How many times?

What was you?

are yaar...koi baat nahi.......x² wale graph ko y axis pe higher kar do by 1.......n plot it..u will get the points........u can find the third point by plotting the graph dude...on a big paper..just try i guess u can plot it.....just shift mah x² grph above by +1 on y axis..n u will get x²+1.

How will find out third point?

P.S
Graph shd b between 2^x and x²+1

thin graph is 2^x............even i wwas quite shocked wen i plotted the graphs n found that there are 3 points....one is in -ve side.......anoder is 2[that was easy....2^x and x²] are both 2² here............and the third intersection is between x=4 and 8 sumwhere......just try plotting till 5 .u will understand...........

will post in 5 mins..

sorry sorry!!mera graph galat tha.......

Where's the third point of intersection?

That was me :P

In a test I took here, i got this question and gave the ans as two (u can't expect me to draw the graph to scale!!) as 0 and 1 were the only solutions I could think of. Well, thanks everybody :))

Applying numerical technique (which you guys will learn in IIT), the third root comes out to be 4.2575

Any other way to get the ans?

Slope of x²+1 = 2x.

Slope of 2^x = 2^xlog_e2.

At x=1, slope of x²+1 > 2^x.... so x² + 1 is steeper and on the inside.

so seeing the values of 2^xlog_e2 > 2x for higher numbers then they must meet at another point.

I think the another point is around x=4.5

I still dont get it :(

what do u mean "The slopes show that they must meet 3 times" ?