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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?
Options:
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?
Solution:
2512 Upvotes
Verified Answer
The correct answer is:
$f(x)$ is decreasing on $(-1,1)$ and has a local minimum at $x=1$
$f(x)$ is decreasing on $(-1,1)$ and has a local minimum at $x=1$
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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