f(x) \(=\frac{3\mathrm x^2}{4\pi}\) – cos πx
\(\because\) f(x) \(=\frac{3\mathrm x^2}{4\pi}\) – cos πx
And f(–x) \(=\frac{3\mathrm x^2}{4\pi}\) – cos(–πx)
\(=\frac{3\mathrm x^2}{4\pi}\) – cos πx (\(\because\) cos(–θ) = cos θ)
Therefore, f(x) = f(–x)
Hence, f is not one-one function.
\(\therefore\) f is many-one function.
Now, Range of \(\frac{3\mathrm x^2}{4\pi}\) is [0, \(\infty\)]
and Range of cos πx = [–1, 1]
Therefore, the range of f(x) is [–1, \(\infty\)).
\(\therefore\) f(x) never gets negative values less than 1.
Hence, f(x) is not onto (\(\because\) Domain of f(x) is R)
Therefore, f(x) is into function.