Derivative of everything

Calculus Level 4

Given the function $f(x)=x-\dfrac{1}{2}\sin x - \dfrac{m}{2} \ln x +1$ , where $m \in \mathbb R$ is a parameter.

If there exist $x_1,x_2 \in (0,+ \infty), x_1 \neq x_2$ such that $f(x_1)=f(x_2)$ , compare $x_1 x_2$ and $m^2$ .

x_1 x_2 = m^2

x_1 x_2 > m^2

x_1 x_2 < m^2

2 solutions

Saswat Prakash
Mar 4, 2020

By differentiating f(x) for minimization , We find , 1-cosx/2-m/2x=0 Or m/x-1 =1-cosx, => 0<m/x-1<2 ,using range of cosx €[-1,1], Or. m/3<x0<m, x0 is the minima. Thus minima is a point which is smaller tham m, Now consider 2 points x1 ,x2 x1 = x0 -€, x2=∆ + x0 Such that :€,∆<<<1; And f(x1)=f(x2), as x0 is minima we are sure to find such neighbouring points so close to the minima. Now x1.x2= x0^2+(∆-€)x0-∆€; As ∆,€<<<1, m^2/9<X1.x2<m^2 Or ,X1.x2<m^2. .........(1 ) As the inequality is generalized for all x1 , x2 in the question itself and the piont we are studying give us the inequality (1). Hence this is the general inequality for all such x1 ,x2.

Could you use Latex to formulate the result？I’d appreciate it.

Alice Smith - 1 year, 3 months ago

Not Karthik
Feb 26, 2020

There is an intuitive method and an analytical method but I'll just post the intuitive method her because it's a bit simpler. Here it is. Ignore the sine part. The function simply oscillates and does not add much to the magnitude of f(x). Look at x and -(m/2)ln(x). Since ln(x) is negative, the graph of ln(x) is flipped here. The slope becomes zero at some x<1. Let this point be x'. Now f(x1) = f(x2) for and there are infinitely many pairs (x1, x2). Let x1 < x' and x2 > x'. f'(x1) <<< f'(x2) so x1 can become very very small. Therefore x1x2 can become very very small quickly and are always smaller than m^2. Why m^2 I will not explain (I gotta go sleep and I don't know if I'll be able to explain even in full wakefulness) but I have explained my intuition here.

Derivative of everything

2 solutions

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