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Gujarati JEE Mathematics : ગણ, સંબંધ અને વિધેય

Multiple Choice Questions

61. જો f: R →S, f(x) = sin x - $square root of bold 3$ cos + 1 વ્યાપ્ત વિધેય હોય, તો S = .......... .

[0, 3]
[-1, 3]
[-1, 1]
[0, 1]

62.

બે અરિક્ત ગણ X તથા Y માટે, f : X → Y એ એક-એક વિધેય છે જો A ⊂ X તથા B ⊂ Y માટે, f(A) = {f(x) | x ∈ A} અને f^-1(B) = {x ∈ | f(x) ∈ B} હોય, તો .......

f^-1(f(A))=A
f^-1(f(A)) ⊄ A
f(f^-1(N))=B
અપેલ પૈકી એક પણ નહી

63. જો f(x) = cos [ $bold pi to the power of bold 2$ ] x ; જ્યાં [x] એ પૂર્ણાંક ભાગ વિધેય દર્શાવે છે, તો નીચેનામાંથી કયું સત્ય છે ?

$bold f open parentheses bold pi over bold 4 close parentheses bold space bold equals bold space bold 1$
$bold f bold left parenthesis bold minus bold pi bold right parenthesis bold space bold equals bold space bold 0$
$bold f bold left parenthesis bold pi bold right parenthesis bold space bold equals bold space bold 1$
$bold f open parentheses bold pi over bold 2 close parentheses bold space bold equals bold space bold minus bold 1$

64.

જો A = {1, 2, 3, 4},B = {3, 4, 5} હોય, તો A થી B પરના એક-એક વિધેયોની સંખ્યા તથા વ્યાપ્ત વિધેયોની સંખ્યા અનુક્રમે ....... અને ...... મળે. જો A = {3, 4, 5}, B = {1, 2, 3, 4} હોય, તો વ્યાપ્ત વિધેયોની સંખ્યા ...... મળે.

36, 0, 0
0, 0, 36
0, 36, 0
36, 6, 0

65. જો 3^x = 4^x-1 હોય, તો x = .......

$fraction numerator 2 space log subscript 3 space 2 over denominator 2 space log subscript 3 space 2 space minus space 1 end fraction$
$fraction numerator 2 over denominator 2 minus log subscript 2 space 3 end fraction$
$fraction numerator 1 over denominator 1 space minus space log subscript 4 space 3 end fraction$
$fraction numerator 2 space log subscript 2 space 3 over denominator 2 space log subscript 2 space 3 space minus 1 end fraction$

66.

જો f(x) = sin $open curly brackets bold pi over bold 6 bold sin bold space open parentheses bold pi over bold 2 bold space bold sin bold space bold x close parentheses close curly brackets bold space bold semicolon$ x ∈ R તથા g(x) = $bold pi over bold 2 bold space bold sinx bold space bold semicolon$ x ∈ R, (fog) (x) તથા (gof) (x) ને f(g(x)) તથા g(f(x)) થી દર્શાવીએ તો નીચેનામાંથી શું સત્ય બને ?

f નો વિસ્તાર $open square brackets fraction numerator bold minus bold 1 over denominator bold 2 end fraction bold comma bold 1 over bold 2 close square brackets$ છે
fog નો વિસ્તાર $open square brackets fraction numerator bold minus bold 1 over denominator bold 2 end fraction bold comma bold 1 over bold 2 close square brackets$ છે.
$table row cell table row bold lim row cell bold x bold rightwards arrow bold m end cell end table bold space fraction numerator bold f bold left parenthesis bold x bold right parenthesis over denominator bold g bold left parenthesis bold x bold right parenthesis end fraction end cell end table bold space bold equals bold space bold space bold pi over bold 6$
કોઈ એવો મળે કે જેથી

67. વિધેય એ ........ અંતરાલમાં વ્યાખ્યાયિત થાય.

$left parenthesis 0 comma straight pi right parenthesis$
$open parentheses fraction numerator bold minus bold pi over denominator bold 2 end fraction bold comma bold pi over bold 2 close parentheses$
$left square bracket 0 comma space straight pi over 2 right parenthesis$
$left parenthesis 0 comma space straight pi right square bracket$

68. જો X = {1, 2, 3, 4, 5}, Y = {1, 3, 5, 7, 9} હોય, તો નીચેનામાંથી શું સત્ય છે ?

