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Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

71.

સમીકરણ e^x sin x = 1 નાં બે બીજની વચ્ચે સમીકરણ e^x cos x = -1 ને

0< x < $fraction numerator bold 3 bold pi over denominator bold 2 end fraction$ હોય તેવું બીજ મળે.
$fraction numerator bold 3 bold pi over denominator bold 2 end fraction$ < x < 2 હોય તેવું બીજ મળે.
ઓછામાં ઓછું એક વાસ્તવિક બીજ મળે.
એક પણ બીજ પણ ના મળે.

72.

fraction numerator bold d to the power of bold 2 bold x over denominator bold dy to the power of bold 2 end fraction bold space bold equals bold space bold. bold. bold. bold. bold. bold.

$bold space bold space open parentheses fraction numerator bold d to the power of bold 2 bold x over denominator bold dy to the power of bold 2 end fraction close parentheses to the power of bold minus bold 1 end exponent bold space bold space$
$open parentheses fraction numerator bold d to the power of bold 2 bold x over denominator bold dy to the power of bold 2 end fraction close parentheses open parentheses bold dy over bold dx close parentheses to the power of bold minus bold 2 end exponent$
$bold space bold minus open parentheses fraction numerator bold d to the power of bold 2 bold x over denominator bold dy to the power of bold 2 end fraction close parentheses open parentheses bold dy over bold dx close parentheses to the power of bold minus bold 3 end exponent bold space bold space$
$bold minus open parentheses fraction numerator bold d to the power of bold 2 bold x over denominator bold dy to the power of bold 2 end fraction close parentheses to the power of bold minus bold 1 end exponent bold space open parentheses bold dy over bold dx close parentheses to the power of bold minus bold 3 end exponent$

73.

bold lim with bold x bold rightwards arrow bold 0 below bold space bold left parenthesis bold 1 bold space bold minus bold space bold x bold space bold plus bold space bold left square bracket bold x bold space bold minus bold space bold 1 bold right square bracket bold space bold plus bold space bold left square bracket bold 1 bold space bold minus bold space bold x bold right square bracket bold right parenthesis bold space bold. bold. bold. bold. bold. bold. bold. bold. bold. bold જ ્ ય ાં bold space bold left square bracket bold times bold right square bracket bold space bold એ bold space bold મહત ્ તમ bold space bold પ ૂ ર ્ ણ ાં ક bold space bold વ િ ધ ે ય bold space bold છ ે bold. bold space bold space

-1
1
0
લક્ષનું અસ્તિત્વ નથી.

74.

bold lim with bold x bold rightwards arrow bold 0 below bold space fraction numerator bold cos begin display style bold left parenthesis end style begin display style bold sin end style begin display style bold space end style begin display style bold x end style begin display style bold right parenthesis end style begin display style bold space end style begin display style bold minus end style begin display style bold space end style begin display style bold cos end style begin display style bold space end style begin display style bold x end style over denominator bold x to the power of bold 4 end fraction bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold.

$bold 1 over bold 6$
$bold 1 over bold 12$
$bold 2 over bold 3$
$bold 1 over bold 3$

75.

bold જ ો bold space bold space bold x bold space bold equals bold space bold a bold space bold cos bold space bold t bold space square root of bold cos bold space bold 2 bold space bold t end root bold space bold અન ે bold space bold y bold space bold equals bold space bold a bold space bold sin bold space bold t bold space square root of bold cos bold space bold 2 bold space bold t end root bold space bold left parenthesis bold space bold a bold space bold greater than bold space bold 0 bold right parenthesis bold space bold ત ો bold space open vertical bar table row cell fraction numerator bold 1 bold space bold plus bold space open parentheses begin display style bold dy over bold dx end style close parentheses to the power of bold 2 over denominator begin display style fraction numerator bold d to the power of bold 2 bold y over denominator bold dx to the power of bold 2 end fraction end style end fraction end cell end table close vertical bar subscript bold t bold space bold equals bold space bold pi over bold 6 end subscript bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold.

$bold a square root of bold 2 bold space$
$fraction numerator square root of bold 2 over denominator bold 3 bold a end fraction bold space$
$fraction numerator square root of bold 2 bold a end root over denominator bold 3 end fraction$
$fraction numerator begin display style bold a end style over denominator begin display style bold 3 end style end fraction$

76.

