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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

41. Re (z)² = 0 અને |z| = a $square root of bold 2 bold space bold left parenthesis bold a bold space bold greater than bold space bold 0 bold right parenthesis$ સમીકરણને કેટલા ઉકેલ મળે ?

42.

a, b, c, a₁, b₁, c₁ એ શૂન્યેતર સંકર સંખ્યાઓ છે, જ્યાં $bold a over bold a subscript bold 1 bold space bold plus bold space bold b over bold b subscript bold 1 bold space bold plus bold space bold c over bold c subscript bold 1 bold space bold equals bold space bold 1$ અને $bold a subscript bold 1 over bold a bold space bold plus bold space bold b subscript bold 1 over bold b bold space bold plus bold space bold c subscript bold 1 over bold c bold space bold equals bold space bold 0 bold comma$ તો $bold a to the power of bold 2 over bold a subscript bold 1 to the power of bold 2 bold plus bold b to the power of bold 2 over bold b subscript bold 1 to the power of bold 2 bold plus bold c to the power of bold 2 over bold c subscript bold 1 to the power of bold 2 bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold space bold.$

2 + 2i
2i
2
1 + 2i

43. જો (cos θ + i sin θ) (cos2θ + i sin2θ) ... (cos n θ + i sin n θ) = 1 તો θ = .......

$fraction numerator bold 4 bold kπ over denominator bold n bold left parenthesis bold n bold plus bold 1 bold right parenthesis end fraction bold comma bold space bold n bold space bold element of bold space bold Z$
$fraction numerator bold 2 bold pi over denominator bold 2 bold left parenthesis bold n bold left parenthesis bold n bold plus bold 1 bold right parenthesis end fraction bold comma bold space bold k bold space bold element of bold space bold Z$
$fraction numerator bold kπ over denominator bold n bold left parenthesis bold n bold plus bold 1 bold right parenthesis end fraction bold comma bold space bold k bold space bold element of bold space bold Z$
આપેલ પૈકી એક પણ નહી

44.

જો z₁, z₂ અને z₃ ઘડિયાળની વિરુદ્વ દિશામાં દર્શાવેલ $bold increment bold ABC$ નાં શિરોબિંદુઓ છે. જો z₀ એ $bold increment bold ABC$ નું પરિકેન્દ્ર હોય, તો $open parentheses fraction numerator bold z subscript bold 0 bold minus bold z subscript bold 1 over denominator bold z subscript bold 0 bold minus bold z subscript bold 2 end fraction close parentheses fraction numerator bold sin bold 2 bold A over denominator bold sin bold 2 bold B end fraction bold plus bold space open parentheses fraction numerator bold z subscript bold 0 bold minus bold z subscript bold 3 over denominator bold z subscript bold 0 bold minus bold z subscript bold 2 end fraction close parentheses bold space fraction numerator bold sin bold 2 bold space bold C over denominator bold sin bold 2 bold space bold B end fraction$

2
1
-1
0

45.

જો α, β એ u² - 2u + 2 = 0 નાં બીજ હોય તથા cotθ = x + 1 હોય, તો $fraction numerator bold left parenthesis bold x bold plus bold alpha bold right parenthesis to the power of bold n bold minus bold left parenthesis bold x bold plus bold beta bold right parenthesis to the power of bold n over denominator bold alpha bold minus bold beta end fraction$ = ......

$fraction numerator bold sin bold space bold n bold space bold theta over denominator bold n bold space bold sin to the power of bold n bold space bold theta end fraction$
$fraction numerator bold sin bold space bold n bold space bold theta over denominator bold cos to the power of bold n bold space bold theta end fraction$
$fraction numerator bold sin bold space bold n bold space bold theta over denominator bold sin to the power of bold n bold space bold theta end fraction$
$fraction numerator cos space straight n space straight theta over denominator cos to the power of straight n space straight theta end fraction$

fraction numerator bold sin bold space bold n bold space bold theta over denominator bold sin to the power of bold n bold space bold theta end fraction

Tips: -

$bold u to the power of bold 2 bold space bold minus bold space bold 2 bold u bold space bold plus bold space bold 2 bold space bold equals bold space bold 0 bold. bold space$ આથી u = 1 ± i, ધારો કે α = 1 + i, β = 1 - i

