Book Store

Download books and chapters from book store.
Currently only available for.
CBSE

Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

191. વક્ર $bold y bold space bold equals bold space bold x to the power of bold 1 over bold 3 end exponent$ (1 - cos x) ને x = 0 આગળના સ્પર્શકનું સમીકરણ .... થશે.

x = 0
x = 1
y = 0
y = 1

192.

વિધેય $bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space bold x to the power of bold 3 over bold 2 end exponent bold space bold plus bold space bold x to the power of fraction numerator bold minus bold 3 over denominator bold 2 end fraction end exponent bold minus bold 4 bold space open parentheses bold x bold space bold plus bold space bold 1 over bold x close parentheses$ નું ન્યુનતમ મૂલ્ય ....... છે.

-10
10
0
ન મળે.

193.

જો વક્ર y = f(x) ઓ સ્પર્શક જે બિંદુનો x યામ 1 હોય તે બિંદુએ અક્ષની ધન દિશા સામે $bold pi over bold 6$ માપનો ખુણો બનાવે છે તથા જે બિંદુના x - યામ અનુક્રમે 2 તથા 3 હોય તે બિંદુએ $bold pi over 3$ તથા $bold pi over bold 4$ માપના ખૂણા બનાવે છે, તો $table row bold 3 row bold integral row bold 1 end table bold f bold " bold left parenthesis bold x bold right parenthesis bold space bold f bold " bold left parenthesis bold x bold right parenthesis bold space bold dx bold space bold plus bold space table row bold 3 row bold integral row bold 2 end table bold f bold " bold left parenthesis bold x bold right parenthesis bold space bold dx bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold space$

$fraction numerator bold 3 square root of bold 3 over denominator bold 4 end fraction$
$fraction numerator begin display style bold 4 bold minus bold 3 square root of bold 3 end style over denominator begin display style bold 3 end style end fraction$
$fraction numerator bold 4 bold plus square root of bold 3 over denominator bold 3 end fraction$
$fraction numerator bold 4 bold plus bold 3 square root of bold 3 over denominator bold 3 end fraction$

194.

જો વિધેય f(x) = 2x³ - 9ax² + 12a²x + 1 ને x = x₁ આગળ સ્થાનીય મહત્તમ અને x = x₂ આગળ સ્થાનીય ન્યુઅનતમ મળે જ્યાં x₂ = x₁² તો a = ........

0
2
$bold 1 over bold 4$
(A) અથવા (c)

195. વિધેય $bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space fraction numerator bold tan to the power of bold 2 bold x bold space bold minus bold space bold cot to the power of bold 2 bold space bold x bold space bold plus bold space bold 1 over denominator bold tan to the power of bold 2 bold x bold space bold plus bold space bold cot to the power of bold 2 bold x bold minus bold 1 end fraction$ માટે નીચેનામાંથી કયા વિધાન સત્ય છે ?

f(x)નું વૈશ્વિક ન્યુઅનતમ મૂલ્ય -1 છે.
f(x)નું વૈશ્વિક મહત્તમ મુલ્ય $bold 5 over bold 3$ છે.
f(x)ને વૈશ્વિક ન્યુનતમ મૂલ્ય ન મળે.
f(x)ને વૈશ્વિક મહતમ મૂલ્ય ન મળે.

196. R ત્રિજ્યાવાળા ગોલકની અંતર્ગત મહત્તમ ઘનફળ ધરાવતા નળાકારની ઊંચાઈ ....... થશે.

R
$fraction numerator bold 2 bold R over denominator square root of bold 3 end fraction$
$fraction numerator bold 5 bold R over denominator bold 4 end fraction$
$fraction numerator begin display style bold R end style over denominator begin display style square root of bold 3 end style end fraction$

fraction numerator bold 2 bold R over denominator square root of bold 3 end fraction

