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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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Shravani.D.K

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

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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

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

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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

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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Let $\alpha =\frac{-1+i\sqrt{3}}{2}.$ If $a=(1+\alpha )\sum_{k=0}^{100}\alpha ^{2k}$ and $b=\sum_{k=0}^{100}\alpha ^{3k},$ then a and b are the roots of the quadratic equation :
Option: 1 $x^{2}+101x+100=0$
Option: 2 $x^{2}+102x+101=0$
Option: 3 $x^{2}-102x+101=0$
Option: 4 $x^{2}-101x+100=0$

Cube roots of unity

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

Sum of n-term of a GP

Let S_n be the sum of n terms of the G.P. with the first term ‘a’ and common ratio ‘r’. Then

$S_n=a( \frac{r^n-1}{r-1})$

Now,

$\alpha$ = $\omega$ ,

b = 1 + $\omega$ ³ + $\omega$ ⁶ + …… $\omega$ ³⁰⁰= 101

a= (1+ $\omega$ ) (1+ $\omega$ ²+ $\omega$ ⁴+ $\omega$ ⁶.....+ $\omega$ ²⁰⁰)

$a=(1+\omega)\frac{(\omega^{2(101)}-1)}{\omega^2-1}=\frac{1-\omega^2}{1-\omega^2}=1$

Now, equation with roots 1 and 101 is

x² – (1+101)x + 101*1 = 0

x² – 102x + 101 = 0

Correct Option (3)

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vishal kumar

Let $\alpha$ and $\beta$ be the roots of the equation $x^{2}-x-1=0.$ If $P_{k}=\left ( \alpha \right )^{k}+\left ( \beta \right )^{k},k\geq 1,$ then which one of the following statements is not true ?
Option: 1 $P_{1}+P_{2}+P_{3}+P_{_{4}}+P_{5} =26$
Option: 2 $P_{5}=11$
Option: 3 $P_{5}=P_{2}\cdot P_{3}$
Option: 4 $P_{3}=P_{5}- P_{4}$

Polynomial Equation of Higher Degree, Remainder theorem -

$\\\mathrm{An \;equation \;of \;the\; form\; a_0x^n+a_1x^{n-1}+...+a_{n-1}x+a_n=0, } \\\mathrm{where \; a_0, a_1,..., a_n \; are\; constant\; and \; a_0 \; \neq 0}$

Is known as the polynomial equation of degree n which have n and only n roots.

$\\\mathrm{sum \;of\; all\; roots=\sum \alpha_1 = \alpha_1 + \alpha_2 +...+\alpha_n-1+\alpha_n=(-1)\frac{a_1}{a_0}} \\\\\mathrm{sum\; of \; products\; taken \; two \; at \; a \; time\;} \\\mathrm{\;\;\;\sum \alpha_1\alpha_2=\alpha_1\alpha_2 + \alpha_1\alpha_3 + ... + \alpha_1\alpha_n + \alpha_2\alpha_3+...+\alpha_2\alpha_n+...+\alpha_{n-1}\alpha_n=(-1)^2\frac{a_2}{a_0}} \\ \\\mathrm{sum\; of \; products\; taken \; three \; at \; a \; time\;} \\\mathrm{\;\;\; \sum \alpha_1\alpha_2\alpha_3 = (-1)^3\frac{a_3}{a_0}} \\\\\mathrm{product\; of \; all \; roots = \alpha_1\alpha_2...\alpha_n = (-1)^n\frac{a_n}{a_0}}$

For example, suppose n = 3 and ax³ + bx² +cx + d = 0 is polynomial equation with a ≠ 0 and ?, ? and ? are the roots of the equation then :

$\\\mathrm{\alpha+\beta+\gamma = -\frac{b}{a}} \\\mathrm{\sum \alpha \beta =\alpha\beta+\beta\gamma+\gamma\alpha= (-1)^2\frac{c}{a}=\frac{c}{a}} \\\mathrm{\alpha\beta \gamma=(-1)^3\frac{d}{a}=-\frac{d}{a}}$

$\alpha+\beta=1 \text{ and }\alpha\timex\beta=-1$

$\\P_1=\alpha+\beta=1\\P_2=\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=1-2(-1)=3\\P_3=\alpha^3+\beta^3=(\alpha+\beta)(\alpha^2+\beta^2-\alpha\beta)=(1\times(3+1))=4\\P_4=\alpha^4+\beta^4=(\alpha^2+\beta^2)^2-2\alpha^2\beta^2=(P_2)^2-2(-1)^2=7$

$x^2=x+1$

$\\x^2=x+1\Rightarrow x^5=x^4+x^3\\ \alpha^5=\alpha^4+\alpha^3\text{ and }\beta^5=\alpha^4+\alpha^3$

$\alpha^5+\beta^5=P_4+P_3=11$

$\mathrm{P}_{5} \neq \mathrm{P}_{2} \times \mathrm{P}_{3}$

Correct option (3)

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All Questions

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

Posted by

Shravani.D.K

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

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

Posted by

avinash.dongre

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

Posted by

Kuldeep Maurya

JEE Main high-scoring chapters and topics

If the equation, has conjugate complex roots and they satisfy , then: Option: 1 Option: 2 Option: 3 Option: 4

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 $I_{3}$

Let $\alpha =\frac{-1+i\sqrt{3}}{2}.$ If $a=(1+\alpha )\sum_{k=0}^{100}\alpha ^{2k}$ and $b=\sum_{k=0}^{100}\alpha ^{3k},$ then a and b are the roots of the quadratic equation :
Option: 1 $x^{2}+101x+100=0$
Option: 2 $x^{2}+102x+101=0$
Option: 3 $x^{2}-102x+101=0$
Option: 4 $x^{2}-101x+100=0$