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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

$bold arg bold space open parentheses bold z subscript bold 2 over bold z subscript bold 1 close parentheses bold space bold equals bold space bold plus-or-minus bold space bold pi over bold 2$
$bold z subscript bold 1 bold space stack bold z subscript bold 2 with bold bar on top bold space bold plus bold space bold z subscript bold 2 bold space stack bold z subscript bold 1 with bold bar on top bold space bold equals bold space bold 0$
$bold z subscript bold 2 over bold z subscript bold 1$ એ વાસ્તવિક સંખ્યા હોય.
$bold z subscript bold 2 over bold z subscript bold 1$ એ શુદ્વ કાલ્પનિક સંખ્યા છે.

62.

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

63. જો z₁, z₂, .... z_n 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

64. 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 = ......

$bold 3 square root of bold 2$
$2 square root of bold 2$
$4 square root of bold 2$
$5 square root of bold 2$

2 square root of bold 2

Tips: -

ધારો કે z = x + iy અને arg (z - 2) = θ₁ તથા arg (z + 2) = θ₂ .

$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 space bold equals bold space bold arg bold space bold left parenthesis bold z bold space bold minus bold space bold 2 bold right parenthesis bold space bold minus bold space bold arg bold space bold left parenthesis bold z bold space bold plus bold space bold 2 bold right parenthesis bold space bold equals bold space bold pi over bold 4 bold therefore bold space bold theta subscript bold 1 bold space bold minus bold space bold theta subscript bold 2 bold space bold equals bold space bold pi over bold 4 bold therefore 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 minus bold 2 end fraction close parentheses bold minus 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 2 end fraction close parentheses bold space bold equals bold space bold pi over bold 4 bold therefore bold space bold tan to the power of bold minus bold 1 end exponent bold space fraction numerator begin display style fraction numerator bold y over denominator bold x bold minus bold 2 end fraction end style bold minus begin display style fraction numerator bold y over denominator bold x bold plus bold 2 end fraction end style over denominator bold 1 bold plus begin display style fraction numerator bold y to the power of bold 2 over denominator bold x to the power of bold 2 bold minus bold 4 end fraction end style end fraction bold space bold equals bold space bold pi over bold 4 bold therefore bold space fraction numerator bold y bold left parenthesis bold x bold plus bold 2 bold right parenthesis bold minus bold y bold left parenthesis bold x bold minus bold 2 bold right parenthesis over denominator bold x to the power of bold 2 bold minus bold 4 bold plus bold y to the power of bold 2 end fraction bold equals bold 1 bold therefore bold space bold x to the power of bold 2 bold space bold plus bold space bold y to the power of bold 2 bold space bold minus bold space bold 4 bold y bold space bold equals bold space bold 4 bold therefore bold space bold left parenthesis bold x bold minus bold 0 bold right parenthesis to the power of bold 2 bold space bold plus bold space bold left parenthesis bold y bold minus bold 2 bold right parenthesis to the power of bold 2 bold space bold equals bold space bold 8 bold space bold therefore bold space bold vertical line bold x bold space bold plus bold space bold i bold left parenthesis bold y bold minus bold 2 bold right parenthesis bold vertical line to the power of bold 2 bold space bold equals bold space bold 8 bold space bold therefore bold space bold vertical line bold z bold space bold minus bold space bold 2 bold i bold vertical line bold space bold equals bold space bold 2 square root of bold 2$

65. 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

66.

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

67.

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

68. 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.

69. $bold sum from bold p bold equals bold 1 to bold 32 of bold left parenthesis bold 3 bold p bold plus bold 2 bold right parenthesis bold space open parentheses bold sum from bold q bold equals bold 1 to bold 10 of open parentheses bold sin fraction numerator bold 2 bold qπ over denominator bold 11 end fraction bold minus bold space bold i bold space bold cos bold space fraction numerator bold 2 bold qπ over denominator bold 11 end fraction close parentheses close parentheses to the power of bold p bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold space bold.$

48 (1+i)
24(1-i)
48(1-i)
24(1+i)

70. 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.
$bold z bold space bold plus bold space bold z with bold bar on top bold space bold less than bold space bold 0$

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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

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