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CBSE

Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

61.

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

62.

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.

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.

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

1
-1
0
લક્ષનું અસ્તિત્વ નથી.

65.

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 = ..........

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

66.

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}

67. 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

(1 - e^-1)^-1

Tips: -

x < 0 તો -1< x < 0 લેતાં |x| = - x [x] = -1

$bold lim with bold h bold rightwards arrow bold 0 below bold space fraction numerator bold e to the power of bold vertical line bold x bold vertical line bold plus bold left square bracket bold x bold right square bracket end exponent bold minus bold space bold 1 over denominator bold a bold vertical line bold x bold vertical line bold space bold plus bold space bold left square bracket bold x bold right square bracket end fraction bold space bold space bold equals bold space bold lim with bold h bold rightwards arrow bold 0 below bold space fraction numerator bold e to the power of bold minus bold x bold minus bold 1 end exponent bold space bold minus bold space bold 1 over denominator bold minus bold ax bold space bold minus bold space bold 1 end fraction bold space bold equals bold space fraction numerator bold e to the power of bold minus bold 1 end exponent bold space bold minus bold 1 over denominator bold minus bold 1 end fraction bold space bold equals bold space bold 1 bold minus bold 1 over bold e$

x > 0 તો 0< x < 1 લેેેેતાં |x| = x, [x] = 0

$bold lim with bold h bold rightwards arrow bold 0 below bold space fraction numerator bold e to the power of bold vertical line bold x bold vertical line bold plus bold left square bracket bold x bold right square bracket end exponent bold minus bold space bold 1 over denominator bold a bold vertical line bold x bold vertical line bold space bold plus bold space bold left square bracket bold x bold right square bracket end fraction bold space bold equals bold space bold lim with bold h bold rightwards arrow bold 0 below bold space fraction numerator bold e to the power of bold x bold space bold minus bold space bold 1 over denominator bold ax end fraction bold space bold equals bold space bold 1 over bold a bold space$

લક્ષનુંં અસ્તિત્વ હોવાથી , 1 - e^-1 = $bold 1 over bold a$

∴ a = (1 - e^-1)^-1

68.

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.

69.

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

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