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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

31. 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

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

1
2ⁿ
2^n-1
3^n-1

33.

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

34.

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

35.

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

36. 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

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

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

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

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

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

3
0
-3
1

40.

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}

{z | |z| = 1 }

Tips: -

ધારો કે w = a + ib, b ≠ 0 તથા z = x + iy,

w - $top enclose bold w$ z = k (1 - z)

∴ a + ib - (a - ib) (x + iy) = k (1 - x - iy)

∴ a + ib - (ax + ayi - bxi + by) = k (1 - x) - iky

∴ (a - ax - by) + i (b - ay + bx) = k (1 - x) - kyi

વાસ્તવિક અને કાલ્પનિક ભાગને સરખાવતાં,

$bold k bold space bold equals bold space fraction numerator bold a bold space bold minus bold space bold ax bold space bold minus bold space bold by over denominator bold 1 bold space bold minus bold space bold x end fraction bold comma bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold k bold space bold equals bold space fraction numerator bold ay bold space bold minus bold space bold bx bold space bold minus bold space bold b over denominator bold y end fraction$

∴ ay - axy - by² = ay - bx - b - ayx + bx² + bx

∴ -by² = bx² - b

∴ x² + y² = 1 (b ≠ 0)

આથી |z| = 1

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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

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