Question

If the function $f:R\to R$ is defined by $f\left(x\right)=(x^2+1{)}^{35}\forall \in R$, then $f$ is

• A. One-one but not onto
• B. Onto but not one-one
• C. Neither one-one nor onto
• D. Both one-one and onto

Answer 1

Answer: (C)

Neither one-one nor onto

$f\left(x\right)=(x^2+1{)}^{35}$

$f^{\prime }\left(x\right)=35(x^2+1{)}^{34}(2x)=70(x^2+1{)}^{34}x$

So for $x<0,$ $f^{\prime }\left(x\right)<0$ and for $x>0,f^{\prime }\left(x\right)>0$

That means the function is increasing for $x>0$ and decreasing for $x>0$

So the function is many one,

Also the range of $f\left(x\right)$ is $[1,\infty )$ which is not same as codomain

Hence $f$ is into function

Therefore option $C$ is correct choice