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Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

121.

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)

122.

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

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

bold 1 over bold 2

Tips: -

xy + ax + by = 0 નું વિકલન કરતાં $open parentheses bold x bold dy over bold dx bold plus bold space bold y bold space bold plus bold space bold a bold space bold plus bold space bold b bold dy over bold dx close parentheses$ = 0

$bold therefore bold space bold dy over bold dx bold space bold equals bold space open parentheses fraction numerator bold a bold space bold plus bold space bold y over denominator bold b bold space bold plus bold space bold x end fraction close parentheses$

$bold therefore open parentheses bold dy over bold dx close parentheses subscript bold left parenthesis bold 1 bold comma bold 1 bold right parenthesis end subscript bold space bold equals bold space bold minus bold space fraction numerator bold a bold plus bold 1 over denominator bold b bold plus bold 1 end fraction bold equals bold 2$ (આપેલ છે કે સ્પર્શક X-અક્ષ સાથે tan-1 2 માપનો ખૂણો બનાવે છે.)

∴ a + 2b = -3 (1)

વળી, (1, 1) એ વક્ર પરનું બિંદુ છે. આથી 1 + a + b = 0. આથી a + b = -1 (2)

(1) અને (2) ને ઉકેલતાં, a = 1, b = -2

$bold therefore bold space fraction numerator bold a bold space bold plus bold space bold b over denominator bold ab end fraction bold space bold equals bold space bold 1 over bold 2$

124.

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) = ....

125.

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

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

126.

R ત્રિજ્યાવાળા વર્તુળમાં અંતર્ગત ત્રિકોણની બાજુનો શૂન્યેત્તર વૃદ્ધિદર એ તેન સામેની બાજુના ખૂણાના વૃદ્દિદર કરતા Rગણો છે. આ ખૂણાનું માપ ..... થાય.

$bold pi over bold 2$
$bold pi over bold 3$
$bold pi over bold 4$
$bold pi over bold 6$

127.

bold જ ો bold space bold 3 bold space bold f bold left parenthesis bold x bold right parenthesis bold space bold minus bold space bold 2 bold f open parentheses bold 1 over bold x close parentheses bold space bold equals bold space bold x bold comma bold space bold ત ો bold space bold f bold apostrophe bold left parenthesis bold 2 bold right parenthesis bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold space

$bold 1 over bold 2$
$bold 2 over bold 7$
$bold 7 over bold 2$
2

128.

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

129.

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

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

130.

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.

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

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Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

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