જો દ્વિઘાત સમીકરણ (a2 + b2)x2 - 2b(a + c) x + (b2+c2)=0 નાં બીજ સમાન હોય તોa, b, c,  ....... શ્રેણીમાં હોય a, b, c ∈ R.  from Mathematics દ્વિઘાત સમીકરણ

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Gujarati JEE Mathematics : દ્વિઘાત સમીકરણ

Multiple Choice Questions

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21.
જો દ્વિઘાત સમીકરણ (a2 + b2)x2 - 2b(a + c) x + (b2+c2)=0 નાં બીજ સમાન હોય તોa, b, c,  ....... શ્રેણીમાં હોય a, b, c ∈ R. 
  • સમગુણોત્તર 

  • સમાંતર 
  • સ્વરિત 
  • સમાંતર-સમગુણોત્તર

A.

સમગુણોત્તર 

Tips: -

bold left parenthesis bold a to the power of bold 2 bold plus bold b to the power of bold 2 bold right parenthesis bold x to the power of bold 2 bold space bold minus bold space bold 2 bold b bold space bold left parenthesis bold a bold space bold plus bold space bold c bold right parenthesis bold x bold space bold plus bold space bold left parenthesis bold b to the power of bold 2 bold space bold plus bold space bold c to the power of bold 2 bold right parenthesis bold space bold equals bold space bold 0 નાં બીજ સમાન છે.
bold therefore bold space bold increment bold space bold equals bold space bold 0

bold therefore bold space bold 4 bold b to the power of bold 2 bold space bold left parenthesis bold a bold space bold plus bold space bold c bold right parenthesis to the power of bold 2 bold space bold minus bold space bold 4 bold space bold left parenthesis bold a to the power of bold 2 bold space bold plus bold space bold b to the power of bold 2 bold right parenthesis bold space bold left parenthesis bold b to the power of bold 2 bold space bold plus bold space bold c to the power of bold 2 bold right parenthesis bold space bold equals bold space bold 0 bold space

bold therefore bold space bold 4 bold b to the power of bold 2 bold space bold left parenthesis bold a to the power of bold 2 bold space bold plus bold space bold 2 bold ac bold space bold plus bold space bold c to the power of bold 2 bold right parenthesis bold space bold minus bold space bold 4 bold space bold left parenthesis bold a to the power of bold 2 bold b to the power of bold 2 bold space bold plus bold space bold a to the power of bold 2 bold c to the power of bold 2 bold space bold plus bold space bold b to the power of bold 4 bold space bold plus bold space bold b to the power of bold 2 bold c to the power of bold 2 bold right parenthesis bold space bold equals bold space bold 0

bold therefore bold space bold 8 bold ab to the power of bold 2 bold c bold space bold minus bold space bold 4 bold a to the power of bold 2 bold c to the power of bold 2 bold space bold minus bold space bold 4 bold b to the power of bold 4 bold space bold equals bold space bold 0 bold space

bold therefore bold space bold b to the power of bold 4 bold space bold plus bold space bold a to the power of bold 2 bold c to the power of bold 2 bold space bold minus bold space bold 2 bold ab to the power of bold 2 bold c bold space bold equals bold space bold 0 bold space

bold therefore bold space bold left parenthesis bold b to the power of bold 2 bold space bold minus bold space bold ac bold right parenthesis to the power of bold 2 bold space bold equals bold space bold 0 bold space

bold therefore bold space bold b to the power of bold 2 bold space bold equals bold space bold ac bold space

∴ a, b, c સમગુણોત્તર શ્રેણીમાં છે.

