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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

51.

bold sin bold space bold italic alpha bold space bold plus bold space bold italic s bold italic i bold italic n bold space bold italic beta bold space bold plus bold space bold italic s bold italic i bold italic n bold space bold italic gamma bold space bold equals bold space bold 0 bold space bold equals bold space bold italic c bold italic o bold italic s bold space bold italic alpha bold space bold plus bold space bold italic c bold italic o bold italic s bold space bold italic beta bold space bold plus bold space bold italic c bold italic o bold italic s bold space bold italic gamma bold space

જ્યાં

bold minus bold space bold pi bold space bold less than bold space bold alpha bold space bold comma bold space bold beta bold comma bold space bold gamma bold space bold less or equal than bold space bold pi

bold જ ો bold space bold sin to the power of bold 2 bold space bold alpha bold space bold plus bold space bold sin to the power of bold 2 bold space bold beta bold space bold plus bold space bold sin to the power of bold n bold space bold gamma bold space bold equals bold space bold 3 over bold 2 bold space bold તથ ા bold space bold sum bold cos bold space bold left parenthesis bold alpha bold space bold plus bold space bold beta bold right parenthesis bold space bold equals bold space bold lambda bold space bold ત ો bold comma bold space bold left parenthesis bold n bold comma bold space bold lambda bold right parenthesis bold space bold equals bold space bold. bold. bold. bold. bold. bold space

(2, 3)
(0, 3)
(2, 0)
(2, 3)

52. જો સંકર સંખ્યા z માટે |z - 3 - 2i| ≤ 2, તો |2z - 6 + 5i| નું મહત્વ મૂલ્ય ....... છે.

53. z અને w એ સંકર સંખ્યાઓ છે. જો $bold z with bold bar on top bold space bold plus bold space bold i bold space bold w with bold bar on top bold space bold equals bold space bold 0$ અને arg (zw) = $bold pi$ તો argz = ............ .

$fraction numerator bold 5 bold pi over denominator bold 4 end fraction$
$fraction numerator bold minus bold pi over denominator bold 4 end fraction$
$bold pi over bold 4$
$bold pi over 2$

fraction numerator bold minus bold pi over denominator bold 4 end fraction

Tips: -

ધારો કે z = re^iθ આથી arg z + arg w = $straight pi$ . આથી $bold arg bold space bold w bold space bold equals bold space bold pi bold space bold minus bold space bold theta$

$bold therefore bold space bold w bold space bold equals bold space bold r subscript bold 1 bold r to the power of bold i bold left parenthesis bold pi bold minus bold theta bold right parenthesis end exponent$ જ્યાં r₁ = |w|

હવે, $bold z with bold bar on top bold space bold plus bold space bold i bold w with bold bar on top bold space bold equals bold space bold 0$

$bold therefore bold space bold re to the power of bold minus bold iθ end exponent bold space bold plus bold space bold ir subscript bold 1 bold e to the power of bold i bold left parenthesis bold theta bold minus bold pi bold right parenthesis end exponent bold space bold equals bold space bold 0 bold r bold space bold left parenthesis bold cos bold space bold theta bold space bold minus bold space bold i bold space bold sin bold space bold theta bold right parenthesis bold space bold plus bold space bold ir subscript bold 1 bold space bold left square bracket bold cos bold space bold left parenthesis bold pi bold minus bold theta bold right parenthesis bold space bold minus bold space bold i bold space bold sin bold space bold left parenthesis bold pi bold minus bold theta bold right parenthesis bold right square bracket bold space bold equals bold space bold 0 bold space bold therefore bold space bold r bold space bold left parenthesis bold cos bold space bold theta bold space bold minus bold space bold i bold space bold sin bold space bold theta bold right parenthesis bold space bold plus bold space bold r subscript bold 1 bold space bold left square bracket bold minus bold i bold space bold cos bold space bold theta bold space bold plus bold space bold sin bold space bold theta bold right parenthesis bold space bold equals bold space bold 0$

