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Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

51.

f(x) = [x] sin $bold pi$ x, વિધેયનું fનું x = k,k ∈ Z આગળ ડાબી બાજુનું વિકલિત ...... છે.

(-1)^k-1 (k-1) $bold pi$
(-1)^k k $bold pi$
(-1)^k-1 k $bold pi$
(-1)^k (k -1) $bold pi$

52.

ધારો કે f : R →R, f(x+y) = f(x) + f(y); ∀ x, y ∈ R. જો f(x) એ x = 0 આગળ વિકલનીય હોય, તો

f એ R પર સતત નથી.
f(x) એ અમુક બિંદું સિવાય વિકલનીય છે.
f(x) એ (a, b)જ્યાં a < 0 < b અંતરાલમાં વિકલનીય હોય.
f(x) = અચળ

53.

bold જ ો bold space bold lim with bold x bold rightwards arrow bold 0 below bold space bold left square bracket bold 1 bold plus bold xlog bold left parenthesis bold 1 bold plus bold b to the power of bold 2 bold right parenthesis bold right square bracket to the power of bold 1 over bold x end exponent bold space bold equals bold space bold 2 bold space bold b bold space bold sin to the power of bold 2 bold space bold theta bold comma bold space bold b bold space bold greater than bold space bold 0 bold space bold અન ે bold space bold theta bold space bold element of bold space bold left square bracket bold minus bold pi bold comma bold pi bold right square bracket bold space bold ત ો bold space bold theta bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold.

$bold plus-or-minus bold pi over bold 2$
$bold plus-or-minus bold pi over bold 4$
$bold plus-or-minus bold pi over bold 3$
$bold plus-or-minus bold pi over bold 6$

54.

ધારો કે [0,1] માં વિધેય f ના દ્વિતિય વિકલિતનું અસ્તિત્વ છે અને |f"(x)| ≤ 1, ∀ x ∈[0,1]. જો f(0) = f(1) હોય તો, અંતરાલ (0,1) માં

|f'(x)| > 1
|f'(x)|>1
|f''(x)| < 1
|f(x)|<1

55.

જો y = (1+x) (1+x²) (1+x⁴) ... + log (1+x²ⁿ) તો x = 0 આગળની કિંમત ......

1
-1
2
0

56.

bold જ ો bold space bold space bold y bold space bold equals bold space bold cot to the power of bold minus bold 1 end exponent open parentheses fraction numerator bold log open parentheses begin display style bold e over bold x to the power of bold 2 end style close parentheses over denominator bold log bold left parenthesis bold ex to the power of bold 2 bold right parenthesis end fraction close parentheses bold space bold plus bold space bold cot to the power of bold minus bold 1 end exponent bold space open parentheses fraction numerator bold log bold left parenthesis bold ex to the power of bold 4 bold right parenthesis over denominator bold log open parentheses begin display style bold e to the power of bold 2 over bold x to the power of bold 2 end style close parentheses end fraction close parentheses bold comma bold space bold ત ો bold space bold dy over bold dx bold space bold equals bold space bold. bold. bold. bold. bold. bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold 0 bold less than bold space bold logx bold space bold less than bold 1 over bold 2

1
-1
2
0

57.

$bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space bold g bold apostrophe bold left parenthesis bold x bold right parenthesis bold space fraction numerator bold e to the power of begin display style bold a over bold x end style end exponent bold minus bold e to the power of bold minus begin display style bold a over bold x end style end exponent over denominator bold g to the power of begin display style bold a over bold x end style end exponent bold space bold plus bold space bold e to the power of bold minus begin display style bold a over bold x end style end exponent end fraction bold space$ જ્યાં g' એ વિધેય g નું વિકલિત છે તથા સતત વિધેય છે અને a > 0. જો $bold lim with bold x bold rightwards arrow bold 0 below bold space bold f bold left parenthesis bold x bold right parenthesis$ નું અસ્તિત્વ હય તો

g(x) = a₀ + a₁x + a₂x² + ..... + a_nx^-n જ્યાં a_i ∈ R, i = 1,2 ....n તથા a₁ # 0 a_n # 0
g(x) = log (1 + x)
g એવું વિધેય હોય જ્યાં g'(0) = 0
g(x0 = sin x

58.

જો f(x) = (x+1)(x+2)(x+3)....(x + 100) અને g(x) = f(x) - (f(x))2 તો g(x) = 0 ને

એક ઉકેલ મળે.
ઉકેલ નથી.
બે ઉકેલ મળે.
ઓછામાં ઓછા ત્રણ ઉકેલ મળે.

59. નીચેનામાંથી કયું વિધેય અંતરાલ [-1, 1] માં રોલના પ્રમેયની શરતોનું પાલન કરે છે.

f(x) = [x] + [-x]
f(x) = |x| [x]
f(x) = |x| - |sin x|
$bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space open curly brackets table attributes columnalign left end attributes row cell fraction numerator bold sin bold space bold x over denominator bold x end fraction bold comma bold space bold x bold # bold 0 bold space end cell row cell bold 0 bold space bold comma bold space bold x bold space bold equals bold space bold 0 bold space end cell end table close$

60.

bold જ ો bold space bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space open curly brackets table row cell bold minus bold x bold minus bold pi over bold 2 bold comma end cell cell bold x bold less-than or slanted equal to bold minus bold pi over bold 2 end cell row cell bold minus bold cos bold space bold x bold comma end cell cell bold minus bold pi over bold 2 bold less than bold x bold less-than or slanted equal to bold 0 end cell row cell bold x bold minus bold 1 bold comma end cell cell bold 0 bold less than bold x bold less-than or slanted equal to bold 1 end cell row cell bold log bold space bold x bold comma end cell cell bold x bold greater than bold 1 end cell end table close bold space bold ત ો bold space

f(x) એ x = $bold minus bold 3 over bold 2$ , 1આગળ વિકલનીય છે.
f(x) એ x = $bold minus bold pi over bold 2$ આગળ સતત છે.
f(x) એ x = 0 આગળ વિકલનીય નથી.
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Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

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