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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 ?

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

182.

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

3a²
a²
4a²
2a²

183.

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 = ......

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

bold 3 over bold 2 square root of bold 5

Tips: -

$bold dy over bold dx bold space bold equals bold space fraction numerator bold minus bold cosec to the power of bold 2 bold theta over denominator bold 2 bold sec to the power of bold 2 bold theta bold space bold tanθ end fraction bold space bold equals bold space bold minus bold space bold 1 over bold 2 bold space bold cot to the power of bold 3 bold space bold space bold આથ ી bold space open parentheses bold dy over bold dx close parentheses subscript bold theta bold equals bold space bold pi over bold 4 end subscript bold space bold equals bold space bold minus bold 1 over bold 2 bold therefore bold space bold space bold theta bold space bold equals bold space bold pi over bold 4 bold space bold મ ા ટ ે bold space bold x bold space bold equals bold space bold equals bold space bold sec to the power of bold 2 bold space bold pi over bold 4 bold space bold equals bold space bold 2 bold comma bold space bold y bold comma bold space bold equals bold space bold cot bold space bold pi over bold 4 bold space bold equals bold space bold 1$

∴ વક્ર પરનું બિંદુ, બિંદુ P(2,1) થશે.

P આગળ સ્પર્શકનું સમીકરણ, y - 1 = $bold minus bold 1 over bold 2$ (x -2)

∴ x + 2y - 4 = 0 (1)

હવે, x = sec²θ, y = cot θ

sec²θ - tan²θ = 1 નો ઉપયોગ કરતાં θ નો લોપ કરી શકાય. $bold x bold space bold minus bold space bold 1 over bold y to the power of bold 2 bold space bold equals bold space bold 1$

$bold y to the power of bold 2 bold space bold equals bold space fraction numerator bold 1 over denominator bold x bold space bold minus bold space bold 1 end fraction bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold left parenthesis bold 2 bold right parenthesis bold left parenthesis bold 1 bold right parenthesis bold space bold પરથ ી bold comma bold space bold y bold space bold equals bold space fraction numerator bold 4 bold space bold minus bold space bold x over denominator bold 2 end fraction$

છેદ બિંદુ માટે, (2) માં મૂકતાં $open parentheses fraction numerator bold 4 bold minus bold x over denominator bold 2 end fraction close parentheses to the power of bold 2 bold space bold equals bold space fraction numerator bold 1 over denominator bold x bold minus bold 1 end fraction$

∴ (x² - 8x + 16) (x - 1) = 4

∴ x³ - 9x² + 24x - 20 = 0

∴ (x - 2)(x² - 7x + 10) = 0

∴ (x - 2)² (x - 5) = 0

∴ x = 2, 5

જો x = 2 તો y = 1. જો x = 5 તો $bold y bold space bold equals bold space bold minus bold 1 over bold 2$

∴ P(2, 1) (જે આપેલ છે) તેથી Q = $open parentheses bold 5 bold comma bold minus bold 1 over bold 2 close parentheses$

$bold therefore bold space bold PQ bold space bold equals bold space square root of bold left parenthesis bold 5 bold space bold minus bold space bold 2 bold right parenthesis to the power of bold 2 bold space bold plus bold space open parentheses bold 1 bold space bold plus bold space bold 1 over bold 2 close parentheses to the power of bold 2 end root bold space bold equals bold space fraction numerator bold 3 square root of bold 3 over denominator bold 2 end fraction$

184.

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)
$bold left parenthesis square root of bold 8 bold comma bold space bold 0 bold right parenthesis bold space$
$bold left parenthesis bold 0 bold comma bold space square root of bold 18 bold right parenthesis bold space$

185.

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, $bold dy over bold dx bold space bold equals bold space bold 1$

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

186.

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)

187.

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.

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, = .......

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