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CBSE

Gujarati JEE Mathematics : લક્ષ-સાતત્ય અને વિકલન

Multiple Choice Questions

181.

વક્ર ay² - x³, c > 0. પર પ્રથમ ચરણમાં કયા બિંદુએ અભિલંબ બંને અક્ષ પર સમાન લંબાઈના અંતઃખંડ કાપે છે ?

$open parentheses fraction numerator bold 4 bold a over denominator bold 9 end fraction fraction numerator bold minus bold 8 bold a over denominator bold 27 end fraction close parentheses$
$open parentheses bold a over bold 9 bold comma fraction numerator bold 8 bold a over denominator bold 27 end fraction close parentheses$
$open parentheses bold 4 bold a bold comma fraction numerator bold 8 bold a over denominator bold 27 end fraction close parentheses$
$open parentheses fraction numerator bold 4 bold a over denominator bold 9 end fraction bold comma fraction numerator bold 8 bold a over denominator bold 27 end fraction close parentheses$

open parentheses fraction numerator bold 4 bold a over denominator bold 9 end fraction bold comma fraction numerator bold 8 bold a over denominator bold 27 end fraction close parentheses

Tips: -

ધારો કે માંગેલ બિંદુ (x₁, y₁) છે.

ay₁² = x₁³. આથી $bold y subscript bold 1 bold space bold equals bold space bold plus-or-minus bold space square root of begin inline style bold x subscript bold 1 to the power of bold 3 over bold a end style end root$

ay² = x³ નું x પ્રત્યે વિકલન કરતાં,

$bold 2 bold ay bold space bold dy over bold dx bold space bold equals bold space bold 3 bold x to the power of bold 2 bold therefore bold space open parentheses bold dy over bold dx close parentheses subscript bold left parenthesis bold x subscript bold 1 bold comma bold space bold y subscript bold 1 bold right parenthesis end subscript bold space bold equals bold space fraction numerator bold 3 bold x subscript bold 1 to the power of bold 2 over denominator bold 2 bold ay subscript bold 1 end fraction bold space bold equals bold space fraction numerator bold 3 bold x subscript bold 1 to the power of bold 2 over denominator bold 2 bold a square root of begin display style bold x subscript bold 1 to the power of bold 3 over bold a end style end root end fraction bold space bold equals bold space bold plus-or-minus bold space bold 3 over bold 2 bold space square root of bold x subscript bold 1 over bold a end root$

અભિલંબનો ઢાળ $bold equals bold space bold plus-or-minus bold space bold 2 over bold 3 bold space square root of bold a over bold x subscript bold 1 end root$

આપણે જાણીએ છીએ કે જો રેખા બંને અક્ષ પર સમાન લંબાઈનો અંતઃખંડ કાપે તો તેનો ઢાળ = ± 1

$bold therefore bold space bold plus-or-minus bold 2 over bold 3 bold space square root of bold a over bold x subscript bold 1 end root bold space bold equals bold space bold plus-or-minus bold 1 bold. bold space bold space bold space bold space bold space bold આથ ી bold space bold 4 over bold 9 bold space bold a over bold x subscript bold 1 bold space bold equals bold space bold 1 bold. bold space bold space bold space bold space bold space bold space bold space bold આથ ી bold space bold space bold x subscript bold 1 bold space bold equals bold space fraction numerator bold 4 bold a over denominator bold 9 end fraction bold therefore bold space bold y subscript bold 1 bold space bold equals bold space bold plus-or-minus bold space square root of fraction numerator bold 64 bold a to the power of bold 3 over denominator bold 9 to the power of bold 3 bold a end fraction end root bold space bold equals bold space bold plus-or-minus bold space fraction numerator bold 8 bold a over denominator bold 27 end fraction$

પ્રથમ ચરણમાં માંગેલ બિંદુ $open parentheses fraction numerator bold 4 bold a over denominator bold 9 end fraction bold comma fraction numerator bold 8 bold a over denominator bold 27 end fraction close parentheses$ છે.

