Animated Solution for Mathematics - Differentiation: For the function $f(x)=x\cos (1/x),x\ge 1$,

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dissect a function that might look intimidating at first glance, but reveals a beautiful, rhythmic structure once we peel back the layers.

We are looking at $f(x)=x\cos (\frac{1}{x})$ for $x\ge 1$. This isn't just an algebraic exercise; it's a study of how functions behave as they stretch toward infinity.

The First Derivative and the Product Rule

To understand the slope of our function, we need the first derivative, $f^{\prime }(x)$. We have a product of two functions: $u=x$ and $v=\cos (\frac{1}{x})$.

The product rule, $\frac{d}{dx}[uv]=u\frac{dv}{dx}+v\frac{du}{dx}$, is our best friend here. When we differentiate $v=\cos (\frac{1}{x})$, we must use the chain rule.

The derivative of $\cos (\theta )$ is $-\sin (\theta )$, and the derivative of $\frac{1}{x}$ is $-\frac{1}{x^2}$. So, $\frac{dv}{dx}=-\sin (\frac{1}{x})\cdot (-\frac{1}{x^2})=\frac{1}{x^2}\sin (\frac{1}{x})$.

Putting it all together, we get:

Simplifying the first term, $x\cdot \frac{1}{x^2}$ becomes $\frac{1}{x}$. Thus, our derivative is:

The Limit at Infinity

Now, let's see what happens as $x$ grows without bound. As $x\to \infty$, the argument $\frac{1}{x}$ approaches $0$.

We know that $\cos (0)=1$ and $\sin (0)=0$. Therefore, the term $\frac{1}{x}\sin (\frac{1}{x})$ approaches $0\cdot 0=0$.

The derivative $f^{\prime }(x)$ approaches $0+1=1$. This confirms that for very large $x$, the function behaves like a line with a slope of $1$. This validates Option B.

The Second Derivative and Monotonicity

To determine if $f^{\prime }(x)$ is increasing or decreasing, we need $f^{\prime \prime }(x)$. Let's differentiate $f^{\prime }(x)=\cos (\frac{1}{x})+\frac{1}{x}\sin (\frac{1}{x})$.

The derivative of $\cos (\frac{1}{x})$ is $\frac{1}{x^2}\sin (\frac{1}{x})$. For the second term, $\frac{1}{x}\sin (\frac{1}{x})$, we use the product rule again:

When we sum these, the terms $\frac{1}{x^2}\sin (\frac{1}{x})$ and $-\frac{1}{x^2}\sin (\frac{1}{x})$ cancel out! We are left with:

Since $x\ge 1$, $\frac{1}{x}$ is in the first quadrant, where cosine is positive. Thus, $f^{\prime \prime }(x)$ is strictly negative. This means $f^{\prime }(x)$ is strictly decreasing, which proves Option D.

The LMVT Bridge

Finally, we tackle the difference $f(x+2)-f(x)$. The Mean Value Theorem is the perfect tool.

It states that there exists a $c\in (x,x+2)$ such that:

This simplifies to $f(x+2)-f(x)=2f^{\prime }(c)$. Since $f^{\prime }(x)$ is strictly decreasing and approaches $1$ as $x\to \infty$, it must be that $f^{\prime }(x)>1$ for all finite $x$.

Therefore, $f^{\prime }(c)>1$, which implies $2f^{\prime }(c)>2$. Thus, $f(x+2)-f(x)>2$. This confirms Option C.

We have navigated the calculus, simplified the expressions, and used the Mean Value Theorem to bridge the gap. The result? Options B, C, and D are correct.