Question

Let $f:R\to R$ be a continuous onto function satisfying $f\left(x\right)+f(-x)=0,\forall xϵR$. If $f(-3)=2$ and $f\left(5\right)=4$, then the equation $f\left(x\right)=0$ has

• A. Exactly three real roots
• B. Exactly two real roots
• C. At least five real roots
• D. At least three real roots

Answer 1

The correct option is D At least three real roots

$\because f\left(x\right)+f(-x)=0,\forall x\in R$
$\Rightarrow$ $f\left(x\right)=-f(-x),\forall x\in R$
i.e. $f\left(x\right)$ is an odd function.
Now, it is given that $f(-3)=2$ and $f\left(5\right)=4$
$\Rightarrow$ $f\left(3\right)=-2$ and $f(-5)=-4$
$\because f:R\to R$ is a continuous onto function.
So, according to the intermediate value theorem, there should be at least one root each in the intervals $(-5,-3),(-3,3),(3,5).$
Hence, $f\left(x\right)=0$ posses at least three real roots.
Hence,option (D) is correct.