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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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Posted by

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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Posted by

Shravani.D.K

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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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Posted by

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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Posted by

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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Posted by

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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Posted by

Kuldeep Maurya

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The most abundant elements by mass in the body of a healthy human adult are : Oxygen (61.4%); Carbon (22.9%), Hydrogen (10.0%); and Nitrogen (2.6%). The weight (in kg) which a 75 kg person would gain if all ^{1}H atoms are replaced by ^{2}H atoms is :  
Option: 1 7.5
Option: 2 10
Option: 3 15
Option: 4 37.5
 

Given that

Mass of the person = 75 kg

Mass of 1H1 present in person = 10% of 75 kg = 7.5 kg

Since Mass of 1H2 is double the Mass of 1H1

So, Mass of 1H2 will be in person = 2 X 7.5 kg =15 kg

Thus, increase in weight = 15 - 7.5 = 7.5 kg

Therefore, Option (1) is correct

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Posted by

vishal kumar

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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Posted by

Kuldeep Maurya

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The correct order of the atomic radii of C, Cs, Al and S is :
Option: 1 S<C<Cs<Al
Option: 2 C<S<Al<Cs
Option: 3 C<S<Cs<Al
Option: 4 S<C<Al<Cs
 

 

Periodicity of atomic radius and ionic radius in period -

In a period from left to right the effective nuclear charge increases because the next electron fills in the same shell. So the atomic size decrease.

- wherein

Li>Be>B>C>N>O>F

 

 

 

Electronegativity and atomic radius -

The attraction between the outer electrons and the nucleus increases as the atomic radius decreases in a period.

- wherein

Electronegativity\propto\frac{1}{atomic\:radius}

 

 

 

Size of atom and ion in a group -

In a group moving from top to the bottom the number of shell increases.So the atomic size increases.

- wherein

Li<Na<K<Rb<Cs

As we know that

From Left to right in a period size decreases and when going down the group size increases

C< S< Al< Cs

Therefore, Option(2) is correct

  

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Posted by

Ritika Jonwal

The IUPAC symbol for the element with atomic number 119 would be:
Option: 1 uue
Option: 2une
Option: 3 unh
Option: 4 uun

 

Nomenclature of elements with atomic number >100 -

The name is derived directly from the atomic number of the element using the following numerical roots:

0 = nil

1 = un

2 = bi

3 = tri

4 = quad

5 = pent

6 = hex

7 = sept

8 = oct

9 = enn

Eg:

 

Atomic number

Name

Symbol

101

Mendelevium (Unnilunium)

Md (Unu)

102

Nobelium (Unnilbium)

No (Unb)



 

-

uue

1  1  9

Un Un ennium

Therefore, Option(1) is correct.

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Posted by

Ritika Jonwal

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