from Mathematics લક્ષ-સાતત્ય અને વિકલન

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Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

111. જો f(x) = |x|, g(x) = sin x અને h(x) = g(x) f(g(x)) તો ....... 
  • h(x) એ સતત વિધેય તેમજ વિકલનીય વિધેય છે.

  • h(x) એ સતત વિધેય છે અને ફક્ત x = 0 આગળ વિકલનીય છે. 

  • h(x) એ અસતત વિધેય છે.

  • h(x) એ સતત વિધેય છે પરંતુ x = 0 આગળ વિકલનીય નથી. 


112.

વિધેય f : R → R માટે નીચે આપેલ પૈકી કયા વિધન સત્ય (T) કે મિથ્યા (F) છે ?

(1) જો |f(x) - f(y)|≤30 |x - y|, ∀x, y, ∈ R, તો f એ R પર સતત વિધેય છે.

(2) જો |f(x) - f(y)|≤30 |x - y|, ∀x, y, ∈ R, તો f એ R પર વિકલનીય વિધેય છે.
(3) જો |f(x) - f(y)|≤21 |x - y|2, ∀x, y, ∈ R, તો f એ R પર વિકલનીય વિધેય છે.
(4) જો |f(x) - f(y)|≤21 |x - y|2, ∀x, y, ∈ R, તો f એ અચળ વિધેય છે.

  • TFTF 

  •  FTTF

  • TFTT

  • TTTT


113.

f એ વિકલનીય વિધેય છે તથા bold f open parentheses bold x over bold y close parentheses bold space bold equals bold space fraction numerator bold f bold left parenthesis bold x bold right parenthesis over denominator bold f bold left parenthesis bold y bold right parenthesis end fraction bold comma x#0, y#0, f(y) # 0 છે. જો f'(1) = 2 હોય તો f'(1) = 2 હોય તો f'(x) = .......

  • 2x f(x)

  • 2 f(x) 

  • fraction numerator bold 2 bold f bold left parenthesis bold x bold right parenthesis over denominator bold x end fraction
  • fraction numerator bold f bold left parenthesis bold x bold right parenthesis bold space over denominator bold x end fraction

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114. bold જ ો bold space bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space bold lim with bold n bold rightwards arrow bold infinity below bold space open parentheses bold n-th root of bold x bold minus bold 1 end root close parentheses bold space bold n bold space bold ત ો bold space fraction numerator bold 1 over denominator bold f bold apostrophe bold left parenthesis bold 2012 bold right parenthesis end fraction bold comma bold space fraction numerator bold 1 over denominator bold f bold apostrophe bold left parenthesis bold 2013 bold right parenthesis end fraction bold comma bold space fraction numerator bold 1 over denominator bold f bold apostrophe bold left parenthesis bold 2014 bold right parenthesis end fraction bold space bold એ bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold.
  • સ્વરિત શ્રેણિમાં હોય.

  • સમાંતર શ્રેણીમાં હોય. 

  • સમગુણોત્તર શ્રેણીમાં હોય. 

  • એક પણ નહિ.


B.

સમાંતર શ્રેણીમાં હોય. 

Tips: -

Error converting from MathML to accessible text.

bold therefore bold space fraction numerator bold 1 over denominator bold f bold apostrophe bold left parenthesis bold 2012 bold right parenthesis end fraction bold comma bold space fraction numerator bold 1 over denominator bold f bold apostrophe bold 2013 bold right parenthesis end fraction bold comma bold space fraction numerator bold 1 over denominator bold f bold apostrophe bold 2014 bold right parenthesis end fractionએ અનુક્રમે 2012, 2013, 2014 થશે. એ સમાંતર શ્રેણીમાં હોય. 

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115. bold જ ો bold space bold y bold space bold equals bold space square root of bold 2 bold x bold space bold minus bold space bold x to the power of bold 2 end root bold space bold મ ા ટ ે bold comma bold comma bold space bold y to the power of bold 3 bold space fraction numerator bold d to the power of bold 2 bold y over denominator bold dx to the power of bold 2 end fraction bold space bold plus bold space bold k bold space bold equals bold space bold 0 bold comma bold space bold હ ો ય bold space bold ત ો bold space bold k bold space bold equals bold space bold. bold. bold. bold. bold. bold.
  • 4

  • 3

  • 2

  • 1


116.
વિધેય f(x) માટે f'(x) + f(x) = 0, ∀ x અને g(x) = [f(x)]2 + [f'(x)]2 તથા g(3) = 8  તો g(8) = ........ 
  • 0

  • 3

  • 5

  • 8


117. bold જ ો bold space bold y bold space bold equals bold space bold 2 to the power of bold x to the power of bold 2 to the power of bold x end exponent end exponent bold space bold x to the power of bold 2 to the power of bold x end exponent bold space bold ત ો bold space bold dy over bold dx= ......
  • bold 2 to the power of bold x to the power of bold 2 to the power of bold x end exponent end exponent bold space bold left parenthesis bold 1 bold space bold minus bold space bold x bold space bold log subscript bold e bold 2 bold right parenthesis bold space bold 2 to the power of bold x bold space bold log subscript bold e bold 2
  • bold space bold 2 to the power of bold x to the power of bold 2 to the power of bold x end exponent end exponent bold space bold left parenthesis bold space bold 1 bold space bold plus bold space bold log subscript bold e bold 2 bold right parenthesis bold space bold 2 to the power of bold x bold space bold left parenthesis bold log subscript bold e bold 2 bold right parenthesis to the power of bold 2
  • bold 2 to the power of bold x to the power of bold 2 to the power of bold x end exponent end exponent bold space bold left parenthesis bold 1 bold space bold plus bold space bold x bold space bold log subscript bold e bold 2 bold right parenthesis bold space bold 2 to the power of bold x bold space bold log subscript bold e bold 2
  • bold space bold 2 to the power of bold x to the power of bold 2 to the power of bold x end exponent end exponent bold space bold x to the power of bold 2 to the power of bold x end exponent bold space bold log subscript bold e bold 2 bold space open parentheses bold 2 bold x bold space bold log subscript bold e bold space bold xlog subscript bold e bold space bold 2 bold plus bold space bold 2 to the power of bold x over bold x close parentheses

118.
જો g : (-∞, ∞) →open parentheses fraction numerator bold minus bold pi over denominator bold 2 end fraction bold comma bold pi over bold 2 close parentheses g(x) 2 tan-1 (ex)-fraction numerator bold minus bold pi over denominator bold 2 end fraction અને f એ g નું પ્રતિવિધેય હોય તો f'(0) = ........ 
  • 4

  • 3

  • 2

  • 1


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119. f(x) [(1+x)(1+X2)(1+X4) ......... n પદ],તો fopen parentheses bold 1 over bold 2 close parentheses = ........ 
  • 4

  • 3

  • 2

  • 1


120.
જો વિધેય g(x) એ વિધેય f(x)નું પ્રતિવિધેય હોય અને f'(x) = fraction numerator bold 1 over denominator bold 1 bold equals bold x to the power of bold 3 end fractionતો g(x) = ....... 
  • 1+[g(x)]3

  • 1+ g(x) 

  • g(x) 

  • fraction numerator bold 1 over denominator bold 1 bold plus bold left square bracket bold g bold left parenthesis bold x bold right parenthesis bold right square bracket to the power of bold 3 end fraction

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