Explanation:
Exp: (B) : Given,
The function \(\mathrm{f}:[0,3] \rightarrow[1,29]\) defined by \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^3-\) \(15 \mathrm{x}^2+36 \mathrm{x}+1\)
Then, \(f(x)=2 x^3-15 x^2+36 x+1\),
Condition (a) one-one :
For one-one, differentiate the function -
\(f^{\prime}(x)=6 x^2-30 x+36\)
Now, \(\quad \mathrm{f}^{\prime}(\mathrm{x})=0\)
\(6 x^2-30 x+36=0\)
\(x^2-5 x+6=0\)
\((x-3)(x-2)=0\)
\(\Rightarrow \quad x \in\{2,3\} .\)
Then, \(\mathrm{f}(\mathrm{x})\) is increasing for \(\mathrm{x}\lt 2\) or \(\mathrm{x}>3\).
Since, \(\mathrm{f}^{\prime}(\mathrm{x})>0\).
\(\mathrm{f}(\mathrm{x})\) is decreasing for \(2\lt \mathrm{x}\lt 3\).
Since, \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\).
But \(f(x)\) is not strictly increasing or strictly decreasing in the entire domain.
Then, \(\mathrm{f}(\mathrm{x})\) is not one-one.
Condition (b) Onto :
Since, \(\mathrm{f}(\mathrm{x})\) is maximum at \(\mathrm{x}=2\)
\(f(2)=29\) is maximum value of \(f\)
\(f(0)=1, f(3)=28 \text {. }\)
We see that, range of \(f(x)=[1,29]\) which is equal to the co-domain.
So, \(\mathrm{f}(\mathrm{x})\) is onto.
Hence, \(f(x)=2 x^3-15 x^2+36+1\) is onto but not oneone.