Ans: = 4
15 Answers
but this is a graph of 2x = x3
here we have to draw the graph of 2x = x3+1
jagaran's graph is right ....
@maverick.....look the graph is shifted by 1 unit along the x -axis ! that is done !
no of solutions = 4
one intersection is between the x values of (-1,0) as seen from graph ,
other points of intersection are x =0 and x=1
now for x = 2 ,
LHS = 4 , RHS = 9
so LHS > RHS ...(1)
now for x = 100
LHS = 2100 = (24)25 = 1625
RHS = 1003+ 1 = 106 + 1 wich is < 1625
so RHS > LHS (2)
so from (1) and (2)
it is clear that at some value of x ....LHS became = RHS and then after that value of x, LHS always remains < RHS
so 4 points of intersections
@mavarick just post ur graph to justify it
@ jagaran ur graph is pathetic
good work â—‹|~_^|â—‹
but i was looking for a much better solution.......
something that strikes me at the time of the exam......
what maverick???
do u think i am going to carry my laptop to examination hall so that wolfram can draw my graphs for me?????
we can have an idea abt roots by finding f'(x),f"(x) and f"'(x)
where f(x)=2x-(1+x3)
i m still getting 1 pt of intersection .hope graph is correct this time
wat abt x = 0 ???
LHS = 20 = 1
RHS = 03 + 1 = 1 .....
in ur graph the functions dont intersect at x = 0
@ JAGARAN
ur graphs still pathetic.........pls go over the fundamentals again
agreed this is incorrrect. but is is possible to draw a graph as shown by maverick without using a graph plotter???????
i mean if u use fundamentals will u get a better graph.finding pts of intersection and then drawing a graph is not a part of fundamentals
i dont think so we can find all points of intersections by using graph ..i mean for example if a question has has point of intersection at x =101.78
how can u find it by drawing a graph ??
u can find the no of intersections as i hav done above .....putting values in such functions isnt that tough ..lol i hav used x=100 to just elaborate it ..but u can even check it at x =10 ..