Calculus not going to work

Algebra Level 3

If the minimum value of x 1000 + x 900 + x 90 + x 6 + 1996 x x^{1000}+ x^{900}+ x^{90}+ x^{6}+ \dfrac{1996}{x} for x > 0 x > 0 can be expressed as 100 n 100n , where n n is a positive integer, find n n .


The answer is 20.

View wiki
Archit Tripathi by Archit Tripathi , 23, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Archit Tripathi
Sep 17, 2016

Applying A.M > G.M inequality

x 1000 + x 900 + x 90 + x 6 + 1 x + . . . . . . + 1 x 2000 \frac{x^{1000}+x^{900}+x^{90}+x^{6}+\frac{1}{x}+......+\frac{1}{x}}{2000} > 1

hence the minimum value comes out to be 2000

so, n = n = 20 \boxed{20}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

@Archit Tripathi Where do you study and what is the source of your problems?

95% Of your problems are already given to me in fiitjee

Prakhar Bindal - 4 years, 6 months ago

Ha ha.. I also study there, and know u very well.

Archit Tripathi - 4 years, 6 months ago

Log in to reply

OK Which batch??

Prakhar Bindal - 4 years, 6 months ago

By the way, they are the older ones which I have seen in your workshops, however I was not selected for that workshops but A.G sir and NIM sir used to give me those sheets and I posted them here.

Archit Tripathi - 4 years, 6 months ago

NEPA A02 pass out batch

Archit Tripathi - 4 years, 6 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...