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CBSE

Gujarati JEE Mathematics : ગણ, સંબંધ અને વિધેય

Multiple Choice Questions

11. જો X = {4ⁿ - 3n - 1, n ∈ N} અને Y = {9 (n-1); n ∈ N} જ્યાં N = પ્રાકૃતિક સંખ્યાઓનો ગણ હોય તો X ∪ Y = ......... .

X
Y - X
Y
N

12. f : (R -{-1} → R, f(x) = $fraction numerator bold 1 bold minus bold x over denominator bold 1 bold plus bold x end fraction$ હોય તો, $bold f bold space open parentheses fraction numerator bold x bold plus bold y over denominator bold 1 bold plus bold xy end fraction close parentheses$ = ..........

f(x) + f(y)
f(x)•f(y)
(f(x))²
f(x)/(f(y)

13. જો P = {1, 2, 3, 4,} Q = {a, b, c, d} હોય તો નીચેનાં જોડકાં જોડો :

i - c, ii - a, iii - b
i - b, ii - c, ii - a
i - a, ii - b, iii - c
i - a, ii - c, iii - b

14.

જો $bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space fraction numerator bold x bold minus bold 3 over denominator bold x bold plus bold 1 end fraction bold comma bold space bold x bold space bold not equal to bold space bold minus bold 1$ હોય તો f⁽²⁰¹⁶⁾(2015) = ...... જ્યાં f⁽²⁰¹⁶⁾(x) એ f નું f સાથે 2016 વખત સંયોજિત વિધેય દર્શાવે છે.

2017
2014
2016
2015

15. જો f(x) = cos (log x) હોય, તો f(x), f(y) - $bold minus bold 1 over bold 2 bold space open square brackets bold f open parentheses bold x over bold y close parentheses bold plus bold f bold left parenthesis bold xy bold right parenthesis close square brackets bold space bold equals bold space bold. bold. bold. bold. bold space bold.$

16. જો વિધેય f એ સમીકરણ $bold 3 over bold 10 bold space bold f bold left parenthesis bold x bold right parenthesis bold space bold plus bold space bold 2 over bold 10 bold f bold space open parentheses fraction numerator bold x bold plus bold 59 over denominator bold x bold minus bold 1 end fraction close parentheses bold space bold equals bold space bold x bold space bold plus bold space bold 3$ નું સમાધાન કરે (જ્યાં $bold x bold space bold not equal to bold 1$ ) તો f(21) .......

17.

f(x) = ax² + bx + c; x = 1, 2, 3 તથા g(x) = $open curly brackets table attributes columnalign left end attributes row cell bold 3 bold x bold space bold plus bold 1 bold semicolon bold space bold x bold space bold equals bold space bold 2 bold comma bold 3 end cell row cell bold space bold space bold space bold space bold space bold space bold space bold space bold 3 bold semicolon bold space bold x bold space bold equals bold space bold 1 end cell end table close$ હોય તેમજ બંને વિધેયો સમાન હોય, તો નીચેનામાંથી કયું સત્ય બને ?

a = b = c = 1
$straight a space equals space minus 1 half comma space straight b space equals space 11 over 2 comma space straight c space equals space minus 2$
$straight a space equals 11 over 2 comma space straight b space equals space minus 2 comma space straight c space equals 1 half$
$straight a space equals space 1 half comma space straight b space equals space 2 comma space straight c space equals space 11 over 2$

18. વાસ્તવિક વિધેય f(x) = $square root of bold x to the power of bold 2 bold space bold plus bold space bold 6 bold x bold space bold plus bold space bold 10 end root$ નો વિસ્તાર ...... મળે.

(-∞, 1)
[1, ∞]
R
(1, ∞)

[1, ∞]

Tips: -

f(x) = $square root of bold x to the power of bold 2 bold space bold plus bold space bold 6 bold x bold space bold plus bold space bold 10 end root bold space bold equals bold space square root of bold left parenthesis bold x to the power of bold 2 bold space bold plus bold space bold 6 bold x bold space bold plus bold space bold 9 bold right parenthesis bold space bold plus bold space bold 1 end root bold space bold equals bold space square root of bold left parenthesis bold x bold space bold plus bold space bold 3 bold right parenthesis to the power of bold 2 bold space bold plus bold space bold 1 end root$

∴ f(x) ≥ $square root of bold 1$ કારણ કે (x + 3)² ≥ 0

વળી, x = -3 લેતાં ન્યુનતમ કિંમત 1 મળે.

∴ વિધેય f નો વિસ્તાર [1, ∞] થાય.

19. જો f(x)= $square root of bold log bold space bold left parenthesis bold sin bold space bold x bold right parenthesis end root$ હોય, તો વિધેય f નો મહત્તમ પ્રદેશ ......... હોય.

$left curly bracket left parenthesis 4 straight k space plus space 3 space right parenthesis space straight pi over 2 vertical line space straight k space element of space straight Z right curly bracket$
$left curly bracket space left parenthesis 4 straight k space minus 1 right parenthesis space straight pi over 2 vertical line space straight k space element of space straight Z right curly bracket$
$left curly bracket left parenthesis 4 straight k space plus space 1 right parenthesis space straight pi over 2 vertical line space straight k element of straight Z right curly bracket space$
$left curly bracket left parenthesis 4 straight k space plus space 3 space right parenthesis space straight pi over 4 vertical line space straight k space element of space straight Z right curly bracket$