જો |z| = 1 અને z2n + 1 ≠ 0 તો  from Mathematics સંકર સંખ્યાઓ

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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

1. જો ચતુર્ઘાત સમીકરણ x4 + ax3 + bx2 + cx + d = 0 (a, b, c, d ∈ R) નું કોઈપણ બીજ વાસ્તવિક સંખ્યા ન હોય તથા બે બીજનો સરવાળો 3 + 4 bold i with bold hat on top હોય અને બાકીનાં બે બીજનો ગુણાકાર 13 + i હોય, તો b = ......... . 
  • 51

  • 15

  • 6

  • 30


2.
bold log subscript begin inline style bold 1 over bold 2 end style end subscript bold space open parentheses fraction numerator bold vertical line bold z bold minus bold 1 bold vertical line bold plus bold 4 over denominator bold 3 bold vertical line bold z bold space bold minus bold space bold 1 bold vertical line bold minus bold 1 end fraction close parentheses bold space bold greater than bold space bold 1 (જ્યાં bold vertical line bold z bold space bold minus bold space bold 1 bold vertical line bold space bold not equal to bold space bold 2 over bold 3 bold right parenthesis અસમતાનું પાલન કરતી સંકર સંખ્યાનો બિંદુગણ ......
  • વર્તુળનો બહારનો ભાગ. 

  • વર્તુળ છે. 

  • વર્તુળની અંદરનો ભાગ. 

  • રેખા છે.


3. જો x2 + x + 1 = 0 નાં બે બીજ a અને b હોય, તો જેનાં બીજ a19 અને b7 હોય, તેવું સમીકરણ = .........
  • x2 + x + 1 = 0

  • x2 + x - 1 = 0

  • x2 - x - 1 = 0

  • x2 - x + 1 = 0


4.
z1 અને z2 ભિન્ન સંકર સંખ્યાઓ છે તથા |z1| = |z2|. જો z1 નો વાસ્તવિક ભાગ ધન સંખ્યા હોય તથા z2 નો કાલ્પનિક ભાગ ઋણ સંખ્યા હોય, તો fraction numerator bold z subscript bold 1 bold space bold plus bold space bold z subscript bold 2 over denominator bold z subscript bold 2 bold space bold minus bold space bold z subscript bold 2 end fraction એ ...... છે.          open parentheses fraction numerator bold z subscript bold 1 bold space bold plus bold space bold z subscript bold 2 over denominator bold z subscript bold 1 bold space bold minus bold space bold z subscript bold 2 end fraction bold space bold not equal to bold space bold 0 close parentheses
  • વાસ્તવિક અને ધન

  • શુદ્વ કાલ્પનિક સંખ્યા 

  • વાસ્તવિક અને ઋણ 

  • શૂન્ય સંખ્યા


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5. જો |z| = 1 અને z2n + 1 ≠ 0 તો fraction numerator bold z to the power of bold n over denominator bold z to the power of bold 2 bold n end exponent bold space bold plus bold space bold 1 end fraction bold minus bold space fraction numerator bold x to the power of bold minus bold n end exponent over denominator open parentheses begin display style bold z with bold minus on top end style close parentheses to the power of bold 2 bold n end exponent bold space bold plus bold space bold 1 end fraction bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold. bold space bold.
  • 1

  • i

  • 0

  • 3


C.

0

Tips: -

ધારો કે z = cos θ + i sin θ


∴  fraction numerator bold z to the power of bold n over denominator bold z to the power of bold 2 bold n end exponent bold space bold plus bold space bold 1 end fraction bold minus fraction numerator bold z to the power of bold minus bold n end exponent over denominator open parentheses begin display style bold z with bold minus on top end style close parentheses to the power of bold 2 bold n end exponent bold space bold plus bold space bold 1 end fraction bold equals fraction numerator bold cos bold space bold nθ bold space bold plus bold space bold i bold space bold sin bold space bold n bold space bold theta bold space over denominator bold 1 bold space bold plus bold space bold space bold cos bold space bold 2 bold n bold space bold space bold plus bold space bold i bold space bold sin bold space bold 2 bold space bold n bold space bold theta end fraction bold minus fraction numerator bold cos bold space bold n bold space bold space bold minus bold space bold i bold space bold sin bold space bold n bold space bold theta over denominator bold 1 bold space bold plus bold space bold cos bold space bold n bold space bold space bold left parenthesis bold cos bold space bold n bold space bold theta bold space bold minus bold space bold i bold space bold sin bold space bold n bold space bold theta bold space bold right parenthesis end fraction

