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CBSE

Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

71.

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

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

72.

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.

73.

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.

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

74.

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.

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

75.

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.

76.

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

77.

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.

ઓછામાં ઓછું એક વાસ્તવિક બીજ મળે.

Tips: -

ધારો કે, f(x) = e^-x - sin x અને α, β (α < β) એ f(x) નાં બે શૂન્ય છે. f(x) એ [α, β] પર સતત અને વિકલનીય છે તથા f(α) = f(β) = 0

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

∴ f'(x) = -e^-x - cos x

f'(x) = 0 ⇒ - e^-k -cos k = 0 ⇒ -e^-k (1 + e^k cos k) = 0 ⇒ e^k cos k = - 1

∴ k એ e^x cos x = - 1 નું બીજ થાય.