Tips: -

જો નળાકાળની ત્રિજ્યા r અને ઊંચાઈ h હોય તો

$open parentheses bold h over bold 2 close parentheses to the power of bold 2 bold space bold plus bold space bold r to the power of bold 2 bold space bold equals bold space bold R to the power of bold 2 bold therefore bold space bold h to the power of bold 2 over bold 4 bold space bold plus bold space bold r to the power of bold 2 bold space bold equals bold space bold R to the power of bold 2 bold r to the power of bold 2 bold space bold plus bold space bold R to the power of bold 2 bold space bold minus bold space bold h to the power of bold 2 over bold 4$

નળાકરનું ઘનફળ V = $bold πr to the power of bold 2 bold h$

$bold equals bold space bold pi bold space open parentheses bold R to the power of bold 2 bold minus bold space bold h to the power of bold 2 over bold 4 close parentheses bold h bold space bold equals bold space bold pi open parentheses bold R to the power of bold 2 bold h bold minus bold h to the power of bold 3 over bold 4 close parentheses bold dV over bold dh bold space bold equals bold space bold pi bold space open parentheses bold R to the power of bold 2 bold minus fraction numerator bold 3 bold h to the power of bold 2 over denominator bold 4 end fraction close parentheses bold space bold equals bold space bold 0 bold space bold rightwards double arrow bold space bold pi open parentheses bold R to the power of bold 2 bold minus fraction numerator bold 3 bold h to the power of bold 2 over denominator bold 4 end fraction close parentheses bold equals bold space bold 0 bold space bold space bold space bold આથ ી bold space bold R bold 2 bold space bold equals bold space bold 3 over bold 4 bold h to the power of bold 2 bold. bold space bold આથ ી bold space bold h bold space bold equals bold space fraction numerator bold 2 over denominator square root of bold 3 end fraction bold space bold R bold dV over bold dh bold space bold equals bold space bold pi bold space open parentheses bold 0 bold space bold minus bold space fraction numerator bold 6 bold h over denominator bold 4 end fraction close parentheses bold space bold equals bold space bold 3 over bold 2 bold πh bold. bold space bold space bold space bold space bold space bold space bold space bold space bold h bold space bold equals bold space fraction numerator bold 2 over denominator square root of bold 3 end fraction bold space bold R bold space bold space bold space bold space bold space bold space bold space bold space bold space bold હ ો વ ા થ ી bold space fraction numerator bold d to the power of bold 2 bold V over denominator bold dh to the power of bold 2 end fraction bold less than bold 0 bold space$

h = $fraction numerator bold 2 over denominator square root of bold 3 end fraction$ R હોય ત્યારે નળાકારનું ઘનફળ મહત્તમ થશે.

197.

ધારો કે f(x) = ax³ + bx² + cx + d જ્યાં a, b, c, d ∈Rઅને 3b² < c² એ વધતું વિધેય છે અને g(x) = af (x) + bf"(x) + c². જો G(x) = g(t) dt, α ∈R તો α < x < α + 1 માટે

G(x) એ વધતું વિધેય છે.
G(x) એ ઘટતું વિધેય છે.
G(x) એ એક-એક વિધેય છે.
G(x) એ વધતું કે ઘટતું વિધેય નથી.

198. જો બિંદુ (0, 3) અને (5, -2)ને જોડતી રેખા એ વક્ર $bold y bold space bold equals bold space fraction numerator bold c over denominator bold x bold space bold plus bold space bold 1 end fraction$ નો સ્પર્શક હોય, તો c = .......

199.

વક્ર $bold y bold space bold equals bold space bold x bold space bold tan bold space bold alpha bold space bold minus bold space bold 1 over bold 2 bold space fraction numerator bold x to the power of bold 2 over denominator bold u to the power of bold 2 bold space bold cos to the power of bold 2 bold space bold alpha end fraction bold comma bold space bold alpha bold space bold element of bold space open parentheses bold 0 bold comma bold pi over bold 2 close parentheses$ ના બિંદુ P આગળનો સ્પર્શક રેખા y = x + 5 ને સમાંતર છે. જો બિંદુ P નો y - યામ $bold u to the power of bold 2 over bold 4$ હોય તો α = ......