Match List-I with List-II

Consider the function f(x) = 2x³ - 21x² + 36x + 80, x∈[0, 6]. Then

List-I	List-II
(A) one of its critical points is at x =	(I) -28
(B) Its absolute maximum value is	(II) -42
(C) Its absolute minimum value is	(III) 97
(D) Its second derivative at x = 0 is	(IV) 6

Choose the correct answer from the options given below:

Given: $f(x) = 2x³ - 21x² + 36x + 80$ on the interval $[0, 6]$

---

Finding the critical points:

$f'(x) = 6x² - 42x + 36$

Setting $f'(x) = 0$:

$6x² - 42x + 36 = 0$

$x² - 7x + 6 = 0$

$(x - 1)(x - 6) = 0$

Critical points: $x = 1$ and $x = 6$

Therefore, (A) matches with (IV) 6

---

To find the absolute maximum and minimum, evaluate $f(x)$ at the endpoints and critical points:

At $x = 0$:

$f(0) = 2(0)³ - 21(0)² + 36(0) + 80$

$f(0) = 80$

At $x = 1$:

$f(1) = 2(1)³ - 21(1)² + 36(1) + 80$

$f(1) = 2 - 21 + 36 + 80$

$f(1) = 97$

At $x = 6$:

$f(6) = 2(216) - 21(36) + 36(6) + 80$

$f(6) = 432 - 756 + 216 + 80$

$f(6) = -28$

Comparing values: $f(0) = 80$, $f(1) = 97$, $f(6) = -28$

The absolute maximum value is $97$.

Therefore, (B) matches with (III) 97

The absolute minimum value is $-28$.

Therefore, (C) matches with (I) -28

---

Finding the second derivative at $x = 0$:

$f'(x) = 6x² - 42x + 36$

$f''(x) = 12x - 42$

At $x = 0$:

$f''(0) = 12(0) - 42$

$f''(0) = -42$

Therefore, (D) matches with (II) -42

---

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