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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

11. જો z એ વાસ્તવિક ન હોય તેવી સંકર સંખ્યા વર્તુળ |z| = 1 પર આવેલ છે, તો z = ...... .

$fraction numerator 1 plus itan left parenthesis arg space straight z right parenthesis over denominator 1 minus itan space left parenthesis arg space straight z right parenthesis end fraction$
$fraction numerator 1 space minus space itan space left parenthesis arg space straight z right parenthesis over denominator 1 space plus space itan space left parenthesis arg space straight z right parenthesis end fraction$
$fraction numerator bold 1 bold plus bold itan bold space open parentheses begin display style fraction numerator bold arg bold space bold z over denominator bold 2 end fraction end style close parentheses over denominator bold space bold 1 bold space bold minus bold space bold itan bold space open parentheses begin display style fraction numerator bold arg bold space bold z over denominator bold 2 end fraction end style close parentheses end fraction$
$fraction numerator bold 1 bold minus bold itan bold space open parentheses begin display style fraction numerator bold arg bold space bold z over denominator bold 2 end fraction end style close parentheses over denominator bold space bold 1 bold space bold plus bold space bold itan bold space open parentheses begin display style fraction numerator bold arg bold space bold z over denominator bold 2 end fraction end style close parentheses end fraction$

12. 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|

13.

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 ?

14. 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)

15.

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

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

16.

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
$fraction numerator bold n to the power of bold 2 bold left parenthesis bold n bold plus bold 1 bold right parenthesis to the power of bold 2 over denominator bold 4 end fraction$
$fraction numerator bold n to the power of bold 2 bold left parenthesis bold n bold plus bold 1 bold right parenthesis to the power of bold 2 over denominator bold 4 end fraction bold minus bold n$
$fraction numerator bold n to the power of bold 2 bold left parenthesis bold n bold plus bold 1 bold right parenthesis to the power of bold 2 over denominator bold 2 end fraction bold plus bold n$

17. 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.

18.

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

19. 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)

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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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

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