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Gujarati JEE Mathematics : ગણ, સંબંધ અને વિધેય

Multiple Choice Questions

41. sin^-1 $open square brackets bold log subscript bold 3 open parentheses bold x over bold 3 close parentheses close square brackets$ નો મહત્તમ પ્રદેશ ............ છે.

[1, 9]
(1, 9)
[9, 1]
[-1, 9]

42. જો S = {(1, 3), (4, 2), (2, 3), (3, 1)} એ ગણ A = {1, 2, 3, 4} પરનો સંબંધ દર્શાવે તો S એ .........

સંમિત સંબંધ નથી
સ્વવાચક સંબંધ છે.
પરંપરિત સંબંધ છે.
કોઈ સંબંધ નથી.

43. f : [1, ∞] → [1, ∞) f(x) = 2^x(x-1) હોય, તો f^-1(x) = ........

1 + $square root of 4 space log subscript 2 space straight x end root$
$1 half plus 1 half square root of 1 plus 4 log subscript 2 x end root$
$square root of 1 plus 4 log subscript 2 straight g end root$
અસ્તિત્વ ધરાવતું નથી

44. f: N→Z, દ્વારા, વ્યાખ્યાયિત વિધેય $bold f bold left parenthesis bold n bold right parenthesis bold space bold equals bold space open curly brackets table attributes columnalign left end attributes row cell fraction numerator bold n bold minus bold 1 over denominator bold 2 end fraction end cell row cell fraction numerator bold minus bold n over denominator bold 2 end fraction end cell end table close$

એક-એક નથી તથા વ્યાપ્ત વિધેય પણ નથી.
એક-એક વિધેય છે પરંતુ વ્યાપ્ત વિધેય નથી.
વ્યાપ્ત વિધેય છે પરંતુ એક-એક વિધેય નથી.
એક-એક તથા વ્યાપ્ત વિધેય છે.

45. $bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space fraction numerator bold sin to the power of bold minus bold 1 end exponent bold left parenthesis bold x bold minus bold 3 bold right parenthesis over denominator square root of bold 9 bold minus bold x to the power of bold 2 end root end fraction$ નો મહત્તમ પ્રદેશ ....... હોય.

[3, 2]
[2, 1]
[2, 3)
[1, 2]

46. વિધેય f(x) = log $open parentheses bold x bold plus square root of bold x to the power of bold 2 bold space bold plus bold space bold 1 end root close parentheses$ એ .........

અયુગ્મ વિધેય છે.
યુગ્મ કે અયુગ્મ વિધેય નથી.
યુગ્મ વિધેય છે.
આવર્તી વિધેય છે.

47.

bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space fraction numerator bold x to the power of bold 2 bold space bold plus bold space bold x bold space bold plus bold space bold 2 over denominator bold x to the power of bold 2 bold space bold plus bold space bold x bold space bold plus bold space bold 1 end fraction bold comma bold space bold x bold space bold element of bold space bold R

નો વિસ્તાર ....... હોય.

(1, ∞)
$left parenthesis 1 comma 7 over 3 right square bracket$
$left parenthesis 1 comma 7 over 5 right square bracket$
$left parenthesis 1 comma space 1 over 7 right parenthesis$

left parenthesis 1 comma 7 over 3 right square bracket

Tips: -

$bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space fraction numerator bold x to the power of bold 21 bold space bold plus bold space bold x bold space bold plus bold space bold 2 over denominator bold x to the power of bold 2 bold space bold plus bold space bold x bold space bold plus bold space bold 1 end fraction bold space bold equals bold space fraction numerator bold left parenthesis bold x to the power of bold 2 bold space bold plus bold space bold x bold space bold plus bold space bold 1 bold right parenthesis bold space bold plus bold 1 over denominator bold left parenthesis bold x to the power of bold 2 bold space bold plus bold space bold x bold space bold plus bold space bold 1 bold right parenthesis end fraction$

