Question

$f\left(x\right)$ is continuous function on [1, 3] and $f\left(1\right)=2,f\left(3\right)=-2,$ then which of the following not necessarily hold good?

• A. $f\left(2\right)⩾0$
• B. $x^{2}f\left(x\right)=0\text{has a root in}(1,3)$
• C. $-2\le f\left(x\right)\le 2\forall x\in [1,3]$
• D. $f\left(x\right)-x^2=0\text{has a root in}(1,3)$

Answer 1

Answer: (A), (C)

The correct options are
A $f\left(2\right)⩾0$
C $-2\le f\left(x\right)\le 2\forall x\in [1,3]$

$f\left(2\right)⩾0$ is not necessarily true as from the figure we can see, f(2) may be negative.

$Letg\left(x\right)=x^{2}f\left(x\right)$
$g\left(1\right)=f\left(1\right)=2g\left(3\right)=9\times f\left(3\right)=-18g\left(1\right)g\left(3\right)<0$
$x^{2}f\left(x\right)=0\text{has a root in}(1,3)$

Graph of $f\left(x\right)$:

It is clear from the above figure, $f\left(x\right)$ may be greater than 2 or less than -2.

$Leth\left(x\right)=f\left(x\right)-x^2$
$h\left(1\right)=2-1=1h\left(3\right)=-2-9=-11h\left(1\right)h\left(3\right)<0$
So $f\left(x\right)-x^2$ has a root in
(1, 3)