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Let $f(x)=\cos \left(\frac{\pi}{x}\right), x \neq 0,$ then assuming $k$ as an integer,
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Verified Answer
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$f(x)$ increases in the interval $\left(\frac{1}{2 k+1}, \frac{1}{2 k}\right)$, $f(x)$ decreases in the interval $\left(\frac{1}{2 k+2}, \frac{1}{2 k+1}\right)$
$f(x)=\cos \left(\frac{\pi}{x}\right)$
$\Rightarrow \quad f'(x)=-\sin \left(\frac{\pi}{x}\right)\left(\frac{-\pi}{x^{2}}\right)=\frac{\pi}{x^{2}} \sin \frac{\pi}{x}$
For increasing function, $f'(x)>0$ $\Rightarrow \sin \left(\frac{\pi}{x}\right)>0 \Rightarrow 2 k \pi < \frac{\pi}{x} < (2 k+1) \pi$
$\Rightarrow \quad \frac{1}{2 k}>x>\frac{1}{2 k+1}$
For decreasing function, $f(x) < 0$ $\Rightarrow \quad \sin \left(\frac{\pi}{x}\right) < 0$
$\Rightarrow \frac{\pi}{x} \in[(2 k+1) \pi,(2 k+2) \pi] \Rightarrow x \in\left(\frac{1}{2 k+2}, \frac{1}{2 k+1}\right)$
$\Rightarrow \quad f'(x)=-\sin \left(\frac{\pi}{x}\right)\left(\frac{-\pi}{x^{2}}\right)=\frac{\pi}{x^{2}} \sin \frac{\pi}{x}$
For increasing function, $f'(x)>0$ $\Rightarrow \sin \left(\frac{\pi}{x}\right)>0 \Rightarrow 2 k \pi < \frac{\pi}{x} < (2 k+1) \pi$
$\Rightarrow \quad \frac{1}{2 k}>x>\frac{1}{2 k+1}$
For decreasing function, $f(x) < 0$ $\Rightarrow \quad \sin \left(\frac{\pi}{x}\right) < 0$
$\Rightarrow \frac{\pi}{x} \in[(2 k+1) \pi,(2 k+2) \pi] \Rightarrow x \in\left(\frac{1}{2 k+2}, \frac{1}{2 k+1}\right)$
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