Mithapur, Patna, India - 800001.
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Hindi Mother Tongue (Native)
English Proficient
VTU Belgaun 2007
Bachelor of Engineering (B.E.)
Mithapur, Patna, India - 800001
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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, CBSE, ISC/ICSE
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 Advanced Coaching, IIT JEE Integrated Coaching, IIT JEE Mains Coaching, IIT JEE Crash Course, 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. "
1. Which school boards of Class 12 do you teach for?
State, CBSE and ISC/ICSE
2. Have you ever taught in any School or College?
Yes
3. Which classes do you teach?
I teach Class 10 Tuition, Class 11 Tuition, Class 12 Tuition, Class 9 Tuition, Engineering Entrance Coaching and UGC NET Exam Coaching Classes.
4. Do you provide a demo class?
Yes, I provide a free demo class.
5. How many years of experience do you have?
I have been teaching for 2 years.
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
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, CBSE, ISC/ICSE
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 Advanced Coaching, IIT JEE Integrated Coaching, IIT JEE Mains Coaching, IIT JEE Crash Course, 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) 
Uttam Nagar, Delhi 
