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Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

11.

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

12. 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, ∞)

13.

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.

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

15. 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
$bold 10 over bold 3$
$bold 16 over bold 5$

Tips: -

જો g(x) = વિકલનીય હોય, તો g એ x = 3 આગળ સતત પણ હોય.

∴ $bold lim with bold x bold rightwards arrow bold 3 to the power of bold plus below$ g(x) = g(3)

∴3m + 2 = 2k

g(x) વિકલનીય છે. ∴ g'(3+) = g'(3-)

∴ $bold m bold space bold equals bold space bold k over bold 4$ . આથી k = 4m આથી 3m + 2 = 8m

$bold m bold space bold equals bold space bold 2 over bold 5 bold comma bold space bold k bold space bold equals bold space bold 8 over bold 5$ . આથી k + m = 2

16.

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

17.

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.

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.

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

20.

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

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Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

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