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CBSE

Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

141.

વક્ર $bold x bold space bold equals bold space bold a bold space square root of bold cos bold 2 bold theta end root bold space bold cosθ bold comma bold space bold y bold space bold equals bold space bold a square root of bold cos bold 2 bold theta end root bold space bold sinθ bold space bold ન ો bold space bold theta bold space bold equals bold pi over bold 6 bold space$ આગળનો સ્પર્શક .........

રેખા y = x ને સમાંતર છે.
રેખા x + y = 1 ને સમાંતર છે.
X-અક્ષને સમાંતર છે.
Y-અક્ષને સમાંતર છે.

142.

bold વ િ ધ ે ય bold space bold g bold left parenthesis bold x bold right parenthesis bold space bold equals bold space bold tanx bold. bold space bold ex bold space bold e to the power of bold x to the power of bold 2 end exponent bold space fraction numerator bold log bold left parenthesis bold pi bold space bold plus bold space bold x bold right parenthesis over denominator bold log bold left parenthesis bold e bold space bold plus bold space bold x bold right parenthesis bold space end fraction bold left parenthesis bold x bold greater-than or slanted equal to bold 0 bold right parenthesis bold space bold એ

$open square brackets 0 comma straight pi over e close square brackets$ પર વધતું તથા $open square brackets bold pi over bold e bold comma bold infinity close square brackets$ પર ઘટતું વિધેય છે.
$open square brackets bold 0 bold comma bold pi over bold e close square brackets$ પર ઘટતું તથા $open square brackets bold pi over bold e bold comma bold infinity close square brackets$ પર વધતું વિધેય છે.
[0, ∞]પર વધતું વિધેય છે.
[0, ∞]પર ઘટતું વિધેય છે.

143.

જો f(x)= અને g(x) = , 0<x<1, તો આ અંતરાલમાં

f(x) વધતું વિધેય છે અને g(x) ઘટતું વિધેય છે.
g(x) વધતું વિધેય છે અને f(x) ઘટતું વિધેય છે.
f(x) અને g(x) બંને વધતાં વિધેય છે.
f(x) અને g(x) બંને ઘટતાં વિધેય છે.

144. $bold વ િ ધ ે ય bold space bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space open curly brackets table row cell bold left parenthesis bold x bold space bold plus bold space bold 1 bold right parenthesis bold 3 bold comma bold space end cell row cell bold x to the power of bold 2 over bold 3 end exponent bold minus bold 1 bold comma bold space end cell row cell bold minus bold left parenthesis bold x bold space bold minus bold 1 bold right parenthesis end cell end table close bold space bold space bold space table attributes columnalign left end attributes row cell bold minus bold 2 bold space bold less than bold x bold space bold less-than or slanted equal to bold space bold minus bold space bold 1 end cell row cell bold minus bold 1 bold space bold less than bold space bold x bold space bold less-than or slanted equal to bold space bold 1 end cell row cell bold 1 bold space bold less than bold space bold X bold less than bold space bold 2 end cell end table$

ને કેટલાં બિંદુઓએ સ્થાનીય મહત્તમ ન્યુનત્તમ મૂલ્યોનું અસ્તિત્વ હોય ?

145.

વક્ર y = [| sinx | + | cosx |] અને x² + y² = 5 વચ્ચેના ખૂણાનું માપ ........ છે. જ્યાં [.] એ પૂર્ણાંક ભાગ વિધેય છે.

$bold cos to the power of bold minus bold 1 end exponent open parentheses fraction numerator bold 2 over denominator square root of bold 5 end fraction close parentheses$
$bold tan to the power of bold minus bold 1 end exponent bold space bold 3 bold space$
$bold pi over bold 6 bold space$
$bold pi over bold 2$

146.

જો a, b > 0 તો y = $fraction numerator bold b to the power of bold 2 over denominator bold a bold minus bold x end fraction bold space bold plus bold space bold a to the power of bold 2 over bold x$ , 0 < x < a નું ન્યુનતમ મૂલ્ય ...... થાય.

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

147.

વક્ર 4x² + a²y² = 4a², 4 < a² < 8 પરનું ......... બિંદુ એ (0, -2) થી સૌથી દૂરનું બિંદુ છે.

(0, 2)
(0, -2)
(0, 1)
(0, -1)

148.

O કેન્દ્રવાળો ઉપવલય છે $bold x to the power of bold 2 over bold y to the power of bold 2 bold space bold plus bold space bold y to the power of bold 2 over bold b to the power of bold 2 bold space bold equals bold space bold 1$ પર P કોઈ એક બિંદુ છે. કેન્દ્ર 0 માંથી દોરેલ P પરના સ્પર્શક પર દોરેલ લંબનો લંબપાદ N છે. ધારો કે A_max એ ∆OPN નું મહત્તમ ક્ષેત્રફળ છે. $fraction numerator bold a to the power of bold 2 bold space bold minus bold space bold b to the power of bold 2 over denominator bold A subscript bold max end fraction$ ની કિંમત .....

Tips: -

બિંદુ P ના યામ (acosθ, bsinθ) લો.

