }^{ 2 }{ e }^{ { x }^{ 3 }+5y })\)
= 2y cos(xy²)+y²(-sin(xy²)2xy) + 3x² \({ e }^{ { x }^{ 3 }+5y }\) 5
= 2y cos(xy²)+2xy³ sin(xy²)+15 x² \({ e }^{ { x }^{ 3 }+5y }\)
Finally,
\(\frac { { \partial }^{ 2 }f }{ { \partial x\partial y } } =\frac { \partial }{ \partial y } \left( \frac { \partial f }{ \partial x } \right) =\frac { \partial }{ \partial y } (cos(x{ y }^{ 2 })2xy+5{ e }^{ { x }^{ 3 }+5y })\)
= -sin(xy²)y²2xy+cos(xy²)2y+5\({ e }^{ { x }^{ 3 }+5y }\) 3x²
= 2y cos(xy²)- 2xy³ sin(xy²)+15x²\({ e }^{ { x }^{ 3 }+5y }\)
Note that we have first used sum rule, then in the next step we have used chain rule. In the third step, product rule is used. Also, we see that f_xy = f_yx Is it a coincidence? or is it always true? Actually, there are functions for which f_xy ≠ f_yz at some points. The following theorem gives conditions under which f_xy = f_yz.

We have,
f'(x) = 4x³+32 = 0 gives x³ = -8
⇒ x = −2
and f′′(x) = 12 x² 
As f''(−2)>0, the function has local minimum at x = −2. The local minimum value is f (−2) = −48 
Therefore, the extreme point is (−2, −48) .

Let \(g(x)=x^{\frac{1}{1-x}}\). This is an indeterminate of the form \(1^{\infty}\). Taking the logarithm,
\(log \ g(x)=\frac{logx}{1-x}\).
Therefore, \(\underset{x \rightarrow1}{lim \ }log \ g(x) =\underset{x\rightarrow 1}{lim}(\frac{log \ x}{1-x})(\frac{0}{0})\)
An application of l’Hôpital rule,
\(\underset{x \rightarrow1}{lim} \ (\frac{\frac{1}{x}}{-1})=-1\)
But, \(\underset{x\rightarrow 1}{lim}\) log g(x) = log (\(\underset{x\rightarrow 1}{lim}\) g(x)).
Hence on exponentiating, we get
\(\underset{x \rightarrow1}{lim}\quad x^{\frac{1}{1-x}}=e^{-1}=\frac{1}{e}\).

The normal vector to the planes
x + 2y + 3z - 7 = 0, 2x - 3y + 4z = 0 are 
\(\overset { \rightarrow }{ b } =\overset { \wedge }{ i } +2\overset { \wedge }{ j } +3\overset { \wedge }{ k } \) and \(\overset { \rightarrow }{ c } =2\overset { \wedge }{ i } -3\overset { \wedge }{ j } +4\overset { \wedge }{ k } \)
∴ The required planes passes through the point \(\overset { \rightarrow }{ a } =\overset { \wedge }{ i } +\overset { \wedge }{ j } +\overset { \wedge }{ k } \) and parallel to two vector 5 namely \(\overset { \rightarrow }{ b } \) and \(\overset { \rightarrow }{ c } \) 
∴ The Parametric form of vectors equation of the plans is \(\overset { \rightarrow }{ r } =\overset { \rightarrow }{ a } +s\overset { \rightarrow }{ b } +t\overset { \rightarrow }{ c } \) s, t ∈ R
\(\overset { \rightarrow }{ r } =\left( \overset { \rightarrow }{ i } +\overset { \rightarrow }{ j } -\overset { \rightarrow }{ k } \right) +s\left( \overset { \rightarrow }{ i } +2\overset { \rightarrow }{ j } +3\overset { \rightarrow }{ k } \right) +t\left( 2\overset { \rightarrow }{ i } -3\overset { \rightarrow }{ j } +4\overset { \rightarrow }{ k } \right) ,\)
Cartesian equation is \(\left| \begin{matrix} x-{ x }_{ 1 } \\ { b }_{ 1 } \\ { c }_{ 1 } \end{matrix}\begin{matrix} y-{ { y }_{ 1 } } \\ { b }_{ 2 } \\ { c }_{ 2 } \end{matrix}\begin{matrix} z-{ { z }_{ 1 } } \\ { b }_{ 3 } \\ { c }_{ 3 } \end{matrix} \right| =0\)
\(\Rightarrow \left| \begin{matrix} x-1 \\ 1 \\ 2 \end{matrix}\begin{matrix} y-1 \\ 2 \\ -3 \end{matrix}\begin{matrix} z+1 \\ 3 \\ 4 \end{matrix} \right| =0\)
⇒ (x - 1) (8 + 9) - (y - 1)(4 - 6) + (z + 1)(-3 -4) = 0
⇒ 17 (x - 1) +2 (y - 1) -7 (z + 1) = 0
⇒ 17x - 17 + 2y - 2 - 7z - 7 = 0
⇒ 17x + 2y - 7z - 26 = 0

