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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

21. જો x = cos θ + i sin θ અને y = cos ϕ + i sin ϕ તો x^m yⁿ + $fraction numerator bold 1 over denominator bold x to the power of bold m bold space bold y to the power of bold n end fraction bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold space bold.$

2cos (mθ-nϕ)
2cos(mθ + nϕ)
cos(mθ - nϕ)
cos(mθ + nϕ)

22.

જો (1+x)ⁿ ના દ્વિપદી વિસ્તરણના સહગુણકો c₀, c₂, ..., c_n હોય, તો નીચે આપેલ વિકલ્પમાંથી કયો વિકલ્પ સત્ય ના બને ?

$bold c subscript bold 1 bold space bold plus bold space bold c subscript bold 5 bold space bold plus bold space bold c subscript bold 9 bold space bold plus bold space bold c subscript bold 0 bold space bold plus bold space bold. bold. bold. bold space bold equals bold space open parentheses bold 2 to the power of bold n bold minus bold 1 end exponent bold space bold plus bold space bold 2 to the power of begin inline style bold n over bold 2 end style end exponent bold space bold sin bold space bold pi close parentheses bold 4$
$Error converting from MathML to accessible text.$
$bold c subscript bold 1 bold space bold minus bold space bold c subscript bold 3 bold space bold minus bold space bold c subscript bold 5 bold space bold minus bold space bold. bold. bold. bold space bold equals bold space bold 2 to the power of begin inline style bold n over bold 2 end style end exponent bold space bold sin bold space bold nπ over bold 4$
$bold c subscript bold 0 bold space end subscript bold minus bold space bold c subscript bold 2 bold space bold plus bold space bold c subscript bold 4 bold space bold minus bold space bold c subscript bold 6 bold space bold plus bold space bold. bold. bold. bold space bold 2 to the power of begin inline style bold n over bold 2 end style end exponent bold space bold cos bold space bold nπ over bold 4$

23.

જો સંકર સંખ્યા z (z ≠ 2)એ સમીકરણ z² = 4z + |z²| + $fraction numerator bold 16 over denominator bold vertical line bold z bold vertical line to the power of bold 3 end fraction$ નું સમાધાન કરે તો |z|⁴ની કિંમત ...... થાય.

24. z₁ અને z₂ એવી સંકર સંખ્યાઓ છે. જ્યાં $open vertical bar fraction numerator bold z subscript bold 1 bold minus bold 2 bold z subscript bold 2 over denominator bold 2 bold minus bold z subscript bold 1 bold space bold z with bold bar on top subscript bold 2 end fraction close vertical bar bold space bold equals bold space bold 1$ તથા |z₂| ≠ 1. બિંદુ એ

2 ત્રિજ્યાવાળા વર્તુળ પર હોય.
4 ત્રિજ્યાવાળા વર્તુળ પર હોય.
X-અક્ષને સમાંતર રેખા પર હોય.
Y-અક્ષને સમાંતર રેખા પર હોય.

25. |z-i| + |z+i| ≤ 4 એ આર્ગન્ડ સમતલમાં કયો પ્રદેશ દર્શાવશે ?

ઉપવલયની અંદરનો ભાગ
વર્તુળની બહારનો ભાગ
ઉપવલય ઉપર તથા તેની અંદરનો ભાગ
વર્તુળ ઉપર તથા તેની અંદરનો ભાગ

26. જો |z| < 1, |v| < 1 અને z = $fraction numerator bold u bold minus bold v over denominator bold 1 bold minus bold uv end fraction$ તો |z| ની ન્યુનતમ કિંમત ....... થાય.

$fraction numerator bold vertical line bold u bold vertical line bold plus bold vertical line bold v bold vertical line over denominator bold 1 bold minus bold vertical line bold u bold vertical line bold vertical line bold v bold vertical line end fraction$
$fraction numerator open vertical bar bold vertical line bold u bold vertical line bold minus bold vertical line bold v bold vertical line close vertical bar over denominator bold 1 bold minus bold vertical line bold u bold vertical line bold vertical line bold v bold vertical line end fraction$
$fraction numerator bold vertical line bold u bold vertical line bold minus bold vertical line bold v bold vertical line over denominator bold 1 bold plus bold vertical line bold u bold vertical line bold vertical line bold v bold vertical line end fraction$
$fraction numerator bold vertical line bold u bold vertical line bold plus bold vertical line bold v bold vertical line over denominator bold 1 bold minus bold vertical line bold u bold vertical line bold vertical line bold v bold vertical line end fraction$

27. જો |z²-1| = |z|² + 1, તો z એ ................ .

કાલ્પનિક અક્ષ પર હોય.
ઉપવલય પર હોય.
વર્તુળ પર હોય.
વાસ્તવિક અક્ષ પર હોય.

