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If $\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{xy}{x^{2}+y^{2}};y(1)=1;$ then a value of $x$ satisfying $y(x)=e$ is :
Option: 1 $\sqrt{3}\: e$

Option: 2 $\frac{1}{2}\sqrt{3}\: e$

Option: 3 $\sqrt{2}\: e$

Option: 4 $\frac{e}{\sqrt{2}}$

√3e

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Nalla mahalakshmi

If $z$ be a complex number satisfying $\left | Re\left ( z \right ) \right |+\left | Im(z) \right |=4,$ then $\left | z \right |$ cannot be :
Option: 1 $\sqrt{7}$

Option: 2 $\sqrt{\frac{17}{2}}$

Option: 3 $\sqrt{10}$

Option: 4 $\sqrt{8}$

Option d

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Let $a,b\epsilon \textbf{R},a\neq 0$ be such that the equation, $ax^{2}-2bx+5=0$ has a repeated root $\alpha$ , which is also a root of the equation, $x^{2}-2bx-10=0$ . If $\beta$ is the other root of this equation, then $\alpha ^{2}+\beta ^{2}$ is equal to:
Option: 1 $24$
Option: 2 $25$
Option: 3 $26$
Option: 4 $28$

Nature of Roots -

Let the quadratic equation is ax² + bx + c = 0

D is the discriminant of the equation.

iii) if roots D = 0, then roots will be real and equal, then

$\\\mathrm{x_1=x_2 = \frac{-b}{2a} } \\\mathrm{Then, \;\; ax^2+bx +c =a(x-x_1)^2 }$

ax² – 2bx + 5 = 0 having equal roots or $D=0$ and $\alpha=\frac{b}{a}$

$(2b)^2=4\times5\times a\;\;\Rightarrow \;\;b^2=5a$

Put $\alpha=\frac{b}{a}$ in the second equation

${x^{2}-2 b x-10=0} \\ {\Rightarrow b^{2}-2 a b^{2}-10 a^{2}=0}$

$\\\Rightarrow 5 a-10 a^{2}-10 a^{2}=0 \\ \Rightarrow 20 a^{2}=5 a \\ \Rightarrow a=\frac{1}{4} \text { and } \mathrm{b}^{2}=\frac{5}{4} \\ \alpha^{2}= 20 \text { and } \beta^{2}=5 \\ \alpha^{2}+\beta^{2} \\ =5+20 \\ =25$

Correct Option 2

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avinash.dongre

The number of real roots of the equation, $e^{4x}+e^{3x}-4e^{2x}+e^{x}+1=0$ is :
Option: 1 $3$
Option: 2 4
Option: 3 1
Option: 4 2

Transcendental function -

Transcendental functions: the functions which are not algebraic are called transcendental functions. Exponential, logarithmic, trigonometric and inverse trigonometric functions are transcendental functions.

Exponential Function: function f(x) such that $\mathrm{f(x)=a^x}$ is known as an exponential function.

$\\\mathrm{base:\;\;a>0,a\neq1}\\\mathrm{domain:x\in \mathbb{R}}\\\mathrm{range:f(x)>0}$

Logarithmic function: function f(x) such that $f\left ( x \right )= \log\: _{a}x$ is called logarithmic function

$\\\mathrm{base:\;\;a>0,a\neq1}\\\mathrm{domain:x>0}\\\mathrm{range:f(x)\in\mathbb{R}}$

If a > 1 If a < 1

Properties of Logarithmic Function

$\\\mathrm1.\;{\log_e(ab)=\log_ea+\log_eb}\\\mathrm{2.\;\log_e\left ( \frac{a}{b} \right )=\log_ea-\log_e b}\\\mathrm{3.\;\log_ea^m=m\log_ea}\\\mathrm{4.\;\log_aa=1}\\\mathrm{5.\;\log_{b^m}a=\frac{1}{m}\log_ba}\\\mathrm{6.\;\log_ba=\frac{1}{\log_ab}}\\\mathrm{7.\;\log_ba=\frac{\log_ma}{\log_mb}}\\\mathrm{8.\;a^{\log_am}=m}\\\mathrm{9.\;a^{\log_cb}=b^{\log_ca}}\\\mathrm{10.\;\log_ma=b\Rightarrow a=m^b}$

Quadratic Equation -

The root of the quadratic equation is given by the formula:

