અને  બંને શરતનું પાલન કરતી બધી સંકર સંખ્યાઓના કાલ્પનિક ભાગનો સરવાળો ....... થાય. from Mathematics સંકર સંખ્યાઓ

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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

11. જો open vertical bar bold z bold minus bold 4 over bold z close vertical bar bold space bold equals bold space bold 2 હોય, તો |z| નાં મહત્તમ તથા ન્યુનતમ મૂલ્યો વચ્ચેનો તફાવત ......... છે. (z≠0) 
  • 4

  • 1

  • 2

  • 3


12. જો z એ વાસ્તવિક ન હોય તેવી સંકર સંખ્યા વર્તુળ |z| = 1 પર આવેલ છે, તો z = ...... .  
  • fraction numerator 1 plus itan left parenthesis arg space straight z right parenthesis over denominator 1 minus itan space left parenthesis arg space straight z right parenthesis end fraction
  • fraction numerator 1 space minus space itan space left parenthesis arg space straight z right parenthesis over denominator 1 space plus space itan space left parenthesis arg space straight z right parenthesis end fraction
  • fraction numerator bold 1 bold plus bold itan bold space open parentheses begin display style fraction numerator bold arg bold space bold z over denominator bold 2 end fraction end style close parentheses over denominator bold space bold 1 bold space bold minus bold space bold itan bold space open parentheses begin display style fraction numerator bold arg bold space bold z over denominator bold 2 end fraction end style close parentheses end fraction
  • fraction numerator bold 1 bold minus bold itan bold space open parentheses begin display style fraction numerator bold arg bold space bold z over denominator bold 2 end fraction end style close parentheses over denominator bold space bold 1 bold space bold plus bold space bold itan bold space open parentheses begin display style fraction numerator bold arg bold space bold z over denominator bold 2 end fraction end style close parentheses end fraction

13.
વાસ્તવિક સહગુણકવાળી બહુપદી f(x) = x4 + ax3 + bx3 + cx + d માટે f(2i) = f(2+i) = 0 હોય તો a + b + c + d = ....... 
  • 10

  • 9

  • 4

  • 1


14. જો z એ સંકર સંખ્યા હોય તથા bold vertical line bold z bold vertical line bold space bold greater or equal than bold space bold 2 bold space bold ત ો bold space open vertical bar bold z bold plus bold 1 over bold 2 close vertical bar  તો ની ન્યુનતમ કિંમત
  • અંતરાલ (1, 2) માં છે,

  • 5/2 થી વધુ હોય. 
  • 3/2 થી વધુ તથા 5/2 થી ઓછી હોય. 
  • 5/2 હોય.

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15.
જો w એ 1 નું ઘનમૂળ હોય તો 2 (1 + w) (1 + w2) + 3 (2w + 1) + ... + (n+1) (nw2+1) = ......... (w ≠1) 
  • fraction numerator bold n to the power of bold 2 bold left parenthesis bold n bold plus bold 1 bold right parenthesis to the power of bold 2 over denominator bold 4 end fraction+n
  • fraction numerator bold n to the power of bold 2 bold left parenthesis bold n bold plus bold 1 bold right parenthesis to the power of bold 2 over denominator bold 4 end fraction
  • fraction numerator bold n to the power of bold 2 bold left parenthesis bold n bold plus bold 1 bold right parenthesis to the power of bold 2 over denominator bold 4 end fraction bold minus bold n
  • fraction numerator bold n to the power of bold 2 bold left parenthesis bold n bold plus bold 1 bold right parenthesis to the power of bold 2 over denominator bold 2 end fraction bold plus bold n

16. જો z ≠ 0, |zi| ની મહત્તમ સીમા ........ થશે.
  • bold e to the power of bold pi
  • bold e to the power of bold minus bold pi end exponent
  • 1

  • |z|


17. bold z with bold minus on top bold space bold equals bold space bold italic z to the power of bold 2 શરતનું પાલન કરતી કેટલી સંકર સંખ્યાઓ મળે ? 
  • 4

  • 3

  • 2

  • 1


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18.
open vertical bar fraction numerator bold z bold minus bold 12 over denominator bold z bold minus bold 8 bold i end fraction close vertical bar bold space bold equals bold space bold 5 over bold 3અને open vertical bar fraction numerator bold z bold minus bold 4 over denominator bold z bold minus bold 8 end fraction close vertical bar bold space bold equals bold space bold 1 બંને શરતનું પાલન કરતી બધી સંકર સંખ્યાઓના કાલ્પનિક ભાગનો સરવાળો ....... થાય.
  • 35

  • 28

  • 25

  • 28


C.

25

Tips: -

સહેલાઇથી જોઈ શકાય છે કે પ્રથમ સમીકરણ વર્તુળ દર્શાવે છે.

બીજું સમીકરણ એ (4, 0) તથા (8, 0) નો લંબદ્વિભાજક દર્શાવે છે.

∴ z નું સ્વરૂપ 6 + yi પ્રકારનું હોય.

open vertical bar fraction numerator bold z bold minus bold 12 over denominator bold z bold minus bold 8 bold i end fraction close vertical bar bold space bold equals bold space bold 5 over bold 3

bold therefore bold space open vertical bar fraction numerator bold minus bold 6 bold plus bold y bold i over denominator bold 6 bold plus bold left parenthesis bold y bold minus bold 8 bold right parenthesis bold i end fraction close vertical bar bold equals bold 5 over bold 3

bold therefore bold space bold 9 bold space bold left parenthesis bold 6 to the power of bold 2 bold space bold plus bold space bold italic y to the power of bold 2 bold right parenthesis bold space bold equals bold space bold 25 bold space bold left parenthesis bold 6 to the power of bold 2 bold space bold plus bold space bold left parenthesis bold italic y bold space bold minus bold space bold 8 bold right parenthesis to the power of bold 2 bold right parenthesis

bold therefore bold space bold 16 bold space bold italic y to the power of bold 2 bold space bold minus bold space bold 400 bold italic y bold space bold plus bold space bold 2176 bold space bold equals bold space bold 0 bold space

bold therefore bold space bold italic y to the power of bold 2 bold space bold minus bold space bold 25 bold space bold plus bold space bold 136 bold space bold equals bold space bold 0 bold space


∴ y = 8 અથવા y = 17

∴ z = 6 + 8i અથવા 6 + 7i

∴ માંગેલ સરવાળો = 8 + 17 = 25 


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19. cos y sin y + cos2 y sin 2y + cos3y sin3y + ...... n પદ ......
  • tan y (1 - cosn y cosny)
  • cot y (1 - cosn cosny)
  • cot y (1 - sinn y sinny)
  • tan y (1 - sinn y sin n y)

20. શૂન્યેતર ભિન્ન સંકર સંખ્યાઓ z અને w માટે જો  |z|2 w-|w|2 z = z - w તો ...... 
  • bold zw bold space bold equals bold space bold 1
  • bold z bold w with bold minus on top bold space bold equals bold space bold 1
  • bold z bold space bold equals bold space bold w with bold minus on top
  • z = -w


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