Important Questions of લક્ષ-સાતત્ય અને વિકલન for JEE Mathematics | Zigya

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Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

11. bold lim with bold x bold rightwards arrow bold 0 below bold space fraction numerator bold left parenthesis bold 1 bold minus bold cos bold 2 bold x bold right parenthesis bold left parenthesis bold 3 bold plus bold cosx bold right parenthesis over denominator bold x bold space bold tan bold 4 bold x end fraction bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold space
  • 4

  • 3

  • 2

  • bold 1 over bold 2

12.
જો વિધેય f એ a આગળ ડાબી બાજુ તથા જમણી બાજુ વિકલનીય હોય, તો f એ ...... 
  • a આગળ સતત નથી.

  • a આગળ સતત હોય. 

  • a આગળ વિકલનીય હોય. 

  • 0 આગળ વિકલનીય હોય. 


13.

જો xm yn = (x + y)m+n તો bold dy over bold dx bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold. bold.

  • bold x over bold y
  • bold y over bold x
  • fraction numerator bold x bold plus bold y over denominator bold xy end fraction
  • xy


14. bold lim with bold n bold rightwards arrow bold infinity below bold space fraction numerator bold 1 bold plus bold 2 to the power of bold 4 bold space bold plus bold space bold 3 to the power of bold 4 bold space bold plus bold space bold. bold. bold. bold. bold space bold plus bold space bold n to the power of bold 4 over denominator bold n to the power of bold 5 end fraction bold space bold minus bold space bold lim with bold n bold rightwards arrow bold infinity below bold space fraction numerator bold 1 bold plus bold 2 to the power of bold 3 bold space bold plus bold space bold 3 to the power of bold 3 bold space bold plus bold space bold. bold. bold. bold. bold. bold space bold plus bold space bold n to the power of bold 3 over denominator bold n to the power of bold 5 end fraction bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold space
  • 0

  • bold 1 over bold 4
  • bold 1 over bold 30
  • fraction numerator begin display style bold 1 end style over denominator begin display style bold 5 end style end fraction

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15.

જો f(x) = (1 + x)n, તો f(0) + f(0) +bold 1 over bold 2 fn(0) + ..... +fraction numerator bold 1 over denominator bold n bold factorial end fraction fn(0) = ......

  • 1

  • 2n 

  • 2n-1


16. bold lim with bold x bold rightwards arrow bold 2 below bold space fraction numerator root index bold 3 of bold 3 bold x bold plus bold space bold 2 end root begin display style bold minus end style begin display style bold 2 end style over denominator root index bold 5 of bold x bold space bold plus bold space bold 30 end root begin display style bold minus end style begin display style bold 2 end style end fraction bold equals bold space bold. bold. bold. bold. bold.
  • 10

  • 20

  • 30

  • 40


17. f(x) = |3 - |3 - |x| ||એ કેટલા બિંદુ આગળ વિકલનીય નથી ? 
  • 2

  • 3

  • 5

  • 6


18.
જો α, β એ દ્વિઘાત સમીકરણ ax2 + bx + c = 0 નાં ભિન્ન વાસ્તવિક બીજ હોય, તો bold lim with bold x bold rightwards arrow bold alpha below bold space fraction numerator bold 1 bold minus bold cos bold left parenthesis bold ax bold 2 bold space bold plus bold space bold bx bold space bold plus bold space bold c bold right parenthesis bold space over denominator bold left parenthesis bold x bold space bold minus bold space bold alpha bold right parenthesis to the power of bold 2 end fraction bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. 
  • fraction numerator begin display style bold b to the power of bold 2 bold space bold minus bold space bold 4 bold ac end style over denominator begin display style bold 2 end style end fraction
  • fraction numerator bold b to the power of bold 2 bold space bold plus bold space bold 4 bold ac over denominator bold 2 end fraction bold space
  • 1

  • 0


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19. નીચેનામાંથી કયા બિંદુગણમાં f(x) = fraction numerator bold x over denominator bold 1 bold plus bold 1 bold plus bold 1 end fractionવિકલનીય થશે ? 
  • (-∞, ∞)

  • (0, ∞)

  • (-∞, 0) ∪ (0, ∞) 

  • (-∞, -1) ∪ (-1, ∞) 


20. જો g(x) =open curly brackets table row cell bold k square root of bold x bold plus bold 1 end root bold comma bold space end cell cell bold 0 bold space bold less or equal than bold space bold x bold less or equal than bold space bold 3 end cell row cell bold mx bold space bold plus bold space bold 2 bold comma bold space end cell cell bold 3 bold space bold less or equal than bold space bold x bold space bold less or equal than bold space bold 5 end cell end table close વિકલનીય હોય, તો k + m = ........ 
  • 2

  • 4

  • bold 10 over bold 3
  • bold 16 over bold 5

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