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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

31. જો z₁ = 24 + 7i અને |z₂| = 6 તો |z₁+z₂| નું મહત્તમ તથા ન્યુનતમ મૂલ્ય અનુક્રમે ........ થાય.

24, 7
25, 19
31, 29
31, 19

32.

bold e to the power of bold i bold space bold left parenthesis bold 2 bold k bold space bold cot to the power of bold minus bold 1 bold space bold m bold right parenthesis end exponent end exponent bold space open parentheses fraction numerator bold mi bold space bold plus bold space bold 1 over denominator bold mi bold space bold minus bold space bold 1 end fraction close parentheses to the power of bold k bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold space bold space bold જ ્ ય ાં bold space bold k bold space bold element of bold space bold Z bold comma bold space bold m bold space bold greater than bold space bold 0

1
-1
2
0

33. જો z₁, z₂, z₃ ∈ C, θ₁ $bold sum bold z subscript bold 1 bold space bold Im bold space bold left parenthesis stack bold z subscript bold 2 with bold bar on top bold space bold z subscript bold 3 bold right parenthesis bold space bold equals bold space bold. bold. bold. bold. bold. bold space$

$fraction numerator bold 1 over denominator bold 2 bold space bold i end fraction bold space bold left parenthesis bold z subscript bold 1 bold space bold plus bold space bold z subscript bold 2 bold space bold plus bold space bold z subscript bold 3 bold right parenthesis$
z₁, z₂, z₃
z₁ + z₂ + z₃
0

34. જો cos (1 - i) = a + ib જ્યાં, a, b ∈ R તો a = ......... b = .........

$bold 1 over bold 2 open parentheses bold e bold minus bold 1 over bold e close parentheses bold space bold cos bold space bold 1 bold comma bold space bold 1 over bold 2 bold space open parentheses bold e bold minus bold 1 over bold e close parentheses bold space bold sin bold space bold 1$
$bold 1 over bold 2 open parentheses bold e bold minus bold 1 over bold e close parentheses bold space bold cos bold space bold 1 bold comma bold space bold 1 over bold 2 bold space open parentheses bold e bold plus bold 1 over bold e close parentheses bold space bold sin bold space bold 1$
$bold 1 over bold 2 open parentheses bold e bold plus bold 1 over bold e close parentheses bold space bold cos bold space bold 1 bold comma bold space bold 1 over bold 2 bold space open parentheses bold e bold minus bold 1 over bold e close parentheses bold space bold sin bold space bold 1$
$bold 1 over bold 2 open parentheses bold e bold minus bold 1 over bold e close parentheses bold space bold cos bold space bold 1 bold comma bold space bold 1 over bold 2 bold space open parentheses bold e bold plus bold 1 over bold e close parentheses bold space bold sin bold space bold 1$

35.

સમીકરણ az² + z + 1 = 0 નાં બીજ શુદ્વ કાલ્પનિક સંખ્યા છે, જ્યાં a = cos θ + i sin θ નીચેનામાંથી કયું વિધાન f(x) = x³ - 3x² + 3 (1 + cos θ) x + 5 માટે સત્ય છે ?

f(x) = 0 ને એક ઋણ વાસ્તવિક બીજ મળે.
f(x) = 0 ને ત્રણ વાસ્તવિક બીજ મળે.
f(x) =0 ને એક ધન વાસ્તવિક બીજ મળે.
f(x) = 0 ને ત્રણ બીજ મળે છે પણ બધા જ ભિન્ન ન હોય

36.

જો સમીકરણ 8x² - 10x + 3 = 0 બીજ α અને β² જ્યાં β² > $bold 1 over bold 2$ હોય, તો જેના બીજ (α + iβ)¹⁰⁰ અને (α - iβ)¹⁰⁰ હોય તેવું સમીકરણ .......

x² + x - 1 = 0
x² - x - 1 = 0
x² - x + 1 = 0
x² + x + 1 = 0

37. સંકર સંખ્યા a માટે |a| = 1 છે. જો az² + z + 1 = 0 ને એક શુદ્વ કાલ્પનિક બીજ હોય, તો a ની કિંમત ......

$bold a bold space bold equals bold space bold cos bold space bold theta bold space bold plus bold space bold sin bold space bold theta bold comma bold space bold theta bold space bold equals bold space bold cos to the power of bold minus bold 1 end exponent bold space open parentheses fraction numerator square root of bold 5 bold minus bold 1 over denominator bold 2 end fraction close parentheses$
$bold a bold space bold equals bold space bold sin bold space bold theta bold space bold plus bold space bold i bold space bold cos bold space bold theta bold comma bold space bold theta bold space bold equals bold space bold cos to the power of bold minus bold 1 end exponent bold space open parentheses fraction numerator square root of bold 5 bold minus bold 1 over denominator bold 2 end fraction close parentheses$
$bold a bold space bold equals bold space bold cos bold space bold theta bold space bold plus bold space bold i bold space bold sin bold space bold theta bold comma bold space bold theta bold space bold equals bold space bold cos to the power of bold minus bold 1 end exponent bold space open parentheses fraction numerator square root of bold 5 bold minus bold 1 over denominator bold 4 end fraction close parentheses$
$bold a bold space bold equals bold space bold sin bold space bold theta bold space bold plus bold space bold i bold space bold cos bold space bold theta bold comma bold space bold theta bold space bold equals bold space bold cos to the power of bold minus bold 1 end exponent bold space open parentheses fraction numerator square root of bold 5 bold minus bold 1 over denominator bold 4 end fraction close parentheses$

