Mithapur, Patna, India - 800001
2
Details verified of Rakesh Kumar Singh✕
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2 Students
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
2
Board
State, ISC/ICSE, CBSE
ISC/ICSE Subjects taught
Mathematics
CBSE Subjects taught
Mathematics
Taught in School or College
Yes
State Syllabus Subjects taught
Mathematics
2 Students
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 11 Tuition
2
Board
CBSE
CBSE Subjects taught
Mathematics
Taught in School or College
Yes
2 Students
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
2
Board
CBSE
CBSE Subjects taught
Mathematics
Taught in School or College
Yes
2 Students
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
2
Board
CBSE
CBSE Subjects taught
Mathematics
Taught in School or College
Yes
2 Students
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Engineering Entrance Coaching classes
2
Engineering Entrance Exams
IIT JEE Coaching Classes
IITJEE Coaching
IIT JEE Crash Course, IIT JEE Integrated Coaching, IIT JEE Mains Coaching, IIT JEE Advanced Coaching, IIT JEE Foundation Course
Type of class
Regular Classes, Crash Course
IIT-JEE Subjects
Maths
2 Students
Class Location
Online (video chat via skype, google hangout etc)
I am willing to Travel
Tutor's Home
Years of Experience in UGC NET Exam Coaching classes
2
UGC_NET_Paper_I_Subjects
Logical Reasoning, Data Interpretation
UGC_NET_Papers
Paper I
5 out of 5 2 reviews
Levi Ackerman
"One of the best teachers. Very skilled in Teaching, always prove every formula with great understanding."
Adarsh Srivastava
"He is a genius teacher of maths but he talk about all the formula or equation of maths used in our daily life or other subjects. "
Answered on 06/09/2019 Learn CBSE/Class 12/Mathematics/Continuity and Differentiability/NCERT Solutions/Exercise 5.7
,
,
Answered on 06/09/2019 Learn CBSE/Class 12/Mathematics/Unit I: Relations and Functions/NCERT Solutions/Exercise 1.3
Answered on 05/09/2019 Learn CBSE/Class 12/Mathematics/Differential Equations/NCERT Solutions/Exercise 9.4
For the differential equation 
find the solution curve passing through the point (1, –1).
xydy = (x+2)(y+2)dx
ydy/(y+2) = (1 + 2/x) dx
((Y+2) -2) dy / (y+2) = (1+ 2/x) dx
(1 - 2/(y+2)) dy. = (1 + 2/x) dx
∫ (1- 2/(y+2)) dy. =. ∫ (1 + 2/x) dx + C
y -2 Ln(y+2) = x + 2 Ln x +C
y - x = 2 Ln x(y+2) + C
It must satisfy point (1,-1),. Therefore,
-1 - 1 = 2 Ln 1(-1+2) + C
C = - 2 Ln 1 - 2 = -2
Required equation is,
y = x +2 Ln x(y+2) -2
y + 2 = x + Ln (x(y+2))^2
Answered on 05/09/2019 Learn CBSE/Class 12/Mathematics/Differential Equations/NCERT Solutions/Exercise 9.5
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, then fof(x) is
(B) x3 (C) x (D) (3 − x3)
Rajajinagar,
Bangalore
