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Gujarati JEE Mathematics : ગણ, સંબંધ અને વિધેય

Multiple Choice Questions

31.

બહુપદી વિધેય f(x) = xⁿ + 1 એ f(x) f $open parentheses bold 1 over bold x close parentheses bold space bold equals bold space bold f bold left parenthesis bold x bold right parenthesis bold space bold plus bold space bold f open parentheses bold 1 over bold x close parentheses$ શરત સંતોષે છે. જો f(12) = 1729 હોય, તો f(15) = ............ .x ∈ R

3357
3376
2075
1001

3376

Tips: -

બહુપદી વિધેય f(x) એ f(x) f $open parentheses bold 1 over bold x close parentheses bold space bold equals bold space bold f bold left parenthesis bold x bold right parenthesis bold space bold plus bold space bold f open parentheses bold 1 over bold x close parentheses$ શરત સંતોષે છે.

f(x)0 = xⁿ + 1 હોવાથી $bold f open parentheses bold 1 over bold x close parentheses bold space bold equals bold space bold 1 over bold x to the power of bold n bold space bold plus bold space bold 1$

$bold therefore bold space bold f bold left parenthesis bold x bold right parenthesis bold space bold plus bold space bold f bold space open parentheses bold 1 over bold x close parentheses bold space bold equals bold space bold x to the power of bold n bold space bold plus bold space bold 1 bold space bold plus bold space bold 1 over bold x to the power of bold n bold space bold plus bold space bold 1$

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

... (1)

$bold therefore bold space bold f bold left parenthesis bold x bold right parenthesis bold space bold. bold space bold f open parentheses bold 1 over bold x close parentheses bold space bold equals bold space bold left parenthesis bold x to the power of bold n bold space bold plus bold space bold 1 bold right parenthesis bold space open parentheses bold 1 bold plus bold 1 over bold x to the power of bold n close parentheses$

$fraction numerator bold left parenthesis bold x to the power of bold n bold plus bold 1 bold right parenthesis bold space bold left parenthesis bold x to the power of bold n bold plus bold 1 bold right parenthesis to the power of bold 2 over denominator bold x to the power of bold n end fraction bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold. bold. bold. bold space bold left parenthesis bold 2 bold right parenthesis$

ટુંકમાં f(x) = xⁿ + 1 હોય,તો $bold f bold left parenthesis bold x bold right parenthesis bold space bold f open parentheses bold 1 over bold x close parentheses bold space bold equals bold space bold f bold left parenthesis bold x bold right parenthesis bold space bold plus bold space bold f bold space open parentheses bold 1 over bold x close parentheses$ મળે. ((1) અને(2) પરથી)
આથી, f(x) xⁿ + 1 લેતાં,

f(12) = 12ⁿ + 1 = 1729

∴ f(12) = 12ⁿ = 1728 = (12)³
∴ n = 3 મળે.

∴ f(x) = x³ + 1 મળે.

x = 15 લેતાં, f(15) = (15)³ + 1 = 3376 મળે.

32.

f:R-[-1, ∞] પર f(x) = (x + 1)²-1 દ્વારા વ્યાખ્યાયિત વિધેય માટે જો S = {x| f(x) = f^-1(x)} હોત, તો x = .......... મળે.

{0, 1, -1}
{0, -1}
$up diagonal strike 0$
$open curly brackets 0 comma space 1 comma space fraction numerator negative square root of 3 plus square root of 31 over denominator 2 end fraction comma space fraction numerator negative square root of 3 minus square root of 31 over denominator 2 end fraction close curly brackets$

33. વિધેય f(x) = 9^x - 3^x + 1 નો વિસ્તાર ......... હોય.

(0, ∞)
(-∞, 0)
(-∞, ∞)
$open square brackets 3 over 4 comma space infinity close square brackets$

34.

$bold f bold space bold colon bold space bold R bold space bold rightwards arrow bold space bold left curly bracket bold y bold space bold vertical line bold space bold vertical line bold y bold vertical line bold less than bold 1 bold comma bold space bold y bold element of bold R bold right curly bracket bold comma bold space bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space fraction numerator bold 10 to the power of bold x bold minus bold 10 to the power of bold minus bold x end exponent over denominator bold 10 to the power of bold x bold space bold plus bold space bold 10 to the power of bold minus bold x end exponent end fraction$ હોય, તો f^-1(x) = ......... (સ્વીકારી લો કે f^-1 નું અસ્તિત્વ છે.)

$1 half space log subscript 10 space open parentheses fraction numerator 1 plus straight x over denominator 1 minus straight x end fraction close parentheses$
$1 fourth space log subscript 10 space open parentheses fraction numerator 2 x over denominator 2 minus x end fraction close parentheses$
$1 half space log subscript 10 space left parenthesis 2 x minus 1 right parenthesis$
$log subscript 10 space left parenthesis 2 minus straight x right parenthesis$

35. જો d: R →R, f(x) = sin x અને g : (1, ∞) →R, g(x) = $square root of bold x to the power of bold 2 bold space bold minus bold space bold 1 end root$ હોય, તો (gof) (x) = ......

$cos space x$
$sin space square root of x squared space minus space 1 end root$
$square root of sin space left parenthesis straight x squared space minus space 1 right parenthesis end root$
અવ્યાખ્યાયિત

36.

જો f: R - {-3} →R - {1} પર વ્યાખ્યાયિત વિધેય, f(x) = $fraction numerator bold x bold plus bold 2 over denominator bold x bold plus bold 3 end fraction$ નું પ્રતિવિધેય અસ્તિત્વ ધરાવતો x ∈ ........... માટે, f^-1(x) > 0 મળે.

$open parentheses 2 over 3 comma space 1 close parentheses$
$open parentheses negative infinity comma 2 over 3 close parentheses$
$left parenthesis 1 comma space infinity right parenthesis$
શક્ય નથી.

37. {-2, 2} → {1, 3} પર એક-એક સંગતતા ધરાવતા બે ભિન્ન વિધેયો નીચેનામાંથી કયા હોઈ શકે ? $straight x over 2 plus 2$

y = ±
y = ±x + 4
y = x ± 2
y = ± x ± 2

38. વિધેય f(x) = log₄ [log₅ {log₃, (18x - x² - 77}] નો મહત્તમ પ્રદેશ ........ મળે.

(8, 10)
(0, 10)
(10, 12)
(4, 5)

39.

વિધેય f : A → R, f(x) = $bold log subscript begin inline style fraction numerator bold x bold minus bold 2 over denominator bold x bold plus bold 3 end fraction end style end subscript bold 2$ તથા g : B → R, g(x) = $fraction numerator bold 1 over denominator square root of bold x to the power of bold 2 bold space bold minus bold 9 end root end fraction$ હોય, તો x ની કઈ કિંમતો માટે વિધેય $bold f over bold g$ અસ્તિત્વ ધરાવે ?

(-3, 2)
(-∞, -3) ∪ (3, ∞)
[2,3]
[-3, -2]

40. જો f(x) = sin² x + sin² $open parentheses bold x bold plus bold pi over bold 3 close parentheses bold space bold plus bold space bold italic c bold italic o bold italic s bold space bold italic x bold space bold. bold space bold italic c bold italic o bold italic s bold space open parentheses bold x bold plus bold pi over bold 3 close parentheses bold space$ તથા g $bold 5 over bold 4$ = 1 હોય તો (gof) (x) = ............ .

1
$4 over 5$
$5 over 4$
અવ્યાખ્યાયિત

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Gujarati JEE Mathematics : ગણ, સંબંધ અને વિધેય

Multiple Choice Questions

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