MHT CET · Maths · Application of Derivatives
For every value of \(x\), the function \(f(x)=\frac{1}{a^{x}}, a>0\) is
- A decreasing
- B increasing
- C Constant
- D Neither increasing nor decreasing
Answer & Solution
Correct Answer
(A) decreasing
Step-by-step Solution
Detailed explanation
(D)
\(\begin{aligned}
\mathrm{f}(\mathrm{x})=& \frac{1}{\mathrm{a}^{\mathrm{x}}}=\mathrm{a}^{-\mathrm{x}} \\
\therefore \quad \mathrm{f}^{\prime}(\mathrm{x}) &=-\mathrm{a}^{-\mathrm{x}} \cdot \log _{e} \mathrm{a}=-\frac{\log _{\mathrm{e}} \mathrm{a}}{\mathrm{a}^{\mathrm{x}}}= < 0
\end{aligned}\)
So \(\mathrm{f}(\mathrm{x})\) is decreasing for all \(\mathrm{x}\).
\(\begin{aligned}
\mathrm{f}(\mathrm{x})=& \frac{1}{\mathrm{a}^{\mathrm{x}}}=\mathrm{a}^{-\mathrm{x}} \\
\therefore \quad \mathrm{f}^{\prime}(\mathrm{x}) &=-\mathrm{a}^{-\mathrm{x}} \cdot \log _{e} \mathrm{a}=-\frac{\log _{\mathrm{e}} \mathrm{a}}{\mathrm{a}^{\mathrm{x}}}= < 0
\end{aligned}\)
So \(\mathrm{f}(\mathrm{x})\) is decreasing for all \(\mathrm{x}\).
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