Question:
The minimum value of \(\frac{logx}{x}\) in the interval \((2,\infty )\) is
The minimum value of \(\frac{logx}{x}\) in the interval \((2,\infty )\) is
Show Hint
Differentiate: the function has a maximum at x = e and decreases afterwards.
Updated On: Oct 1, 2026
- \(0\)
- \(e\)
- \(\frac{1}{e}\)
- Does not exist
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
Let \(f(x) = \dfrac{\log x}{x}\). Its stationary points come from \(f'(x) = 0\).
Step 2: Differentiate:
\[ f'(x) = \frac{1 - \log x}{x^2} \]
\(f' = 0\) at \(x = e\). For \(x < e\), \(f' > 0\) (increasing). For \(x > e\), \(f' < 0\) (decreasing).
Step 3: Analyse the interval (2, infinity):
\(e \approx 2.718\) lies in \((2, \infty)\), and it gives a maximum value \(\dfrac1e\), not a minimum.
After \(x = e\) the function keeps decreasing and approaches 0 as \(x \to \infty\), but never reaches 0, because \(\log x > 0\) there.
Step 4: Conclusion:
There is no smallest value attained. So the minimum does not exist, option (D). Option (C) \(1/e\) is the maximum.
Final Answer:
The function peaks at x = e and has no minimum. \[ \boxed{\text{(D) }\text{Does not exist}} \]
Let \(f(x) = \dfrac{\log x}{x}\). Its stationary points come from \(f'(x) = 0\).
Step 2: Differentiate:
\[ f'(x) = \frac{1 - \log x}{x^2} \]
\(f' = 0\) at \(x = e\). For \(x < e\), \(f' > 0\) (increasing). For \(x > e\), \(f' < 0\) (decreasing).
Step 3: Analyse the interval (2, infinity):
\(e \approx 2.718\) lies in \((2, \infty)\), and it gives a maximum value \(\dfrac1e\), not a minimum.
After \(x = e\) the function keeps decreasing and approaches 0 as \(x \to \infty\), but never reaches 0, because \(\log x > 0\) there.
Step 4: Conclusion:
There is no smallest value attained. So the minimum does not exist, option (D). Option (C) \(1/e\) is the maximum.
Final Answer:
The function peaks at x = e and has no minimum. \[ \boxed{\text{(D) }\text{Does not exist}} \]
Was this answer helpful?
0
0
Top MHT CET Mathematics Questions
- The line $5x + y - 1 = 0$ coincides with one of the lines given by $5x^2 + xy - kx - 2y + 2 = 0 $ then the value of k is
- If $\int\frac{f\left(x\right)}{log \left(sin\,x\right)}dx = log\left[log\,sin\,x\right]+c$ then $f\left(x\right)=$
- If $\int\limits^{K}_0 \frac{dx}{2 + 18 x^2} = \frac{\pi}{24}$, then the value of K is
- If $\int^{\pi/2}_{0} \log\cos x dx =\frac{\pi}{2} \log\left(\frac{1}{2}\right)$ then $ \int^{\pi/2}_{0} \log\sec x dx = $
- The point on the curve $y = \sqrt{x - 1}$ where the tangent is perpendicular to the line $2x + y - 5 = 0 $ is
View More Questions
Top MHT CET Maxima and Minima Questions
- The maximum value of $2x + y$ subject to $3x + 5y \leq 26$ and $5x + 3y \leq 30, x \geq 0, y \geq 0$ is
- Find the approximate value of , where .
- Order of is:
- The integrating factor of the differential equation is, :
- Calculate the area under the curve and included between the lines .
View More Questions
Top MHT CET Questions
- During r- DNA technology, which one of the following enzymes is used for cleaving DNA molecule ?
- In non uniform circular motion, the ratio of tangential to radial acceleration is (r = radius of circle, $v =$ speed of the particle, $\alpha =$ angular acceleration)
- The temperature of $32^{\circ}C$ is equivalent to
- The line $5x + y - 1 = 0$ coincides with one of the lines given by $5x^2 + xy - kx - 2y + 2 = 0 $ then the value of k is
- The heat of formation of water is $ 260\, kJ $ . How much $ H_2O $ is decomposed by $ 130\, kJ $ of heat ?
View More Questions