Practicing Success
For the function $f(x)=x \cos \frac{1}{x}, x \geq 1$, which one of the following is incorrect ? |
for at least one x in the interval $[1, \infty), f(x+2)-f(x)<2$ $\lim\limits_{x \rightarrow \infty} f^{\prime}(x)=1$ for all x in the interval $[1, \infty), f(x+2)-f(x)>2$ f'(x) is strictly decreasing in the interval $[1, \infty)$ |
for at least one x in the interval $[1, \infty), f(x+2)-f(x)<2$ |
We have, $f(x)=x \cos \frac{1}{x}, x \geq 1$ $\Rightarrow f'(x)=\cos \frac{1}{x}+\frac{1}{x} \sin \frac{1}{x}$ and $f''(x)=-\frac{1}{x^3} \cos \frac{1}{x}$ $\Rightarrow \lim\limits_{x \rightarrow \infty} f'(x)=1$ and f''(x) < 0 for all $x \geq 1$ ⇒ f'(x) is strictly decreasing in the interval $[1, \infty)$ Let g(x) = f(x + 2) - f(x). Then, $g'(x)=f'(x+2)-f'(x)<0$ [∵ f'(x) is strictly decreasing] ⇒ g(x) is strictly decreasing on $[1, \infty)$ Now, $\lim\limits_{x \rightarrow \infty} g(x)=\lim\limits_{x \rightarrow \infty} f(x+2)-f(x)$ $=\lim\limits_{x \rightarrow \infty}(x+2) \cos \frac{1}{x+2}-x \cos \frac{1}{x}$ $=\lim\limits_{x \rightarrow \infty} x\left(\cos \frac{1}{x+2}-\cos \frac{1}{x}\right)+2 \cos \frac{1}{x+2}$ $=\lim\limits_{x \rightarrow \infty} 2 x \sin \frac{x+1}{x^2+2 x} \sin \frac{1}{x^2+2 x}+2 \cos \frac{1}{x+2}$ $=2 \lim\limits_{x \rightarrow \infty} \frac{\sin \left(\frac{x+1}{x^2+2 x}\right)}{\left(\frac{x+1}{x^2+2 x}\right)} \times \frac{\sin \left(\frac{1}{x^2+2 x}\right)}{\left(\frac{1}{x^2+2 x}\right)} \times \frac{x(x+1)}{\left(x^2+2 x\right)^2}+2 \cos \frac{1}{x+2}$ $=2 \times 1 \times 1 \times 0+2=2$ As g(x) is decreasing on $[1, \infty)$ and $\lim\limits_{x \rightarrow \infty} g(x)=2$ ∴ $g(x)>2$ for all $x \in[1, \infty)$ $\Rightarrow f(x+2)-f(x)>2$ for all $x \in[1, \infty)$. Hence, option (a) is incorrect. |