સંકર સંખ્યા z માટે 4 arg(z-5i) = 2arg(z+3)=cos-1(-1) તો from Mathematics સંકર સંખ્યાઓ

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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

61. જો z1, z2 બે શુન્યેતર સંકર સંખ્યાઓ છે, જ્યાં |2z1- 3z2|2 = |2z1|2 + |3z2|2 તો
  • bold arg bold space open parentheses bold z subscript bold 2 over bold z subscript bold 1 close parentheses bold space bold equals bold space bold plus-or-minus bold space bold pi over bold 2
  • bold z subscript bold 1 bold space stack bold z subscript bold 2 with bold bar on top bold space bold plus bold space bold z subscript bold 2 bold space stack bold z subscript bold 1 with bold bar on top bold space bold equals bold space bold 0
  • bold z subscript bold 2 over bold z subscript bold 1એ વાસ્તવિક સંખ્યા હોય.
  • bold z subscript bold 2 over bold z subscript bold 1 એ શુદ્વ કાલ્પનિક સંખ્યા છે.

62. ધારો કે bold z subscript bold k bold space bold equals bold space bold cos bold space open parentheses fraction numerator bold 2 bold k bold space bold pi over denominator bold 10 end fraction close parentheses bold space bold plus bold space bold i bold space bold sin bold space open parentheses fraction numerator bold 2 bold k bold space bold pi over denominator bold 10 end fraction close parentheses bold comma bold space bold k bold space bold equals bold space bold 1 bold comma bold space bold 2 bold comma bold space bold. bold. bold. bold space bold 9

  • P-1, Q-2, R-4, S-3

  • P-2, Q-1, R-3, S-4

  • P-2, Q-1, R-4, S-3

  • P-1, Q-2, R-3, S-4


63. P(x) અને Q (x) એ બહુપદી છે. ધારો કે x2 + x + 1 એ f(x) = P(x3) + xQ(x3) નો અવયવ છે, તો 
  • (x-1)એ P(x) તથા Q(x) બંનેનો અવયવ હોય.

  • (x-1)એ f(x)નો અવયવ હોય.
  • (x-1) એ P(x) નો અવયવ હોય પરંતુ Q(x) નો ન હોય. 
  • (x-1)એ Q(x) નો અવયવ હોય પરંતુ P(x)નો ન હોય. 

64.
  • P-i, Q-ii, R-iii, S-iv

  • P-ii, Q-ii, R-iv, S-iii

  • P-iv, Q-iii, R-ii, S-i

  • P-iv, Q-i, R-ii, S-iii


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65.
  • P-iii, Q-ii, R-iv, S-i

  • P-i, Q-ii, R-iii, S-iv

  • P-ii, Q-iii, R-iv, S-i

  • P-iv, Q-iii, R-ii, S-i


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66. સંકર સંખ્યા z માટે 4 arg(z-5i) = 2arg(z+3)=cos-1(-1) તો
  • bold z with bold bar on top bold space bold equals bold space bold 2 bold italic i bold space bold plus bold space bold 3
  • z એ x+y+1 = 0 પર છે.
  • |z| એ x2 + y2 + 2x - 4y - 8 = 0 ની ત્રિજ્યા છે.
  • bold z bold space bold plus bold space bold z with bold bar on top bold space bold less than bold space bold 0

A.

bold z with bold bar on top bold space bold equals bold space bold 2 bold italic i bold space bold plus bold space bold 3

B.

z એ x+y+1 = 0 પર છે.

C.

|z| એ x2 + y2 + 2x - 4y - 8 = 0 ની ત્રિજ્યા છે.