LHS = \(\left( \frac { i+\sqrt { 3 } }{ -i+\sqrt { 3 } } \right) ^{ 2\omega }+\left( \frac { i-\sqrt { 3 } }{ i+\sqrt { 3 } } \right) ^{ 2\omega }\)
= \(\left( \frac { \sqrt { 3 } +i }{ \sqrt { 3 } -i } \times \frac { \sqrt { 3 } +i }{ \sqrt { 3 } +i } \right) ^{ 2\omega }+\left( \frac { -\sqrt { 3 } +i }{ \sqrt { 3 } +i } \times \frac { \sqrt { 3 } -i }{ \sqrt { 3 } -1 } \right) ^{ 2\omega }\)
= \(\left( \frac { 3-1+2\sqrt { 3 } i }{ 3+1 } \right) ^{ 2\omega }+\left( \frac { -3+1+2\sqrt { 3 } i }{ 3+1 } \right) ^{ 2\omega }\)
= \(\left( \frac { 2+2\sqrt { 3 } i }{ 4 } \right) ^{ 2\omega }+\left( \frac { -2+2\sqrt { 3 } i }{ 4 } \right) ^{ 2\omega }\)
= \(\left( \frac { 1+\sqrt { 3 } i }{ 2 } \right) ^{ 2\omega }+\left( \frac { -2+2\sqrt { 3 } i }{ 2 } \right) ^{ 2\omega }\)
=\(\left[ -\left( \frac { -1-\sqrt { 3 } i }{ 2 } \right) \right] ^{ 2\omega }+\left[ \frac { -1+\sqrt { 3 } i }{ 2 } \right] ^{ 2\omega }\)
= (-ω²)^2ω+(ω)^2ω
[∴ ω = \(\frac { -1+i\sqrt { 3 } }{ 2 } \), ω² = \(\frac { -1-i\sqrt { 3 } }{ 2 } \)]
= ω^4ω+ω^2ω
= (ω³)¹³³. ω¹ + (ω³)⁶⁶.ω²
= 1.ω+1.ω²  [∴ 1+ω+ω² = 0 & ω³ = 1]
= ω + ω²
= -1 = RHS

Equation of the ellipse is \(\frac { { x }^{ 2 } }{ 25 } +\frac { y^{ 2 } }{ 9 } =1\)
∴ a² = 25, b² = 9
∴ e = \(\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 9 }{ 25 } } =\sqrt { \frac { 16 }{ 25 } } =\frac { 4 }{ 5 } \)
Focus is (ae, 0) = \(\left( 5\times \frac { 4 }{ 5 } \right) \) = (4, 0)
Since the focus of the hyperbola coincides with the focus of the ellipse, foci of the hyperbola are (±4,0).
Let A be the length of the semi-transverse axis
∴ Ae - 4 ⇒ 2A = \(\frac { 4 }{ e } =\frac { 4 }{ 2 } =2\) [∵ e = 2]
Let B b th length of the semi conjugate axis
B² = A²(e² - 1) = 4(4 - 1) = 12
Equation of the hyperbola is \(\frac { { x }^{ 2 } }{ { A }^{ 2 } } -\frac { { y }^{ 2 } }{ { B }^{ 2 } } =1\)
⇒ \(\frac { { x }^{ 2 } }{ 4 } -\frac { { y }^{ 2 } }{ 12 } =1\)

