Question

Assertion(A): The range of the function $\frac{x^2}{1+x^2}$ is $[0,1)$

Reason $\left(R\right)$ : lf $f\left(x\right)\le g\left(x\right)$ then the range of $\frac{f\left(x\right)}{g\left(x\right)},g\left(x\right)\ne 0$ is $[0,1)$

• A. Both A and R are true and R is the correct explanation of A
• B. Both A and R are true but R is not correct explanation of A
• C. A is true but R is false
• D. A is false but R is true

Answer 1

The correct option is C A is true but R is false

$f\left(x\right)=\frac{x^2}{1+x^2}=1-\frac{1}{1+x^2}$
$\frac{1}{1+x^2}\le 1$ for all $x$
$\therefore f\left(x\right)\in [0,1)$ for all $x$
Reason is false as $f\left(x\right)$ need not have an existing zero. Then $\frac{f\left(x\right)}{g\left(x\right)}$ will not be zero anywhere.