JEE MAIN - Mathematics (2024 - 31st January Evening Shift - No. 11)
Let $$f: \rightarrow \mathbb{R} \rightarrow(0, \infty)$$ be strictly increasing function such that $$\lim _\limits{x \rightarrow \infty} \frac{f(7 x)}{f(x)}=1$$. Then, the value of $$\lim _\limits{x \rightarrow \infty}\left[\frac{f(5 x)}{f(x)}-1\right]$$ is equal to
0
4
1
7/5
Explanation
$$\begin{aligned} & f: R \rightarrow(0, \infty) \\ & \lim _{x \rightarrow \infty} \frac{f(7 x)}{f(x)}=1 \end{aligned}$$
$$\because \mathrm{f}$$ is increasing
$$\begin{aligned} & \therefore \mathrm{f}(\mathrm{x})<\mathrm{f}(5 \mathrm{x})<\mathrm{f}(7 \mathrm{x}) \\ & \because \frac{\mathrm{f}(\mathrm{x})}{\mathrm{f}(\mathrm{x})}<\frac{\mathrm{f}(5 \mathrm{x})}{\mathrm{f}(\mathrm{x})}<\frac{\mathrm{f}(7 \mathrm{x})}{\mathrm{f}(\mathrm{x})} \\ & 1<\lim _{\mathrm{x} \rightarrow \infty} \frac{\mathrm{f}(5 \mathrm{x})}{\mathrm{f}(\mathrm{x})}<1 \\ & \therefore\left[\frac{\mathrm{f}(5 \mathrm{x})}{\mathrm{f}(\mathrm{x})}-1\right] \\ & \Rightarrow 1-1=0 \end{aligned}$$
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