LHS =\({ tan }^{ -1 }\left( \frac { 1-x }{ 1+x } \right) -{ tan }^{ -1 }\left( \frac { 1-y }{ 1+y } \right) \)
= tan^-1(1) - tan^-1 (x) - (tan^-1(1) - tan^-1(y)
\(\left[ \because { tan }^{ -1 }(\frac { x-y }{ 1+xy } )={ tan }^{ -1 }x-{ tan }^{ -1 }y \right] \)
= tan^-1(1) - tan^-1 (x) - tan^-1(1) + tan^-1(y)
= tan^-1(y) - tan^-1(x)
= \({ tan }^{ -1 }\left( \frac { y-x }{ 1+xy } \right) \)
= \({ sin }^{ -1 }\left( \frac { y-x }{ \sqrt { 1+\left( yx \right) ^{ 2 }+\left( y-x \right) ^{ 2 } } } \right) \)
= \({ sin }^{ -1 }\left( \frac { y-x }{ \sqrt { (1+{ x }^{ 2 })(1+{ x }^{ 2 }) } } \right) \)
RHS

Let the required numbers be x, y and z
By the given data,
x + y + z = 20  ....(1)
2z + x = 23 ⇒ x + 2z = 23...(2)
y + z + 3x = 46 ⇒  3x + y + z = 46..(3)
Δ = \(\left| \begin{matrix} 1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1 \end{matrix} \right| =1\left| \begin{matrix} 0 & 2 \\ 0 & 1 \end{matrix} \right| -1\left| \begin{matrix} 1 & 2 \\ 1 & 1 \end{matrix} \right| +1\left| \begin{matrix} 1 & 0 \\ 3 & 1 \end{matrix} \right| \)
= -2 + 5 + 1 =4
Δ₁ = \(\left| \begin{matrix} 20 & 1 & 1 \\ 23 & 0 & 2 \\ 46 & 1 & 1 \end{matrix} \right| =20\left| \begin{matrix} 0 & 2 \\ 1 & 1 \end{matrix} \right| -1\left| \begin{matrix} 23 & 2 \\ 46 & 1 \end{matrix} \right| +1\left| \begin{matrix} 23 & 0 \\ 46 & 1 \end{matrix} \right| \)
= -40 + 69 + 23 = 52
Δ₂ = \(\left| \begin{matrix} 1 & 20 & 1 \\ 1 & 23 & 2 \\ 3 & 46 & 1 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 23 & 2 \\ 46 & 1 \end{matrix} \right| -20\left| \begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix} \right| +1\left| \begin{matrix} 1 & 23 \\ 3 & 46 \end{matrix} \right| \)
= -69 + 100 - 23 = 8
Δ₃ = \(\left| \begin{matrix} 1 & 1 & 20 \\ 1 & 2 & 23 \\ 3 & 1 & 46 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 0 & 23 \\ 1 & 46 \end{matrix} \right| -1\left| \begin{matrix} 1 & 23 \\ 3 & 46 \end{matrix} \right| +20\left| \begin{matrix} 1 & 0 \\ 3 & 1 \end{matrix} \right| \)
= -23 + 23 + 20 = 20
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 52 }{ 4 } \) = 13
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 8 }{ 4 } \) = 2 and z =\(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { 20 }{ 4 } \) = 5
Hence the required numbers are 13, 2 and 5.

