AITS

The eqn-- e^x-ax-b=0 has
A. 1 real root if a<0
B. 1 real root if b>0 & a<=0
C. 2 real roots if a>0 & a.ln a>=(a-b)
D. No real root if a ln a<a-b

Multiple Ans correct

6 Answers

i guess A&D

AM I RITE I SAW DIS QUESN IN FIITJEE AITS

oh! sorry i didnt see it was multiple

D is ans too..

ya A & D is right..can u show u me ur working guys..

see.. e^x=ax+b inLHS dy/dx= e^x>0(alwys) (increasing)
inRHS dy/dx=a
now if RHs is decreasin then the two curves must meeet a some point hence a<0

e^x-ax-b=0

thus

e^x = ax + b

draw the graph of both sides on a paper and then decide :)

that is the easiest and best way

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