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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(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.

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.

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

65. 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 = ......

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

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

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

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

48(1-i)

Tips: -

$Error converting from MathML to accessible text.$

$bold equals bold space bold minus bold i bold space open square brackets bold e to the power of begin inline style fraction numerator bold i bold space bold 2 bold space bold pi over denominator bold 11 end fraction end style end exponent bold space bold plus bold space bold e to the power of begin inline style fraction numerator bold i bold space bold 4 bold space bold pi over denominator bold 11 end fraction end style end exponent bold space bold plus bold space bold. bold. bold. bold space bold plus bold space bold e to the power of begin inline style fraction numerator bold i bold space bold 20 bold space bold pi over denominator bold 11 end fraction end style end exponent close square brackets bold space bold equals bold space bold minus bold i bold space bold left parenthesis bold minus bold 1 bold right parenthesis bold space bold equals bold space bold i bold space$

(zⁿ - 1 = 0 નાં બીજ $bold cos bold space fraction numerator bold 2 bold kπ over denominator bold 11 end fraction bold plus bold space bold isin bold space fraction numerator bold 2 bold kπ over denominator bold 11 end fraction bold space bold left parenthesis bold i bold space bold equals bold space bold 0 bold comma bold space bold 1 bold comma bold space bold 2 bold comma bold space bold 3 bold comma bold space bold. bold. bold. bold. bold. bold. bold space bold 10 bold right parenthesis$ છે.

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

જ્યાં $bold S subscript bold 1 bold space bold equals bold space bold sum from bold p bold equals bold 1 to bold 32 of bold space bold pi to the power of bold p bold minus end exponent bold comma bold space bold S subscript bold 2 bold space bold equals bold space bold sum from bold p bold equals bold 1 to bold 32 of bold i to the power of bold p$

$bold S subscript bold 1 bold space bold equals bold space bold i bold space bold plus bold space bold 2 bold i to the power of bold 2 bold space bold plus bold space bold 3 bold i to the power of bold 3 bold space bold plus bold space bold. bold. bold. bold space bold plus bold space bold 32 bold i to the power of bold 32 bold i bold space bold S subscript bold 1 bold space bold equals bold space bold i to the power of bold 2 bold space bold plus bold space bold 2 bold i to the power of bold 3 bold space bold plus bold space bold 3 bold i to the power of bold 4 bold space bold plus bold space bold. bold. bold. bold space bold plus bold space bold 32 bold i to the power of bold 33 bold S subscript bold 1 bold space bold left parenthesis bold 1 bold space bold minus bold space bold i bold right parenthesis bold space bold equals bold space bold i bold space bold plus bold space bold i to the power of bold 2 bold space bold plus bold space bold i to the power of bold 3 bold space bold plus bold space bold. bold. bold. bold space bold plus bold space bold i to the power of bold 32 bold space bold minus bold space bold 32 bold space bold i to the power of bold 33 bold space bold equals bold space bold 8 bold left parenthesis bold i bold space bold plus bold space bold i to the power of bold 3 bold space bold plus bold space bold i to the power of bold 4 bold right parenthesis bold space bold minus bold space bold 32 bold space bold i bold space bold. bold space bold i to the power of bold 32 bold S subscript bold 1 bold space bold equals bold space fraction numerator bold minus bold 32 bold i over denominator bold 1 bold minus bold i end fraction bold equals bold space bold 16 bold space bold left parenthesis bold 1 bold minus bold i bold right parenthesis bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold left square bracket bold left parenthesis bold 1 bold minus bold i bold right parenthesis to the power of bold 2 bold space bold equals bold space bold minus bold 2 bold i bold right square bracket bold space bold S subscript bold 2 bold space bold equals bold space bold 0 bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold left parenthesis bold i bold space bold plus bold space bold i to the power of bold 2 bold space bold plus bold space bold. bold. bold. bold space bold plus bold space bold i to the power of bold 32 bold space bold equals bold space bold 0 bold comma bold space bold i to the power of bold 33 bold space bold equals bold space bold i bold right parenthesis$

∴ માંગેલ કિંમત 3 × 16 (1-i) + 0 = 48 (1-i)

69. 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

70.

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

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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

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