Question

Suppose $f$ is defined from $R\to [-1,1]$ as $f(x)=\frac{x^2-1}{x^2+1}$ where $R$ is the set of real number. Then the statement which does not hold is

• A. f is many one onto
• B. f increases for x > 0 and decrease for x < 0
• C. minimum value is not attained even though f is bounded
• D. the area included by the curve y = f (x) and the line y = 1 is $\pi$ sq. units

Answer 1

Answer: (A), (C), (D)

$y=f(x)=\frac{x^2-1}{x^2+1}=1-\frac{2}{x^2+1}$
$f^{\prime }(x)=\frac{4x}{{\left(x^2+1\right)}^2}$
$x>0$, fis increasing and $x<0$ fis decreasing $\Rightarrow (B)$ is true range is $[-1,1)\Rightarrow$ into $\Rightarrow$ (A) is false;
minimum value occurs at $x=0$ and $f(0)=-1\Rightarrow$
(C) is false
$A=2\underset{0}{\overset{\infty }{\int }}\left(1-\frac{x^2-1}{x^2+1}\right)dx=4\underset{0}{\overset{\infty }{\int }}\frac{dx}{x^2+1}={4\cdot {\tan }^{-1}|}_0^{\infty }=4\cdot \frac{\pi }{2}=2\pi \Rightarrow$(D) is false