11thank you bhaiya
IF f(x)=e-ln x and f(2m2 + m +1) < f(3m2 - 4m + 1) then m belongs to
A. (0,5)
B. (-7,-3)
C. (3,17)
D. none of these.
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5 Answers
1none of these...just solving it might come out (0,5) but actually the values (1/3,1) is not satisfying it....so (d).....how much xpecting dis tym from aits??
62- ln x = ln (1/x)
e-ln x = 1/x
so we have f(x)=1/x (as long as ln x is defined.. ie x >0
So we want to solve for 2m2+m+1>0 and 3m2-4m+1>0 and 2m2 + m +1>3m2 - 4m + 1
to hold simultaneously.
1According to question,
e-ln(2m^2+m+1) < e-ln(3m^2 -4m+1)
=> 1/(2m2+m+1) < 1/(3m2-4m+1)...........................(1)
Cross multiplying both sides of (1) by 2m2+m+1 without changing the direction of the inequality, {Since 2m2+m+1 is always positive on account of Discriminant <0} we have,
(2m2+m+1)/(3m2-4m+1) -1 > 0
=> (-m2+5m)/(3m2-4m+1) > 0
=> m(m-5)/(3m2-4m+1) < 0 {Since the inequality has been multiplied by -1, sign is reversed} => m(m-5)/(3m-1)(m-1) < 0 .....................(2)
And now, the solution of (2) gives the range of m:-
m belongs to (0,1/3) U (1,5)
1http://www.goiit.com/posts/list/differential-calculus-if-f-x-e-ln-x-and-f-2m-2-m-1-f-3m-2-4m-1006749.htm