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Gujarati JEE Mathematics : શ્રેણિક

Multiple Choice Questions

51. જો $bold space bold A bold space bold equals bold space open square brackets table row bold 1 bold 2 row cell bold minus bold 2 end cell cell bold minus bold 1 end cell end table close square brackets$ નો લંબ શ્રેણિક ....... થાય.

$open square brackets table row bold 0 bold 1 row cell bold minus bold 1 end cell cell bold minus bold 1 end cell end table close square brackets$
$bold space open square brackets table row bold 1 bold 1 row cell bold minus bold 1 end cell bold 0 end table close square brackets$
$bold space open square brackets table row bold 0 bold 1 row bold 1 bold 0 end table close square brackets$
$open square brackets table row bold 1 bold 0 row bold 0 bold 1 end table close square brackets$

52.

જો a,b,c શુન્યેત્તર ધન વાસ્તવિક સંખ્યાઓ હોય, તો x, y, z ની સમીકરણ સંહતિ
$bold x to the power of bold 2 over bold y to the power of bold 2 bold minus bold y to the power of bold 2 over bold b to the power of bold 2 bold minus bold z to the power of bold 2 over bold c to the power of bold 2 bold space bold equals bold space bold 1 bold comma bold minus bold space fraction numerator begin display style bold x to the power of bold 2 end style over denominator begin display style bold y to the power of bold 2 end style end fraction bold plus fraction numerator begin display style bold y to the power of bold 2 end style over denominator begin display style bold b to the power of bold 2 end style end fraction bold minus fraction numerator begin display style bold z to the power of bold 2 end style over denominator begin display style bold c to the power of bold 2 end style end fraction bold equals bold 1 bold comma bold space bold space bold minus fraction numerator begin display style bold x to the power of bold 2 end style over denominator begin display style bold y to the power of bold 2 end style end fraction bold minus fraction numerator begin display style bold y to the power of bold 2 end style over denominator begin display style bold b to the power of bold 2 end style end fraction bold plus fraction numerator begin display style bold z to the power of bold 2 end style over denominator begin display style bold c to the power of bold 2 end style end fraction bold equals bold space bold 1 bold comma$

અનંત ઉકેલો છે.
એકથી વધુ સાન્ત ઉકેલો છે.
અનન્ય ઉકેલ છે.
ઉકેલ નથી.

53.

જે સમીકરણો x + 3y + z = 0, 2x - y - z = 0, kx + 2y + 3z = 0 નો શુન્યેતર ઉકેલ મળે નહિ તો k = ..... (x#0, y#0, z#0)

$fraction numerator begin display style bold 9 end style over denominator begin display style bold 2 end style end fraction$
$bold 13 over bold 2$
$bold minus bold 13 over bold 2$
$bold minus fraction numerator begin display style bold 15 end style over denominator begin display style bold 2 end style end fraction$

bold minus fraction numerator begin display style bold 15 end style over denominator begin display style bold 2 end style end fraction

Tips: -

$open square brackets table row bold 1 bold 3 bold 1 row bold 2 cell bold minus bold 1 end cell cell bold minus bold 1 end cell row bold k bold 2 bold 3 end table close square brackets bold space open square brackets table row bold x row bold y row bold z end table close square brackets bold space bold equals bold space open square brackets table row bold 0 row bold 0 row bold 0 end table close square brackets bold space bold rightwards double arrow bold space bold AX bold space bold equals bold space bold D bold space bold space bold space bold જ ્ ય ાં bold space bold A bold space bold equals bold space open square brackets table row bold 1 bold 3 bold 1 row bold 2 cell bold minus bold 1 end cell cell bold minus bold 1 end cell row bold k bold 2 bold 3 end table close square brackets bold શ ુ ન ્ ય ે તર bold space bold ઉક ે લ bold space bold મ ા ટ ે bold space bold equals bold space bold A bold space bold equals bold space open square brackets table row bold 1 bold 3 bold 1 row bold 2 cell bold minus bold 1 end cell cell bold minus bold 1 end cell row bold k bold 2 bold 3 end table close square brackets bold space bold equals bold space bold 0 bold space$

1[-3 + 2] -3 [6 + k] + 1 [4 + k] = 0

-1 - 18 - 3k + 4 + k = 0

-15 -2k = 0

$bold therefore bold space bold k bold space bold equals bold space bold minus bold 15 over bold 2$

54. જો $bold A bold space bold equals bold space open square brackets table row bold 3 bold 1 row cell bold minus bold 9 end cell cell bold minus bold 1 end cell end table close square brackets$ તો I + 2A + 3A2 + 4A3 + ........ ∞ = ..........