નીચેનામાંથી કયું વિધેય અંતરાલ [0,1] પર મધ્યકમાન પ્રમેયની શરતોનું પાલન કરતું નથી ?

f(x) = (x² - 3x + 2) |x - 1|
f(x) = |2x - 1|
$bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space open curly brackets table attributes columnalign left end attributes row cell bold x bold comma bold space end cell row cell bold 1 over bold 2 open parentheses bold 1 over bold 2 bold plus bold x close parentheses bold comma end cell end table close table row cell bold x bold less than bold 1 over bold 2 end cell row cell bold x bold greater-than or slanted equal to bold 1 over bold 2 end cell end table$
$bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space open curly brackets table attributes columnalign left end attributes row cell fraction numerator bold sin bold space bold x over denominator bold x end fraction bold comma end cell row cell bold 1 bold comma end cell end table close table row cell bold x bold # bold 0 end cell row cell bold x bold equals bold 0 end cell end table$

77.

bold જ ો bold space bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space bold product from bold 2 bold equals bold 1 to bold 18 of bold space bold left parenthesis bold space bold x bold space bold minus bold space bold r bold right parenthesis to the power of bold r to the power of bold 2 bold left parenthesis bold 2016 bold minus bold r bold right parenthesis end exponent bold comma bold space bold ત ો bold space fraction numerator bold f bold apostrophe bold left parenthesis bold 2016 bold right parenthesis over denominator bold f bold apostrophe bold left parenthesis bold 2016 bold right parenthesis end fraction bold space bold equals bold space bold. bold. bold. bold. bold space

0
અવિભાજ્ય સંખ્યા
અયુગ્મ વિભાજ્ય સંખ્યા
અસંમેય સંખ્યા

78.

bold જ ો bold space bold D bold space bold left parenthesis bold x bold right parenthesis bold space bold equals bold space open vertical bar table row cell bold tan bold space bold x end cell cell bold tan bold left parenthesis bold x bold space bold plus bold space bold h bold right parenthesis end cell cell bold tan bold left parenthesis bold x bold space bold plus bold space bold 2 bold h bold right parenthesis end cell row cell bold tan bold left parenthesis bold x bold space bold plus bold space bold 2 bold h bold right parenthesis end cell cell bold tan bold space bold x end cell cell bold tan bold left parenthesis bold x bold space bold plus bold space bold h bold right parenthesis end cell row cell bold tan bold left parenthesis bold x bold space bold plus bold space bold h bold right parenthesis end cell cell bold tan bold left parenthesis bold x bold space bold plus bold space bold 2 bold h bold right parenthesis end cell cell bold tan bold space bold x end cell end table close vertical bar bold space bold ત ો bold space bold lim with bold h bold rightwards arrow bold 0 below bold space fraction numerator bold D open parentheses begin display style bold pi over bold 3 end style close parentheses over denominator bold h to the power of bold 2 square root of bold 3 end fraction bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold space

79.

f : R → R એ n ઘાતવાળું બહુપદી વિધેય છે. જ્યાં f(x) = 0 ને n ભિન્ન વાસ્તવિક ઉકેલો છે. સમીકરણ [f'(x)]² - f(x0 f"(x) = 0 ને કેટલા વાસ્તવિક ઉકેલ મળે ?

1
2
n
એક પણ નહિ.

80. જો a + b + c હોય તો (0, 1) અંતરાલમાં સમીકરણ 3ax² + 2bx + c = 0 ને ......

એક વસ્તવિક અને એક સંકર બીજ મળે.
એક પણ બીજ ન મળે.
બે અવાસ્તવિક સંકર બીજ મળે.
ઓછામાં ઓછું એક વાસ્તવિક બીજ મળે.

ઓછામાં ઓછું એક વાસ્તવિક બીજ મળે.

Tips: -

f(x) = ax³ + bx² + cx અંતરાલ [0, 1] પર સતત અને વિકલનીય છે.

f'(x) = 3ax² + 2bx + c

વળી, f(0) = 0, f(1) = a + b + c = 0. આથી f(0) = f(1)

રોલના પ્રમેય પરથી ઓછામાં ઓછો એક k ∈ (0, 1) મળે કે, જેથી f'(k) = 0 થાય.

3ax² + 2bx + c = 0

k એ ax² = 2bx + c નું બીજ થાય.

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Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

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