JEE Advanced · Mathematics · 25. AOD
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Consider the function \(f:(-\infty, \infty) \rightarrow(-\infty, \infty)\) defined by \(f(x)=\frac{x^2-a x+1}{x^2+a x+1} ; 0 < a < 2\)Question:
Which of the following is true?
- A
\(f(x)\) is decreasing on \((-1,1)\) and has a local minimum at \(x=1\)
- B
\(f(x)\) is increasing on \((-1,1)\) and has a local maximum at \(x=1\)
- C
\(f(x)\) is increasing on \((-1,1)\) but has neither a local maximum nor a local minimum at \(x=1\)
- D
\(f(x)\) is decreasing on \((-1,1)\) but has neither a local maximum nor a local minimum at \(x=1\)
Answer & Solution
Correct Answer
(A)
\(f(x)\) is decreasing on \((-1,1)\) and has a local minimum at \(x=1\)
Step-by-step Solution
Detailed explanation
When \(x \in(-1,1)\)
\[
x^2 < 1 \Rightarrow x^2-1 < 0
\]
\(\therefore \quad f^{\prime}(x) < 0 \Rightarrow f(x)\) is decreasing.
Also, at \(x=1, f^{\prime \prime}(1)=\frac{4 a}{(a+2)^2>0}\)
\([\because 0 < a < 2]\)
\(\therefore f(x)\) has a local minimum at \(x=1\).
\[
x^2 < 1 \Rightarrow x^2-1 < 0
\]
\(\therefore \quad f^{\prime}(x) < 0 \Rightarrow f(x)\) is decreasing.
Also, at \(x=1, f^{\prime \prime}(1)=\frac{4 a}{(a+2)^2>0}\)
\([\because 0 < a < 2]\)
\(\therefore f(x)\) has a local minimum at \(x=1\).
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