$open square brackets table row bold 7 bold 2 row cell bold minus bold 18 end cell cell bold minus bold 5 end cell end table close square brackets bold space$
$open square brackets table row bold 4 bold 1 row cell bold minus bold 9 end cell cell bold minus bold 1 end cell end table close square brackets bold space$
$open square brackets table row bold 7 bold 2 row cell bold minus bold 5 end cell cell bold minus bold 18 end cell end table close square brackets$
$bold space open square brackets table row bold 9 bold 1 row cell bold minus bold 9 end cell bold 0 end table close square brackets$

55.

bold જ ો bold space bold A bold space bold equals bold space open square brackets table row bold 1 bold 1 row bold 1 bold 1 end table close square brackets bold space bold ત ો bold space bold space bold A to the power of bold 100 bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold space

I
2¹⁰⁰A
2⁹⁹A
2¹⁰¹A

56.

શ્રેણિકોના ગુણાકારને સાપેક્ષ સમૂહ $bold M bold space bold equals bold space open curly brackets open square brackets table row bold x bold x bold x row bold x bold x bold x row bold x bold x bold x end table close square brackets bold comma bold space bold x bold element of bold R bold comma bold x bold # bold 0 close curly brackets$ નો એકમ ઘટક ....... છે.

$open square brackets table row bold 1 bold 1 bold 0 row bold 0 bold 1 bold 0 row bold 0 bold 0 bold 1 end table close square brackets$
$open square brackets table row bold 1 bold 1 bold 1 row bold 1 bold 1 bold 1 row bold 1 bold 1 bold 1 end table close square brackets$
$open square brackets table row cell bold 1 over bold 3 end cell cell bold 1 over bold 3 end cell cell bold 1 over bold 3 end cell row cell bold 1 over bold 3 end cell cell bold 1 over bold 3 end cell cell bold 1 over bold 3 end cell row cell bold 1 over bold 3 end cell cell bold 1 over bold 3 end cell cell bold 1 over bold 3 end cell end table close square brackets bold space bold space$
$open square brackets table row cell bold 1 over bold 2 end cell cell bold 1 over bold 2 end cell cell bold 1 over bold 2 end cell row cell bold 1 over bold 2 end cell cell bold 1 over bold 2 end cell cell bold 1 over bold 2 end cell row cell bold 1 over bold 2 end cell cell bold 1 over bold 2 end cell cell bold 1 over bold 2 end cell end table close square brackets$

57.

જો [x] એ x નું મહત્તમ પૂર્ણાંક ભાગ વિધેય દર્શાવે તો નીચેના શ્રેણિકના નિશ્ચાયકનું મૂલ્ય .......... છે.

$open square brackets table row cell bold left square bracket bold e bold right square bracket end cell cell bold left square bracket bold pi bold right square bracket end cell cell bold left square bracket bold pi to the power of bold 2 bold minus bold 6 bold right square bracket end cell row cell bold left square bracket bold pi bold right square bracket end cell cell bold left square bracket bold pi bold 2 bold minus bold 6 bold right square bracket end cell cell bold left square bracket bold e bold right square bracket end cell row cell bold left square bracket bold pi to the power of bold 2 bold minus bold 6 bold right square bracket end cell cell bold left square bracket bold e bold right square bracket end cell cell bold left square bracket bold pi bold right square bracket end cell end table close square brackets$

-8
8
1
0

58.

જો સમીકરણો y + z = - ax,z + x = -by, x + y = - cz નો શુન્યોતર ઉકેલ શક્ય હોય, તો $fraction numerator bold 1 over denominator bold 1 bold minus bold a end fraction bold plus fraction numerator bold 1 over denominator bold 1 bold minus bold b end fraction bold plus fraction numerator bold 1 over denominator bold 1 bold minus bold c end fraction bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold left parenthesis bold x bold space bold plus bold space bold y bold space bold plus bold space bold z bold space bold # bold space bold 0 bold right parenthesis bold space$

1
•2
-1
-2

59.

bold જ ો bold space bold A bold space bold equals bold space open square brackets table row bold 1 bold tanx row cell bold minus bold tanx end cell bold 1 end table close square brackets bold space bold ત ો bold space bold A to the power of bold T bold A to the power of bold minus bold 1 end exponent bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold space

$open square brackets table row cell bold son bold 2 bold x end cell cell bold cos bold 2 bold x end cell row cell bold cos bold 2 bold x end cell cell bold sin bold 2 bold x end cell end table close square brackets bold space$
$open square brackets table row bold tanx bold 1 row cell bold minus bold 1 end cell bold tanx end table close square brackets bold space$
$open square brackets table row cell bold cos bold 2 bold x end cell cell bold minus bold sin bold 2 bold x end cell row cell bold sin bold 2 bold x end cell cell bold cos bold 2 bold x end cell end table close square brackets$
$open square brackets table row cell bold minus bold cos bold 2 bold x end cell cell bold sin bold 2 bold x end cell row cell bold minus bold sin bold 2 bold x end cell cell bold cos bold 2 bold x end cell end table close square brackets$

60.

3×3 શ્રેણિક A એ સમીકરણ A² - 5A + 71 = 0 નુ સમાધાન કરે છે. જો A⁵ = aA + bI તો 2a = 3b નું મુલ્ય .......... છે.

1453
1435
4135
3135

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Gujarati JEE Mathematics : શ્રેણિક

Multiple Choice Questions

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