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Gujarati JEE Mathematics : ગણ, સંબંધ અને વિધેય

Multiple Choice Questions

31.

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

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

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

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

33.

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

34. 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

35.

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.

36.

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

{0, -1}

Tips: -

x₁, x₂ ∈ R માટે, જો f(x₁) = f(x₂) લઈએ, તો

(x₁ + 1)² = (x₂ + 1)^2-1 (જ્યાં x₁, x₂ ≥ -1 )

∴ (x₁ + 1)² = (x₂ + 1)²

∴ x₁ + 1 = x₂ + 1 અથવા x₁ + 1 = -x₂ - 1

∴ x₁ = x₂ અથવા x₁ + x₂ = -2 પરંતુ x₁x₂ ≥ - 1 હોવાથી x₁ + x₂ = -2 થવા માટે, x₁ = x₂ = -1 મળે.

આમ, f(x₁) = f(x₂) ⇒ x₁ = x₂ હોવાથી, f એક-એક વિધેય છે.

હવે, y, y ≥ -1, f(x) = (x+1)² - 1 = y લેતાં,

(x+1)² = y + 1

∴ x + 1 = $square root of bold y bold space bold plus bold 1 end root$ (x ≥-1)

∴ x = -1 + $square root of bold y bold plus bold 1 end root$ મળે. ... (1)

∴ f(x) = f(-1 + $square root of bold left parenthesis bold y bold plus bold 1 bold right parenthesis end root bold space bold equals bold space bold left square bracket bold minus bold 1 bold space bold plus bold space square root of bold left parenthesis bold y bold plus bold 1 bold right parenthesis end root bold space bold 1 bold right square bracket to the power of bold 2 bold space bold minus bold space bold 1 bold space bold equals bold space bold italic y bold space$

આમ, f વ્યાપ્ત વિધેય હોવાથી, f^-1(x) અસ્તિત્વ ધરાવે.

હવે, f(x) = f^-1(x) હોવાથી,

(x+1)² -1 = -1 + $square root of bold x bold plus bold 1 end root$ (f(x) = y ⇒ x = f^-1(y))

∴ (x+1)² = $square root of bold x bold plus bold 1 end root$

∴ (x+1)² = $square root of bold x bold plus bold 1 end root$ = 0

∴ $square root of bold x bold plus bold 1 end root bold space open square brackets bold left parenthesis bold x bold plus bold 1 bold right parenthesis to the power of begin inline style bold 3 over bold 2 end style end exponent bold minus bold 2 close square brackets bold space bold equals bold space bold 0$
∴ $square root of bold x bold plus bold 1 end root bold space bold equals bold space bold 0$ અથવા $bold left parenthesis bold x bold plus bold 1 bold right parenthesis to the power of begin inline style bold 3 over bold 2 end style end exponent bold space bold equals bold space bold 1 bold space$

∴ x + 1 = 0 અથવા (x + 1)³ = 1 ⇒ x = 0

આમ x = 0 , અથવા - 1 મળે.

S { x | f(x) = f^-1(x)} હોય, તો S = {0, -1} મળે.

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

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

38. 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

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. {-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

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Gujarati JEE Mathematics : ગણ, સંબંધ અને વિધેય

Multiple Choice Questions

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