વાસ્તવિક ભાગ તથા કાલ્પનિક ભાગ સરખાવતાં,

r cos θ + r₁ sin θ = 0, -r sin θ - r₁ cos θ = 0

$bold therefore bold space bold r over bold r subscript bold 1 bold space bold equals bold space bold minus bold tan bold space bold theta bold space bold equals bold space bold r subscript bold 1 over bold r bold.$ આથી r² = r₁² આથી r = r₁

$bold therefore bold space bold tan bold space bold theta bold space bold equals bold space bold minus bold 1$ આથી $bold argz bold space bold equals bold space fraction numerator bold minus bold pi over denominator bold 4 end fraction$

54. z અને w એ શૂન્યેતર સંકર સંખ્યાઓ છે. જો |zw| = 1 અને argz - argw = $bold pi over bold 2$ તો $bold z with bold bar on top bold space bold w bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold space$

-1
-i
i
1

55. $bold sum from bold k bold equals bold 1 to bold 6 of bold space open parentheses bold sin open parentheses fraction numerator bold 2 bold kπ over denominator bold 7 end fraction close parentheses bold plus bold space bold i bold space bold cos bold space open parentheses fraction numerator bold 2 bold kπ over denominator bold 7 end fraction close parentheses close parentheses bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold space bold.$

-i
1
0
-1

56. જો |z| = 1 અને z ≠ ± 1 તો કોઈ પણ સંકર સંખ્યા z માટે $fraction numerator bold z over denominator bold 1 bold space bold minus bold space bold z to the power of bold 2 end fraction$ એ = ....... .

Y-અક્ષ પર હોય.
ઉંગમબિંદુમાંથી પસાર ન થતી રેખા પર હોય.
X-અક્ષ પર હોય.
$bold vertical line bold z bold vertical line bold space bold equals bold space square root of bold 2$

57.

જો સંકર સંખ્યા $bold alpha bold space bold left parenthesis bold space bold alpha bold space bold not equal to bold space bold 1 bold right parenthesis$ એ સમીકરણ $bold z to the power of bold p bold plus bold q end exponent bold space bold minus bold space bold z to the power of bold minus bold space bold minus bold space bold z to the power of bold q bold space bold plus bold space bold 1 bold space bold equals bold space bold 0$ જ્યાં p, q ભિન્ન અવિભાજ્ય સંખ્યા છે, નું બીજ હોય તો

1 + α + α² + ... + α^p = 0
1 + α + α² + ... + α^q = 0 અને 1 + α + a² + ... + a^q-1 = 0
1 + α + a² + ... α^p-1 = 0 અને 1 + α + a² + ... + a^q-1 = 0
પરંતુ બંને એક સાથે ન ઉદભવે

58. જો |z| = 1 અને arf (z) = તો arg $open parentheses fraction numerator bold 1 bold plus bold z over denominator bold 1 bold plus bold z end fraction close parentheses bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold space bold.$

θ
$bold pi bold minus bold theta$
-θ
$bold pi over bold 2 bold minus bold theta$

59. જો f(z) = z⁴ - 8z³ + 4z² + 4z + 39 અને f (3 + 2i) = a + ib તો a:b = .........

-1 : 8
1 : 8
1 : 4
-1 : 4

60. જો $bold alpha bold space bold equals bold space bold cos bold space open parentheses fraction numerator bold 8 bold pi over denominator bold 11 end fraction close parentheses bold space bold plus bold space bold i bold space bold sin bold space open parentheses fraction numerator bold 8 bold pi over denominator bold 11 end fraction close parentheses$ તો, $bold Re bold space bold left parenthesis bold space bold alpha bold space bold plus bold space bold alpha to the power of bold 2 bold space bold plus bold space bold alpha to the power of bold 3 bold space bold plus bold space bold alpha to the power of bold 4 bold space bold plus bold space bold alpha to the power of bold 5 bold right parenthesis bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold space bold.$

1
0
$bold minus bold 1 over bold 2$
$bold 1 over bold 2$

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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

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