$\\\mathrm{x = \frac{-b \pm \sqrt{D}}{2a}}\\\\\mathrm{or} \\\mathrm{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}$

Where D is called the discriminant of the quadratic equation, given by $D = b^2 - 4ac$ ,

Let $e^{x}=t \in(0, \infty)$

Now the equation

$\begin{array}{l}{t^{4}+t^{3}-4 t^{2}+t+1=0} \\ {t^{2}+t-4+\frac{1}{t}+\frac{1}{t^{2}}=0} \\ {\left(t^{2}+\frac{1}{t^{2}}\right)+\left(t+\frac{1}{t}\right)-4=0}\end{array}$

Let $\mathrm{t}+\frac{1}{\mathrm{t}}=\alpha$

$\begin{array}{l}{\left(\alpha^{2}-2\right)+\alpha-4=0} \\ {\alpha^{2}+\alpha-6=0} \\ {\alpha^{2}+\alpha-6=0}\end{array}$

$\alpha=-3,2$

Only positive value possible so $\alpha=2 \Rightarrow \quad \mathrm{e}^{x}+\mathrm{e}^{-\mathrm{x}}=2$

x=0 is the only solution.

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avinash.dongre

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The least positive value of 'a' for which the equation, $2x^{2}+(a-10)x+\frac{33}{2}=2a$ has real roots is
Option: 1 8
Option: 2 6
Option: 3 4
Option: 42

Nature of Roots -

Let the quadratic equation is ax² + bx + c = 0

D is the discriminant of the equation.

ii) If D > 0, then roots will be real and distinct.

$\\\mathrm{x_1 = \frac{-b + \sqrt{D}}{2a} } \;\mathrm{and \;\;x_2 = \frac{-b - \sqrt{D}}{2a} } \\\\\mathrm{Then,\;\; ax^2+bx +c =a(x-x_1)(x-x_2) }$

iii) if roots D = 0, then roots will be real and equal, then

$\\\mathrm{x_1=x_2 = \frac{-b}{2a} } \\\mathrm{Then, \;\; ax^2+bx +c =a(x-x_1)^2 }$

${D \geqslant 0} \\\\ {(a-10)^{2}-8\left(\frac{33}{2}-2 a\right) \geq 0} \\\\ {a^{2}+100-20 a-132+16 a \geq 0}$

$\\ {a^{2}-4 a-32 \geqslant 0} \\\\ {a^{2}-8 a+4 a-32 \geq 0} \\\\ {(a+4)(a-8) \geq 0}$

$a \leq -4 \ \text{ or }\ a \geq \ 8$

least positive value is 8.

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Kuldeep Maurya

If the equation, $x^{2}+bx+45=0(b\epsilon R)$ has conjugate complex roots and they satisfy $\left | z+1 \right |=2\sqrt{10}$ , then:

Option: 1 $b^{2}+b=12$
Option: 2 $b^{2}-b=42$
Option: 3 $b^{2}-b=30$
Option: 4 $b^{2}+b=72$

Nature of Roots -

Let the quadratic equation is ax² + bx + c = 0

D is the discriminant of the equation.

i) if D < 0, then root are in the form of complex number,

If a,b,c ∈ R (real number) then roots will be conjugate of each other, means if p + iq is one of

the roots then other root will be p - iq

Let $z=\alpha\pm i\beta$ be roots of the equation

so $2 \alpha=-b \text { and } \alpha^{2}+\beta^{2}=45,(\alpha+1)^{2}+\beta^{2}=40$

So, $(\alpha+1)^{2}-\alpha^{2}=-5$

$\begin{array}{l}{\Rightarrow \quad 2 \alpha+1=-5 \quad \Rightarrow \quad 2 \alpha=-6} \\ {\text { so } \mathrm{b}=6}\end{array}\\Hence,\;\;b^2-b=30$

Correct Option 3

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Kuldeep Maurya

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Let $y=y(x)$ be a solution of the differential equation, $\sqrt{1-x^{2}}\frac{dy}{dx}+\sqrt{1-y^{2}}=0,\left | x \right |<1.$ If $y\left ( \frac{1}{2} \right )=\frac{\sqrt{3}}{2},$ then $y\left ( \frac{-1}{\sqrt{2}} \right )$ is equal to :
Option: 1 $-\frac{1}{\sqrt{2}}$
Option: 2 $-\frac{\sqrt{3}}{2}$
Option: 3 $\frac{1}{\sqrt{2}}$
Option: 4 $\frac{\sqrt{3}}{2}$

Formation of Differential Equation and Solutions of a Differential Equation -

This is the general solution of the differential equation (2), which represents the family of the parabola (when a = 1) and one member of the family of parabola is given in Eq (1).

