Question

The local maximum value \(l\) and local minimum value \(m\) of \(f(x) = \frac{x^2+2x+2}{x+1}\) in \(\mathbb{R} - \{-1\}\) exist at \(\alpha, \beta\) respectively, then \(\frac{l+m}{\alpha+\beta} =\)

• A. 0
• B. -4
• C. -2
• D. 2

Hint:

For functions of the form \(x + \frac{1}{x}\), the local minimum is 2 (at \(x=1\)) and local maximum is -2 (at \(x=-1\)). Shifting \(x\) to \(x+1\) simply shifts the location of these extrema.

Answer 1

The Correct Option is A

Step 1: Rewrite the Function:
\(f(x) = \frac{(x^2+2x+1)+1}{x+1} = \frac{(x+1)^2+1}{x+1} = (x+1) + \frac{1}{x+1}\). Let \(t = x+1\). Then \(g(t) = t + \frac{1}{t}\) for \(t \neq 0\).
Step 2: Find Local Extrema:
For \(t>0\) (i.e., \(x>-1\)): By AM-GM, \(t + \frac{1}{t} \ge 2\). Minimum value is 2 at \(t=1\). So local minimum \(m = 2\) at \(x+1=1 \implies x = 0\). Thus \(\beta = 0\). For \(t<0\) (i.e., \(x<-1\)): \(t + \frac{1}{t} \le -2\). Maximum value is -2 at \(t=-1\). So local maximum \(l = -2\) at \(x+1=-1 \implies x = -2\). Thus \(\alpha = -2\).
Step 3: Calculate Required Value:
We have \(l = -2, m = 2, \alpha = -2, \beta = 0\). \[ \frac{l+m}{\alpha+\beta} = \frac{-2 + 2}{-2 + 0} = \frac{0}{-2} = 0 \]