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 cubic equation is 2x³-6x²+3x+k = 0
Here, a = 2, b = -6, c = 3, d = k
Let ∝, β, ૪ be the roots
Given ∝ = 2(β+૪) ⇒ \(\frac{\alpha}{2}\) = β+૪ ...(1)
Now, \(\alpha +\beta +\gamma =\frac { -b }{ a } =-\frac { (-6) }{ 2 } =3\)
\(\frac { \alpha }{ 2 } +\alpha =3\Rightarrow \frac { \alpha +2\alpha }{ 2 } =3\Rightarrow \frac { 3\alpha }{ 2 } =3\)
\(\Rightarrow \alpha =2\)
\(\alpha \beta \gamma =\frac { -d }{ a } =\frac { -k }{ 2 } \Rightarrow 2.\beta \gamma =\frac { -k }{ 2 } \)
\(\beta \gamma =\frac { -k }{ 4 } ...(2)\)
Also, \(\alpha \beta +\beta \gamma +\gamma \alpha =\frac { c }{ a } \) 
\(2\beta +\beta \gamma +2\gamma =\frac { 3 }{ 2 } \)
\(2(\beta +\gamma )+\beta \gamma =\frac { 3 }{ 2 } \)
\(\alpha \frac { -k }{ 4 } =\frac { 3 }{ 2 } \quad [from(1)\& (2)]\)
Also, \(2-\frac { k }{ 4 } =\frac { 3 }{ 2 } [\because \alpha =2]\)
\(2-\frac { 3 }{ 2 } =\frac { k }{ 4 } \Rightarrow \frac { 1 }{ 2 } =\frac { k }{ 4 } \)
\(\\ k=\frac { 4 }{ 2 } \Rightarrow k=2\)
From(2), \(\beta \gamma =\frac { -k }{ 4 } =\frac { -2 }{ 4 } =\frac { -1 }{ 2 } \)\(\Rightarrow \gamma =\frac { -1 }{ 2\beta } \)
From \((1),\beta +\gamma =\frac { \alpha }{ 2 } =\frac { 2 }{ 2 } =1\)
Substituting \(\gamma =\frac { -1 }{ 2\beta } \) We get
\(\beta -\frac { 1 }{ 2\beta } =1\Rightarrow 2{ \beta }^{ 2 }-1=2\beta \Rightarrow 2\beta -2\beta -1=0\)
\(\beta =\frac { 2\pm \sqrt { 4-4(2)(-1) } }{ 4 } =\frac { 2\pm \sqrt { 4+8 } }{ 4 } \)
\(=\frac { 2\pm \sqrt { 12 } }{ 4 } =\frac { 2\pm 2\sqrt { 3 } }{ 4 } \)
\(\beta =\frac { 1\pm \sqrt { 3 } }{ 2 } \)
Hence the roots are \(2,\frac { 1+\sqrt { 3 } }{ 2 } ,\frac { 1-\sqrt { 3 } }{ 2 } \) and k = 2

\(\frac { \pi }{ 4 } \)

The equation of the family of ellipses having centre at the origin & foci on the y-axis, is given
\(\frac { { x }^{ 2 } }{ { b }^{ 2 } } +\frac { { y }^{ 2 } }{ { a }^{ 2 } } =1\) ...(1)
where b >a & a, b are the parameters or a,b are arbitrary constant.
Differentiating equation (1) twice successively, because we have two arbitrary constant) we get
\( \frac{2 x}{a^{2}}+\frac{2 y}{b^{2}} \frac{d y}{d x} =0 \)
\(2\left(\frac{x}{a^{2}}+\frac{y}{b^{2}} \frac{d y}{d x}\right) =0 \)
\(\frac{x}{a^{2}}+\frac{y}{b^{2}} \frac{d y}{d x} =0\) ............(2)
Again differentiating equation (2)
\(\frac{1}{a^{2}}+\frac{y}{b^{2}} \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x} \frac{d y}{d x b^{2}}=0\)
\(\frac{1}{a^{2}}+\frac{y}{b^{2}} \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2} \frac{1}{b^{2}}=0\)
multiply by x
\(\frac{x}{a^{2}}+\frac{x y}{b^{2}}\left(\frac{d^{2} y}{d x^{2}}\right)+\left(\frac{d y}{d x}\right)^{2} \frac{x}{b^{2}}=0\) .........(3) 
Equation (3)-(2)
\( \frac{x}{a^{2}}+\frac{x y}{b^{2}}\left(\frac{d^{2} y}{d x^{2}}\right)+\left(\frac{d y}{d x}\right)^{2}\left(\frac{x}{b^{2}}\right) -\left(\frac{x}{a^{2}}+\frac{y}{b^{2}} \frac{d y}{d x}\right) =0 \)
\(\frac{x y}{b^{2}}\left(\frac{d^{2} y}{d x^{2}}\right)+\left(\frac{d y}{d x}\right)^{2} \frac{x}{b^{2}}-\frac{y}{b^{2}} \frac{d y}{d x} =0 \)
Taking \(\frac{1}{b^{2}}\) outside, we get
\( \frac{1}{b^{2}}\left[x y \frac{d^{2} y}{d x^{2}}+x\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}\right]=0 \\ x y \frac{d^{2} y}{d x^{2}}+x\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0 \)
is the required differential equation.