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22.
દ્વિઘાત સમીકરણનો ઉકેલ શોધતી વખતે બે વિદ્યાર્થીઓ પૈકી એક વિદ્યાર્થી સમીકરણનું અચળ પદ ખોટું લખે છે અને સમીકરણનાં સાચાં બીજનો સરવાળો 3 મળે છે જ્યારે બીજો વિદ્યાર્થી x2 નો સહગુણક તથા અચળ પદ સાચાં લખે છે જે અનુક્રમે 1 તથા -18 છે તો મળતા દ્વિઘાત સમીકરણનાં સાચાં બીજ ...... હોય.
  • -6, 3
  • 6, -3
  • -3, -6
  • 3, 6

23. જો log2 x + logx 2 = begin inline style bold 10 over bold 3 end style = log2y + logy2 તથા x ≠ y હોય તો x+y = ......   
  • bold 8 bold minus bold 2 to the power of begin inline style bold 1 over bold 3 end style end exponent
  • bold 2 to the power of begin inline style bold 1 over bold 3 end style end exponent
  • bold 8 bold plus bold 2 to the power of begin inline style bold 1 over bold 3 end style end exponent
  • 8

24. જો x, y, z ભિન્ન અને વાસ્તવિક હોય, તો x2 + 4y2 + 9z2 - 6yz - 3zx - 2xy હંમેશાં ....... હોય. 
  • 0

  • ઋણ 
  • અનૃણ
  • સંકર સંખ્યા

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25. સમીકરણ open parentheses bold 5 bold plus bold 2 square root of bold 6 close parentheses to the power of bold x bold minus bold 3 end exponent bold plus open parentheses bold 5 bold minus bold 2 square root of bold 6 close parentheses to the power of bold x bold minus bold 3 end exponent bold space bold equals bold space bold 10 નું એક બીજ ....... હોય.
  • 3

  • 8

  • 4

  • 2


26.
જો α ≠ β તથા α2 = 5α - 3 તેમજ β2 = 5β - 3 હોય, તો bold alpha over bold beta અને bold beta over bold alpha બીજ ધરાવતું દ્વિઘાત સમીકરણ ....... હોય.
  • x2 + 19x -3 = 0
  • 3x2 - 19x + 3 = 0
  • 3x2-16x + 1 = 0
  • 3x2 -- 19x - 3 = 0

27.
જો 3,3 એ દ્વિઘાત સમીકરણ x2 + ax + β = 0 નાં બીજ હોય તથા -8, 2 એ દ્વિઘાત સમીકરણ x2 + αx + b = 0 નાં બીજ હોય, તો સમીકરણ x2 + ax + b = 0 નાં બીજ ...... હોય.
  • -8, 2
  • 8, -2
  • -2, -8
  • 2, 8

28. સમીકરણ square root of bold x bold minus bold 2 end root bold left parenthesis bold italic x to the power of bold 2 bold space bold minus bold space bold 4 bold italic x bold space bold plus bold space bold 3 bold right parenthesis bold space bold equals bold space bold 0 નાં વાસ્તવિક બીજ ...... હોય. 
  • 1,2,3

  • 2, 3

  • 1, 3

  • 1, 2


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29. જો bold x bold space bold equals bold space square root of bold 12 bold minus square root of bold 140 end root હોય, તો bold x bold space bold plus bold space bold 1 over bold x bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold.
  • fraction numerator square root of bold 7 bold minus bold 3 square root of bold 5 over denominator bold 2 end fraction
  • fraction numerator square root of bold 7 bold minus square root of bold 5 over denominator bold 2 end fraction
  • square root of bold 7 bold plus square root of bold 5
  • fraction numerator bold 3 square root of bold 7 bold minus square root of bold 5 over denominator bold 2 end fraction

30. સમીકરણ fraction numerator bold log bold space bold 3 bold space bold plus bold space bold log bold space bold left parenthesis bold x to the power of bold 2 bold space bold plus bold space bold 2 bold right parenthesis over denominator bold log bold space bold left parenthesis bold x bold space bold minus bold space bold 2 bold right parenthesis end fraction bold space bold equals bold space bold 2 ના ....... ઉકેલ મળે.
  • 0

  • 1

  • 2

  • 3


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