$bold therefore bold space open parentheses bold dy over bold dx close parentheses subscript bold p bold space bold equals bold space fraction numerator bold bcosθ over denominator bold minus bold a bold space bold sinθ end fraction bold space bold equals bold space fraction numerator bold minus bold b over denominator bold a end fraction bold cotθ$

P બિંદુએ સ્પર્શકનું સમીકરણ

bxcosθ + aysinθ = ab

કાટકોણ ∆OPNમાં

PN² - OP² - ON²

$bold equals bold space bold a bold 2 bold cos bold 2 bold theta bold space bold plus bold space bold b bold 2 bold sin bold 2 bold theta bold space bold minus bold space fraction numerator bold a to the power of bold 2 bold b to the power of bold 2 over denominator bold a to the power of bold 2 bold sinθ bold space bold plus bold space bold b to the power of bold 2 bold cos to the power of bold 2 bold theta end fraction bold space bold equals bold space fraction numerator bold a to the power of bold 4 bold sin to the power of bold 2 bold θcos to the power of bold 2 bold theta bold plus bold b to the power of bold 4 bold sin to the power of bold 2 bold θcos to the power of bold 2 bold theta bold space bold plus bold space bold a to the power of bold 2 bold b to the power of bold 2 bold left parenthesis bold sin to the power of bold 4 bold theta bold space bold plus bold space bold cos to the power of bold 4 bold theta bold right parenthesis bold space bold minus bold space bold a to the power of bold 2 bold b to the power of bold 2 bold left parenthesis bold sin to the power of bold 2 bold theta bold space bold plus bold space bold cos to the power of bold 2 bold theta bold right parenthesis to the power of bold 2 over denominator bold a to the power of bold 2 bold sin to the power of bold 2 bold theta bold space bold plus bold space bold b to the power of bold 2 bold cos to the power of bold 2 bold theta end fraction bold equals bold space fraction numerator open parentheses bold a to the power of bold 2 bold space bold minus bold space bold b to the power of bold 2 close parentheses to the power of bold 2 bold space bold sin to the power of bold 2 bold θcos to the power of bold 2 bold theta over denominator bold a to the power of bold 2 bold sin to the power of bold 2 bold theta bold space bold plus bold space bold b to the power of bold 2 bold cos to the power of bold 2 bold theta end fraction bold therefore bold space bold PN bold space bold equals bold space fraction numerator open parentheses bold a to the power of bold 2 bold space bold minus bold space bold b to the power of bold 2 close parentheses to the power of bold 2 bold space bold sin to the power of bold 2 bold θcos to the power of bold 2 bold theta over denominator square root of bold a to the power of bold 2 bold sin to the power of bold 2 bold theta bold space bold plus bold space bold b to the power of bold 2 bold cos to the power of bold 2 bold theta end root end fraction bold space bold A bold space bold equals bold space bold 1 over bold 2 bold space bold PN bold space bold cross times bold space bold ON bold space bold equals bold space fraction numerator bold ab bold left parenthesis bold a to the power of bold 2 bold space bold minus bold space bold b to the power of bold 2 bold right parenthesis bold space bold sinθcosθ over denominator bold 2 bold left parenthesis bold a to the power of bold 2 bold tanθ bold space bold plus bold space bold b to the power of bold 2 bold cotθ bold right parenthesis end fraction bold equals bold space fraction numerator bold ab bold left parenthesis bold a to the power of bold 2 bold space bold minus bold space bold b to the power of bold 2 bold right parenthesis over denominator bold 2 open parentheses bold a to the power of bold 2 bold tanθ bold space bold plus bold space bold b to the power of bold 2 bold cotθ close parentheses end fraction$

A ને મહત્તમ થવા માટે, f(0) = a²tanθ + b²cotθ ન્યુનત્તમ થાય.

$fraction numerator bold d bold left parenthesis bold f bold left parenthesis bold theta bold right parenthesis bold right parenthesis over denominator bold dθ end fraction$ = a²sec²-b²cosec²θ, f(θ) = 2b²cosec²a cotθ

$fraction numerator bold d bold left parenthesis bold f bold left parenthesis bold theta bold right parenthesis bold right parenthesis over denominator bold dθ end fraction$ 0 ⇒ tanθ = ± $bold b over bold a$

$right enclose fraction numerator bold d bold left parenthesis bold f bold left parenthesis bold theta bold right parenthesis bold right parenthesis over denominator bold dθ end fraction end enclose bold space subscript bold tanθ bold space bold b over bold a bold space bold greater than bold space bold 0 end subscript$

f(θ) એ tanθ = $bold b over bold a$ હોય ત્યારે ન્યુનત્તમ થાય. અથી ન્યુનત્તમ f(θ) = $bold a to the power of bold 2 bold space bold b over bold a bold space bold plus bold space bold b to the power of bold 2 bold space bold a over bold b bold space bold equals bold space bold 2 bold ab$

$bold therefore bold space bold A subscript bold max bold space bold equals bold space fraction numerator bold ab open parentheses bold a to the power of bold 2 bold space bold minus bold space bold b to the power of bold 2 close parentheses over denominator bold 2 bold times bold 2 bold ab end fraction bold space bold equals bold space fraction numerator bold a to the power of bold 2 bold space bold minus bold space bold b to the power of bold 2 over denominator bold 4 end fraction bold space bold therefore bold space fraction numerator bold a to the power of bold 2 bold space bold minus bold space bold b to the power of bold 2 over denominator bold A subscript bold max end fraction bold equals bold space bold 4 bold space$