Paragraph: Consider the function f:(-∞, ∞) arrow(-∞, ∞) defined by f(x)=(x^2-a x+1)/(x^2+a x+1) ; 0 Question: Which of the following is true?

Correct answer: A. f(x) is decreasing on (-1,1) and has a local minimum at x=1 When x ∈(-1,1) x² < 1 x²-1 < 0 f^ (x) 0 f(x) is decreasing. Also, at x=1, f^ (1)=(4 a)/((a+2)^2 0)[ 0 a 2] f(x) has a local minimum at x=1 .

JEE Advanced 2008Paper 2

Exam: JEE Advanced
Year: 2008
Paper: Paper 2
Subject: Mathematics
Chapter: Application of Derivatives

JEE Advanced 2008 · Paper 2 · Mathematics · Application of Derivatives

Paragraph: Consider the function f:(-∞, ∞) arrow(-∞, ∞) defined by...

Paragraph: 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

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 $x=1$

Correct answer

A. $f(x)$ is decreasing on $(-1,1)$ and has a local minimum at $x=1$

Step-by-step solution

When

x \in(-1,1)

x² < 1 \Rightarrow x²-1 < 0

\therefore \quad f^{\prime}(x) \lt 0 \Rightarrow f(x)

is decreasing. Also, at

x=1, f^{\prime \prime}(1)=\frac{4 a}{(a+2)^2 \gt 0}[\because 0 \lt a \lt 2]\therefore f(x)

has a local minimum at

x=1

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