38.

w(I_m(w)≠0) એ સંકર સંખ્યા છે. સમીકરણ $bold w bold space bold minus bold space top enclose bold w bold space bold z bold space bold equals bold space bold k bold space bold left parenthesis bold 1 bold space bold minus bold space bold z bold right parenthesis bold comma bold space bold k bold space bold element of bold space bold R$ ના બિંદુગણનું સવરૂપ ....... થશે.

{z | |z| = 1, z ≠1 }
{z | |z| = 1 }
{z |z = z }
{z | z ≠ 1}

39. જો (1+ir)³ = s (1 + i) જ્યાં, r, s ∈ R તો r ની શક્ય કિંમતોનો સરવાળો ..... થાય.

3
0
-3
1

Tips: -

જો $bold cos bold space bold theta bold space bold equals bold space fraction numerator bold 1 over denominator square root of bold 1 bold plus bold r to the power of bold 2 end root end fraction$ અને $bold sin bold space bold theta bold space bold equals bold space fraction numerator bold r over denominator square root of bold 1 bold plus bold r to the power of bold 2 end root end fraction$ લઈએ તો,

આપેલ સમીકરણ પરથી $open square brackets square root of bold 1 bold plus bold r to the power of bold 2 end root bold space bold left parenthesis bold cos bold space bold theta bold space bold plus bold space bold i bold space bold sin bold space bold theta bold right parenthesis close square brackets to the power of bold 3 bold space bold equals bold space bold italic s bold space bold left parenthesis bold 1 bold space bold plus bold space bold italic i bold right parenthesis$

$bold therefore bold space open parentheses square root of bold 1 bold plus bold r to the power of bold 2 end root close parentheses to the power of bold 3 bold space bold left parenthesis bold cos bold space bold 3 bold theta bold space bold plus bold space bold i bold space bold sin bold space bold 3 bold theta bold right parenthesis bold space bold equals bold space bold s bold space bold left parenthesis bold 1 bold space bold plus bold space bold i bold right parenthesis open parentheses bold 1 bold plus bold r to the power of bold 2 close parentheses to the power of begin inline style bold 3 over bold 2 end style end exponent bold space bold cos bold space bold 3 bold theta bold space bold equals bold space bold s bold comma bold space open parentheses bold 1 bold plus bold r to the power of bold 2 close parentheses to the power of begin inline style bold 3 over bold 2 end style end exponent bold space bold sin bold space bold 3 bold theta bold therefore bold space bold cos bold space bold 3 bold theta bold space bold equals bold space bold sin bold space bold 3 bold theta bold therefore bold space bold 4 bold space bold cos to the power of bold 3 bold theta bold space bold minus bold space bold 3 bold space bold cos bold space bold theta bold space bold equals bold space bold 3 bold space bold sin bold space bold theta bold space bold minus bold space bold 4 bold space bold sin to the power of bold 3 bold space bold theta bold space bold therefore bold space bold 4 over open parentheses square root of bold 1 bold plus bold r to the power of bold 2 end root close parentheses to the power of bold 3 bold space bold minus bold space fraction numerator bold 3 over denominator square root of bold 1 bold plus bold r to the power of bold 2 end root end fraction bold space bold equals bold space fraction numerator bold 3 bold r over denominator square root of bold 1 bold plus bold r to the power of bold 2 end root end fraction bold space bold minus bold space fraction numerator bold 4 bold r to the power of bold 3 over denominator open parentheses square root of bold 1 bold plus bold r to the power of bold 2 end root close parentheses end fraction bold therefore bold space bold 4 bold space bold minus bold space bold 3 bold left parenthesis bold 1 bold space bold plus bold space bold r to the power of bold 2 bold right parenthesis bold space bold equals bold space bold 3 bold r bold space bold left parenthesis bold 1 bold space bold plus bold space bold r to the power of bold 2 bold right parenthesis bold space bold minus bold space bold 4 bold r to the power of bold 2 bold therefore bold space bold r to the power of bold 3 bold space bold minus bold space bold 3 bold r to the power of bold 2 bold space bold minus bold space bold 3 bold r bold space bold plus bold space bold 1 bold space bold equals bold space bold 0 bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space$

$bold therefore bold space bold r$ ની શક્ય કિંમતનો સરવાળો = ઉપરના સમીકરણનાં બીજનો સરવાળો = 3

40. જો (1 + x + x²)ⁿ = a₀ + a₁x + a₂x² + ... + a_2nx²ⁿ, તો a₀ + a₃ + a₆ + ... = .......

1
2ⁿ
2^n-1
3^n-1

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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

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