Given: f(x)=42x3−32x−1
⇒f′(x)=122x2−32
⇒f′(x)=32(4x2−1)
⇒f′(x)=32(2x−1)(2x+1)

So, the function is increasing in x∈(21,1).
Thus, the function will intersect the x−axis only at one point.
Hence, statement I is correct.
Now, putting x=12πin the equation cos3x=4cos312π−3cos12π.
⇒cos4π=4cos312π−3cos12π
⇒42cos312π−32cos12π−1=0
Now, comparing with f(x)=42x3−32x−1 we get.
cos12π is a solution of f(x).
So, statement-II is also correct.
Hence, both the statements are correct.