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Gujarati JEE Mathematics : ગણ, સંબંધ અને વિધેય

Multiple Choice Questions

21. $bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space fraction numerator bold e to the power of bold x bold space bold minus bold space bold e blank presuperscript bold minus bold x end presuperscript over denominator bold 2 end fraction$ તથા f(g(x)) = xહોય, તો g $open parentheses fraction numerator bold e to the power of bold 2016 bold minus bold 1 over denominator bold 2 bold e to the power of bold 1008 end fraction close parentheses bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold.$

2016
252
1008
504

22. જો f(x) = $bold left parenthesis bold a bold minus bold x to the power of bold n bold right parenthesis to the power of begin inline style bold 1 over bold n end style end exponent bold comma bold space bold x bold space bold greater than bold space bold 0$ હોય, તો $bold f bold space bold left parenthesis bold f bold left parenthesis bold x bold right parenthesis bold right parenthesis bold space bold plus bold space bold f bold space open parentheses bold f open parentheses bold 1 over bold x close parentheses bold space close parentheses$ ............ મળે. x > 0

< 2
≥2
0≥0
=10

23.

જો [x] એ મહત્તમ પૂર્ણાંક ભાગ વિધેય હોય અને [x] = x - [x] હોય, તો f(x) = [x] + $bold sum from bold r bold space bold equals bold space bold 1 to bold 1000 of fraction numerator bold left curly bracket bold x bold space bold plus bold space bold r bold right curly bracket over denominator bold 1000 end fraction bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold space bold.$

-x
x²
1/x
x

24. $bold f bold space bold colon bold space bold R bold space bold rightwards arrow bold space bold R bold comma bold space bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space fraction numerator bold x bold space bold minus bold space bold left square bracket bold x bold right square bracket over denominator bold 1 bold space bold plus bold space bold x bold space bold minus bold space bold left square bracket bold x bold right square bracket end fraction$ હોય, તો $bold f bold left parenthesis bold x bold right parenthesis bold space bold element of$ ...... જ્યાં [x] = પૂર્ણાંક ભાગ વિધેય.

$open square brackets 0 comma 1 half close square brackets$
$left parenthesis 0 comma 1 half right square bracket$
આપેલ પૈકી એક પણ નહી

25. ધારો કે f(x) = $fraction numerator bold 9 to the power of bold x over denominator bold 9 to the power of bold x bold space bold plus bold space bold 3 end fraction bold comma bold space bold x bold space bold element of bold space bold R$ થી વિધેય વ્યાખ્યાયિત હોય, તો
$bold f bold space open parentheses bold 1 over bold 2017 close parentheses bold space bold plus bold space bold f open parentheses bold 2 over bold 2017 close parentheses bold space bold plus bold space bold f bold space open parentheses bold 3 over bold 2017 close parentheses bold space bold plus bold space bold. bold. bold. bold space bold plus bold space bold f open parentheses bold 2016 over bold 2017 close parentheses bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold space$

504
4032
2016
1008

1008

Tips: -

અહીં $bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space fraction numerator bold 9 to the power of bold x over denominator bold 9 to the power of bold x bold plus bold 3 end fraction bold semicolon bold space bold x bold element of bold R$ આપેલ છે. આથી f(1-x) = $fraction numerator bold 9 to the power of bold 1 bold minus bold x end exponent over denominator bold 9 to the power of bold 1 bold minus bold x end exponent bold space bold plus bold space bold 3 end fraction$ મળે.

વળી, f(x) + f(1-x) = $fraction numerator bold 9 to the power of bold x over denominator bold 9 to the power of bold x bold space bold plus bold space bold 3 end fraction bold space bold plus bold space fraction numerator bold 9 to the power of bold 1 bold minus bold x end exponent over denominator bold 9 to the power of bold 1 bold minus bold x end exponent bold space bold plus bold space bold 3 end fraction$

$bold equals bold space fraction numerator bold 9 to the power of bold x over denominator bold 9 to the power of bold x bold space bold plus bold space bold 3 end fraction bold space bold plus bold space fraction numerator bold 9 over bold 9 to the power of bold x over denominator bold 9 over bold 9 to the power of bold x bold plus bold 3 end fraction bold equals bold space fraction numerator bold 9 to the power of bold x over denominator bold 9 to the power of bold x bold space bold plus bold space bold 3 end fraction bold space bold plus bold space fraction numerator bold 9 over denominator bold 9 bold space bold plus bold space bold 3 to the power of bold x end fraction bold equals bold space fraction numerator bold 9 to the power of bold x over denominator bold 9 to the power of bold x bold plus bold 3 end fraction bold space bold plus bold space fraction numerator bold 3 over denominator bold 3 bold plus bold 9 to the power of bold x end fraction bold equals bold space fraction numerator bold 9 to the power of bold x bold space bold plus bold space bold 3 over denominator bold 9 to the power of bold x bold space bold plus bold space bold 3 end fraction bold space$

= 1 મળે
આમ, f(x) + f(1-x) = 1 ... (1)

