If the corner points of the bounded feasible region of an LPP are (0,2), (3,0), (6,0), (6,8) and (0,5), then the minimum value of objective function…

A function $f(x) = |g(x)|$f(x)=∣g(x)∣ is not differentiable where $g(x) = 0$g(x)=0, because the absolute value creates a sharp corner at that point.

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$$f(x) = |x| + 1$$f(x)=∣x∣+1

The expression inside the absolute value is $x$x.

$$x = 0$$x=0

Not differentiable at $x = 0$x=0 only → (IV)

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$$f(x) = |x - 3|$$f(x)=∣x−3∣

The expression inside the absolute value is $x - 3$x−3.

$$x - 3 = 0$$x−3=0

$$x = 3$$x=3

Not differentiable at $x = 3$x=3 only → (I)

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$$f(x) = |x + 3|$$f(x)=∣x+3∣

The expression inside the absolute value is $x + 3$x+3.

$$x + 3 = 0$$x+3=0

$$x = -3$$x=−3

Not differentiable at $x = -3$x=−3 only → (II)

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$$f(x) = |x^2 - 9|$$f(x)=∣x2−9∣

The expression inside the absolute value is $x^2 - 9$x2−9.

$$x^2 - 9 = (x - 3)(x + 3) = 0$$x2−9=(x−3)(x+3)=0

$x = 3$x=3 or $x = -3$x=−3

Not differentiable at $x = 3, -3$x=3,−3 only → (III)

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(A) → (IV)
(B) → (I)
(C) → (II)
(D) → (III)

Therefore, the correct answer is (A)-(IV), (B)-(I), (C)-(II), (D)-(III).