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CBSE

Gujarati JEE Mathematics : ગણ, સંબંધ અને વિધેય

Multiple Choice Questions

31. 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

32.

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

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

33.

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]

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.

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

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

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

(8, 10)

Tips: -

f(x) = log₄ [log₅ {log₃, (18x - x² - 77}]આપેલ છે.

log વિધેયનો પ્રદેશ R⁺ હોવાથી, log₅ {log₃ (18x - x² - 77)} > 0 જોઈએ.

$bold therefore bold space bold log subscript bold 3 bold space bold left parenthesis bold minus bold x to the power of bold 2 bold space bold plus bold space bold 18 bold x bold space bold minus bold space bold 77 bold right parenthesis bold space bold greater than bold space bold 5 bold degree bold space bold equals bold space bold 1 bold space bold therefore bold space bold minus bold x to the power of bold 2 bold space bold plus bold space bold 18 bold x bold space bold minus bold space bold 88 bold space bold greater than bold space bold 3 bold space bold therefore bold space bold x to the power of bold 2 bold space bold minus bold space bold 18 bold x bold space bold plus bold space bold 77 bold space bold less than bold space bold minus bold space bold 3 bold space bold therefore bold space bold x to the power of bold 2 bold space bold minus bold space bold 18 bold x bold space bold plus bold space bold 80 bold space bold less than bold space bold 0 bold space bold therefore bold space bold left parenthesis bold x bold minus bold 8 bold right parenthesis bold space bold left parenthesis bold x bold minus bold 1 bold right parenthesis bold space bold less than bold 0 bold space$

∴ (x-8) < 0 તથા (x-10) > 0 હોય અથવા (x-8) > 0 અને (x-10) < 0

હોય, તો જ (x-8) (x-10) < 0 મળે.

∴ (i) x < 8 તથા x > 10 મળે અથવા (iii) x > 8 તથા x < 10 મળે.

પરંતુ x < 8 તથા x > 10 શક્ય નથી.
∴ x ∈ (8, 10) માટે f(x) વ્યાખ્યાયિત થાય.

38.

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.