$bold therefore bold space bold f bold space open parentheses bold 1 over bold 2017 close parentheses bold space bold plus bold f bold space open parentheses bold 2 over bold 2017 close parentheses bold space bold plus bold f bold space open parentheses bold 3 over bold 2017 close parentheses bold space bold plus bold f bold space open parentheses bold 4 over bold 2017 close parentheses bold space bold plus bold space bold. bold. bold. bold space bold f bold space open parentheses bold 2014 over bold 2017 close parentheses bold f bold space open parentheses bold 2015 over bold 2017 close parentheses bold plus bold f bold space open parentheses bold 2016 over bold 2017 close parentheses bold equals open square brackets bold f bold space open parentheses bold 1 over bold 2017 close parentheses bold plus bold space bold f bold space open parentheses bold 2016 over bold 2017 close parentheses close square brackets bold space bold plus bold space open square brackets bold f bold space open parentheses bold 2 over bold 2017 close parentheses bold plus bold space bold f bold space open parentheses bold 2015 over bold 2017 close parentheses close square brackets bold space bold plus bold space bold. bold. bold. bold space bold plus bold space open square brackets bold f bold space open parentheses bold 1008 over bold 2017 close parentheses bold plus bold space bold f bold space open parentheses bold 1009 over bold 2017 close parentheses close square brackets bold equals bold space open square brackets bold f bold space open parentheses bold 1 over bold 2017 close parentheses bold plus bold space bold f bold space open parentheses fraction numerator bold 2017 bold space bold minus bold space bold 1 over denominator bold 2017 end fraction close parentheses close square brackets bold space bold plus bold space open square brackets bold f bold space open parentheses bold 2 over bold 2017 close parentheses bold plus bold space bold f bold space open parentheses fraction numerator bold 2017 bold space bold minus bold space bold 2 over denominator bold 2017 end fraction close parentheses close square brackets bold plus bold space bold. bold. bold. bold space open square brackets bold f bold space open parentheses bold 1008 over bold 2017 close parentheses bold plus bold space bold f bold space open parentheses fraction numerator bold 2017 bold minus bold 1008 over denominator bold 2017 end fraction close parentheses close square brackets bold equals bold space open square brackets bold f bold space open parentheses bold 1 over bold 2017 close parentheses bold plus bold space bold f bold space open parentheses bold 1 over bold 2017 close parentheses close square brackets bold space bold plus bold space open square brackets bold f bold space open parentheses bold 2 over bold 2017 close parentheses bold plus bold space bold f bold space open parentheses bold 1 bold minus bold 2 over bold 2017 close parentheses close square brackets bold space bold plus bold space bold. bold. bold. bold space bold plus bold space open square brackets bold f bold space open parentheses bold 1008 over bold 2017 close parentheses bold plus bold space bold f bold space open parentheses bold 1 bold minus bold 1008 over bold 2017 close parentheses close square brackets$

= 1 + 1 + ... + (108) વખત ((1)પરથી)

= 1008

26.

bold f bold colon bold R bold space bold rightwards arrow bold space bold R bold comma bold space bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space fraction numerator bold e to the power of bold vertical line bold x bold vertical line end exponent bold minus bold x to the power of bold minus bold x end exponent over denominator bold e to the power of bold e bold plus bold e to the power of bold minus bold x end exponent end fraction

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

27. વિધેય f:A→(1, ∞); f(x) = 1 + x³ એ એક-એક વિધેય હોય, તો A = ......... શક્ય બને.

[1, ∞]
R⁺
R
[0, ∞]

28. f:R →R, f(x) = x-1 હોય તો { f^-1 (-2)} ∪ {f^-1 (17) } = ......

{-1, 18}
{0}
{±2(3)}
0

29. જો $bold e to the power of bold f bold left parenthesis bold x bold right parenthesis end exponent bold space bold equals bold space fraction numerator bold 10 bold plus bold x over denominator bold 10 bold minus bold x end fraction bold apostrophe bold space bold e bold element of bold space bold left parenthesis bold minus bold 10 bold comma bold space bold 10 bold right parenthesis$ તથા f(x) = kf $open parentheses fraction numerator bold 200 bold space bold x over denominator bold 100 bold space bold plus bold space bold x to the power of bold 2 end fraction close parentheses$ હોય, તો k = ...... .

$1 half$
$3 over 5$
$7 over 10$
$4 over 5$

30.

f : (- ∞, 64) → R, f(x) = - $square root of bold 64 bold space bold minus bold space bold x end root$ તથા g : [0, 2 $square root of bold 2 bold right square bracket bold space bold rightwards arrow bold space bold R bold comma bold space bold g bold left parenthesis bold x bold right parenthesis bold space bold equals bold space bold 8 bold x to the power of bold 2$ હોય, તો સંયોજિત વિધેય નો મહત્તમ પ્રદેશ ......તેમજ વિસ્તાર ......... મળે.

$bold left square bracket bold minus bold 2 square root of bold 2 bold comma bold space bold 2 square root of bold 2 bold right square bracket bold comma bold space bold left square bracket bold minus bold 8 bold comma bold space bold 0 bold right square bracket$

$bold left parenthesis bold minus bold 2 square root of bold 2 bold comma bold 2 square root of bold 2 bold right parenthesis bold left parenthesis bold 8 bold comma bold space bold 0 bold right parenthesis$
$bold left square bracket bold 0 bold comma bold space bold 2 square root of bold 2 bold right square bracket bold comma bold space bold left parenthesis bold minus bold 8 bold comma bold space bold 0 bold right parenthesis bold space$
$bold minus bold 2 square root of bold 2 bold comma bold space bold 2 square root of bold 2 bold right parenthesis bold comma bold space bold left parenthesis bold 0 bold comma bold space bold 8 bold right parenthesis$

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Gujarati JEE Mathematics : ગણ, સંબંધ અને વિધેય

Multiple Choice Questions

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