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Gujarati JEE Mathematics : ગણ, સંબંધ અને વિધેય

Multiple Choice Questions

21.

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.

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

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

Tips: -

(fog) (x) = f(g(x)) = f(8x²)

$bold equals bold minus bold space square root of bold 64 bold space bold minus bold space bold 8 bold x to the power of bold 2 end root bold equals bold minus bold space square root of bold 8 bold left parenthesis bold 8 bold minus bold x to the power of bold 2 bold right parenthesis end root bold equals bold space bold minus bold 2 bold space square root of bold 2 bold space square root of bold 8 bold minus bold x to the power of bold 2 end root$

વળી, 8 - x² ≥ 0 જરૂરી છે. આથી 8 ≥ x² એટલે કે x² ≤ 8 થાય.

∴ -2 $square root of bold 2$ ≤ x ≤ 2 $square root of bold 2$ મળે. (અહીંથી જ જવાબ (A) મળી ગયો. )

આથી, x ∈ [-2 $square root of bold 2$ , 2 $square root of bold 2$ ]લેતાં, સંયોજિત વિધેય (fog) નો મહત્તમ પ્રદેશ [-2 $square root of bold 2 bold comma bold space bold 2 square root of bold 2 bold right square bracket$ થાય.

(વળી, સંયોજિત વિધેય fog ના અસ્તિત્વ માટે, R_g ⊂ D_f હોય જ.)

હવે x ∈ [ -2 $square root of bold 2$ , 2 $square root of bold 2$ ] લેતાં, વિધેય ffogની મહત્તમ કિંમત 0 મળે.
જ્યારે x = 0 લેતાં, ન્યુનતમ કિંમત $bold minus square root of bold 64 bold minus bold 0 end root bold space bold equals bold space bold minus bold 8$ મળે.

આથી, સંયોજિત વિધેયનો વિસ્તાર [-8, 0] મળે.

22. $bold f bold space bold colon bold space bold R bold space bold rightwards arrow bold space bold R bold comma bold space bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space fraction numerator bold x bold space bold minus bold space bold left square bracket bold x bold right square bracket over denominator bold 1 bold space bold plus bold space bold x bold space bold minus bold space bold left square bracket bold x bold right square bracket end fraction$ 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

23. $bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space fraction numerator bold e to the power of bold x bold space bold minus bold space bold e blank presuperscript bold minus bold x end presuperscript over denominator bold 2 end fraction$ 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

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

{-1, 18}
{0}
{±2(3)}
0

25. 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 = ...... .

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

26.

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

27.

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.

28. 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

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

[1, ∞]
R⁺
R
[0, ∞]

30. 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

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Gujarati JEE Mathematics : ગણ, સંબંધ અને વિધેય

Multiple Choice Questions

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