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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

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

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

32.

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

f(x) = 0 ને એક ઋણ વાસ્તવિક બીજ મળે.

Tips: -

az² + z + 1 = 0 જ્યાં a = cosθ + i sin θ

ધારો કે શુદ્વ કાલ્પનિક સંખ્યા z₁ અને z₂ એ આ સમીકરણનાં બીજ છે.

$bold therefore bold space bold z subscript bold 1 bold space bold equals bold space bold minus stack bold z subscript bold 1 with bold bar on top$ અને $bold z subscript bold 2 bold space bold equals bold minus stack bold z subscript bold 2 with bold bar on top$

બંને બીજનો સરવાળો $bold z subscript bold 1 bold space bold plus bold space bold z subscript bold 2 bold space bold equals bold space bold minus bold 1 over bold a bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold. bold. bold. bold space bold left parenthesis bold 1 bold right parenthesis$
$bold therefore bold space stack bold z subscript bold 1 with bold bar on top bold space bold plus bold space stack bold z subscript bold 2 with bold bar on top bold space bold equals bold space bold 1 over bold a bold therefore bold space stack bold z subscript bold 1 with bold bar on top bold space bold plus bold space bold space stack bold z subscript bold 2 with bold bar on top bold space bold equals bold space bold minus bold 1 over bold a$

$bold therefore bold space bold minus bold 1 over bold a bold space bold equals bold space bold 1 over bold a bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold therefore bold space bold 1 over bold a bold space bold plus bold space bold 1 over bold a bold space bold equals bold space bold 0 bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold therefore bold space bold cos bold space bold theta bold space bold equals bold space bold 0 bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold left parenthesis bold a bold space bold equals bold space bold cos bold space bold theta bold space bold plus bold space bold i bold space bold sin bold space bold theta bold right parenthesis bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold therefore bold space bold theta bold space bold equals bold space bold pi over bold 2$

હવે f(x) = 0

$bold therefore bold space bold x to the power of bold 3 bold space bold minus bold space bold 3 bold x to the power of bold 2 bold space bold plus bold space bold 3 bold space bold left parenthesis bold 1 bold space bold plus bold space bold cos bold pi over bold 2 bold right parenthesis bold space bold x bold space bold plus bold space bold 5 bold space bold equals bold space bold 0 bold space bold therefore bold space bold x to the power of bold 3 bold space bold minus bold space bold 3 bold x to the power of bold 2 bold space bold plus bold space bold 3 bold x bold space bold plus bold space bold 5 bold space bold equals bold space bold 0 bold space bold therefore bold space bold left parenthesis bold x bold space bold minus bold space bold 1 bold right parenthesis to the power of bold 3 bold space bold plus bold space bold 6 bold space bold equals bold space bold 0 bold space bold space bold આથ ી bold space bold x bold space bold equals bold space bold 1 bold space bold plus bold space bold left parenthesis bold minus bold 6 bold right parenthesis to the power of begin inline style bold 1 over bold 3 end style end exponent$

$bold therefore bold space bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space bold 0 bold space$ ને એક ઋણ વાસ્તવિક બીજ મળે.

33.

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

34. 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

35.

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}

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

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