JEE Advanced · Mathematics · 22. Functions
The function \(f:[0,3] \rightarrow[1,29]\), defined by \(f(x)=2 x^{3}-15 x^{2}+36 x+1\), is
- A one-one and onto
- B onto but not one-one
- C one-one but not onto
- D neither one-one nor onto
Answer & Solution
Correct Answer
(B) onto but not one-one
Step-by-step Solution
Detailed explanation
Given : \(f(x)=2 x^{3}-15 x^{2}+36 x+1\)
\(\Rightarrow f^{\prime}(x)=6 x^{2}-30 x+36\)
\(=6\left(x^{2}-5 x+6\right)=6(x-2)(x-3)\)
\(\because f^{\prime}(x)>0 \forall x \in[0,2)\) and \(f^{\prime}(x) < 0 \forall x \in(2,3)\)
\(\therefore f(x)\) is increasing on \([0,2)\) and decreasing
on \((2,3)\)
\(\therefore \quad f(x)\) is many one on \([0,3]\)
Also \(f(0)=1, f(2)=29, f(3)=28\)
\(\therefore \quad\) Absolute min \(=1\) and Absolute max \(=29\)
\(\therefore \quad\) Range of \(f=[1,29]=\) codomain
Hence \(f\) is onto.
\(\Rightarrow f^{\prime}(x)=6 x^{2}-30 x+36\)
\(=6\left(x^{2}-5 x+6\right)=6(x-2)(x-3)\)
\(\because f^{\prime}(x)>0 \forall x \in[0,2)\) and \(f^{\prime}(x) < 0 \forall x \in(2,3)\)
\(\therefore f(x)\) is increasing on \([0,2)\) and decreasing
on \((2,3)\)
\(\therefore \quad f(x)\) is many one on \([0,3]\)
Also \(f(0)=1, f(2)=29, f(3)=28\)
\(\therefore \quad\) Absolute min \(=1\) and Absolute max \(=29\)
\(\therefore \quad\) Range of \(f=[1,29]=\) codomain
Hence \(f\) is onto.
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