Given  \(sin\left( { sin }^{ -1 }\frac { 1 }{ 5 } +{ cos }^{ -1 }x \right) =1=sin\frac { \pi }{ 2 } \)
\(\left[ \because sin\frac { \pi }{ 2 } =1 \right] \)
\(\Rightarrow { sin }^{ -1 }\frac { 1 }{ 5 } +{ cos }^{ -1 }x={ sin }^{ -1 }\left( sin\left( \frac { \pi }{ 2 } \right) \right) \)
\(\Rightarrow { sin }^{ -1 }\frac { 1 }{ 5 } +{ cos }^{ -1 }x=\frac { \pi }{ 2 } \)
\({ sin }^{ -1 }\frac { 1 }{ 5 } =\frac { \pi }{ 2 } -{ cos }^{ -1 }x{ sin }^{ -1 }x\)
\(\left[ \because { sin }^{ -1 }x+{ cos }^{ -1 }x=\frac { \pi }{ 2 } \right] \)
\(\Rightarrow x=\frac { 1 }{ 5 } \)

L. H. S = \(\left[ \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } ,\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } ,\overset { \rightarrow }{ c } \right] \)
\(=\left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \right) .\left\{ \left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) \times \overset { \rightarrow }{ c } \right\} \)
\(=\left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \right) .\left\{ \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \times \overset { \rightarrow }{ c } \right\} \)
\(=\left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \right) .\left\{ \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right\} \ \left[ \because \overset { \rightarrow }{ c } \times \overset { \rightarrow }{ c } =\overset { \rightarrow }{ 0 } \right] \)
\(=\overset { \rightarrow }{ a } .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) +\overset { \rightarrow }{ b } .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) +\overset { \rightarrow }{ c } .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) \)
\(=\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] +0+0\)
\(\left[ \because \left[ \overset { \rightarrow }{ b } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] =\left[ \overset { \rightarrow }{ c } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] =0 \right] \)
\(=\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] \)= R. H. S
Hence proved

\(\Delta = \left| \begin{matrix} t & 3 & -1 \\ 1 & 2 & 1 \\ -t & 1 & 2 \end{matrix} \right| =t\left| \begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix} \right| -3\left| \begin{matrix} 1 & 1 \\ -t & 2 \end{matrix} \right| -\left| \begin{matrix} 1 & 2 \\ -t & 1 \end{matrix} \right| \)
= t(4 - 1) -3 (2 + t) -1(1 + 2t)
= 3t - 6i - 3t - 1 - 2t = - 7 - 2t
The system will fail to have unique solution if
Δ = 0 ⇒ -7-2t = 0 ⇒ -2t = 7 ⇒ t = \(\frac { -7 }{ 2 } \)
∴ t = \(\frac { -7 }{ 2 } \).

The given equation can be written as
\({ 2 }^{ z }\left( 1+\frac { 1 }{ 2 } +\frac { 1 }{ 4 } \right) ={ 7 }^{ x }\left( 1+\frac { 1 }{ 7 } +\frac { 1 }{ 49 } \right) \)
\(\Rightarrow { 2 }^{ x }\left( \frac { 8+4+2 }{ 8 } \right) ={ 7 }^{ x }\left( \frac { 49+7+1 }{ 49 } \right) \)
\(\Rightarrow { 2 }^{ x }\left( \frac { 7 }{ 4 } \right) ={ 7 }^{ x }\left( \frac { 57 }{ 49 } \right) \Rightarrow \frac { 7 }{ 4 } \times \frac { 49 }{ 57 } =\frac { { 7 }^{ x } }{ { 2 }^{ x } } \)
\(\Rightarrow \frac { 7 }{ 4 } \times \frac { 49 }{ 57 } ={ \left( \frac { 7 }{ 4 } \right) }^{ x }\Rightarrow \frac { { 7 }^{ 3 } }{ 4\times 57 } ={ \left( \frac { 7 }{ 4 } \right) }^{ x }\)
\(\Rightarrow xlog\left( \frac { 7 }{ 4 } \right) =3log\ 7-log4-log57\)
\(\Rightarrow x=\frac { 3log7-log4-log57 }{ log\left( \frac { 7 }{ 2 } \right) } \)