જો f(x) = ax2 + bx જ્યાં f(1) = 2, f(i2)=0 અને f(z+z1) = 0 તો |z| = ......  from Mathematics સંકર સંખ્યાઓ

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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

71. નીચેના ડાબી બાજુના સમીકરણને જમણી બાજુ દર્શાવતા વક્ર સાથે જોડો.
  • P-iv, Q-ii, R-i, S-i

  • P-iv, Q-ii, R-iii, S-i

  • P-iv, Q-iii, R-ii, S-i

  • P-i, Q-iv, R-ii, S-iii


72.
  • P-iv, Q-iii, R-ii, S-i
  • P-iv, Q-ii, R-iii, S-i
  • P-i, Q-iv, R-ii, S-iii
  • P-iv, Q-ii, R-i, S-iii

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73. જો f(x) = ax2 + bx જ્યાં f(1) = 2, f(i2)=0 અને f(z+z1) = 0 તો |z| = ...... 
  • 2

  • 1

  • 0

  • 1/2


B.

1

Tips: -

f(x) = ax2 + bx                 ...(1)

f(1) = 2. આથી a + b = 2          ... (2)

(1) અને (2) પરથી, a = 1, b = 1 

∴ f(x) = x2 + x

f(z+z-1) = 0,    આથી, open parentheses bold z bold plus bold 1 over bold 2 close parentheses to the power of bold 2 bold space bold plus bold space open parentheses bold z bold plus bold 1 over bold 2 close parentheses bold space bold equals bold space bold 0 bold space

                               bold therefore bold space bold z to the power of bold 4 bold space bold plus bold space bold z to the power of bold 3 bold space bold plus bold space bold 2 bold z to the power of bold 2 bold space bold plus bold space bold z bold space bold plus bold space bold 1 bold space bold equals bold space bold 0 bold space

bold therefore bold space bold left parenthesis bold z to the power of bold 2 bold space bold plus bold space bold 1 bold right parenthesis bold space bold left parenthesis bold z to the power of bold 2 bold space bold plus bold space bold z bold space bold plus bold space bold 1 bold space bold right parenthesis bold space bold equals bold space bold 0 bold space

bold therefore bold space bold z bold space bold equals bold space bold plus-or-minus bold space bold i bold comma bold space bold w bold comma bold space bold w to the power of bold 2 bold space

bold therefore bold space bold vertical line bold z bold vertical line bold space bold equals bold space bold 1 bold space 

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