Search for: The maximum value of the function f(x)=3×3−18×2+27x−40 on the set S=x∈R:x2+30≤11x is The maximum value of the function f(x)=3x3−18x2+27x−40 on the set S=x∈R:x2+30≤11x is Fill Out the Form for Expert Academic Guidance!l Grade ---Class 1Class 2Class 3Class 4Class 5Class 6Class 7Class 8Class 9Class 10Class 11Class 12 Target Exam JEENEETCBSE +91 Preferred time slot for the call ---9 am10 am11 am12 pm1 pm2 pm3 pm4 pm5 pm6 pm7 pm8pm9 pm10pm Please indicate your interest Live ClassesBooksTest SeriesSelf Learning Language ---EnglishHindiMarathiTamilTeluguMalayalam Are you a Sri Chaitanya student? NoYes Verify OTP Code (required) I agree to the terms and conditions and privacy policy. Solution:Here, x2−11x+30≤0⇒x2−5x−6x+30≤0 ⇒(x−5)(x−6)≤0⇒5≤x≤6 ∴S={x∈R,5≤x≤6}Now, f(x)=3x3−18x2+27x−40∴ f′(x)=9x2−36x+27=9(x−1)(x−3),which is positive in [5, 6]. So So,f(x) is increasing in [5,6]Hence, maximum value of f(x)=f(6)=122 Related content Nucleus Diagram for Class 9 Neuron Diagram Class 9 Human Heart Diagram Class 10 CBSE Class 7 Science Chapter 1 Nutrition in Plants MCQ Questions with Answers CBSE Class 9 English Worksheet – The Road not Taken 50+ GK Questions and Answers for Class 8 “Kabir Ki Sakhiyan” MCQ Questions for Class 9 Hindi with Answers Kshitij, Kritika, Sparsh, Sanchayan Bhag 1 CBSE Class 7 English Honeycomb Chapter 1 Three Questions MCQ Questions with Answers CBSE Class 9 English Beehive Book Poem 1 The Road Not Taken MCQ Questions with Answers MCQs Questions with Answers for Class 9 English Chapter 1 The Fun They Had