                                     bold equals bold space fraction numerator bold cos bold space bold n bold space bold theta bold space bold plus bold space bold i bold space bold sin bold space bold n bold space bold theta over denominator bold 2 bold space bold cos bold space bold n bold space bold left parenthesis bold cos bold space bold n bold space bold space bold plus bold space bold i bold space bold sin bold space bold n bold space bold right parenthesis end fraction bold space bold minus bold space fraction numerator bold cos bold space bold n bold space bold space bold minus bold space bold i bold space bold sin bold space bold n bold space bold theta over denominator bold 2 bold space bold cos bold space bold n bold space bold space bold left parenthesis bold cos bold space bold n bold space bold space bold minus bold space bold i bold space bold sin bold space bold right parenthesis end fraction

bold equals bold space fraction numerator bold i over denominator bold 2 bold space bold cos bold space bold m bold space bold theta end fraction bold space bold minus bold space fraction numerator bold 1 over denominator bold 2 bold space bold cos bold space bold n bold space bold theta end fraction bold space bold equals bold space bold 0

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6. જો bold x open parentheses bold minus bold space bold x square root of bold 3 close parentheses bold space bold plus bold space bold 1 bold space bold equals bold space bold 0 bold comma bold spaceતો bold sum from bold n bold equals bold 1 to bold 36 of open parentheses bold x to the power of bold n bold minus bold 1 over bold x to the power of bold n close parentheses to the power of bold 2 = ........... . 
  • 72

  • 36

  • -72

  • 0


7.
f(z)ને z-i વડે ભાગીએ તો શેષ i મળે છે તથા જો z + i વડે ભાગીએ તો શેષ 1 + i મળે છે. જો f(z) ને z2 + 1 વડે ભાંગીએ તો મળતી શેષ ......
  • 0

  • bold 1 over bold 2 bold left parenthesis bold iz bold plus bold 1 bold plus bold 2 bold i bold right parenthesis
  • bold 1 over bold 2 bold left parenthesis bold italic i bold italic z bold plus bold 1 bold right parenthesis
  • iz+1+i


8. જો α, β  એ x2-2x + 2 = 0 નાં બીજ હોય, તો αn + βn = ......... .
  • bold 2 to the power of begin inline style bold n over bold 2 end style end exponent bold space bold cos bold nπ over bold 4
  • bold 2 to the power of begin inline style bold n over bold 2 end style bold minus bold 1 end exponent bold space bold cos bold nπ over bold 4
  • 0

  • bold 2 to the power of begin inline style bold n over bold 2 end style bold plus bold 1 end exponent bold space bold cos bold space bold nπ over bold 4

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9. જો, open vertical bar fraction numerator bold 1 bold minus bold minus bold iz over denominator bold z bold minus bold i end fraction close vertical bar bold space bold equals bold space bold 1 bold space તો....... 
  • z એ શુદ્વ કાલ્પનિક સંખ્યા હોય.

  • P(z) એ બીજા ચરણમાં હોય. 

  • P(z) એ ત્રીજા ચરણમાં હોય.

  • z એ વાસ્તવિક સંખ્યા હોય. 


10.
ધારો કે bold a bold space bold equals bold space fraction numerator bold 12 bold pi over denominator bold e to the power of bold 13 end fraction bold. bold space bold alpha bold space bold plus bold space bold a bold space bold plus bold space bold a to the power of bold 3 bold space bold plus bold space bold a to the power of bold 4 bold space bold plus bold space bold a to the power of bold minus bold 4 end exponent bold space bold space bold plus bold space bold a to the power of bold minus bold 3 end exponent bold space bold plus bold space bold a to the power of bold minus bold 1 end exponent તથા bold beta bold space bold equals bold space bold a to the power of bold 2 bold space bold plus bold space bold a to the power of bold 2 bold space bold plus bold space bold a to the power of bold 6 bold space bold plus bold space bold a to the power of bold minus bold 6 end exponent bold space bold plus bold space bold a to the power of bold minus bold 5 end exponent bold space bold plus bold space bold a to the power of bold minus bold 2 end exponent જેનાં બીજ હોય તેવું દ્વિઘાત સમીકરણ ......... છે.
  • x2 + x + 3 = 0

  • x2 - x + 2 = 0

  • x2 + x - 3 = 0

  • x3 - x - 3 = 0


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