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Anshuman Misra Engineering Entrance trainer in Bangalore/>

Anshuman Misra

Tutor

Electronic City, Bangalore, India - 560100.

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Overview

Anshuman Misra describes himself as Tutor. He conducts classes in Engineering Entrance Coaching. Anshuman is located in Electronic City, Bangalore. Anshuman takes at students Home and Regular Classes- at his Home. He has 1 years of teaching experience . Anshuman has completed Bachelor of Technology (B.Tech.) from Sardar Vallabhbhai Institute of Technology Surat in 2017. He is well versed in English and Hindi.

Languages Spoken

English

Hindi

Education

Sardar Vallabhbhai Institute of Technology Surat 2017

Bachelor of Technology (B.Tech.)

Udacity 2018

Nano degree

Address

Electronic City, Bangalore, India - 560100

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Teaches

Engineering Entrance Coaching classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Engineering Entrance Coaching classes

1

Engineering Entrance Exams

IIT JEE Coaching Classes

IITJEE Coaching

IIT JEE Crash Course, IIT JEE Foundation Course, IIT JEE Advanced Coaching, IIT JEE Mains Coaching

Type of class

Crash Course, Regular Classes

IIT-JEE Subjects

Chemistry

Reviews

No Reviews yet!

FAQs

1. Which classes do you teach?

I teach Engineering Entrance Coaching Class.

2. Do you provide a demo class?

Yes, I provide a free demo class.

3. How many years of experience do you have?

I have been teaching for 1 year.

Contact

Answers by Anshuman Misra (2)

Answered on 29/06/2017 Learn Tuition/Class XI-XII Tuition (PUC) +1 CBSE/Class 12/Mathematics

Find the maximum value of (x^4-x^2)/(x^6+2x^3-1).

the solution is to differentiate the above equation and find the soln to the equation at which the differentiated equation gives the answer zero. let us say the answer is "z" . Now double differentiate the given equation and use the value "z" instead of x . if the value comes out to be negative its... ...more
the solution is to differentiate the above equation and find the soln to the equation at which the differentiated equation gives the answer zero. let us say the answer is "z" . Now double differentiate the given equation and use the value "z" instead of x . if the value comes out to be negative its maximum other wise its minimum. Its important to check because certain functions have both maxima and minima. Which means you will get a quadratic equation or a higher order function. Some times you can have a local maxima or minima. putting the value"z" in the given equation will give you the final answer of your maxima or minima
Answers 49 Comments
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Answered on 29/06/2017 Learn Exam Coaching/Engineering Entrance Coaching/IIT JEE Coaching +1 Exam Coaching/Engineering Entrance Coaching/IIT JEE Coaching/Physics

A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration... ...more
A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration 12 m/s. Find the displacement of the block during the first 0.2 s after the start?

The answer is simple. Initially both the elevator and object are at rest. Since the object isn't attached to any surface there is a possibility that it can deattach from the surface . Since the elevator is descending at a rate greater than that of gravitational acceleration the lift will fall at a higher... ...more
The answer is simple. Initially both the elevator and object are at rest. Since the object isn't attached to any surface there is a possibility that it can deattach from the surface . Since the elevator is descending at a rate greater than that of gravitational acceleration the lift will fall at a higher velocity (initial velocity of both object and elevator =0 ). So the object and elevator are no longer in contact and object will fall with a rate of 9.8 m/sec^2 had the elevator descending at a rate of<=9.8 m/sec^2 the accelaration of the object would have been same as that of elevator. Now all you have to do is s= ut + 1/2 at^2. I hope you are familiar with the symbols.
Answers 25 Comments
Dislike Bookmark

Contact

Teaches

Engineering Entrance Coaching classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Engineering Entrance Coaching classes

1

Engineering Entrance Exams

IIT JEE Coaching Classes

IITJEE Coaching

IIT JEE Crash Course, IIT JEE Foundation Course, IIT JEE Advanced Coaching, IIT JEE Mains Coaching

Type of class

Crash Course, Regular Classes

IIT-JEE Subjects

Chemistry

No Reviews yet!

Answers by Anshuman Misra (2)

Answered on 29/06/2017 Learn Tuition/Class XI-XII Tuition (PUC) +1 CBSE/Class 12/Mathematics

Find the maximum value of (x^4-x^2)/(x^6+2x^3-1).

the solution is to differentiate the above equation and find the soln to the equation at which the differentiated equation gives the answer zero. let us say the answer is "z" . Now double differentiate the given equation and use the value "z" instead of x . if the value comes out to be negative its... ...more
the solution is to differentiate the above equation and find the soln to the equation at which the differentiated equation gives the answer zero. let us say the answer is "z" . Now double differentiate the given equation and use the value "z" instead of x . if the value comes out to be negative its maximum other wise its minimum. Its important to check because certain functions have both maxima and minima. Which means you will get a quadratic equation or a higher order function. Some times you can have a local maxima or minima. putting the value"z" in the given equation will give you the final answer of your maxima or minima
Answers 49 Comments
Dislike Bookmark

Answered on 29/06/2017 Learn Exam Coaching/Engineering Entrance Coaching/IIT JEE Coaching +1 Exam Coaching/Engineering Entrance Coaching/IIT JEE Coaching/Physics

A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration... ...more
A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration 12 m/s. Find the displacement of the block during the first 0.2 s after the start?

The answer is simple. Initially both the elevator and object are at rest. Since the object isn't attached to any surface there is a possibility that it can deattach from the surface . Since the elevator is descending at a rate greater than that of gravitational acceleration the lift will fall at a higher... ...more
The answer is simple. Initially both the elevator and object are at rest. Since the object isn't attached to any surface there is a possibility that it can deattach from the surface . Since the elevator is descending at a rate greater than that of gravitational acceleration the lift will fall at a higher velocity (initial velocity of both object and elevator =0 ). So the object and elevator are no longer in contact and object will fall with a rate of 9.8 m/sec^2 had the elevator descending at a rate of<=9.8 m/sec^2 the accelaration of the object would have been same as that of elevator. Now all you have to do is s= ut + 1/2 at^2. I hope you are familiar with the symbols.
Answers 25 Comments
Dislike Bookmark

Contact

Anshuman Misra describes himself as Tutor. He conducts classes in Engineering Entrance Coaching. Anshuman is located in Electronic City, Bangalore. Anshuman takes at students Home and Regular Classes- at his Home. He has 1 years of teaching experience . Anshuman has completed Bachelor of Technology (B.Tech.) from Sardar Vallabhbhai Institute of Technology Surat in 2017. He is well versed in English and Hindi.

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