Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B2

Chapter

Calculus

Question:

Let $f: R→R$ be defined such that

$f(x) = 16x^2 -16x+12 $

(A) Maximum value of f(x) is 8

(B) Minimum value of f(x) is 8

(C) Minimum value of f(x) is 16

(D) No maximum value of f(x)

Choose the correct answer from the options given below :

Options:

(A), (C) Only

(B), (D) Only

(A), (B), (C) Only

(A), (B) Only

Correct Answer:

(B), (D) Only