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4 solutions
I still don't get it. Can you explain a little bit more elaborately.
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There are several steps to my solution... which one(s) do you want me to elaborate on?
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All of them. Actually I am not able to understand the question itself. I tried solving it by taking third order differential to get one solution ln(6).why is it wrong and what your answer is saying exposing every single step. Further what do you mean by convex, LHS>RHS and one time RHS>LHS
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@Puneet Pinku – I will use "concave" and "decreasing" in the sense of strictly concave and decreasing, resp.
Concave means that all the chords (line segments connecting two points on the graph) are below the graph (except for the end points, of course). For a differentiable function, concave means that the derivative is decreasing.
LHS and RHS stand for "Left hand side" and "right hand side".
Now to our problem: We have to count the positive solutions x of the equation x = 4 ln ( x ) . (Are you clear on the negative case?)
It is not hard to see that a concave graph can intersect a straight line at most twice; thus we have at most two solutions.
Let f ( x ) = x and g ( x ) = 4 ln ( x ) . Now g ( 1 ) = 0 < 1 = f ( 1 ) but g ( 4 ) = 4 ln ( 4 ) > 4 = f ( 4 ) , so that the equation g ( x ) = f ( x ) must have a solution between 1 and 4 as the graphs cross. Then g ( e 4 ) = 1 6 < e 4 = f ( e 4 ) so that there must be another solution between 4 and e 4 as the graphs cross again.
Together with the negative solution, this makes for a grand total of 3 solutions.
Please let me know whether this makes any sense
My approach as we could easily see from graph there will be exactly one solution for x<0.
for x> 0 consider the function y = e^x / x^4
Differentiate it dy/dx = e^x x^4 -4 e^x*x^3 / x^8
for x>4 function will increasing and for 4>x>0 it will decrease.
as x =0 function will tend to infinity and so on .
Now consider the line y = 1
for seeing the points of intersection with graph consider value at x= 4
y = e^4 / 4^4 <1 .
Hence y= 1 will cut the graph at two points.
so one solution for x<0 and two for x>0 and hence a total of 3 solutions
Nice q Bro!
@Prakhar Bindal can you explain it more clearly as i often get wrong answers in such questions.
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in which part do you have problem?
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kK! i understand now actually i misread your solution thanks for replying @Prakhar Bindal and @Aniket Sanghi , plz keep supporting.
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@A Former Brilliant Member – I guess brilliant also wants me to help you, as your every doubt to anyone is showed to me in my notification :P
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@Aniket Sanghi – wow that's great !!
If you want to to understand in an easy way it can be like
Infinity ^a can't beat e^infinity , so at end e^x will again overtake x^4 . You can check this like take limit of (e^x/x^4 when x tends to infinity . You will get infinity!
This way will be useful in JEE! And for complete solution , check bro Prakhar's Sol!
a good one when we draw a rough graph of 2 we get to know they intersect at only one pt on the negative side as graph of e^(x) is increasing and that of x^(4) is decreasing.
now to find no of intersect on positive side
x=ln(x)4
x/ln(x)=4 (differentiating) we find that function is decreasing before e and increasing afterward so drawing rough sketch we get our ans to 2. making a total of 3
I also got 2
Good question ......and a nice solution comrade ....! Otto bretscher
There is one negative solution x since x 4 is decreasing and e x is increasing.
For positive x we can write the equation as x = 4 ln ( x ) . This equation has at most two solutions x > 0 since ln ( x ) is (strictly) convex. Now RHS>LHS when x = 4 but LHS>RHS when x = 1 and x = e 4 , so that we have exactly two more solutions, for a total of 3 .