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Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

211.

ધારો કે f : [0, 1] એ R પરનું વિધેય છે તથા તે દ્વિતિય વિકલિત ધરાવે છે. વળી, f(0) = f(1) = 0 તથા f"(x) - 2 f'(x) + f(x) ≥ e^x જો વિધેય e^-x f(x) એ અંતરાલ [0, 1] માં x = $bold 1 over bold 4$ આગળ ન્યુનત્તમ હોય, તો નીચેનમાંથી કયું વિધાન સાચું છે.

$bold f bold left parenthesis bold x bold right parenthesis bold space bold less than bold space bold f bold left parenthesis bold x bold right parenthesis bold comma bold space bold 3 over bold 4 bold space bold less than bold space bold x bold space bold less than bold space bold 1$
$bold f bold left parenthesis bold x bold right parenthesis bold space bold less than bold space bold f bold left parenthesis bold x bold right parenthesis bold comma bold space bold 1 over bold 4 bold space bold less than bold space bold x bold space bold less than bold space bold 3 over bold 4$
$bold f bold left parenthesis bold x bold right parenthesis bold space bold greater than bold space bold f bold left parenthesis bold x bold right parenthesis bold comma bold space bold 0 bold space bold less than bold space bold x bold space bold less than bold space bold 1 over bold 4$
$bold f bold left parenthesis bold x bold right parenthesis bold space bold less than bold space bold f bold left parenthesis bold x bold right parenthesis bold comma bold space bold 0 bold space bold less than bold space bold x bold space bold less than bold space bold 1 over bold 4$

212.

a ને કઈ કિંમત માટે f(x) = x5 - 3x + log 5, b ∈ R એ R પર ઘટતું વિધેય થાય ?

(2, 7)
(-1, 2)
(0, 7)
[-7, 1]

213. $bold y bold space bold equals bold space fraction numerator bold b to the power of bold 2 over denominator bold a bold space bold minus bold space bold x end fraction bold space bold plus bold space bold a to the power of bold 2 over bold x bold comma bold space bold 0 bold space bold less than bold space bold x bold space bold less than bold space bold a bold comma bold space bold a bold comma bold space bold b bold space bold greater than bold space bold 0 bold space$ નું ન્યુનતમ મૂલ્ય ....... છે.

$fraction numerator bold left parenthesis bold a bold space bold plus bold space bold b bold right parenthesis to the power of bold 2 over denominator bold a end fraction$
$fraction numerator begin display style bold a bold space bold plus bold space bold b end style over denominator begin display style bold a end style end fraction$
$fraction numerator begin display style bold ab end style over denominator begin display style bold a bold space bold plus bold space bold b bold space end style end fraction$
$fraction numerator begin display style bold 1 end style over denominator begin display style bold a end style end fraction bold space bold plus bold space fraction numerator begin display style bold 1 end style over denominator begin display style bold b end style end fraction bold space$

214.

ધારો કે P(x) એ ચાર ઘાતવાળી બહુપદી છે તથા $bold lim with bold x bold rightwards arrow bold 0 below bold space open parentheses bold 1 bold space bold plus bold space fraction numerator bold P bold left parenthesis bold x bold right parenthesis over denominator bold x to the power of bold 2 end fraction close parentheses$ જો P(x) ને x = 1,2 આગળ મહત્તમ કે ન્યુનતમ અસ્તિત્વ ધરાવે તો P(2) = .......

Tips: -

$bold lim with bold x bold rightwards arrow bold 0 below bold space open parentheses bold 1 bold space bold plus bold space fraction numerator bold P bold left parenthesis bold x bold right parenthesis over denominator bold x to the power of bold 2 end fraction close parentheses bold space bold equals bold space bold 2 bold. bold space bold આથ ી bold space bold lim with bold x bold rightwards arrow bold 0 below bold space fraction numerator bold P bold left parenthesis bold x bold right parenthesis over denominator bold x to the power of bold 2 end fraction bold space bold equals bold 1$

ધારો કે P(x) = a₀x⁴ + a₁x³ + a₂x² + a₃x + a₄

$bold lim with bold x bold rightwards arrow bold 0 below bold space bold space fraction numerator bold P bold left parenthesis bold x bold right parenthesis over denominator bold x to the power of bold 2 end fraction$ = 1 હોવાથી, a₃ = a₄ = 0 તથા a₂ = 1

P(x) = a₀x⁴ + a₁x³ + x²

વળી, P(x) ને x = 1,2 આગળ મહત્તમ કે ન્યુનતમ મળે છે. આથી P'(1) = P'(2) = 0

P(x) = 4a₀x³ + 3a₁x² + 2x

4a₀ + 3a₁ = - 2 અને 32a₀ + 12a = - 4

આ બંને સમીકરણો ઉકેલતાં $bold a subscript bold 0 bold space bold equals bold space bold 1 over bold 4 bold comma bold space bold a subscript bold 1 bold space bold equals bold space bold minus bold 1$

