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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. 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.

23. |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

24. 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.

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

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

Tips: -

bold z bold space bold equals bold space fraction numerator bold u bold minus bold v over denominator bold 1 bold minus bold uv end fraction bold vertical line bold i bold minus bold v bold vertical line bold space bold greater than bold space bold vertical line bold thin space bold vertical line bold u bold vertical line bold space bold minus bold space bold v bold vertical line bold vertical line 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 space bold space bold space bold space bold space bold space bold space bold. bold. bold. bold space bold left parenthesis bold 1 bold right parenthesis

ધારો કે $bold u bold space bold equals bold space bold re to the power of bold iθ bold comma bold space bold v bold space bold equals bold space bold ρe to the power of bold iϕ bold semicolon bold space bold r bold comma bold space bold rho bold space bold less than bold space bold 1$
∴ |u| = r, |v| = ρ

|1-uv|² = |1-r ρ r^i(θ+ϕ)|² = |1-rρcos (θ + ϕ) - irρsin(θ+ϕ)|²

= 1 + r²ρ² cos²(θ+ϕ) + r² ρ² sin² (θ+ϕ) -2rρ cos (θ+ϕ)

= 1 + r²ρ² - 2rρ cos (θ+ϕ)

≤ (1-rρ)² = (1-|u| |v|)² (cos(θ+ϕ) ≤1))

∴ |1-uv| ≤ (1-|u||v|)

(1) અને (2) પરથી

$bold vertical line bold z bold vertical line bold space bold equals bold space open vertical bar fraction numerator bold u bold minus bold v over denominator bold 1 bold minus bold uv end fraction close vertical bar bold space bold greater or equal than bold space fraction numerator open vertical bar bold vertical line bold u bold vertical line bold minus bold vertical line bold v bold vertical line close vertical bar over denominator bold 1 bold minus bold vertical line bold u bold vertical line bold vertical line bold v bold vertical line end fraction$

25.

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

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

26.

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.

27. 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.

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

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

29. 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.

30. 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

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

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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

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