| M.Phil | punjab university chandigarh | 1977 |
25 Years
I have been teaching Physics, Mathematics to the students of class VII - X of all boards and Mathematics to the students of class XI - XII of all boards since last thirty five years. My teaching style is unique it is beneficial for boards as well as competitive examination like WBJEE, IIT JEE preparation, AIEEE.
I Can Manage Both
my approach tutoring in such a way that it show the students the basics, and then give them a variety of practice exercises and urge them figure out the answers themselves. If in doubt, I would keep prompting them to find the right answers thereby helping them to grasp the right concepts.
1000.00
| Class 9 - 10 | Mathematics, All Boards, English Medium | INR 1000.00 /hour |
| Class 11 - 12 | Mathematics, All Boards, English Medium | INR 1000.00 /hour |
| Engineering Entrance & IITJEE | Mathematics, IIT JEE Mains, WBJEE, AIEEE, IIT JEE Advanced | INR 1500.00 /hour |
| Class 6 - 8 | Mathematics, All Boards | INR 800.00 /hour |
Integration for class 12th. It covers all type of questions.
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♢ = denote ♢r/r *100 = 1% . To find ♢V/V*100 = ??? V= 4/3 pie r^3 ♢V/♢r = 4 pie r^2 (approximate) ♢V/V*100 = (4 pie r^2 ♢r)*100)÷V = (4 pie r^2 ♢r)*100)÷4/3 pie r^3 = 3 * ♢r/r *100 = 3 * 1 = 3%
♢ = denote ♢r/r *100 = 1% . To find ♢V/V*100 = ??? V= 4/3 pie r^3 ♢V/♢r = 4 pie r^2 (approximate) ♢V/V*100 = (4 pie r^2 ♢r)*100)÷V = (4 pie r^2 ♢r)*100)÷4/3 pie r^3 = 3 * ♢r/r *100 = 3 * 1 = 3%
distance = time * speed distance in down = t (x + 4) distance in up = t (x - 4) t (x + 4) = 3 * t (x - 4) ➡ x = 8 km /hr
Reminder P (2) = Q (2) 8-24+2a-6 = 8 + (5-a) 4 +24 - Solving we get a = 11
P (x) and Q (x) becomes zero at x = 8 p(x -8) is a factor of both (x) and Q (x)
P (x) and Q (x) becomes zero at x = 8 p(x -8) is a factor of both P(x) and Q (x)
= x^3 -15x^2+66x-80
Let A(at1^2, 2at1) and B(at2^2, 2at2) be two distinct points on a parabola y^2=4x Since the value of a=1 for the given parabola coordinates of A and B will be (t1^2, 2t1) and (t2^2, 2t2) Since the axis of the parabola touches a circle of radius r having AB as the diameter, A and B are also lying on the circle and must satisfy circle equation. Coordinates for the centre of the circle will be ((t1^2+t2^2)/2, t1+t2)) Since the axis of the parabola touches the circle, radius of the circle will be y coordinate of the centre of the circle ∴ t1+t2=±r Slope of the line joining AB will be (2t2-2t1)/(t2^2-t1^2)=2(t2-t1)/(t2-t1)(t2+t1) =2/(t2+t1)= ±2/r
X^2+(Y-3)^2=9
X^2+Y^2-6Y=0
