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CBSE

Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

121.

વિધેય f : (0, ∞) → (0, ∞) માટે,

(1) f(ab) = f(a) f(b) અને
(2) $bold lim with bold x bold rightwards arrow bold infinity below$ f(x) = c, (જ્યાં ક # 0) પ્રકારનું છે. f(4) = ....

122.

f : R → R વિકલનીય વિધેય છે. જો f(y) f(x - y) = f(x), ∀x, y ∈ R અને f'(0) = p, f'(5) = q, p, q # 0 તો f'(-5) = .......

q
$bold p over bold q bold space$
$bold p to the power of bold 2 over bold q bold space$
$bold q over bold p$

bold p to the power of bold 2 over bold q bold space

Tips: -

f(y) f(x - y) = f(x) નું x પ્રત્યે વિકલન કરતાં, f(y) f'(x - y) = f'(x)

x = y મૂકતાં, f(x) f'(0) = f'(x)

∴ f'(x) = p f(x) (1)

$bold therefore bold space bold integral bold space fraction numerator bold f bold apostrophe bold left parenthesis bold x bold right parenthesis over denominator bold f bold left parenthesis bold x bold right parenthesis end fraction bold space bold dx bold space bold equals bold space bold integral bold space bold p bold space bold dx bold space bold plus bold space bold c$

∴ log^ef(x) = px + c (2)

સમીકરણ (1) માં x = 0 મૂકતાં, f'(0) = p f(0). આથી p = pf(0). આથી f(0) = 1

∴ log_e f(0) = c

સમીકરણ (2) પરથી, આથી log_e 1 = c. તેથી c = 0

∴ log_ef(x) = px

∴ f(x) = e^px. આથી f'(x) = pe^px

∴ f'(5) = pe^5p = q. આથી e^5p =

∴ f'(-5) = pe^-5p = $fraction numerator bold p over denominator bold q bold divided by bold p end fraction bold space bold equals bold space bold p to the power of bold 2 over bold q$

123.

જો વક્ર xy + ax + by = 0 ને (1, 1) આગળનો સ્પર્શક X-અક્ષ સાથે tan-1 2 માપનો ખૂણો બનાવે, તો $fraction numerator bold a bold space bold plus bold space bold b bold space over denominator bold ab bold space end fraction bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold space$

0
1
$bold 1 over bold 2$
$fraction numerator bold minus bold 1 over denominator bold 2 end fraction$

124.

bold જ ો bold space bold v bold space bold e to the power of bold u over bold v to the power of bold 3 end exponent bold space bold equals bold space bold 1 bold space bold ત ો bold space bold space bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold.

$bold v bold space fraction numerator bold d to the power of bold 2 bold u over denominator bold dv to the power of bold 2 end fraction bold space bold plus bold space bold 2 bold space bold du over bold dv bold space bold plus bold space bold 3 bold v to the power of bold 2 bold space bold equals bold space bold 0 bold space$
$bold v bold space fraction numerator bold d to the power of bold 2 bold u over denominator bold dv to the power of bold 2 end fraction bold space bold plus bold space bold 2 bold space bold du over bold dv bold space bold equals bold space bold space bold 3 bold v to the power of bold 2 bold space$
$bold v bold space fraction numerator bold d to the power of bold 2 bold u over denominator bold dv to the power of bold 2 end fraction bold space bold minus bold space bold 2 bold space bold du over bold dv bold space bold plus bold space bold 3 bold v to the power of bold 2 bold space bold equals bold space bold 0 bold space$
$fraction numerator bold d to the power of bold 2 bold u over denominator bold dv to the power of bold 2 end fraction bold space bold minus bold space bold 2 bold space bold du over bold dv bold space bold equals bold space bold 3 bold v to the power of bold 2 bold space$

125.

અરિક્ત ગણ A, B માટે f : A → B અને g : B → A એવાં વિધેય છે જ્યાં f(g(x)) = x, ∀ x ∈ B. નીચેનામાંથી કયા વિધાન સત્ય (T) અને મિથ્યા (F) છે ?

(1) વિધેય f એક-એક વિધેય છે.
(2) વિધેય f વ્યાત્પ વિધેય છે.
(3) વિધેય g એક-એક વિધેય છે.
(4) વિધેય g વ્યાપ્ત વિધેય છે.

FTFT
TTFF
TFTT
FTTF

126.

જો a < b < c, f(x) એ (a, c) પર ચુસ્ત રીતે વધતું વિધેય હોય અને f(x) એ [a, c] પર સતત હોય તો .......

(b - c) f(a) + (c - b) f(b) > (c - a) f(c)
(b - a) f(c) + (c - b) f(a) > (c - a) f(b)
(b - a) f(c) + (c - b) f(a) < (c - a) f(b)
(b - c) f(a) + (c - b) f(b) > (c - a) f(c)

127.

bold lim with bold n bold rightwards arrow bold infinity below bold space open parentheses root index bold 3 of bold n to the power of bold 2 bold space bold minus bold space bold n to the power of bold 3 bold space end root bold plus bold space bold n close parentheses bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold space

$bold minus bold 1 over bold 3$
$bold 2 over bold 3$
$bold minus bold 2 over bold 3$
$fraction numerator begin display style bold 1 end style over denominator begin display style bold 3 end style end fraction$

128.

f(x) = sin x + cos x, 0 ≤ x ≤ 2 $bold pi$ એ ...... અંતરાલમાં ચુસ્ત ઘટતું વિધેય છે.

$open parentheses fraction numerator bold 5 bold pi over denominator bold 4 end fraction bold comma bold 2 bold pi close parentheses$
$bold left parenthesis bold 0 bold comma bold space bold 2 bold pi bold right parenthesis$
$open parentheses bold 0 bold comma bold pi over bold 4 close parentheses$
$open parentheses bold pi over bold 4 bold comma bold pi over bold 4 close parentheses$