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CBSE

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.

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.
આપેલ તમામ

53.

$bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space bold g bold apostrophe bold left parenthesis bold x bold right parenthesis bold space fraction numerator bold e to the power of begin display style bold a over bold x end style end exponent bold minus bold e to the power of bold minus begin display style bold a over bold x end style end exponent over denominator bold g to the power of begin display style bold a over bold x end style end exponent bold space bold plus bold space bold e to the power of bold minus begin display style bold a over bold x end style end exponent end fraction bold space$ 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

54. 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|
$bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space open curly brackets table attributes columnalign left end attributes row cell fraction numerator bold sin bold space bold x over denominator bold x end fraction bold comma bold space bold x bold # bold 0 bold space end cell row cell bold 0 bold space bold comma bold space bold x bold space bold equals bold space bold 0 bold space end cell end table close$

55.

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

56.

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

1
-1
2
0

Tips: -

y = (1+x) (1+x²) (1+x⁴) ... + log (1+x²ⁿ)

log y = log (1+x)+ log (1+x²) + log (1+x⁴) .... + log (1+x²ⁿ)

વિકલન કરતાં,
$bold 1 over bold y bold space bold dy over bold dx bold space bold equals bold space open square brackets fraction numerator bold 1 over denominator bold 1 bold plus bold x end fraction bold plus fraction numerator bold 2 bold x over denominator bold 1 bold plus bold x to the power of bold 2 end fraction bold plus fraction numerator bold 4 bold x to the power of bold 3 over denominator bold 1 bold plus bold x to the power of bold 3 end fraction bold space bold plus bold space bold. bold. bold. bold plus bold space fraction numerator bold 2 bold nx to the power of bold 2 bold n bold minus bold 1 end exponent over denominator bold left parenthesis bold 1 bold plus bold x bold right parenthesis to the power of bold 2 bold n end exponent end fraction close square brackets bold therefore bold space bold dy over bold dx bold space bold equals bold space bold y bold space open square brackets fraction numerator begin display style bold 1 end style over denominator begin display style bold 1 bold plus bold x end style end fraction bold plus fraction numerator begin display style bold 2 bold x end style over denominator begin display style bold 1 bold plus bold x to the power of bold 2 end style end fraction bold plus fraction numerator begin display style bold 4 bold x to the power of bold 3 end style over denominator begin display style bold 1 bold plus bold x to the power of bold 3 end style end fraction bold space bold plus bold space bold. bold. bold. bold plus bold space fraction numerator begin display style bold 2 bold nx to the power of bold 2 bold n bold minus bold 1 end exponent end style over denominator begin display style bold left parenthesis bold 1 bold plus bold x bold right parenthesis to the power of bold 2 bold n end exponent end style end fraction close square brackets bold space bold આથ ી bold space open parentheses bold dy over bold dx close parentheses subscript bold x bold equals bold 0 end subscript bold space bold equals bold space bold y bold left parenthesis bold 0 bold right parenthesis bold space bold times bold 1 bold space bold equals bold space bold 1$