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Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

91. જો 5 f(x) + 3f $open parentheses bold 1 over bold x close parentheses$ = x + 2 અને y = x f(x), તો $open parentheses bold dy over bold dx close parentheses subscript bold x bold equals bold 1 end subscript bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold. bold.$

$bold 7 over bold 8$
14
-1
1

92.

bold જ ો bold space bold lim with bold x bold rightwards arrow bold infinity below bold space open parentheses square root of bold x to the power of bold 6 bold space bold plus bold space bold ax to the power of bold 5 bold space bold equals bold space bold bx to the power of bold 3 bold space bold plus bold space bold cx bold space bold plus bold space bold d end root bold space bold minus bold space square root of bold x to the power of bold 6 bold space bold minus bold space bold 2 bold x to the power of bold 5 bold space bold plus bold space bold x to the power of bold 3 bold space bold minus bold space bold x bold space bold plus bold space bold 1 bold space end root close parentheses bold space bold equals bold space bold 2 bold comma bold comma bold space bold ત ો bold space

a = -2
b = -2
b = - 5
a = 2

a = -2

Tips: -

$bold આપ ે લ bold space bold લક ્ ષ bold space bold equals bold space bold lim with bold x bold rightwards arrow bold infinity below bold space fraction numerator bold left parenthesis bold a bold space bold plus bold space bold 2 bold right parenthesis bold x to the power of bold 5 bold space bold plus bold space bold left parenthesis bold b bold space bold plus bold space bold 1 bold right parenthesis bold x to the power of bold 3 bold space bold plus bold space bold left parenthesis bold c bold space bold plus bold space bold 1 bold right parenthesis bold x bold space bold plus bold d bold minus bold 1 over denominator square root of bold x to the power of bold 6 bold space bold plus bold space bold ax to the power of bold 5 bold space bold plus bold space bold bx to the power of bold 3 bold space bold plus bold space bold cx bold space bold plus bold space bold d bold space bold plus bold space end root square root of bold x to the power of bold 6 bold space bold minus bold space bold 2 bold x to the power of bold 5 bold space bold plus bold space bold x to the power of bold 3 bold space bold minus bold space bold x bold space bold plus bold space bold 1 end root end fraction bold therefore bold space bold 2 bold space bold equals bold space bold space bold lim with bold x bold rightwards arrow bold infinity below bold space fraction numerator bold left parenthesis bold a bold space bold plus bold space bold 2 bold right parenthesis bold x to the power of bold 2 bold space bold plus bold space bold left parenthesis bold b bold space bold minus bold space bold 1 bold right parenthesis bold x bold space bold plus bold space open parentheses begin display style fraction numerator bold c bold plus bold 1 over denominator bold x to the power of bold 2 end fraction end style close parentheses open parentheses begin display style fraction numerator bold d bold minus bold 1 over denominator bold x to the power of bold 3 end fraction end style close parentheses over denominator open parentheses square root of bold 1 bold plus begin display style bold a over bold x end style begin display style bold b over bold x to the power of bold 3 end style begin display style bold c over bold x to the power of bold 5 end style begin display style bold d over bold x to the power of bold 6 end style end root bold plus bold space square root of bold 1 bold minus begin display style bold 2 over bold x end style begin display style bold 1 over bold x to the power of bold 3 end style begin display style bold 1 over bold x to the power of bold 4 end style begin display style bold 1 over bold x to the power of bold 6 end style end root close parentheses end fraction$

લક્ષનું અસ્તિત્વ હોવાથી a + 2 = 0, $fraction numerator bold b bold minus bold 1 over denominator bold 2 end fraction$ = 2 c, d ∈ R

∴ a = -2, b = 5, c, d ∈ R

93.

bold lim with bold e to the power of bold x bold rightwards arrow bold 1 end exponent below bold space open parentheses bold 4 over bold pi bold tan to the power of bold minus bold 1 end exponent bold x bold minus bold 1 close parentheses to the power of fraction numerator bold 1 over denominator bold x to the power of bold 2 bold minus bold 1 end fraction bold space end exponent bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold space

$bold e to the power of bold n bold space$
$bold e to the power of bold 2 over bold pi end exponent$
$bold e to the power of bold 1 over bold pi end exponent$
$bold space bold e to the power of fraction numerator bold minus bold 1 over denominator bold pi end fraction end exponent$

94.

