in this circle centered at (1, -2) with radius= root(24)........... wat does y/x representis it the slope of the tangent...???
yup thts the thng whenever u face a prob of y/x i mark 0/1 without solving although it actually goes from infinity to -infinity bt always there is sum or the othr condition making it most likely zero or 1 ....:D :D
yup thts the thng whenever u face a prob of y/x i mark 0/1 without solving although it actually goes from infinity to -infinity bt always there is sum or the othr condition making it most likely zero or 1 ....
lekin infinity to answer nahi hona chahiye... kuchh to number aayega i guess... i m not sure if my concept is clear abt slope of tangent and all ......... wait for GODs to come...
Well, obviously if we put x=0 then y/x = infinity..
If we can't put x = 0
Then, maximum of y can be at the lowest point of the circle if sign(+ or -) is not taken into consideration otherwise it's the uppermost point in a circle of radius, R = 2_/6
and centre ( 1, -2)
We can calculate the coordinates of uppermost and lowermost point easily since we know R=2_/6
Oter than this I don't know anything about it's solution :splat:
Well, obviously if we put x=0 then y/x = infinity..If we can't put x = 0Then, maximum of y can be at the lowest point of the circle if sign(+ or -) is not taken into consideration otherwise it's the uppermost point in a circle of radius, R = 2_/6 and centre ( 1, -2)We can calculate the coordinates of uppermost and lowermost point easily since we know R=2_/6Oter than this I don't know anything about it's solution
sir, the highest point of the circle is at x = 1, where y is approx = 3
y/x = 3 (approx)
and dy/dx >0 for 0
so as x becomes less than 1, y is still > 1, so y /x should go on increasing.
sir, the highest point of the circle is at x = 1, where y is approx = 3y/x = 3 (approx)and dy/dx >0 for 0 so as x becomes less than 1, y is still > 1, so y /x should go on increasing.is it possible to find the maximum value then?
Actually, I really don't think that we can find a global maxima of the function y/x with the given domain!
At least I can't find a solution to it 😞 Others may well!
Out of 120 students in a school, 87 plays hockey, 76 plays football, 65 plays tennis and 19 plays carrom. What can be minimum number of students who play at least two of the three games?
Out of 120 students in a school, 87 plays hockey, 76 plays football, 65 plays tennis and 19 plays carrom.What can be minimum number of students who play at least two of the three games?don't have the OA
Out of 120 students in a school, 87 plays hockey, 76 plays football, 65 plays tennis and 19 plays carrom.What can be minimum number of students who play at least two of the three games?don't have the OA
I + II + III + IV= 120
I + 2II + 3III +4IV = 247
Minimise (II + III + IV) OR Maximise I
Also, II + 2III + 3IV = 127
Put, II = 0
WE get, III = 2 and IV = 41
Hence minimum number of students playing at least 2 games = 0 + 2 + 41 = 43 ?..
Actually, I really don't think that we can find a global maxima of the function y/x with the given domain! At least I can't find a solution to it Others may well!(x-1)^2 + (y-2)^2 = 24Suppose y/x = k(1-1/x)^2 + (k-2/x)^2 = 24/x^2k = [ 2 + _/(24 - (x-1)^2 ] / xTo maximise k, put limit: x->0Even then I couldn't solve it! May be @bodhi_vriksha sir would help!
......... wait for GODs to come...