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Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

161.

aની કઈ કિંમત માટે f(x) = x³ + 3(a-7)x² + 3(a² - 9) x - 1 નું સ્થાનીય મહત્તમ મૂલ્ય ધન સંખ્યા માટે મળે ?

a < -3
-1 , a < 1
3 < a < 4
5 < a < 6

162.

ધારો કે p(x) = a₀ + a₁x² + a₂x⁴ + ... + a_nx²ⁿ એ વાસ્તવિક સંખ્યા x માં બહુપદી છે, જ્યાં 0 < a₀ < a₁ ..... < a_n વિધેય P(x0 ને

n મહત્તમ તથા n ન્યુનત્તમ મૂલ્ય મળે.
મહત્તમ કે ન્યુનતમ મુલ્ય ન મળે.
ફક્ત એક મહત્તમ મૂલ્ય મળે.
ફક્ત એક ન્યુનતમ મૂલ્ય મળે.

163. વિધેય f(x) = |3 - x| + |2 + x| + |5 - x| નું ન્યુનતમ મૂલ્ય ........ છે.

164.

bold જ ો bold space bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space open curly brackets table attributes columnalign left columnspacing 1.4ex end attributes row cell bold 3 bold x to the power of bold 2 bold space bold plus bold space bold 12 bold x bold space bold minus bold space bold 1 bold comma bold space end cell cell bold minus bold 1 bold space bold less-than or slanted equal to bold space bold x bold space bold less than bold space bold 2 end cell row cell bold 37 bold space bold minus bold space bold x end cell cell bold 2 bold space bold less-than or slanted equal to bold space bold x bold space bold less-than or slanted equal to bold space bold 3 end cell end table close bold ત ો bold space

f ને x = 2 આગળ મહત્તમ કિંમત મળે.
f એ (10, 2) માં વધતું વિધેય છે.
f એ [-1, 3] પર સતત વિધેય છે.
f'(2)નું અસ્તિત્વ નથી.

165.

f(x) એ ત્રણ ઘાતવાળું બહુપદી વિધેય છે, જ્યાં f(0) + 2 = 0 અને f"'(0) = 6. જો f(x) = 0 નાં ત્રણેય બીજ ધન પૂર્ણાંક હોય, તો

f'(1) = 0
2f(0) + f"(0) = 0
2f(0) + f"(0) = 2
f(2) = 0

166.

bold lim with bold n bold rightwards arrow bold infinity below bold space fraction numerator bold minus bold 3 bold n bold space bold plus bold space bold left parenthesis bold minus bold 1 bold right parenthesis to the power of bold n bold space over denominator bold 4 bold n bold space bold minus bold space bold left parenthesis bold minus bold 1 bold right parenthesis to the power of bold n end fraction bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold space

જો n અયુગ્મ સંખ્યા હોય તો $fraction numerator bold minus bold 3 over denominator bold 4 end fraction$
જો n યુગ્મ સંખ્યા હોય તો $fraction numerator bold minus bold 3 over denominator bold 4 end fraction$
$fraction numerator bold minus bold 3 over denominator bold 4 end fraction$
લક્ષનું અસ્તિત્વ નથી.

167.

h ઉંચાઈ તથા અર્ધશીર્ષકોણ વાળા શંકુમાં અંતર્ગત મહત્તમ ઘનફળવાળા નળકારનું ઘનફળ ..... છે.

$bold 4 over bold 27 bold space bold πh to the power of bold 3 bold space bold tan to the power of bold 3 bold alpha bold space$
$bold 1 over bold 27 bold space bold h to the power of bold 3 bold space bold tan to the power of bold 3 bold space bold alpha$
$bold 4 over bold 27 bold space bold h to the power of bold 2 bold space bold tan to the power of bold 2 bold space bold alpha$
$bold 4 bold πh to the power of bold 3 bold space bold tan to the power of bold 3 bold space bold alpha bold space$

bold 4 over bold 27 bold space bold πh to the power of bold 3 bold space bold tan to the power of bold 3 bold alpha bold space

Tips: -

નળાકાળ ની ત્રિજ્યા x તથા શંકુનો અર્ધશિર્ષકોણ હોય, તો tan α $bold equals bold space bold x over bold VN$

