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Gujarati JEE Mathematics : ગણ, સંબંધ અને વિધેય

Multiple Choice Questions

31. {-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

32.

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.

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

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

34. 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

35.

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

36.

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

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

37. 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

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

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

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]

(-∞, -3) ∪ (3, ∞)

Tips: -

$bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space bold log subscript fraction numerator bold x bold minus bold 2 over denominator bold x bold plus bold 3 end fraction end subscript bold 2 bold space bold equals bold space fraction numerator bold log bold space bold 2 over denominator bold log bold space open parentheses begin display style fraction numerator bold x bold minus bold 2 over denominator bold x bold plus bold 3 end fraction end style close parentheses end fraction$

હવે log વિધેયના અસ્તિત્વ માટે, $fraction numerator bold x bold minus bold 2 over denominator bold x bold plus bold 3 end fraction bold space bold greater than bold 0$ જરૂરી છે.

$bold therefore bold x bold space bold greater than bold space bold 2$ હોય અથવા x < -3 હોય, તો $fraction numerator bold x bold minus bold 2 over denominator bold x bold plus bold 3 end fraction bold space bold greater than bold space bold 0$ મળે.

∴ D_f = (-∞, -3) ∪ (2, ∞)

વળી, g(x) = $fraction numerator bold 1 over denominator square root of bold x to the power of bold 2 bold space bold minus bold space bold 9 end root end fraction$ ના અસ્તિત્વ માટે, x² - 9 > 0 જરૂરી છે.

∴ x² > 9 એટલે કે, |x| > 3

∴ x ∈ (-∞, -3) ∪ (3, ∞)

∴ D_f ∩ D_g = (-∞, -3) ∪ (3, ∞)

40.

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
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Gujarati JEE Mathematics : ગણ, સંબંધ અને વિધેય

Multiple Choice Questions

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