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CBSE

Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

41.

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.

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

42. 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.

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 |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
$fraction numerator square root of bold 2 over denominator bold vertical line bold z bold plus bold 1 bold vertical line to the power of bold 2 end fraction$
$fraction numerator bold 1 over denominator bold vertical line bold z bold plus bold 1 bold vertical line to the power of bold 2 end fraction$

Tips: -

$bold w bold space bold equals bold space fraction numerator bold z bold minus bold 1 over denominator bold z bold plus bold 1 end fraction$

$bold therefore bold space bold wz bold space bold plus bold space bold w bold space bold equals bold space bold z bold space bold minus bold space bold 1 bold space bold therefore bold space bold w bold space bold plus bold space bold 1 bold space bold equals bold space bold z bold space bold left parenthesis bold 1 bold space bold minus bold space bold w bold right parenthesis bold therefore bold space bold z bold space bold equals bold space fraction numerator bold w bold plus bold 1 over denominator bold 1 bold minus bold w end fraction bold therefore bold space bold vertical line bold z bold vertical line bold space bold equals bold space open vertical bar fraction numerator bold w bold plus bold 1 over denominator bold 1 bold minus bold w end fraction close vertical bar bold therefore bold space open vertical bar fraction numerator bold w bold plus bold 1 over denominator bold 1 bold minus bold w end fraction close vertical bar bold space bold equals bold space bold 1 bold space bold space bold space bold space bold space bold space bold space bold space bold left parenthesis bold vertical line bold z bold vertical line bold space bold equals bold space bold 1 bold right parenthesis$

∴ |w+1| = |w-1|

∴ w એ (1, 0) તથા (-1, 0) થી સમાન અંતરે આવેલ છે.

∴ w એ (1, 0) તથા (-1, 0) ના લંબદ્વિભાજક પર આવેલ છે.

∴ w એ કાલ્પનિક અક્ષ પર આવેલ બિંદુ છે.

∴ Re (w) = 0

45.

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

46.

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

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

47.

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

48.

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 ......

49. 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

50. 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 ?

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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

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