TS EAMCET · Maths · Application of Derivatives
The local maximum value \(l\) and local minimum value \(m\) of \(f(x)=\frac{x^2+2 x+2}{x+1}\) in \(\mathbb{R}-\{-1\}\) exist at \(\alpha, \beta\) respectively, then \(\frac{l+m}{\alpha+\beta}=\)
- A \(0\)
- B \(-4\)
- C \(-2\)
- D \(2\)
Answer & Solution
Correct Answer
(A) \(0\)
Step-by-step Solution
Detailed explanation
\(f(x) = x+1+\frac{1}{x+1}\) \(f'(x) = 1-\frac{1}{(x+1)^2}\) \(f'(x)=0 \Rightarrow (x+1)^2=1 \Rightarrow x+1=\pm 1 \Rightarrow x=0, x=-2\) \(f''(x) = \frac{2}{(x+1)^3}\) \(f''(-2) = -2 \(f''(0) = 2 > 0 \Rightarrow \beta = 0\), \(m = f(0) = 2\)…
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