$bold equals bold space bold 1 bold space bold plus bold space fraction numerator bold 1 over denominator bold x to the power of bold 2 bold space bold plus bold space bold x bold space bold plus bold space bold 1 end fraction bold equals bold space bold 1 bold space bold plus bold space fraction numerator bold 1 over denominator bold x to the power of bold 2 bold space bold plus bold space bold x bold space bold plus bold space begin display style bold 1 over bold 4 end style bold plus begin display style bold 3 over bold 4 end style end fraction bold equals bold space bold 1 bold space bold plus bold space fraction numerator bold 1 over denominator open parentheses bold x bold plus bold space begin display style bold 1 over bold 2 end style close parentheses to the power of bold 2 bold space bold plus bold space begin display style bold 3 over bold 4 end style end fraction$

હવે f(x) એ $open parentheses bold x bold plus bold 1 over bold 2 close parentheses to the power of bold 2 bold space bold plus bold space bold 3 over bold 4$ ન્યુનતમ કિંમત ધારણ કરે ત્યારે મહત્તમ બને. આથી, જ્યારે $bold x bold plus bold 1 over bold 2 bold space bold equals bold space bold 0$
હોય, તો f(x) મહત્તમ ધારણ કરે. વળી, $fraction numerator bold 1 over denominator open parentheses bold x bold plus begin display style bold 1 over bold 2 end style close parentheses to the power of bold 2 bold space bold plus bold space begin display style bold 3 over bold 4 end style end fraction bold space bold not equal to$ આથી f(x) > 1જ થાય.

$bold therefore bold f bold left parenthesis bold x bold right parenthesis bold space bold less than bold space bold 1 bold space bold plus bold space fraction numerator bold 1 over denominator bold 0 bold plus bold 3 over bold 4 end fraction bold space bold equals bold space bold 1 bold plus bold 4 over bold 3 bold space bold equals bold space bold 7 over bold 3 bold therefore bold space bold R subscript bold f bold space bold equals bold space bold left parenthesis bold 1 bold comma bold 7 over bold 3 bold right square bracket$

48. f: [0, ∞) → [0, ∞) દ્વારા વ્યાખ્યાયિત વિધેય f(x) = $fraction numerator bold x over denominator bold 1 bold plus bold x end fraction$ હોય, તો f એ ......... .

એક-એક તથા વ્યાપ્ત વિધેય છે.
એક-એક વિધેય નથી પરંતુ વ્યાપ્ત વિધેય છે.
એક-એક નથી તથા વ્યાપ્ત વિધેય પણ નથી
એક-એક વિધેય છે પરંતુ વ્યાપ્ત વિધેય નથી

49. જો g (f(x) = |sin x| તથા f(g(x) = (sin $square root of bold x$ )² હોય, તો ......

$straight f left parenthesis straight x right parenthesis space equals space sin squared straight x comma space straight g left parenthesis straight x right parenthesis space equals space square root of straight x$
$straight f left parenthesis straight x right parenthesis space equals space sin space straight x comma space straight g left parenthesis straight x right parenthesis space equals space vertical line straight x vertical line$
$bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space bold x to the power of bold 2 bold comma bold space bold g bold left parenthesis bold x bold right parenthesis bold space bold equals bold space bold sin bold space square root of bold x$
f અને g નિશ્વિત ન કરી શકાય.

50. f(x) sin x + cos x; g(x) = x²-1 હોય, તો g(f(x)) એ x ∈ ....... માટે પ્રતિવિધેય ધરાવે.

$left square bracket 0 comma space straight pi right square bracket$
$open square brackets negative straight pi over 4 comma straight pi over 4 close square brackets$
$open square brackets negative straight pi over 2 comma pi over 2 close square brackets$
$open square brackets 0 comma pi over 4 close square brackets$

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Gujarati JEE Mathematics : ગણ, સંબંધ અને વિધેય

Multiple Choice Questions

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