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QJEEM24Maa6b6: JEE Main 2024 Mathematics - Class 12 JEEM24Maa6b6 Solution: Let be strictly increasing function such that lim x → ∞ f ( 7 x …

Let f:→R→(0,∞)f: \rightarrow \mathbb{R} \rightarrow(0, \infty) be strictly increasing function such that limx→∞f(7x)f(x)=1. Then, the value of limx→∞[f(5x)f(x)−1] is equal to

Multiple Choice Options

  • A 0
  • B 4
  • C 1
  • D `7/5

Step-by-Step Verified Solution

amp;f:R→(0,∞)amp;lim⁡x→∞f(7x)f(x)=1\begin{aligned}& f: R \rightarrow(0, \infty) \\& \lim _{x \rightarrow \infty} \frac{f(7 x)}{f(x)}=1\end{aligned}

∵f\because \mathrm{f} is increasing

amp;∴f(x)lt;f(5x)lt;f(7x)amp;∵f(x)f(x)lt;f(5x)f(x)lt;f(7x)f(x)amp;1lt;lim⁡x→∞f(5x)f(x)lt;1amp;∴[f(5x)f(x)−1]amp;⇒1−1=0\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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