નોંધ y₁ = $bold minus square root of bold x to the power of bold 3 over bold a end root$ હોય તો.

182.

જો રેખા (3 - a)x + ay + a² = 1 = 0 એ વક્ર xy = 1 નો કોઈક બિંદુએ અભિલંબ હોય તો a ∈ ......

(-∞, 0) ∪ (3, ∞)
(0, 3)
(0, ∞)
(0, 1)

183.

જો P₁ અને P₂ એ અનુક્રમે ઉગમબિંદુથી વક્ર પરના સ્પર્શક તથા અભિલંબની લંબાઈ હોય, તો 4P₁² + P₂² = ....

3a²
a²
4a²
2a²

184.

f(x) = $bold x to the power of bold 3 over bold 3 bold space bold plus bold space fraction numerator bold 3 bold x to the power of bold 2 over denominator bold 2 end fraction$ + ax + b. જો F(-2) = 0 હોય, તો એવી કેટલી ક્રમયુક્ત હોડ (a, b) શક્ય બને જ્યાં વિધેય F એ P(-2, 0) આગળ ન્યુનતમ હોય.

185.

વક્ર x = sec²θ, y = cot θ ના બિંદુ $bold P open parentheses bold pi over bold 4 close parentheses$ આગળનો સ્પર્શક જો વક્રને ફરીથી બિંદુ Q આગલ મળે, તો PQ = ......

$bold 2 square root of bold 15$
$square root of bold 15$
$bold 3 over bold 2 square root of bold 5$
$bold 1 over bold 2 square root of bold 15$

186.

વક્ર y = $table row bold x row bold integral row bold 0 end table$ 2|t|dt માટે પ્રથમ ચરણની દ્વિભજક રેખા એટૅલે કે y = x ને સમાંતર હોવાથી, $bold dy over bold dx bold space bold equals bold space bold 1$

$bold y bold space bold equals bold space bold x bold space bold plus bold 3 over bold 4 bold space bold comma bold space bold y bold space bold equals bold space bold x bold space bold minus bold space bold 1 over bold 4 bold space bold space$
$bold y bold space bold equals bold space bold x bold plus-or-minus bold 1 over bold 2$
$bold y bold space bold equals bold space bold x bold space bold plus-or-minus bold 1 bold space$
$bold space bold y bold space bold equals bold space bold x bold plus-or-minus bold space bold 3 over bold 4$

187.

ઉપવલય $bold x to the power of bold 2 over bold 8 bold space bold plus bold space bold y to the power of bold 2 over bold 18 bold space bold equals bold space bold 1$ ના બિંદુ P (x, y) આગળનો સ્પર્શક અક્ષોને બિંદુ A તથા B માં છેદે છે. જો OAB નું ક્ષેત્રફળ ન્યુનતમ હોય તો બિંદુ P એ ......

$open parentheses bold 1 bold comma bold space fraction numerator square root of bold 63 over denominator bold 2 end fraction close parentheses$
(2, 3)
$bold left parenthesis square root of bold 8 bold comma bold space bold 0 bold right parenthesis bold space$
$bold left parenthesis bold 0 bold comma bold space square root of bold 18 bold right parenthesis bold space$

188.

વક્ર y = ax³ + bx³ + cx + 5 એ X - અક્ષને બિંદું P(-2, 0) એ સ્પર્શે છે. આ વક્ર Y-અક્ષને Q-બિંદુએ છેદે છે. ત્યાં સ્પર્શકનો ઢાળ 3 છે. a, b, c, = .......

$bold 3 bold comma bold space bold minus bold 1 over bold 2 bold comma bold space bold minus bold 3 over bold 2$
$bold minus bold 3 over bold 4 bold comma bold space bold minus bold 1 over bold 2 bold comma bold 3$
$bold minus bold space bold 1 over bold 2 bold comma bold space bold minus bold 3 over bold 4 bold comma bold space bold 3 bold space$
$bold 1 over bold 2 bold comma bold 3 over bold 2 bold comma bold minus bold 3$