Arrange \(k, \alpha, \beta\) and \(\mu\) in descending order :
A. If \(y=\displaystyle\frac{1}{x}\) then \(\left[\displaystyle\frac{d^2y}{dx^2}\right]_{x=\frac{1}{2}}=k\)
B. If \(y=\displaystyle\frac{2}{x}\) then \(\left[\displaystyle\frac{d^2y}{dx^2}\right]_{x=2}=\alpha\)
C. If \(y=\displaystyle\frac{1}{x+1}\) then \(\left[\displaystyle\frac{d^2y}{dx^2}\right]_{x=1}=\beta\)
D. If \(y=\log_e(x+1)\) then \(\left[\displaystyle\frac{d^2y}{dx^2}\right]_{x=1}=\mu\)
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(k\gt\alpha\gt\beta\gt\mu\)
Each of k, α, β, μ is a second derivative at a point.
(A) \(y=x^{-1}\):
\( \displaystyle y''=\frac{2}{x^3},\quad k=\frac{2}{(1/2)^3}=16 \)
(B) \(y=2x^{-1}\):
\( \displaystyle y''=\frac{4}{x^3},\quad \alpha=\frac{4}{8}=\frac{1}{2} \)
(C) \(y=(x+1)^{-1}\):
\( \displaystyle y''=\frac{2}{(x+1)^3},\quad \beta=\frac{2}{8}=\frac{1}{4} \)
(D) \(y=\log_e(x+1)\):
\( \displaystyle y'=\frac{1}{x+1},\quad y''=-\frac{1}{(x+1)^2},\quad \mu=-\frac{1}{4} \)
So
\( \displaystyle 16\gt\frac{1}{2}\gt\frac{1}{4}\gt-\frac{1}{4} \), that is \(k\gt\alpha\gt\beta\gt\mu\)
Option (1) treats \(\beta=\mu\), but they differ in sign (the second derivative of the logarithm is negative).