Aspire's Library
A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
If the function $f(x)=\left(\dfrac1x\right)^{2x}, x>0$ attains its maximum at $x=\dfrac1e$, then:
Solution
Take log: $y = \left(\frac{1}{x}\right)^{2x}$$\ln y = 2x \ln\left(\frac{1}{x}\right) = -2x \ln x$
Differentiate: $\frac{1}{y}\frac{dy}{dx} = -2(\ln x + 1)$
For extrema: $\ln x + 1 = 0 \Rightarrow x = \frac{1}{e}$
Now value at maximum: $f\left(\frac{1}{e}\right) = \left(e\right)^{2/e} = e^{2/e}$
Hence: Maximum value $= e^{2/e}$
Now compare options using logs: Take $\ln$ on both sides
Option (1):
$e^\pi < \pi^e$
$\Rightarrow \frac{\pi}{e} < \ln \pi$ (false, since $\frac{\pi}{e} > \ln \pi$)
$\Rightarrow \frac{\pi}{e} < \ln \pi$ (false, since $\frac{\pi}{e} > \ln \pi$)
Option (2):
$e^{2\pi} < (2\pi)^e$
$\Rightarrow \frac{2\pi}{e} < \ln(2\pi)$ (false)
$\Rightarrow \frac{2\pi}{e} < \ln(2\pi)$ (false)
Option (3):
$(2e)^\pi > \pi^{2e}$
$\Rightarrow \frac{\pi}{2e} > \frac{\ln \pi}{\ln(2e)}$ (true comparison holds)
$\Rightarrow \frac{\pi}{2e} > \frac{\ln \pi}{\ln(2e)}$ (true comparison holds)
Option (4):
$e^\pi > \pi^e$
$\Rightarrow \frac{\pi}{e} > \ln \pi$ (true)
$\Rightarrow \frac{\pi}{e} > \ln \pi$ (true)
Final Answer: $ \boxed{\text{Option (4)}} $
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Ask Your Question or Put Your Review.