28. જો $fraction numerator bold z bold minus bold 1 over denominator bold e to the power of bold iθ end fraction bold space bold plus bold space fraction numerator bold e to the power of bold iθ over denominator bold z bold minus bold 1 end fraction$ કાલ્પનિક ભાગ શુન્ય હોય, તથા $fraction numerator bold z bold minus bold 1 over denominator bold e to the power of bold iθ end fraction$ વાસ્તવિક ન હોય તો z એ

રેખા પર હોય.
વર્તુળ હોય.
પરવલય પર હોય.
ઉપવલય પર હોય.

29. સંકર સંખ્યા z₁ = x₁ + iy₁ અને z₂ = x₂ + iy₂ માટે જો x₁ ≤ x₂ અને y₁ ≤ y₂ તો આપણે z₁ ∩ z₂ વડે દર્શાવીએ.
ધારો કે z એ સંકર સંખ્યા છે જ્યાં 1 ∩ z, તો

$fraction numerator bold 1 bold minus bold z over denominator bold 1 bold plus bold z end fraction bold intersection bold 0$
$fraction numerator bold 1 bold plus bold z over denominator bold 1 bold minus bold z end fraction bold intersection bold 0$
$fraction numerator bold 1 bold minus bold z over denominator bold 1 bold plus bold z end fraction bold intersection bold minus bold i$

30. સમીકરણ zⁿ = (z+1)ⁿ નાં બીજ .....

1/2 ત્રિજ્યાવાળા વર્તુળ પર આવેલ છે.
2n બાજુવાળા નિયમિત બહુકોણ પર આવેલ છે.
n બાજુવાળા નિયમિત બહુકોણ પર આવેલ છે.
2x + 1 = 0રેખા પર આવેલ છે.

2x + 1 = 0રેખા પર આવેલ છે.

Tips: -

$bold z to the power of bold n bold space bold equals bold space bold left parenthesis bold z bold plus bold 1 bold right parenthesis to the power of bold n bold therefore bold space open parentheses bold 1 bold plus bold 1 over bold z close parentheses to the power of bold n bold space bold equals bold space bold 1 bold space bold therefore bold space bold 1 bold space bold plus bold space bold 1 over bold z bold space bold space bold equals bold space bold cos bold space fraction numerator bold 2 bold kπ over denominator bold n end fraction bold space bold plus bold space bold i bold space bold sin bold space fraction numerator bold 2 bold kπ over denominator bold n end fraction bold comma bold space bold k bold space bold equals bold space bold 0 bold comma bold space bold 1 bold comma bold space bold 2 bold comma bold space bold. bold. bold. bold. bold space bold n bold space bold minus bold space bold 1 bold space$

$bold therefore bold space bold k bold space bold equals bold space bold 0$ માટે $bold 1 over bold z bold space bold equals bold space bold 0$ શક્ય નથી.

bold therefore bold space bold z bold space bold equals bold space fraction numerator bold 1 over denominator bold minus bold 1 bold plus bold cos begin display style fraction numerator bold 2 bold kπ over denominator bold n end fraction end style bold plus bold i bold space bold sin bold space begin display style fraction numerator bold 2 bold kπ over denominator bold n end fraction end style end fraction 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. bold comma bold space bold n bold space bold minus bold space bold 1 bold space

bold equals bold space fraction numerator bold minus bold 1 bold plus bold cos begin display style fraction numerator bold 2 bold kπ over denominator bold n end fraction end style bold minus bold i bold space bold sin bold space begin display style fraction numerator bold 2 bold kπ over denominator bold n end fraction end style over denominator open parentheses bold minus bold 1 bold plus bold cos bold space begin display style fraction numerator bold 2 bold kπ over denominator bold n end fraction end style close parentheses bold plus bold sin to the power of bold 2 bold space begin display style fraction numerator bold 2 bold kπ over denominator bold n end fraction end style end fraction bold equals bold space fraction numerator bold minus open parentheses bold 1 bold minus bold cos begin display style fraction numerator bold 2 bold kπ over denominator bold n end fraction end style close parentheses bold minus bold i bold space bold sin bold space begin display style fraction numerator bold 2 bold kπ over denominator bold n end fraction end style over denominator blank end fraction

bold equals bold space fraction numerator bold minus bold 1 over denominator bold 2 end fraction bold space bold minus bold space fraction numerator bold i bold space bold sin bold space begin display style fraction numerator bold 2 bold kπ over denominator bold n end fraction end style over denominator bold 2 open parentheses bold 1 bold minus bold cos begin display style fraction numerator bold 2 bold kπ over denominator bold n end fraction end style close parentheses end fraction bold equals bold space fraction numerator bold minus bold 1 over denominator bold 2 end fraction bold space bold minus bold space open parentheses fraction numerator bold minus bold 1 over denominator bold 2 end fraction bold space bold cot bold kπ over bold n close parentheses

bold therefore bold space bold Re bold space bold left parenthesis bold z bold right parenthesis bold space bold equals bold space fraction numerator bold minus bold 1 over denominator bold 2 end fraction bold.

આથી $bold x bold space bold equals bold space fraction numerator bold minus bold 1 over denominator bold 2 end fraction$

સમીકરણ zⁿ = (z+1)ⁿ ના બીજ રેખા 2x + 1 = 0 પર આવેલ છે.

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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

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