Question 12 - NCERT Exemplar - MCQs - Chapter 6 Class 12 Application of Derivatives

Last updated at August 8, 2026 by

The function f (x) = 2x 3 – 3x 2 – 12x + 4, has

(A) two points of local maximum

(B) two points of local minimum

(C) one maxima and one minima

(D) no maxima or minima

The function f (x) = 2x3 – 3x2 – 12x + 4, has - Class 12 MCQ [Teachoo] - NCERT Exemplar - MCQs

part 2 - Question 12 - NCERT Exemplar - MCQs - Serial order wise - Chapter 6 Class 12 Application of Derivatives
part 3 - Question 12 - NCERT Exemplar - MCQs - Serial order wise - Chapter 6 Class 12 Application of Derivatives

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Question 12 The function f (x) = 2x3 ā€“ 3x2 ā€“ 12x + 4, has two points of local maximum (B) two points of local minimum (C) one maxima and one minima (D) no maxima or minima f (š‘„) = 2š‘„3 ā€“ 3š‘„2 ā€“ 12š‘„ + 4 Finding fā€™ (š’™) fā€™ (š’™) = 6š‘„2 ā€“ 6š‘„ ā€“ 12 = 6 (š‘„"2 ā€“" š‘„" ā€“ 2" ) = 6 (š‘„"2 ā€“ 2" š‘„ "+ " š‘„"ā€“ 2 " ) = 6 (š‘„(š‘„" ā€“ 2" )+1(š‘„ "ā€“ 2" )) = 6 (š’™" + " šŸ) (š’™ "ā€“" šŸ) Putting fā€™ (š’™) = 0 6 (š‘„+1) (š‘„āˆ’2) = 0 āˆ“ š’™ = āˆ’1, 2 For maxima or minima Finding fā€ (š’™) fā€ (š’™) = 12š‘„ āˆ’ 6 For š’™ = āˆ’1 fā€ (āˆ’1) = 12 (āˆ’1) āˆ’6 = āˆ’12 āˆ’ 6 = āˆ’18 < 0 āˆ“ f has local maxima at x = āˆ’1 For š’™ = 2 fā€ (2) = 12 (2) āˆ’6 = 24 āˆ’ 6 = 18 > 0 āˆ“ f has local minima at x = 2 Hence, š‘“ has one maxima and one minima. So, the correct answer is (C)

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