For the function f(x) = 2x³ − 9x² + 12x − 5, x ∈ [0,3], match List-I with List-II:

The correct option is: (A)- (IV), (B)- (III), (C)- (I), (D)- (II).

By analyzing the provided image, we see a table that lists several properties of a function under List-I, and corresponding values or points under List-II:
List-I:
(A) Absolute maximum value
(B) Absolute minimum value
(C) Point of maxima
(D) Point of minima

List-II:
(I) 3
(II) 0
(III) -5
(IV) 4

To determine the correct match, we analyze the function:

$f\left(x\right)=2x^3-9x^2+12x-5$

over the closed interval:

$x\in [0,3]$

Step 1: Find the critical points of the function
We first compute the first derivative of the function with respect to $x$:

$f^{\text{'}}\left(x\right)=\frac{d}{dx}(2x^3-9x^2+12x-5)$

$f^{\text{'}}\left(x\right)=6x^2-18x+12$

Next, we set the derivative equal to zero to find the critical points:

$6x^2-18x+12=0$

Divide the entire equation by 6:

$x^2-3x+2=0$

Factor the quadratic equation:

$(x-1)(x-2)=0$

This gives the critical points:

$x=1$

and

$x=2$

Both of these critical points lie within the specified closed interval $[0,3]$.

Step 2: Evaluate the function at the critical points and the boundaries
We evaluate the value of $f\left(x\right)$ at the boundaries $x=0$ and $x=3$, as well as at the critical points $x=1$ and $x=2$:

1. At the boundary $x=0$:
$f\left(0\right)=2(0{)}^3-9(0{)}^2+12\left(0\right)-5=-5$

2. At the critical point $x=1$:
$f\left(1\right)=2(1{)}^3-9(1{)}^2+12\left(1\right)-5=2-9+12-5=0$

3. At the critical point $x=2$:
$f\left(2\right)=2(2{)}^3-9(2{)}^2+12\left(2\right)-5=16-36+24-5=-1$

4. At the boundary $x=3$:
$f\left(3\right)=2(3{)}^3-9(3{)}^2+12\left(3\right)-5=54-81+36-5=4$

Step 3: Determine the matches for List-I and List-II

(A) Absolute maximum value:
Comparing the function values $-5,0,-1,4$, the largest value is $4$.
Thus, the absolute maximum value is $4$, which corresponds to (IV).

(B) Absolute minimum value:
Comparing the function values, the smallest value is $-5$.
Thus, the absolute minimum value is $-5$, which corresponds to (III).

(C) Point of maxima (Absolute Maxima):
The absolute maximum value of $4$ is achieved at $x=3$.
Thus, the point of maxima is $3$, which corresponds to (I).

(D) Point of minima (Absolute Minima):
The absolute minimum value of $-5$ is achieved at $x=0$.
Thus, the point of minima is $0$, which corresponds to (II).

Combining all of the matches, we get:
(A) - (IV), (B) - (III), (C) - (I), (D) - (II)