f(x) [(1+x)(1+X2)(1+X4) ......... n પદ],તો f = ........  from Mathematics લક્ષ-સાતત્ય અને વિકલન

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Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

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111. f(x) [(1+x)(1+X2)(1+X4) ......... n પદ],તો fopen parentheses bold 1 over bold 2 close parentheses = ........ 
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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 curly brackets fraction numerator bold left parenthesis bold 1 bold minus bold x bold right parenthesis bold left parenthesis bold 1 bold plus bold x bold right parenthesis bold left parenthesis bold 1 bold plus bold x to the power of bold 2 bold right parenthesis bold left parenthesis bold 1 bold plus bold x to the power of bold 4 bold right parenthesis bold. bold. bold n over denominator bold 1 bold space bold minus bold space bold x end fraction close curly brackets

bold lim with bold n bold rightwards arrow bold infinity below bold space fraction numerator bold 1 bold space bold minus bold space bold x to the power of bold 2 bold n end exponent over denominator bold 1 bold minus bold x end fraction bold space bold equals bold space fraction numerator bold 1 over denominator bold 1 bold space bold minus bold space bold x end fraction bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space open parentheses bold lim with bold n bold rightwards arrow bold infinity below bold space bold x to the power of bold n bold equals bold space bold 0 bold comma bold space bold vertical line bold x bold vertical line bold space bold less than bold space bold 1 close parentheses bold space

bold therefore bold space bold f open parentheses bold 1 over bold 2 close parentheses bold space bold equals bold space fraction numerator bold 1 over denominator bold 1 bold minus begin display style bold 1 over bold 2 end style end fraction bold space bold equals bold space bold 2

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112. 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. 


113. 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.


114.
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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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.
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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) = ........ 
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117.
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) = ........ 
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118.

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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119.

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


120. 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

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