number of solutions

Ans: = 4

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......

http://www.wolframalpha.com/input/?i=(2^x)%3D(x^3)%2B1

what maverick???
do u think i am going to carry my laptop to examination hall so that wolfram can draw my graphs for me?????

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

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 ..