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CBSE

Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

71.

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.

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

72.

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

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.

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.

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