IIT JEE , AIEEE Forum - Mathematics , Algebra

sourab_MCA

the no. of real roots of eq

e^(x -1) + x - 2 = 0 is

magiclko

the answer is 1....

draw the graphs of y=e^(x-1) and y=2-x ...and find the Point of Intersection

see the attached picture...

sourab_MCA

how to solve without graphing?

sourab_MCA

and secondly is the graph of e^(x-1) correct???

feynmann

let us denote x-1 by y

then the eqn becomes

e^( y ) +y -1 = 0

let us denote the LHS by f(y )

now f'( y ) = e^y + 1 >= 0 for all y

so it is an increasing fn

So it may have atmost one zero .......................... ( 1 )

Agian we see that f ( - inf ) = -inf and f( inf ) = inf

so f( y ) has at least one zero for real y .................. ( 2 )

Then from ( 1 ) & ( 2 ) it follows that f( y ) = 0 has exactly one soln .

elastiboysai

its of d form e^t=1-t
lhs is an increasing function
rhs is a decreasing function
clearly only one real root is possible
and thats at t=0