S₁ = {(x,a)|a=x+2, x ∈ X, a ∈ Y} સંબંધ દર્શાવે પરંતુ X થી Y પરનું વિધેય નથી.
S₂ = {(1,1), (2,1),(3,3),(4,3),(5,5)} એ X થી Y પરનો સંબંધ દર્શાવે તથા વિધેય છે.
S₃ = {(1,1), (1,3), (3,5), (3,7), (5,7)} એ X થી Y પરનો સંબધ દર્શાવે તથા પરંતુ વિધેય નથી.
S₄ = {(1,3), (2,5), (4,7), (5,9), (3, 1) } એ X થી Y પરનો સંબંધ દર્શાવે તથા વિધેય છે.

69. કોઈ રિક્ત ગણ માટે, n[P{P{P(P( $up diagonal strike bold 0$ ))}}] = .......

70. f:(-1, 1) → B, f(x) = tan^-1 $fraction numerator bold 2 bold x over denominator bold 1 bold minus bold x to the power of bold 2 end fraction$ એ એક-એક તથા વ્યાપ્ત વિધેય હોય તો B =

$open square brackets 0 comma space straight pi over 2 close square brackets$
$open parentheses fraction numerator negative straight pi over denominator 2 end fraction comma straight pi over 2 close parentheses$
$open parentheses 0 comma space pi over 2 close parentheses$
$open square brackets fraction numerator italic minus pi over denominator 2 end fraction comma pi over 2 close square brackets$

open parentheses fraction numerator negative straight pi over denominator 2 end fraction comma straight pi over 2 close parentheses

Tips: -

f(x) = tan^-1 $fraction numerator bold 2 bold x over denominator bold 1 bold minus bold x to the power of bold 2 end fraction bold comma bold space bold italic x bold space bold element of bold space bold left parenthesis bold minus bold 1 bold comma bold space bold 1 bold right parenthesis bold space$ આપેલ છે.

x = tan θ જ્યાં, θ ∈ $open parentheses fraction numerator bold minus bold pi over denominator 2 end fraction bold comma bold pi over 2 close parentheses$ ⇒ θ = tan^-1 x

પરંતું x ∈ (-1, 1) હોવાથી, θ ∈ $open parentheses fraction numerator bold minus bold pi over denominator 4 end fraction bold comma bold pi over 4 close parentheses$ મળે.

હવે f(x) = tan^-1 $open parentheses fraction numerator bold 2 bold tanθ over denominator bold 1 bold minus bold tan to the power of bold 2 bold theta end fraction close parentheses$

= tan^-1 (tan 2θ)

વળી, $fraction numerator bold minus bold pi over denominator bold 4 end fraction bold space bold less than bold space bold theta bold space bold less than bold space bold pi over bold 4 bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold rightwards double arrow bold space fraction numerator bold minus bold pi over denominator bold 2 end fraction bold space bold less than bold space bold 2 bold theta bold space bold less than bold space bold pi over bold 2$

$bold therefore bold space bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space bold 2 bold theta bold space bold equals bold space bold 2 bold space bold tan to the power of bold minus bold 1 end exponent bold space bold x bold therefore bold space bold R subscript bold f bold space bold equals bold space bold left parenthesis bold 2 bold space bold tan to the power of bold minus bold 1 end exponent bold space bold left parenthesis bold minus bold 1 bold right parenthesis bold comma bold space bold 2 bold tan to the power of bold minus bold 1 end exponent bold space bold 1 bold right parenthesis bold space bold equals bold space open parentheses bold 2 open parentheses fraction numerator bold minus bold pi over denominator bold 4 end fraction close parentheses bold comma bold space bold 2 bold space open parentheses bold pi over bold 4 close parentheses close parentheses bold space bold equals bold space open parentheses fraction numerator bold minus bold pi over denominator bold 2 end fraction bold comma bold pi over bold 2 close parentheses bold therefore bold space bold B bold space bold equals bold space open parentheses fraction numerator bold minus bold pi over denominator bold 2 end fraction bold comma bold pi over bold 2 close parentheses bold space bold મળ ે bold.$

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Gujarati JEE Mathematics : ગણ, સંબંધ અને વિધેય

Multiple Choice Questions

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