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Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

61.

bold જ ો bold space bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space bold sum from bold r bold equals bold 1 to bold n of bold tan to the power of bold minus bold 1 end exponent bold space open parentheses fraction numerator bold 1 over denominator bold x to the power of bold 2 bold space bold plus bold space bold left parenthesis bold 2 bold r bold space bold minus bold space bold 1 bold right parenthesis bold space bold x bold space bold plus bold space bold left parenthesis bold r to the power of bold 2 bold space bold minus bold space bold r bold space bold plus bold space bold 1 bold right parenthesis end fraction close parentheses bold space bold ત ો bold space bold lim with bold n bold rightwards arrow bold infinity below bold space bold f bold apostrophe bold left parenthesis bold 0 bold right parenthesis bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold space bold space bold space bold space bold space bold space bold left parenthesis bold x bold greater than bold 0 bold right parenthesis

e
-1
$bold 3 over bold 2 bold space$
$bold minus bold e over bold 2$

62.

$Error converting from MathML to accessible text.$ જ્યાં [•] એ મહત્તમ પૂર્ણાંક ભાગ વિધેય છે.

1
-1
0
લક્ષનું અસ્તિત્વ નથી.

63.

ધારો કે, f(x) અને g(x) એ R પર વિકલિત વિધેયો છે. જો f(2) = 0, f(4) = 10 અને g(4)= 8 તો

કોઈક x ∈(2, 4) માટે g'(x) = 4f'(x)
g'(x) > 4 f'(x), ∀x ∈(2, 4)
g(x) > f(x), ∀x ∈ (2, 4)
કોઈક x ∈(2, 4) માટે 3 g'(x) = 4f'(x)

64.

bold જ ો bold space bold space bold lim with bold x bold rightwards arrow bold 0 below bold space open square brackets fraction numerator bold left square bracket bold x bold right square bracket bold space bold plus bold space bold left square bracket bold x to the power of bold 2 bold right square bracket bold space bold plus bold space bold left square bracket bold x to the power of bold 3 bold right square bracket bold space bold plus bold space bold. bold. bold. bold. bold plus bold space bold left square bracket bold x to the power of bold 2 bold n bold plus bold 1 end exponent bold right square bracket bold plus bold n bold plus bold 1 over denominator bold 1 bold plus bold left square bracket bold x bold right square bracket bold plus bold vertical line bold x bold vertical line bold space bold plus bold space bold 2 bold x end fraction close square brackets bold space bold equals bold space bold. bold. bold. bold.

1
-1
0
લક્ષનું અસ્તિત્વ નથી.

65.

bold જ ો bold space bold f bold " bold left parenthesis bold x bold right parenthesis bold space bold equals bold space fraction numerator bold cos bold left parenthesis bold log bold space bold x bold right parenthesis over denominator bold x end fraction bold comma bold space bold f bold apostrophe bold left parenthesis bold 1 bold right parenthesis bold space bold equals bold space bold 0 bold space bold space bold અન ે bold space bold y bold space bold equals bold space bold f open parentheses fraction numerator bold 2 bold x bold space bold plus bold space bold 3 over denominator bold 3 bold minus bold 2 bold x end fraction close parentheses bold space bold ત ો bold space bold dy over bold dx bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold. bold space

$Error converting from MathML to accessible text.$
$bold space fraction numerator bold sin bold left parenthesis bold logx bold right parenthesis bold space over denominator bold cos bold space bold x end fraction$
$fraction numerator bold 1 over denominator bold left parenthesis bold 3 bold space bold minus bold space bold 2 bold x bold right parenthesis to the power of bold 2 end fraction bold space bold sin bold space open parentheses bold log open parentheses fraction numerator begin display style bold 2 bold x bold plus bold 3 end style over denominator begin display style bold 3 bold minus bold 2 bold x end style end fraction close parentheses close parentheses$
$bold space bold sin bold space open parentheses bold log open parentheses fraction numerator bold 2 bold x bold plus bold 3 over denominator bold 3 bold minus bold 2 bold x end fraction close parentheses close parentheses bold space$

