Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Applications of Derivatives

Question:

For finding absolute maximum and minimum values of a function f gives by

$f(x)=2x^3-15x^2+36x + 1 $ on [1, 5].

A. the critical points in (1, 5) are 2 and 3.

B. the absolute maximum value of f on [1, 5] is 29.

C. the absolute minimum value of f on [1, 5] is 24.

D. the absolute minimum values of f on [1, 5] is 20.

E. we evaluate the value of f ate critical points and at the end points of the interval [1, 5].

Choose the correct answer from the options given below :

Options:

A, C, E only

D, E only

B, C, D only

C, D, E only

Correct Answer:

A, C, E only

Explanation:

The correct answer is Option (1) → A, C, E only

$f(x)=2x^3-15x^2+36x + 1$,  $x∈[1, 5]$

$f'(x)=6x^2-30x+36=0$

$x^2-5x+6=0$

$(x-2)(x-3)=0⇒x=2,3$ critical points

so $f(1)=2-15+36+1=24$

$f(2)=2×2^3-15×2^2+36×2+1=29$

$f(3)=2×3^3-15×3^2+36×3+1=28$

$f(5)=2×5^3-15×5^2+36×5+1=56$