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Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

11.

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) = ......

n
1
2n
2^n-1

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.

a આગળ સતત હોય.

Tips: -

$bold lim with bold h bold rightwards arrow bold 0 bold plus below bold space bold left parenthesis bold f bold left parenthesis bold a bold right parenthesis bold space bold minus bold space bold f bold left parenthesis bold a bold space bold minus bold space bold h bold right parenthesis bold space bold equals bold space bold space bold lim with bold h bold rightwards arrow bold 0 bold plus below bold space fraction numerator bold f bold left parenthesis bold a bold right parenthesis bold space bold minus bold space bold f bold left parenthesis bold a bold space bold minus bold space bold h bold right parenthesis over denominator bold h end fraction bold h bold space bold equals bold space bold space bold f bold apostrophe bold left parenthesis bold a to the power of bold minus bold right parenthesis bold space bold times bold space bold h bold space bold equals bold space bold 0 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 left parenthesis bold f bold apostrophe bold left parenthesis bold a to the power of bold minus bold right parenthesis bold ન ું bold space bold અસ ્ ત િ ત ્ વ bold space bold છ ે bold. bold right parenthesis bold ત ે bold space bold જ bold space bold ર ી ત ે bold space bold space bold lim with bold h bold rightwards arrow bold 0 bold plus below bold space bold left parenthesis bold f bold left parenthesis bold a bold space bold plus bold space bold h bold right parenthesis bold right parenthesis bold space bold minus bold space bold f bold left parenthesis bold a bold right parenthesis bold space bold equals bold space bold 0 bold space bold therefore bold space bold lim with bold h bold rightwards arrow bold 0 bold plus below bold space bold f bold left parenthesis bold a bold space bold minus bold space bold h bold right parenthesis bold right parenthesis bold space bold equals bold space bold lim with bold h bold rightwards arrow bold 0 bold plus below bold space bold f bold left parenthesis bold a bold space bold plus bold space bold h bold right parenthesis bold space bold equals bold space bold f bold left parenthesis bold a bold right parenthesis bold space$

∴ f એ a આગળ સતત છે.

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

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

15.

જો x^m 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

16. 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
$bold 10 over bold 3$
$bold 16 over bold 5$

17.

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

18.

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
$bold 1 over bold 2$

19. નીચેનામાંથી કયા બિંદુગણમાં f(x) = $fraction numerator bold x over denominator bold 1 bold plus bold 1 bold plus bold 1 end fraction$ વિકલનીય થશે ?

(-∞, ∞)
(0, ∞)
(-∞, 0) ∪ (0, ∞)
(-∞, -1) ∪ (-1, ∞)

20.

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.

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Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

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