Also, Eq (1) is a particular solution of the differential equation (2).

The solution of the differential equation is a relation between the variables of the equation not containing the derivatives, but satisfying the given differential equation.

A general solution of a differential equation is a relation between the variables (not involving the derivatives) which contains the same number of the arbitrary constants as the order of the differential equation.

Particular solution of the differential equation obtained from the general solution by assigning particular values to the arbitrary constant in the general solution.

$\sqrt{1-x^{2}} \frac{d y}{d x}=-\sqrt{1-y^{2}}$

${\frac{d y}{\sqrt{1-y^{2}}}=-\frac{d x}{\sqrt{1-x^{2}}} \Rightarrow \sin ^{-1} y=-\sin ^{-1} x+c} \\ {x=\frac{1}{2}, y=\frac{\sqrt{3}}{2} \Rightarrow \frac{\pi}{2}=-\frac{\pi}{6}-c}$

$\begin{aligned} \sin ^{-1} y &=\frac{\pi}{2}-\sin ^{-1} x \\ &=\cos ^{-1} x \Rightarrow y=\sin (\cos^{-1}x) \end{aligned}$

$y\left ( \frac{1}{\sqrt{2}} \right ) =\frac{1}{\sqrt2}$

Correct Option (3)

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Kuldeep Maurya

Let $\alpha$ and $\beta$ be two real roots of the equation $(k+1)\tan ^{2}x-\sqrt{2}\cdot\lambda \tan x=(1-k),$ where, $k(\neq-1 )$ and $\lambda$ are real numbers. If $\tan^2 (\alpha +\beta )=50,$ then a value of $\lambda$ is :
Option: 1 $5\sqrt{2}$
Option: 2 $10\sqrt{2}$
Option: 3 $10$
Option: 4 $5$

As we have learnt,

Sum of roots:

$\\\mathrm{\alpha + \beta =\frac{-b}{a}}$

Product of roots:

$\alpha \cdot \beta = \frac{c}{a}$

Trigonometric Ratio for Compound Angles (Part 2)

$\\\mathrm{\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}}\\\\\mathrm{\tan (\alpha-\beta)=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \tan \beta}}$

Now,

$\\\tan \alpha + \tan{\beta } = \frac{\sqrt{2} \lambda}{1+k}\\\\\tan \alpha \times \tan{\beta } = \frac{k-1}{1+k}$

Since $\tan \alpha \& \tan \beta$ are the roots of the given equation

$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan{\beta }}{1-\tan{\alpha }\tan{\beta }}$ $= \frac{\frac{\sqrt{2}\lambda }{1+k}}{1- \frac{k-1}{k+1}} =\frac{\lambda}{\sqrt2}$

Now,

$\\ {\tan ^{2}(\alpha+\beta)=\frac{\lambda^{2}}{2}=50} \\ {\lambda=10}$

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Kuldeep Maurya

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Let $\alpha$ be a root of the equation $x^{2}+x+1=0$ and the matrix $A=\frac{1}{\sqrt{3}}\begin{bmatrix} 1 & 1 & 1\\ 1 & \alpha & \alpha ^{2}\\ 1 & \alpha ^{2}& \alpha ^{4} \end{bmatrix},$ Then the matrix $A^{31}$ is equal to :
Option: 1 $A$
Option: 2 $A^{2}$
Option: 3 $A^{3}$
Option: 4 $I_{3}$

Cube roots of unity -

z is a complex number

Let z^₃ = 1

⇒ z^₃ - 1 = 0

⇒ (z - 1)(z^₂ + z + 1) = 0

⇒ z - 1 = 0 or z^₂ + z + 1 = 0

$\\\mathrm{\therefore \;z=1\;\;or\;\;z=\frac{-1\pm\sqrt{(1-4)}}{2}=\frac{-1\pm i\sqrt{3}}{2}} \\\mathrm{Therefore, \;\mathbf{z=1},\;\mathbf{z=\frac{-1+ i\sqrt{3}}{2}}\;\;and\;\;\mathbf{z=\frac{-1- i\sqrt{3}}{2}}}$

If the second root is represented by ?, then the third root will be represented by ?^₂.

