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Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

bold lim with bold alpha bold rightwards arrow bold pi over bold 4 below bold space fraction numerator bold sin begin display style bold alpha end style begin display style bold space end style begin display style bold minus end style begin display style bold space end style begin display style bold cosα end style over denominator bold alpha bold minus begin display style bold pi over bold 4 end style end fraction bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold space

0
1
2
$square root of bold 2$

2. જો $bold space bold f bold left parenthesis bold x bold right parenthesis bold space open curly brackets table attributes columnalign left end attributes row cell fraction numerator bold sin bold left parenthesis bold a bold right parenthesis bold 1 bold right parenthesis bold x bold plus bold sinx over denominator bold x end fraction bold comma end cell row cell bold c bold comma end cell row cell fraction numerator square root of bold x bold plus bold bx to the power of bold 2 end root bold space bold minus bold space square root of bold x over denominator bold bx to the power of begin display style bold 3 over bold 2 end style end exponent end fraction bold space bold comma end cell end table close table row cell bold x bold less than bold 0 end cell row cell bold x bold space bold equals bold space bold 0 bold space end cell row cell bold x bold greater than bold 0 end cell end table$ એ x = 0 આગળ સતત હોય તો ........

$bold a bold space bold equals bold space fraction numerator bold minus bold 3 over denominator bold 2 end fraction bold comma bold space bold b bold space bold element of bold space bold R bold comma bold space bold c bold space bold equals bold space bold 1 over bold 2$
$bold a bold space bold equals bold space fraction numerator bold minus bold 3 over denominator bold 2 end fraction bold comma bold space bold b bold space bold equals bold space bold 0 bold comma bold space bold c bold equals bold 1 over bold 2$
$bold a bold space bold equals bold space bold minus bold 3 over bold 2 bold comma bold space bold b bold space bold equals bold space bold 1 bold comma bold space bold c bold space bold equals bold space bold 1 over bold 2$
$bold a bold space bold equals bold space fraction numerator bold minus bold 3 over denominator bold 2 end fraction bold comma bold space bold b bold space bold element of bold space fraction numerator bold minus bold 1 over denominator bold 2 end fraction$

ધારો કે f(x) = g(x) $fraction numerator bold e to the power of begin display style bold 1 over bold x end style end exponent bold space bold minus bold space bold e to the power of bold 1 over bold x end exponent over denominator bold e to the power of bold 1 over bold x end exponent bold space bold plus bold space bold e to the power of bold 1 over bold x end exponent end fraction$ જ્યાં, g સતત વિધેય છે. જો $bold lim with bold x bold rightwards arrow bold 0 below$ f(x) નું અસ્તિત્વ હોય, તો

g(x) = xh(x) જ્યાં h(x) બહુપદી વિધેય
g(x) એ અચળ વિધેય છે.
g(x) = x + 2
g(x) = x² + 4

જો $bold lim with bold x bold rightwards arrow bold 0 below bold space fraction numerator bold log begin display style bold space end style begin display style bold left parenthesis end style begin display style bold space end style begin display style bold 3 end style begin display style bold space end style begin display style bold plus end style begin display style bold space end style begin display style bold x end style begin display style bold right parenthesis end style begin display style bold space end style begin display style bold minus end style begin display style bold space end style begin display style bold log end style begin display style bold left parenthesis end style begin display style bold 3 end style begin display style bold space end style begin display style bold minus end style begin display style bold space end style begin display style bold x end style begin display style bold right parenthesis end style over denominator bold x end fraction bold space bold equals bold space bold k bold space bold ત ો bold space bold k bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold space$

0
$fraction numerator bold minus bold 1 over denominator bold 3 end fraction$
$fraction numerator bold minus bold 2 over denominator bold 3 end fraction$
$bold 2 over bold 3$

bold lim with bold alpha bold rightwards arrow bold 0 below bold space fraction numerator square root of bold 1 bold minus bold cos bold 2 square root of bold x end root over denominator bold 2 square root of bold x end fraction bold space bold equals bold space bold. bold. bold. bold. bold. bold.

$fraction numerator bold 1 over denominator square root of bold 2 end fraction$
$bold 1 over bold 2$
1
લક્ષનું અસ્તિત્વ નથી.

bold lim with bold n bold rightwards arrow bold infinity below bold space open curly brackets bold tan bold x over bold 2 bold secx bold plus bold tan bold x over bold 2 to the power of bold 2 bold sec bold x over bold 2 bold plus bold space bold. bold. bold. bold. bold space bold plus bold space bold tan open parentheses bold x over bold 2 to the power of bold n close parentheses bold sec open parentheses bold x over bold 2 to the power of bold n bold minus bold 1 end exponent close parentheses close curly brackets

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tan x
cot x
sec x

bold n bold space bold element of bold space bold N

ની કઈ કિંમત માટે $bold lim with bold x bold rightwards arrow bold 0 below bold space bold equals bold space fraction numerator bold left parenthesis bold cosx bold space bold minus bold space bold 1 bold right parenthesis bold space bold left parenthesis bold cos bold space bold x bold space bold minus bold space bold e to the power of bold x bold right parenthesis over denominator bold x to the power of bold n end fraction$ સાન્ત શુન્યેત્તર સંખ્યા મળે ?

શુન્યેત્તર સતત વિધેય f એ f(x + y) = f(x) f(y0, ∀ x, y ∈R નું પાલન કરે છે. જો f(2) = 9 તો f(3) = .......

વિધેય f સતત છે. જો f(x) f $open parentheses bold 1 over bold x close parentheses$ = f(x) + f $open parentheses bold 1 over bold x close parentheses bold comma bold space bold for all$ x ∈ Df અને f(1) > તો $bold lim with bold x bold rightwards arrow bold 1 below$ f(x) = .......

1
2
3
લક્ષ્યનું અસ્તિત્વ નથી.

Tips: -

વિધેય f સતત હોવાથી $bold lim with bold x bold rightwards arrow bold 1 below$ f(x) = f(1)

આપેલ સમીકરણમાં x = 1 મૂકતાં, (f(1))² 2f(1)

f(1) = 0 અથવા f(1) = 2. પરંતુ f(10 >0 હોવાથી, f(1) = 2

$bold lim with bold x bold rightwards arrow bold 1 below$ f(x) = 2

10.

$bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space open curly brackets table attributes columnalign center end attributes row cell fraction numerator bold log bold left parenthesis bold 1 bold plus bold x bold plus bold x to the power of bold 2 bold right parenthesis bold space bold plus bold space bold log bold left parenthesis bold 1 bold minus bold x bold plus bold x to the power of bold 2 bold right parenthesis over denominator bold sec bold space bold x bold space bold minus bold space bold cosx end fraction bold semicolon bold x bold # bold 0 end cell row cell bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold k bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold semicolon bold space bold x bold equals bold space bold 0 bold space end cell end table close$ જો વિધેય f એ x = 0 આગળ સતત હોય, તો k = ..........

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Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

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