f'(x)=e^x -1
f"(x)=e^x,, putting f'(x)=0 gives x=0 and f"(0) =1 which means at x=0 the function has a local minima,, The point of local minima is (0,-1).Since this point lies below the origin,, it must cut the x axis which means f(x) =0 for two real values of x

Neha is right..it's a continuous function and it has a local minima at (0,-1)..just draw the figure once..u'll get it. and u get 1<x<2 and -2<x<-1 through the intermediate value theorem..

I checked by looking at the opposite signs of f(x) at these values.

@vasudha-congratulations saw your name on DSE admission list ,altho i dint gt thru so help me need some tips to crack ISI as its d only option i hv right now :/

thanks anurag. hope u get into isi. it's better than dse in many ways. solving these jrf papers seems to be a good idea. plus u can look at some of the maths questions from DSE option B question papers. just keep solving problems the way u did for the written test. keep in mind what Amit sir posted about the interview. all the best :)

@vasudha- Thnku :) ya i saw you ppl solving jrf quesns so strted this morning only..jst keep posting tips n wtever you practicing for interviews..wts ur prfrnce DSE or ISI ?

Umm, i guess u just put l=y(a/k+m). from here u get y=kl/(a+mk). if v r given a production function we get l in terms of k and y to minimize cost, isnt it? i'm just doing the reverse.

Since a production function is given, we could perhaps calculate the labor demand from this equation (MPL = real wage). This would have to be equal to the labor supply determined in part 1. Substitute this back into the production function and we should find equilibrium output Y. BUt I'm not sure how Investment will figure in this equation, as this is more of a supply side equation :/ . And I think theta determines the Labor share of income?