Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
143 views
in Sets, Relations and Functions by (96.4k points)
closed by
The maximum value of the function f(x) = 3x3 – 18x2 + 27x – 40 on the set S = {x ∈ R : x2 + 30 ≤ 11x} is:
1. -122
2. -222
3. 122
4. 222

1 Answer

0 votes
by (85.7k points)
selected by
 
Best answer
Correct Answer - Option 3 : 122

From question, the function given is:

f(x) = 3x3 – 18x2 + 27x – 40

Now,

⇒ f'(x) = 9x2 – 36x + 27

⇒ f'(x) = 9(x2 – 4x + 3)

⇒ f'(x) = 9(x – 1)(x – 3)

From question,

S = {x ∈ R : x2 + 30 ≤ 11x}

From which,

⇒ x2 + 30 ≤ 11x

⇒ x2 – 11x + 30 ≤ 0

⇒ (x – 5)(x – 6) ≤ 0

∴ x ∈ [5, 6]

∴ S = [5, 6]

Since, f'(x) > 0 when x ∈ [5, 6].

So, f(x) is an increasing function is [5, 6] interval.

Now,

f(x) is maximum when x = 6.

⇒ f(6) = 3(6)3 – 18(6)2 + 27(6) – 40

∴ f(6) = 122

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...