જો $bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space open curly brackets table attributes columnalign left end attributes row cell bold a bold space bold plus bold space bold sin bold minus bold 1 bold space bold left parenthesis bold x bold space bold plus bold space bold b bold right parenthesis bold. bold space end cell row cell bold x bold comma bold space end cell end table close table row cell bold x bold greater than bold 1 end cell row cell bold x bold less than bold 1 end cell end table$ એ x = 1 આગળ વિકલનીય હોય, તો a = ........ , b = .....

1, -1
-1, 1
1,1
-1, -1

95.

bold lim with bold x bold rightwards arrow bold infinity below bold space fraction numerator begin display style bold sum from bold r bold space bold equals bold space bold 1 to bold 2016 of bold left parenthesis bold r bold space bold plus bold space bold x bold right parenthesis to the power of bold 2016 end style over denominator bold x to the power of bold 2016 bold space bold plus bold space bold 2016 to the power of bold 2016 end fraction bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold space

2014
2015
2016
લક્ષનું અસ્તિત્વ નથી.

96.

ધારો કે વિધેય f એ દ્વિતિય વિકલિત ધરાવતું એવું વિધેય છે, જ્યાં f"(x) = -f(x) અને f'(x) = g(x). જો h(x) = (f(x))² + (g(x))² અને h(5) = 3 તો h(10) = .......

97. જો $bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space open curly brackets table attributes columnalign left columnspacing 1.4ex end attributes row cell bold 1 bold plus bold x bold comma bold space end cell cell bold 0 bold less-than or slanted equal to bold x bold less-than or slanted equal to bold 2 end cell row cell bold 3 bold minus bold x bold comma bold space end cell cell bold 2 bold less than bold x bold less-than or slanted equal to bold 3 end cell end table close$ તો વિધેય f0f કેટલા બિંદુએ વિકલન ન થાય ?

98.

Error converting from MathML to accessible text.

1
-1
0
2

99.

bold જ ો bold space bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space bold lim with bold x bold rightwards arrow bold infinity below bold space fraction numerator bold log subscript bold e bold space bold left parenthesis bold 2 bold space bold plus bold space bold x bold right parenthesis bold space bold minus bold x to the power of bold 2 bold n end exponent bold sinx over denominator bold 1 bold space bold plus bold space bold x to the power of bold 2 bold n end exponent end fraction bold comma bold space bold ત ો bold space

$bold lim with bold x bold rightwards arrow bold 1 bold plus below$ f(x) નું અસ્તિત્વ નથી.
f(x) એ x = 1 અગળ સતત થાય.
$bold lim with bold x bold rightwards arrow bold 1 bold plus below$ f(x) = loge3.
$bold lim with bold x bold rightwards arrow bold 1 bold minus below$ f(x) = - sin 1.

100.

વિધેય g(x) એ વિધેય f(x) નું પ્રતિવિધેય છે. જો વિધેય f(x) એ R પર વિકલનીય હોય તો g"(f)x)) = .....

$fraction numerator bold f bold apostrophe bold left parenthesis bold x bold right parenthesis bold space bold f bold " bold left parenthesis bold x bold right parenthesis bold space bold minus bold space bold f bold left parenthesis bold x bold right parenthesis bold right parenthesis to the power of bold 2 over denominator bold f bold apostrophe bold left parenthesis bold x bold right parenthesis end fraction bold space bold space$
$fraction numerator bold f bold apostrophe bold left parenthesis bold x bold right parenthesis bold space bold f bold " bold left parenthesis bold x bold right parenthesis bold space bold minus bold space bold f bold left parenthesis bold x bold right parenthesis bold right parenthesis to the power of bold 2 over denominator bold left parenthesis bold f bold apostrophe bold left parenthesis bold x bold 0 bold right parenthesis to the power of bold 2 end fraction bold space$
$fraction numerator bold space bold f bold " bold left parenthesis bold x bold right parenthesis over denominator bold left parenthesis bold f bold apostrophe bold left parenthesis bold x bold right parenthesis bold right parenthesis to the power of bold 3 end fraction bold space bold space$
$fraction numerator begin display style bold minus bold f bold " bold left parenthesis bold x bold right parenthesis end style over denominator begin display style bold left parenthesis bold f bold apostrophe bold left parenthesis bold x bold right parenthesis bold right parenthesis to the power of bold 3 end style end fraction bold space bold space$

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Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

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