D.

bold z bold space bold plus bold space bold z with bold bar on top bold space bold less than bold space bold 0

Tips: -

bold 4 bold space bold arg bold space bold left parenthesis bold z bold space bold minus bold space bold 5 bold i bold right parenthesis bold space bold equals bold space bold 2 bold arg bold space bold left parenthesis bold z bold space bold plus bold space bold 3 bold right parenthesis bold space bold equals bold space bold cos to the power of bold minus bold 1 end exponent bold space bold left parenthesis bold minus bold 1 bold right parenthesis bold space bold equals bold space bold pi

bold therefore bold space bold tan to the power of bold minus bold 1 end exponent bold space open parentheses fraction numerator bold y bold minus bold 5 over denominator bold x end fraction close parentheses bold space bold equals bold space bold pi over bold 4 bold comma bold space bold tan to the power of bold minus bold 1 end exponent bold space open parentheses fraction numerator bold y over denominator bold x bold plus bold 3 end fraction close parentheses bold space bold equals bold space bold pi over bold 2 bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold left parenthesis bold z bold equals bold x bold plus bold iy bold right parenthesis

∴ y - 5 = x, x + 3 = 0  પરથી, x = -3, y = 2

∴ z = -3 + 2i  આથી bold z with bold bar on top bold space bold equals bold space bold minus bold 3 bold space bold minus bold space bold 2 bold i

બિંદુ (-3, 2) એ x + y + 1 = 0 પર આવેલ છે.
 
∴ z એ x + y + 1 = 0 પર આવેલ છે.

વર્તુળ x2 + y2 + 2x - 4y - 8 = 0  ની ત્રિજ્યા square root of bold 1 bold plus bold 4 bold plus bold 8 end root bold space bold equals bold space square root of bold 13 bold space bold equals bold space bold vertical line bold italic z bold vertical line
 
∴|z| એ વર્તુળ x2 + y2 + 2x - 4y - 8 = 0 ની ત્રિજ્યા છે. 

વળી, z + bold z with bold bar on top = -6 < 0

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67. જો સંકર સંખ્યા z માટે bold arg bold space open parentheses fraction numerator bold z bold minus bold 2 over denominator bold z bold plus bold 2 end fraction close parentheses bold equals bold space bold pi over bold 4 તો |z-2i|  ની કિંમત = ...... 
  • bold 3 square root of bold 2
  • 2 square root of bold 2
  • 4 square root of bold 2
  • 5 square root of bold 2

68. bold sum from bold p bold equals bold 1 to bold 32 of bold left parenthesis bold 3 bold p bold plus bold 2 bold right parenthesis bold space open parentheses bold sum from bold q bold equals bold 1 to bold 10 of open parentheses bold sin fraction numerator bold 2 bold qπ over denominator bold 11 end fraction bold minus bold space bold i bold space bold cos bold space fraction numerator bold 2 bold qπ over denominator bold 11 end fraction close parentheses close parentheses to the power of bold p bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold space bold.
  • 48 (1+i)
  • 24(1-i)
  • 48(1-i)
  • 24(1+i)

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69. જો z1, z2, .... zn એ વર્તુળ |z| = r પર આવેલ હોય તથા હોય તો, bold Re bold space open parentheses bold sum from bold j bold equals bold 1 to bold n of bold space bold sum from bold k bold equals bold 1 to bold n of bold space bold sum from bold z subscript bold k to bold z bold space bold j of close parentheses bold space bold equals bold space bold 0 હોય તો, 
  • bold sum from bold j bold equals bold 1 to bold n of bold space bold z subscript bold j bold space bold equals bold space bold 0
  • open vertical bar bold sum from bold j bold equals bold 1 to bold n of bold space bold z subscript bold j close vertical bar bold space bold equals bold 0
  • bold sum from bold j bold equals bold 1 to bold n of bold space bold z with bold bar on top bold space bold j bold space bold equals bold space bold 0
  • આપેલ પૈકી એક પણ નહી

70.
  • P-ii, Q-iii, R-i, S-iv

  • P-i, Q-ii, R-iii, S-iv

  • P-iv, Q-iii, R-ii, S-i

  • P-ii, Q-i, R-iv, S-iiii


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