Solution using $f\left( x \right) = x\cos \frac{1}{x},x \ge 1$

Question

If $f\left( x \right) = x\cos \frac{1}{x},x \ge 1$. Then which of the following is the correct option for the domain $x \ge 1$

(A) $f(x+2)-f(x)<2$

(B) $f(x+2)-f(x)>2$

My approach is as follow

$T\left( x \right) = f\left( {x + 2} \right) - f\left( x \right) = \left( {x + 2} \right)\cos \frac{1}{{x + 2}} - x\cos \frac{1}{x}$

$T\left( x \right) = x\left( {\cos \frac{1}{{x + 2}} - \cos \frac{1}{x}} \right) + 2\cos \frac{1}{{x + 2}} \Rightarrow T\left( x \right) = x\left( {2\sin \left( {\frac{{\frac{1}{{x + 2}} + \frac{1}{x}}}{2}} \right)\sin \left( {\frac{{\frac{1}{x} - \frac{1}{{x + 2}}}}{2}} \right)} \right) + 2\cos \frac{1}{{x + 2}}$

$T\left( x \right) = x\left( {2\sin \left( {\frac{{2x + 2}}{{2x\left( {x + 2} \right)}}} \right)\sin \left( {\frac{2}{{2x\left( {x + 2} \right)}}} \right)} \right) + 2\cos \frac{1}{{x + 2}} \Rightarrow T\left( x \right) = x\left( {2\sin \left( {\frac{{x + 1}}{{x\left( {x + 2} \right)}}} \right)\sin \left( {\frac{1}{{x\left( {x + 2} \right)}}} \right)} \right) + 2\cos \frac{1}{{x + 2}}$

$T\left( x \right) = 2\left( {\left( x{\sin \left( {\frac{{x + 1}}{{x\left( {x + 2} \right)}}} \right)\sin \left( {\frac{1}{{x\left( {x + 2} \right)}}} \right)} \right) + \cos \frac{1}{{x + 2}}} \right)$

From here onward how do I proceed

Answer 1

From your expression $T(x)$, now we find roots for

$$ T(x) - 2 = 0 \\ x \sin\left( \frac{x+1}{x(x+2)} \right) \sin\left( \frac{1}{x(x+2)} \right) + \cos \left(\frac{1}{x+2}\right) - 1 = 0 $$

Comparing denominators and numerators inside every sin and cos expresion:

$$ x+1 < x(x+2)\\ 1 < x(x+2)\\ 1 < x+2 $$

So sin and cos expresions are monotonic. Check that:

$$ x \sin\left( \frac{x+1}{x(x+2)} \right) \sin\left( \frac{1}{x(x+2)} \right) \to 0 \text{ (monotonically increassing)}\\ \cos \left(\frac{1}{x+2}\right) \to 1 \text{ (monotonically decreasing)} $$

Then note that

$$ 1 \sin\left( \frac{1+1}{1(1+2)} \right) \sin\left( \frac{1}{1(1+2)} \right) = 0.20232\dots \\ \cos \left(\frac{1}{1+2}\right) = 0.94495\dots $$

Also $$ T(1) - 2 = 0.20232 + 0.94495 - 1 = 0.14727 > 0\\ T(x) - 2 \to 0 \text{(monotonically)} $$

Thus

$T(x) > 2$

Not that rigorous but hope it helps, Bests!