$bold therefore bold space bold P bold left parenthesis bold x bold right parenthesis bold space bold equals bold space bold 1 over bold 4 bold x to the power of bold 4 bold space bold minus bold space bold x to the power of bold 3 bold space bold plus bold space bold x to the power of bold 2 bold. bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold આથ ી bold space bold space bold P bold left parenthesis bold 2 bold right parenthesis bold space bold equals bold space bold 1 over bold 4 bold space bold 16 bold space bold minus bold space bold 8 bold space bold plus bold space bold 4 bold space bold equals bold space bold 0 bold space$

215.

જો વિધેય y = 1 + a²x - x³ જે બિંદુએ ન્યુનતમ હોય તે અસમતા $fraction numerator bold x to the power of bold 2 bold space end exponent bold plus bold space bold x bold space bold plus bold space bold 2 bold space over denominator bold x to the power of bold 2 bold space bold plus bold space bold 5 bold x bold space bold plus bold space bold 6 bold space end fraction bold less-than or slanted equal to bold space bold 0$ નું સમાધાન કરે તો પ્રચલ a ની કિંમત ...... ગણમાં હોય.

$bold empty set bold space$
$bold left parenthesis bold 2 square root of bold 3 bold space bold comma bold space bold 3 bold space square root of bold 3 bold right parenthesis$
$bold left parenthesis bold minus bold 3 square root of bold 3 bold comma bold space bold minus bold 2 bold space square root of bold 3 bold right parenthesis$
$Error converting from MathML to accessible text.$

216.

વક્ર y² = 8x અને xy = -1 ના સામાન્ય સ્પર્શકનું સમીકરણ ....... છે.

2y = x + 8
3y = 9x + 2
y = x + 2
y = 2x + 1

217. ધારો કે g(x) = 2f $open parentheses bold x over bold 2 close parentheses$ + f(1 - x) અને 0 ≤ x ≤ 1 માટે અને f"(x) 0 તો g(x) એ

$open parentheses bold 2 over bold 3 bold comma bold 1 close parentheses$ માં ઘટતું વિધેય છે.
$open parentheses bold 0 bold comma bold 2 over bold 3 close parentheses$ માં વધતું વિધેય છે.
$bold left square bracket bold 0 bold comma bold space bold 2 over bold 3 bold right parenthesis$ માં ઘટતું વિધેય છે.
$bold left parenthesis bold 2 over bold 3 bold comma bold space bold 0 bold right square bracket$ માં વધતું વિધેય છે.

218.

વિધેય f(x) = 2|x| + |x + 2| - $open vertical bar table row cell bold vertical line bold x bold space bold plus bold space bold 2 bold vertical line bold space bold minus bold space bold 2 bold vertical line bold x bold vertical line end cell end table close vertical bar$ ને x ની કઈ કિંમત માટે સ્થાનિય મહત્તમ કે સ્થાનીય ન્યુનતમ મૂલ્ય મળે ?

2
$bold 2 over bold 3$
-2
$bold minus bold 2 over bold 3$

219.

સંયોજિત વિધેય f1, (f2 (f3 (...(fn (x))) એ ઘટતું વિધેય છે અને n વિધેયોમાંથી r વિધેય છે તથા બાકિના વધતાં વિધેય છે. r(n-r) નું મહત્તમ મૂલ્ય કેટલું થાય ?

$fraction numerator bold n to the power of bold 2 bold space bold minus bold space bold 4 over denominator bold 4 end fraction bold comma bold space bold n bold space bold equals bold space bold 4 bold k bold comma bold space bold k bold space bold element of bold space bold N bold space$
$bold n to the power of bold 2 over bold 4$ યુગ્મ સંખ્યા હોય.
$fraction numerator begin display style bold n to the power of bold 2 bold space bold minus bold space bold 1 end style over denominator begin display style bold 4 end style end fraction bold comma bold space$ અયુગ્મ સંખ્યા હોય.
$fraction numerator begin display style bold n to the power of bold 2 end style over denominator begin display style bold 4 end style end fraction bold comma bold space bold n bold space bold equals bold space bold 4 bold k bold space bold plus bold space bold 2 bold comma bold space bold k bold space bold element of bold space bold N$

220.

જો વક્ર 4x2 + a²y² = 4a², 4 < a² < 8 પરનું બિંદુ (u, v) એ બિંદુ (0, -2) થી સૌથી દૂરનું બિંદુ હોય તો u + v = ..........

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Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

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