$\\\mathrm{\mathbf{\omega=\frac{-1+ i\sqrt{3}}{2}},\;\;\omega^2=\mathbf{\frac{-1- i\sqrt{3}}{2}}}$

Properties of cube roots:

i) 1 + ? + ?² = 0 and ?³ = 1

ii) to find ?ⁿ , first we write ? in multiple of 3 with remainder belonging to 0,1,2 like n=3q + r

Where r is from 0,1,2. Now ?ⁿ = ?^{3q + r} = (?³)^q·?^r = ?^r.

Multiplication of two matrices -

Matrix multiplication:

Two matrices A and B are conformable for the product AB if the number of columns in A and the number of rows in B is equal. Otherwise, these two matrices will be non-conformable for matrix multiplication. So on that basis,

i) AB is defined only if col(A) = row(B)

ii) BA is defined only if col(B) = row(A)

$\\\mathrm{A = \left [ a_{ij} \right ]_{m\times n}} \\\mathrm{\\B=\left [ b_{ij} \right ]_{n\times p}}$

$\\\mathrm{C = AB = \left [ c_{ij} \right ]_{m\times p}} \\\mathrm{Where\;\; c_{ij} = \sum_{j=1}^{n}a_{ij}b_{jk}, 1\leq i\leq m,1\leq k\leq p} \\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=a_{i1}b_{1k} + a_{i2}b_{2k} + a_{i3}b_{3k}+ ... + a_{in}b_{nk}}$

For examples

$\\\mathrm{Suppose,\;two\;matrices\;are\;given}\\\mathrm{A=\begin{bmatrix} a_{11} &a_{12} &a_{13} \\ a_{21} &a_{22} & a_{33} \end{bmatrix}_{2\times3}\;\;\;and\;\;\;B=\begin{bmatrix} b_{11}& b_{12} &b_{13} \\b_{21} &b_{22} &b_{23} \\b_{31} &b_{32} &b_{33} \end{bmatrix}_{3\times3}}\\\\\mathrm{To\:obtain\:the\:entries\:in\:row\:\mathit{i}\:of\:AB,\:we\:multiply\:the\:entries\:in\:row\:\mathit{i}\:of\:A\:by\:}\\\mathrm{column\:\mathit{j}\:in\:B\:and\:add.}\\\mathrm{given\:matrices\:A\:and\:B,\:where\:the\:order\:of\:A\:are\:2\times3\:and\:the\:order\:of\:B\:are\:3\times3,}\\\mathrm{the\:product\:of\:AB\:will\:be\:a\:2\times3\:matrix.}\\\\\mathrm{To\:obtain\:the\:entry\:in\:row\:1,\:column\:1\:of\:AB,\:multiply\:the\:}\\\mathrm{first\:row\:in\:A\:by\:the\:first\:column\:in\:B,and\:add.}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\begin{bmatrix} a_{11} &a_{12} &a_{13} \end{bmatrix}\begin{bmatrix} b_{11}\\b_{21} \\b_{31} \end{bmatrix}=a_{11}\cdot b_{11}+a_{12}\cdot b_{21}+a_{13}\cdot b_{31}}$

$\\\mathrm{To\:obtain\:the\:entry\:in\:row\:1,\:column\:2\:of\:AB,\:multiply\:the\:}\\\mathrm{first\:row\:in\:A\:by\:the\:second\:column\:in\:B,and\:add.}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\begin{bmatrix} a_{11} &a_{12} &a_{13} \end{bmatrix}\begin{bmatrix} b_{12}\\b_{22} \\b_{32} \end{bmatrix}=a_{11}\cdot b_{12}+a_{12}\cdot b_{22}+a_{13}\cdot b_{32}}\\\\\mathrm{To\:obtain\:the\:entry\:in\:row\:1,\:column\:3\:of\:AB,\:multiply\:the\:}\\\mathrm{first\:row\:in\:A\:by\:the\:thired\:column\:in\:B,and\:add.}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\begin{bmatrix} a_{11} &a_{12} &a_{13} \end{bmatrix}\begin{bmatrix} b_{13}\\b_{23} \\b_{33} \end{bmatrix}=a_{11}\cdot b_{13}+a_{12}\cdot b_{23}+a_{13}\cdot b_{33}}\\\\\mathrm{We\:proceed\:the\:same\:way\:to\:obtain\:the\:second\:row\:of\:AB.\:In\:other\:words,\:}\\\mathrm{row\:2\:of\:A\:times\:column\:1\:of\:B;}\\\mathrm{row\:2\:of\:A\:times\:column\:2\:of\:B;}\\\mathrm{row\:2\:of\:A\;times\:column\:3\:of\:B.}$