∴ VN = x cot α

∴ નળાકળની ઉંચાઈ MN = VM - VN = h - x cot α

$bold therefore bold space bold નળ ા ક ા રન ું bold space bold ઘનફળ bold space bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold pi bold space bold x to the power of bold 2 bold left parenthesis bold h bold space bold minus bold space bold x bold space bold cotα bold right parenthesis bold space bold equals bold space bold πx to the power of bold 2 bold h bold space bold minus bold pi bold space bold x bold 3 bold space bold cotα bold space bold equals bold space bold 0 bold space bold f bold apostrophe bold left parenthesis bold x bold right parenthesis bold space bold equals bold space bold 0 bold space bold rightwards double arrow bold space bold 2 bold πhx bold space bold minus bold space bold 3 bold πx to the power of bold 2 bold space bold cotα bold space bold equals bold space bold 0 bold space bold x bold space bold equals bold space bold 2 over bold 3 bold space bold h bold space bold tan bold space bold alpha bold space bold therefore bold space bold f bold " bold left parenthesis bold x bold right parenthesis bold space bold equals bold space bold 2 bold πh bold space bold minus bold space bold 6 bold pi bold space bold x bold space bold cot bold space bold alpha bold f bold " bold left parenthesis bold x bold right parenthesis bold space bold equals bold space bold 2 bold πh bold space bold minus bold space bold 6 bold pi bold space bold x bold space bold cot bold space bold alpha bold f bold " open parentheses bold 2 over bold 3 bold h bold space bold tan bold space bold alpha close parentheses bold space bold equals bold space bold minus bold 2 bold πh bold space bold less than bold space bold 0 bold space bold therefore bold space bold x bold space bold equals bold 2 over bold 3 bold space bold h bold space bold tan bold space bold alpha bold space bold મ ા ટ ે bold space bold નળ ા કરન ું bold space bold ઘનફળ bold space bold મહત ્ તમ bold space bold છ ે bold. bold space bold મહત ્ તમ bold space bold ઘનફળ bold space bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold πx bold 2 bold MN bold space bold equals bold space bold pi bold 4 over bold 9 bold space bold h to the power of bold 2 bold space bold tan to the power of bold 2 bold space bold alpha bold space open parentheses bold h bold space bold minus bold space bold 2 over bold 3 bold h close parentheses bold equals bold space fraction numerator bold 4 bold pi over denominator bold 27 end fraction bold space bold h to the power of bold 3 bold space bold tan to the power of bold 2 bold space bold alpha$

168.

(1, 2) માંથી પસાર થતી કઈ રેખા પ્રથમ ચરણમાં અક્ષો સાથે ન્યુનતમ ક્ષેત્રફળવાઓ ત્રિકોણ બનાવશે ?

x + 2y = 4
2x + y = 2
2x + y = 0
2x + y = 4

169.

જો f'(sin x) < 0 અને f"(sin x) > 0. ∀ x ∈ R, તો g(x) = f(sin x) + f(cos x), x ∈ એ

$open parentheses bold 0 bold comma bold pi over bold 4 close parentheses$ માં ઘટતું વિધેય છે.
$open parentheses bold pi over bold 4 bold apostrophe bold pi over bold 2 close parentheses$ માં વધતું વિધેય છે.
વિધેય g ને x = $bold pi over bold 4$ આગળ સ્થાનીય ન્યુનતમ મળે.
વિધેય g ને x = $bold pi over bold 4$ આગળ સ્થાનીય મહત્તમ મળે.

170. જો y(x) = cos (3 cos^-1x), x∈[-1, 1] x # ± $fraction numerator square root of bold 3 over denominator bold 2 end fraction$ તો $fraction numerator bold 1 over denominator bold y bold left parenthesis bold x bold right parenthesis end fraction bold space open curly brackets bold left parenthesis bold x to the power of bold 2 bold space bold minus bold space bold 1 bold right parenthesis fraction numerator bold d to the power of bold 2 bold y bold left parenthesis bold x bold right parenthesis over denominator bold dx to the power of bold 2 end fraction bold plus bold x fraction numerator bold dy bold left parenthesis bold x bold right parenthesis over denominator bold dx end fraction close curly brackets bold equals bold space bold. bold. bold. bold. bold. bold space$

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Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

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