Ask a Doubt
Request a call back

Join NOW to get access to exclusive study material for best results

JEE Class main Answered

f(x) and g(x) are two functions which satisfy differential equation Image not present = 0 where f(0) = g(0) = 2 and f '(0) = g'(0) = 1, then (1) Both f(x) and g(x) are decreasing functions (2) Both f(x) and g(x) are monotonic functions (3) Area bounded by f(x), g(x) and x = ?n2 is Image not present sq. units. (4) Area bounded by f(x), g(x) and x = ?n2 is Image not present sq. units.
Asked by kuldeepupadhyay530 | 11 Apr, 2020, 21:28: PM
answered-by-expert Expert Answer
f(x) and g(x) satisfy the given differential equation
fraction numerator d squared x over denominator d y squared end fraction times fraction numerator d squared y over denominator d x squared end fraction plus fraction numerator d x over denominator d y end fraction equals 0 space... space left parenthesis i right parenthesis fraction numerator d squared x over denominator d y squared end fraction equals fraction numerator d over denominator d y end fraction open parentheses fraction numerator d x over denominator d. y end fraction close parentheses equals fraction numerator d over denominator d y end fraction open parentheses fraction numerator 1 over denominator begin display style fraction numerator d y over denominator d x end fraction end style end fraction close parentheses equals negative 1 over open parentheses begin display style fraction numerator d y over denominator d x end fraction end style close parentheses squared times fraction numerator d x over denominator d y end fraction fraction numerator d over denominator d x end fraction open parentheses fraction numerator d y over denominator d x end fraction close parentheses equals negative 1 over open parentheses begin display style fraction numerator d y over denominator d x end fraction end style close parentheses cubed times fraction numerator d squared y over denominator d x squared end fraction F r o m space left parenthesis i right parenthesis rightwards double arrow negative 1 over open parentheses begin display style fraction numerator d y over denominator d x end fraction end style close parentheses cubed times fraction numerator d squared y over denominator d x squared end fraction times fraction numerator d squared y over denominator d x squared end fraction plus fraction numerator 1 over denominator begin display style fraction numerator d y over denominator d x end fraction end style end fraction equals 0 M u l t i p l y i n g space a b o v e space e q space w i t h space minus open parentheses fraction numerator d y over denominator d x end fraction close parentheses cubed rightwards double arrow open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses squared minus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared equals 0 rightwards double arrow fraction numerator d squared y over denominator d x squared end fraction minus fraction numerator d y over denominator d x end fraction equals 0 space space space o r space space space fraction numerator d squared y over denominator d x squared end fraction plus fraction numerator d y over denominator d x end fraction equals 0 W h e n space fraction numerator d squared y over denominator d x squared end fraction minus fraction numerator d y over denominator d x end fraction equals 0 comma space t h e space c h a r a c t e r i s t i c space e q space i s space lambda squared minus lambda equals 0 rightwards double arrow lambda equals 0 space o r space 1 rightwards double arrow y subscript 1 equals c subscript 1 plus c subscript 2 e to the power of x F ı n d space c subscript 1 space a n d space c subscript 2 space u s ı n g space t h e space g ı v e n space v a l u e s space F apostrophe open parentheses 0 close parentheses equals G apostrophe open parentheses 0 close parentheses equals 1 space space a n d space space F open parentheses 0 close parentheses equals G open parentheses 0 close parentheses equals 2 W e space g e t space F open parentheses x close parentheses equals 1 plus e to the power of x space a n d space G open parentheses x close parentheses equals 3 minus e to the power of x
Answered by Renu Varma | 19 Apr, 2020, 17:03: PM

Application Videos

Concept Videos

Multiple Choice Questions Practice Test
Answered Unanswered My Questions My Answers
JEE main - Maths
in how many ways a team of 2 boys and 4 girls selected from 5 boys and 6 girls
Asked by pallavibirangal87 | 08 Dec, 2024, 10:18: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Maths
if a chord of length 18 cm is 12 cm away from the centre then find the distance of 24 cm long chord from the centre
Asked by pritisindra6 | 06 Dec, 2024, 21:04: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Maths
The number of ways in which at least one coin can be selected from 3 coins of one rupee, 4 coins of fifty paise, 5 coins of twenty paise and 6 coins of ten paise is-
Asked by kanishthkumawat | 29 Nov, 2024, 00:23: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Maths
what is range of tan^-1 x
Asked by pawandhayal125 | 06 Oct, 2024, 10:03: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Maths
what is 2+2
Asked by srishtipra2008 | 28 Sep, 2024, 10:33: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Maths
n arithmetic means are inserted between the numbers 7 and 49. if the sum of these means is 364 then the sum of their squares is
Asked by yogeshsoni575 | 23 Sep, 2024, 23:01: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Maths
consider a dodecagon then number of ways of selecting two triangles which can be made using the vertices of dodecagon so that they do not have a common side.
Asked by sumit.parsudih | 17 Sep, 2024, 22:22: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Maths
the sum of infinite series 1/2(1/3+1/4)-1/4(1/3^2+1/4^2)+1/6(1/3^3+1/4^3)+
Asked by pritamlovevanshi5 | 15 Sep, 2024, 08:24: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Maths
solve the equation 6x-y+3z=10 , 3x+6y=18 , x+y+5z=-10 by jacobis method with (3,0,-2) as the initial approcimate to the solution
Asked by ganeshbabugottapu7 | 12 Sep, 2024, 18:19: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Maths
find the inverse of a Matrix by using gauss Jordan method -2. 1. 3 A = 0. -1. 1 1. 2. 0
Asked by ganeshbabugottapu7 | 10 Sep, 2024, 07:32: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
My Class Videos Tests ₹ Plans Ask a Doubt