$\\\mathrm{When\:complete,\:the\:product\:matrix\:will\:be}\\\\\mathrm{AB=\begin{bmatrix} a_{11}\cdot b_{11}+a_{12}\cdot b_{21}+a_{13}\cdot b_{31}\;\;& a_{11}\cdot b_{12}+a_{12}\cdot b_{22}+a_{13}\cdot b_{32}\;\; &a_{11}\cdot b_{13}+a_{12}\cdot b_{23}+a_{13}\cdot b_{33} \\ a_{21}\cdot b_{11}+a_{22}\cdot b_{21}+a_{23}\cdot b_{31} \;\;& a_{21}\cdot b_{12}+a_{22}\cdot b_{22}+a_{23}\cdot b_{32} \;\;& a_{21}\cdot b_{13}+a_{22}\cdot b_{23}+a_{23}\cdot b_{33} \end{bmatrix}}$

Given

$\\x^2+x+1=0\\\Rightarrow \alpha\;is\;the\;root\;of\;the\;eq\\\alpha=\omega,\;\;\omega^2\\\text{A}=\frac{1}{\sqrt3}\begin{bmatrix} 1 &1 &1 \\ 1& \omega &\omega^2 \\ 1 &\omega^2 &\omega \end{bmatrix}\\\text{A}^2=\frac{1}{3}\begin{bmatrix} 3 &0 &0 \\ 0& 0 &3 \\ 0 &3 &0 \end{bmatrix}$

$\\\text{A}^4=\frac{1}{3}\begin{bmatrix} 3 &0 &0 \\ 0& 0 &3 \\ 0 &3 &0 \end{bmatrix}\times \frac{1}{3}\begin{bmatrix} 3 &0 &0 \\ 0& 0 &3 \\ 0 &3 &0 \end{bmatrix}\\\text{A}^4=I\\\therefore A^{31}=A^{28}\cdot A^3=A^3$

Correct option (3)

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Ritika Jonwal

If Re $\left [ \frac{z-1}{2z+i} \right ]=1,$ where $z=x+iy,$ then the point $(x,y)$ lies on a :

Option: 1 circle whose centre is at $\left [ -\frac{1}{2},-\frac{3}{2} \right ].$

Option: 2 straight line whose slope is $\frac{3}{2}.$

Option: 3 circle whose diameter is $\frac{\sqrt{5}}{2}.$

Option: 4 straight line whose slope is $-\frac{2}{3}.$

Conjugate of complex numbers and their properties -

The complex conjugate of a complex number a + ib (a, b are real numbers and b ≠ 0) is a − ib.

It is denoted as $\bar{z}$ .

i.e. if z = a + ib, then its conjugate is $\bar{z}$ = a - ib.

Conjugate of complex numbers is obtained by changing the sign of the imaginary part of the complex number. The real part of the number is left unchanged.

Note:

When a complex number is added to its complex conjugate, the result is a real number. i.e. z = a + ib, $\bar{z}$ = a - ib

Then the sum, z + $\bar{z}$ = a + ib + a - ib = 2a (which is real)

When a complex number is multiplied by its complex conjugate, the result is a real number i.e. z = a + ib, $\bar{z}$ = a - ib

Then the product, z? $\bar{z}$ = (a + ib)?(a - ib) = a² - (ib)²

= a² + b²(which is real)

Circle(Definition) -

General Form:

The equation of a circle with centre at (h,k) and radius r is

$\\ {\Rightarrow(x-h)^{2}+(y-k)^{2}=r^{2}} \\ {\Rightarrow x^{2}+y^{2}-2 h x-2 k y+h^{2}+k^{2}-r^{2}=0\;\;\;\;\;\;\;\;\;\;\;\ldots(i)} \\ {\text { Which is of the form : }} \\ {\mathbf{x}^{2}+\mathbf{y}^{2}+2 \mathbf{g} \mathbf{x}+2 \mathbf{f} \mathbf{y}+\mathbf{c}=\mathbf{0}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(ii)}$

This is known as the general equation of the circle.

