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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

21.

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

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

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

Tips: -

(1 + x)ⁿ = c₀ + c₁ + c₂x² + c₃x³ + ... (1)

(1) માં x = i મૂકતાં

(1+i)n = c₀ + c₁i - c₂ - c₃i + c₄ + ... (2)

સમીકરણ (1) માં x = -i મૂકતાં

(1-i)ⁿ = c₀ - c₁i - c₂ + c₃i + c₄ + ... (3)

(2) અને (3) નો સરવાળો કરતાં,

2 (c₀ - c₂ + c₄ - c₆ + ... ) = (1+i)ⁿ + (1-i)ⁿ

∴ c₀ - c₂ + c₄ - c₆ + ... = $fraction numerator bold left parenthesis bold 1 bold plus bold i bold right parenthesis to the power of bold n bold plus bold left parenthesis bold 1 bold minus bold i bold right parenthesis to the power of bold n over denominator bold 2 end fraction$

$bold equals bold space bold 2 to the power of begin inline style bold n over bold 2 end style end exponent bold space open parentheses fraction numerator open parentheses bold cos begin display style bold pi over bold 4 end style bold space bold plus bold space bold i bold space bold sin bold space begin display style bold pi over bold 4 end style close parentheses to the power of bold n bold space bold plus bold space open parentheses bold cos begin display style bold pi over bold 4 end style bold minus bold space bold i bold space bold sin bold space begin display style bold pi over bold 4 end style close parentheses to the power of bold n over denominator bold 2 end fraction close parentheses bold equals bold space bold 2 to the power of begin inline style bold n over bold 2 end style end exponent bold space open parentheses fraction numerator bold cos begin display style bold nπ over bold 4 end style bold plus bold i bold space bold sin bold space begin display style bold nπ over bold 4 end style bold space bold plus bold space bold cos bold space begin display style bold nπ over bold 4 end style bold space bold minus bold space bold i bold space bold sin bold space begin display style bold nπ over bold 4 end style over denominator bold 2 end fraction close parentheses$

∴ c₀ - c₂ + c₄ - c₆ + ... = $bold 2 to the power of begin inline style bold n over bold 2 end style end exponent bold space bold cos bold space bold nπ over bold 4$ (4)
તે જ રીતે (2) અને (3)ની બાદબાકી કરતાં,

2i (c₁ - c₃ + c₅) = (1+i)ⁿ - (1-i)ⁿ

$bold c subscript bold 1 bold space bold minus bold space bold c subscript bold 3 bold space bold plus bold space bold c subscript bold 5 bold space bold. bold. bold. bold space bold equals bold space bold 2 to the power of begin inline style bold n over bold 2 end style end exponent bold space bold sin bold space bold nπ over bold 4$ (5)

∴ આપણે જાણીએ છીએ કે c₁ + c₃ + c₅ + ... = 2^n-1 (6)

સમીકરણ (5) અને (6) નો સરવાળો કરતાં,

$bold 2 bold space bold left parenthesis bold c subscript bold 1 bold plus bold space bold c subscript bold 5 bold space bold plus bold space bold c subscript bold 9 bold space bold plus bold space bold. bold. bold. bold right parenthesis bold space bold equals bold space bold 2 to the power of bold n bold minus bold 1 end exponent bold space bold plus bold space bold 2 to the power of begin inline style bold n over bold 2 end style end exponent bold space bold sin bold space bold nπ over bold 4 bold therefore bold space bold c subscript bold 1 bold space bold plus bold space bold c subscript bold 5 bold space bold plus bold space bold c subscript bold 9 bold space bold plus bold space bold. bold. bold. bold space bold equals bold space bold 1 over bold 2 bold space open parentheses bold 2 to the power of bold n bold minus bold 1 end exponent bold space bold 2 to the power of begin inline style bold n over bold 2 end style end exponent bold space bold sin bold space bold nπ over bold 4 close parentheses$

22. 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.

23. 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

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

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

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

25. 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.

26. 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)

27. 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.

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

28. 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.

29. |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

30.

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.

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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

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