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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

31.

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

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

3
0
-3
1

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

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

34.

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}

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

1
2ⁿ
2^n-1
3^n-1

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

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

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

Tips: -

az² + z + 1 = 0

ધારો કે z₁ એ આ સમીકરણનું શુદ્વ કાલ્પનિક બીજ છે.

∴ az₁² + z₁ + 1 = 0 ...(1)

$bold therefore bold space top enclose bold a bold space open parentheses top enclose bold z subscript bold 1 end enclose close parentheses to the power of bold 2 bold space bold plus bold space top enclose bold z subscript bold 1 end enclose bold space bold plus bold space bold 1 bold space bold equals bold space bold 0 bold space bold therefore bold space top enclose bold a bold space bold z subscript bold 1 to the power of bold 2 bold space bold minus bold space bold z subscript bold 1 bold space bold plus bold space bold 1 bold space bold equals bold space bold 0 bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold left parenthesis top enclose bold z subscript bold 1 end enclose bold space bold equals bold space bold minus bold z subscript bold 1 bold right parenthesis bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold left parenthesis bold 2 bold right parenthesis$

સમીકરણ (1) અને (2) માંથી (z₁) નો લોપ કરતાં,

$bold therefore bold space fraction numerator bold z subscript bold 1 superscript bold 2 over denominator bold 2 end fraction bold space bold equals bold space fraction numerator bold minus bold z subscript bold 1 over denominator bold a bold minus bold a end fraction bold space bold equals bold space fraction numerator bold 1 over denominator bold minus bold left parenthesis bold a bold plus top enclose bold a bold right parenthesis end fraction bold therefore bold space bold z subscript bold 1 to the power of bold 2 bold space bold equals bold space open parentheses fraction numerator bold a bold space bold minus bold space top enclose bold a over denominator bold a bold space bold plus bold space top enclose bold a end fraction close parentheses to the power of bold 2 bold space bold equals bold space fraction numerator bold minus bold 2 over denominator bold a bold plus top enclose bold a end fraction bold therefore bold space bold left parenthesis top enclose bold a bold space bold minus bold space bold a bold right parenthesis to the power of bold 2 bold space bold plus bold space bold 2 bold space bold left parenthesis bold a bold space bold plus bold space top enclose bold a bold right parenthesis bold space bold equals bold space bold 0$

ધારો કે a = cos θ + i sin θ (|a| = 1)

$bold therefore bold space bold left parenthesis bold minus bold 2 bold space bold i bold space bold sin bold space bold right parenthesis to the power of bold 2 bold space bold plus bold space bold 2 bold space bold left parenthesis bold 2 bold space bold cos bold space bold right parenthesis bold space bold equals bold space bold 0 bold space bold therefore bold space bold minus bold 4 bold left parenthesis bold 1 bold minus bold cos to the power of bold 2 bold space bold right parenthesis bold space bold plus bold space bold 4 bold space bold cos bold space bold space bold equals bold space bold 0 bold therefore bold space bold cos to the power of bold 2 bold space bold space bold plus bold space bold cos bold space bold theta bold space bold minus bold space bold 1 bold space bold equals bold space bold 0 bold therefore bold space bold cos bold space bold theta bold space bold equals bold space fraction numerator bold minus bold 1 bold plus square root of bold 1 bold plus bold 4 end root over denominator bold 2 end fraction bold space bold equals bold space fraction numerator bold minus bold 1 bold plus-or-minus square root of bold 5 over denominator bold 2 end fraction$

∴ cos θ ની કિંમત $fraction numerator bold minus bold 1 bold space bold plus bold space square root of bold 5 over denominator bold 2 end fraction$ ન હોઈ શકે.
$bold therefore bold space bold cos bold space bold theta bold space bold equals bold space fraction numerator square root of bold 5 bold space bold minus bold space bold 1 over denominator bold 2 end fraction bold therefore bold space bold theta bold space bold equals bold space bold cos to the power of bold minus bold 1 end exponent bold space open parentheses fraction numerator square root of bold 5 bold minus bold 1 over denominator bold 2 end fraction close parentheses$

37.

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

38. 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

39. 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

40.

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

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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

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