Correct choice is (b) False
The explanation is: The above statement is false. f is neither one-one nor onto.
For one-one: Consider f(x1)=f(x2)
∴ 5x1^4+2=5x2^4+2
⇒x1=± x2.
Hence, the function is not one – one.
For onto: Consider the real number 1 which lies in co- domain R, and let \(x=(\frac{y-2}{5})^{\frac{1}{4}}\).
Clearly, there is no real value of x which lies in the domain R such that f(x)=y.
Therefore, f is not onto as every element lying in the codomain must have a pre- image in the domain.