66.

bold જ ો bold space bold space bold y bold left parenthesis bold n bold right parenthesis bold space bold equals bold space bold e to the power of bold x bold space bold e to the power of bold x to the power of bold 2 end exponent bold space bold. bold. bold. bold space bold e to the power of bold x to the power of bold n end exponent bold comma bold space bold 0 bold space bold less than bold space bold x bold space bold less than bold space bold 1 bold space bold હ ો ય bold space bold ત ો bold space bold x bold space bold equals bold space bold 1 over bold 2 bold આગળ bold space bold lim with bold x bold rightwards arrow bold infinity below bold space fraction numerator bold dy bold left parenthesis bold n bold right parenthesis over denominator bold dx end fraction bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold space

67.

વિધેય f એ અનૃણ સતત વાસ્તવિક વિધેય છે. જો x ≥ 1 માટે f(x) ≤ p f(x), જ્યાં p > 0 અને f(1) = 0 હોય, તો $bold left square bracket bold f bold left parenthesis square root of bold e bold right parenthesis end root bold space bold plus bold space bold f bold left parenthesis square root of bold pi bold right parenthesis bold right square bracket bold space$

∉R
R^-
R⁺
{0}

{0}

Tips: -

(f'(x) - p f(x)) ≤ 0

$bold therefore bold space bold d over bold dx$ (e^-pxf(x)) ≤ 0

ધારો કે g(x) = e^-pxf(x0

હવે g;(x) ≤ 0 , ∀ x ≥ 1

∴ g(x) ≤ 0. આથી g^-pxf(x) ≤ 0 આથી f(x) ≤ 0, ∀ x ≥ 1

પરંતુ f(x) ≥ 0. આથી f(x) = 0, ∀ x ≥ 1

$bold left square bracket bold f bold left parenthesis square root of bold e bold right parenthesis end root bold space bold plus bold space bold f bold left parenthesis square root of bold pi bold right parenthesis bold right square bracket bold space$

68. જો $bold lim with bold h bold rightwards arrow bold 0 below bold space fraction numerator bold e to the power of bold vertical line bold x bold vertical line bold plus bold left square bracket bold x bold right square bracket end exponent bold minus bold space bold 1 over denominator bold a bold vertical line bold x bold vertical line bold space bold plus bold space bold left square bracket bold x bold right square bracket end fraction bold space$ નું અસ્તિત્વ ન હોય, તો a = ......

1 + e
(1 - e^-1)^-1
1 - e^-1
0

69.

જો $bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space open curly brackets table attributes columnalign left end attributes row cell bold ax bold 2 bold space bold minus bold space bold b bold comma bold space bold vertical line bold x bold vertical line bold greater-than or slanted equal to bold 1 end cell row cell begin inline style fraction numerator bold 1 over denominator bold vertical line bold x bold vertical line end fraction end style bold comma bold space bold space bold vertical line bold x bold vertical line bold greater-than or slanted equal to bold 1 end cell end table close$ એ x = 1 આગળ વિકલનીય હોય, તો a = ....... , b = ..........

$bold 1 over bold 2 bold comma bold 3 over bold 2$
$fraction numerator bold minus bold 1 over denominator bold 2 end fraction bold comma fraction numerator bold minus bold 3 over denominator bold 2 end fraction bold space$
$fraction numerator bold minus bold 1 over denominator bold 2 end fraction bold comma bold 3 over bold 2 bold space$
$bold 1 over bold 2 bold comma fraction numerator bold minus bold 3 over denominator bold 2 end fraction$

70.

જો દ્વિઘાત સમીકરણ f(x) = ax² + bx + c = 0 (a#0) ને બે એકબીજાના વ્યસ્ત ધન બીજ હોય તો

f'(1) = 0
af'(1) < 0
af'(1) > 0
af'(1) વિશે કઈ તારણ ન મળે.

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Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

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