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CBSE

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.

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

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

123.

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.

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

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

Tips: -

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 log bold space bold v bold space bold plus bold space bold u over bold v to the power of bold 3 bold space bold equals bold space bold 0 bold. bold space bold આથ ી bold space bold u bold space bold plus bold space bold v to the power of bold 3 bold space bold log bold space bold v bold space bold equals bold space bold 0 bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold left parenthesis bold 1 bold right parenthesis bold space bold therefore bold space bold du over bold dv bold space bold plus bold space bold v to the power of bold 3 bold space bold log bold space bold v bold space bold equals bold space bold 0 bold space bold therefore bold space bold v bold space bold du over bold dv bold space bold plus bold space bold v to the power of bold 3 bold space end exponent bold plus bold space bold 3 bold v to the power of bold 3 bold space bold log bold space bold v bold space bold equals bold 0 bold space bold therefore bold space bold v bold space bold du over bold dv bold space bold plus bold space bold v to the power of bold 3 bold space bold plus bold space bold 3 bold space bold left parenthesis bold minus bold u bold right parenthesis bold space bold equals bold space bold 0 bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold left square bracket bold left parenthesis bold 1 bold right parenthesis bold space bold પરથ ી bold right square bracket bold therefore 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 du over bold dv bold space bold plus bold space bold v to the power of bold 2 bold space bold minus bold space bold 3 bold space bold space bold du over bold dv bold space bold equals bold space bold 0 bold space bold therefore 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

124.

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

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.

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) = ....

127.

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