JEE Advanced · Mathematics · 25. AOD
For the function \(f(x)=x \cos \frac{1}{x}, x \geq 1\),
- A
for atleast one \(x\) in the interval \([1, \infty)\), \(f(x+2)-f(x) < 2\) - B
\(\lim _{x \rightarrow \infty} f^{\prime}(x)=1\)
- C
for all \(x\) in the interval \([1, \infty)\), \(f(x+2)-f(x)>2\)
- D
\(f^{\prime}(x)\) is strictly decreasing in the interval \([1, \infty)\)
Answer & Solution
Correct Answer
(B)
\(\lim _{x \rightarrow \infty} f^{\prime}(x)=1\)
Step-by-step Solution
Detailed explanation
Given, \(f(x)=x \cos \frac{1}{x}, x \geq 1\)
\[
\begin{aligned}
& \Rightarrow \quad f^{\prime}(x)=\frac{1}{x} \sin \frac{1}{x}+\cos \frac{1}{x} \\
& \Rightarrow \quad f^{\prime \prime}(x)=-\frac{1}{x^3} \cos \left(\frac{1}{x}\right)
\end{aligned}
\]
Now, \(\lim _{x \rightarrow \infty} f^{\prime}(x)=0+1=1\)
\(\Rightarrow\) Option (b) is correct.
Now, \(\quad x \in[1, \infty) \Rightarrow \frac{1}{x} \in(0,1]\)
\(\Rightarrow f^{\prime \prime}(x) < 0 \Rightarrow\) Option (d) is correct.
As \(f^{\prime}(1)=\sin 1+\cos 1>1\)
\(f^{\prime}(x)\) is strictly decreasing and \(\lim _{x \rightarrow \infty} f^{\prime}(x)=1\)
So, graph of \(f^{\prime}(x)\) is shown as below.
Now, in \([x, x+2], x \in[1, \infty), f(x)\) is continuous and differentiable so by LMVT,
\[
f^{\prime}(x)=\frac{f(x+2)-f(x)}{2}
\]
As \(f^{\prime}(x)>1\) for all \(x \in[1, \infty)\)
\[
\begin{aligned}
& \Rightarrow \quad \frac{f(x+2)-f(x)}{2}>1 \\
& \Rightarrow f(x+2)-f(x)>2 \quad \text { for all } x \in[1, \infty)
\end{aligned}
\]
\[
\begin{aligned}
& \Rightarrow \quad f^{\prime}(x)=\frac{1}{x} \sin \frac{1}{x}+\cos \frac{1}{x} \\
& \Rightarrow \quad f^{\prime \prime}(x)=-\frac{1}{x^3} \cos \left(\frac{1}{x}\right)
\end{aligned}
\]
Now, \(\lim _{x \rightarrow \infty} f^{\prime}(x)=0+1=1\)
\(\Rightarrow\) Option (b) is correct.
Now, \(\quad x \in[1, \infty) \Rightarrow \frac{1}{x} \in(0,1]\)
\(\Rightarrow f^{\prime \prime}(x) < 0 \Rightarrow\) Option (d) is correct.
As \(f^{\prime}(1)=\sin 1+\cos 1>1\)
\(f^{\prime}(x)\) is strictly decreasing and \(\lim _{x \rightarrow \infty} f^{\prime}(x)=1\)
So, graph of \(f^{\prime}(x)\) is shown as below.

Now, in \([x, x+2], x \in[1, \infty), f(x)\) is continuous and differentiable so by LMVT,
\[
f^{\prime}(x)=\frac{f(x+2)-f(x)}{2}
\]
As \(f^{\prime}(x)>1\) for all \(x \in[1, \infty)\)
\[
\begin{aligned}
& \Rightarrow \quad \frac{f(x+2)-f(x)}{2}>1 \\
& \Rightarrow f(x+2)-f(x)>2 \quad \text { for all } x \in[1, \infty)
\end{aligned}
\]
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