ડા.બા = $fraction numerator bold left parenthesis bold x bold plus bold alpha bold right parenthesis to the power of bold n bold space bold minus bold space bold left parenthesis bold x bold plus bold beta bold right parenthesis to the power of bold n over denominator bold alpha bold minus bold beta end fraction bold space bold equals bold space fraction numerator bold left square bracket bold cot bold space bold theta bold minus bold 1 bold plus bold 1 bold plus bold i bold right square bracket to the power of bold n bold minus bold left square bracket bold cotθ bold minus bold 1 bold plus bold 1 bold minus bold i bold right square bracket to the power of bold n over denominator bold 2 bold i end fraction$

$bold equals bold space fraction numerator bold left parenthesis bold cos bold space bold theta bold space bold plus bold space bold i bold space bold sin bold space bold theta bold right parenthesis to the power of bold n bold minus bold left parenthesis bold cos bold space bold theta bold space bold minus bold space bold i bold space bold sin bold space bold theta bold right parenthesis to the power of bold n over denominator bold sin to the power of bold n bold theta bold times bold 2 bold i end fraction bold equals bold space fraction numerator bold 2 bold i bold space bold sin bold space bold nθ over denominator bold 2 bold i bold space bold sin bold theta presuperscript bold n end fraction bold space bold equals bold space fraction numerator bold sin bold space bold n bold space bold theta over denominator bold sin bold space bold theta presuperscript bold n end fraction$

46.

ધારો કે z = x + iy જ્યાં, x, y ∈ Z, સમીકરણ $bold z bold space bold z with bold bar on top to the power of bold 3 bold space bold plus bold space bold z with bold bar on top bold space bold z to the power of bold 3 bold space bold equals bold space bold 350$ ના બીજથી રચાતા લંબચોરસનું ક્ષેત્રફળ ......

47. જો |z| = 1 અને $bold w bold space bold equals bold space fraction numerator bold z bold minus bold 1 over denominator bold z bold plus bold 1 end fraction bold space bold left parenthesis bold z bold not equal to bold minus bold 1 bold right parenthesis$ તો Re(w)

$open vertical bar fraction numerator bold 1 over denominator bold z bold plus bold 1 end fraction close vertical bar fraction numerator bold 1 over denominator bold vertical line bold z bold plus bold 1 bold vertical line to the power of bold 2 end fraction$
0
$fraction numerator square root of bold 2 over denominator bold vertical line bold z bold plus bold 1 bold vertical line to the power of bold 2 end fraction$
$fraction numerator bold 1 over denominator bold vertical line bold z bold plus bold 1 bold vertical line to the power of bold 2 end fraction$

48. જો $bold e to the power of bold log bold space bold 2 to the power of begin inline style fraction numerator bold vertical line bold z bold vertical line to the power of bold 2 bold minus bold vertical line bold z bold vertical line bold plus bold 4 over denominator bold vertical line bold z bold vertical line to the power of bold 2 bold plus bold 1 end fraction end style end exponent end exponent bold space bold equals bold space bold log subscript square root of bold 2 end subscript bold left parenthesis bold 16 bold right parenthesis$ તો |z| = .......

$bold 1 over bold 4$
$bold 1 over bold 2$
$bold 1 over bold 3$
1

49. જો શુન્યેતર સંકર સંખ્યાઓ z₁, z₂ માટે $bold z subscript bold 1 over bold z subscript bold 2 bold plus bold z subscript bold 2 over bold z subscript bold 1 bold space bold equals bold space bold 1$ તો O(0), P(z₁) તથા Q(z₂) એ

સમબાજુ ત્રિકોણ બનાવે.
રેખા પરનાં બિંદુઓ છે.
કાટકોણ ત્રિકોણનાં શિરોબિંદુઓ છે.
એકમ વર્તુળ પર આવેલ છે.

50.

વર્તુળ |z| = 2 માં અંતર્ગત સમબાજુ ત્રિકોણનાં શિરોબિંદુઓ z₁, z₂, z₃ છે. z₁ = 1 + $square root of bold 3 bold space bold i$ જો તો z₂, z₃ અનુક્રમે ........ થાય.

$bold minus bold 2 bold comma bold space bold 2 bold space bold plus bold space square root of bold 3 bold space bold i$
$bold minus bold 2 bold comma bold space bold minus bold 1 bold space bold plus bold space square root of bold 3 bold space bold i$
$bold minus bold 2 bold comma bold space bold 1 bold space bold minus bold space square root of bold 3 bold space bold i$
$bold minus bold 1 bold space bold plus bold space square root of bold 3 bold space bold i bold comma bold space bold minus bold 2$

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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

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