Compare eq (i) and eq (ii)

h = -g, k = -h and c=h²+k²-r²

Coordinates of the centre (-g,-f)

Radius =g²+f²-c

$\\Re\left ( \frac{z-1}{2z+i} \right )=1$

$\\\text{put}\;z=x+iy$

$\operatorname{Re}\left(\frac{(x+i y)-1}{2(x+i y)+i}\right)=1$

$\operatorname{Re}\left(\left(\frac{(x-1)+i y}{2 x+i(2 y+1)}\right)\left(\frac{2 x-i(2 y+1}{2 x-i(2 y+1)}\right)\right)=1$

$\Rightarrow 2 x^{2}+2 y^{2}+2 x+3 y+1=0$

$x^{2}+y^{2}+x+\frac{3}{2} y+\frac{1}{2}=0$

$\Rightarrow $ locus is a circle whose$

$\text {Centre is }\left(-\frac{1}{2},-\frac{3}{4}\right)\text { and radius } \frac{\sqrt{5}}{4}.$

$\Rightarrow \text { diameter }=\frac{\sqrt{5}}{2}$

Correct Option (3)

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Ritika Jonwal

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All Questions

If then a value of satisfying is : Option: 1 Option: 2 Option: 3 Option: 4

Posted by

Nalla mahalakshmi

If be a complex number satisfying then cannot be : Option: 1 Option: 2 Option: 3 Option: 4

Posted by

Shravani.D.K

Crack CUET with india's "Best Teachers"

Let be such that the equation, has a repeated root , which is also a root of the equation, . If is the other root of this equation, then is equal to: Option: 1 Option: 2 Option: 3 Option: 4

Posted by

avinash.dongre

The number of real roots of the equation, is : Option: 1 Option: 2 4 Option: 3 1 Option: 4 2

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avinash.dongre

JEE Main high-scoring chapters and topics

The least positive value of 'a' for which the equation, has real roots is Option: 1 8 Option: 2 6 Option: 3 4 Option: 42

If $\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{xy}{x^{2}+y^{2}};y(1)=1;$ then a value of $x$ satisfying $y(x)=e$ is :
Option: 1 $\sqrt{3}\: e$

Option: 2 $\frac{1}{2}\sqrt{3}\: e$

Option: 3 $\sqrt{2}\: e$

Option: 4 $\frac{e}{\sqrt{2}}$

If $z$ be a complex number satisfying $\left | Re\left ( z \right ) \right |+\left | Im(z) \right |=4,$ then $\left | z \right |$ cannot be :
Option: 1 $\sqrt{7}$

Option: 2 $\sqrt{\frac{17}{2}}$

Option: 3 $\sqrt{10}$

Option: 4 $\sqrt{8}$

The number of real roots of the equation, $e^{4x}+e^{3x}-4e^{2x}+e^{x}+1=0$ is :
Option: 1 $3$
Option: 2 4
Option: 3 1
Option: 4 2

The least positive value of 'a' for which the equation, $2x^{2}+(a-10)x+\frac{33}{2}=2a$ has real roots is
Option: 1 8
Option: 2 6
Option: 3 4
Option: 42

If the equation, $x^{2}+bx+45=0(b\epsilon R)$ has conjugate complex roots and they satisfy $\left | z+1 \right |=2\sqrt{10}$ , then:

Option: 1 $b^{2}+b=12$
Option: 2 $b^{2}-b=42$
Option: 3 $b^{2}-b=30$
Option: 4 $b^{2}+b=72$

Let $\alpha$ be a root of the equation $x^{2}+x+1=0$ and the matrix $A=\frac{1}{\sqrt{3}}\begin{bmatrix} 1 & 1 & 1\\ 1 & \alpha & \alpha ^{2}\\ 1 & \alpha ^{2}& \alpha ^{4} \end{bmatrix},$ Then the matrix $A^{31}$ is equal to :
Option: 1 $A$
Option: 2 $A^{2}$
Option: 3 $A^{3}$
Option: 4 $I_{3}$