NCERT Class 12 Maths · Chapter 6 Applications Of Derivatives · Exercise 6.3 Question 5

Question

Find absolute extrema: (i) x³ on [−2,2]; (ii) sin x+cos x on [0,π]; (iii) 4x−x²/2 on [−2,9/2]; (iv) (x−1)²+3 on [−3,1].

Accepted Answer

Final Answer: (1) max $8$, min $-8$; (2) max $\sqrt{2}$, min $-1$; (3) max $8$, min $-10$; (4) max $19$, min $3$ Step 1: Part 1: Step 2: Differentiate: f′=$3 x^{2}$. The interior stationary points in this interval are $0$. Step 3: Evaluate every candidate: f($-2$)=$-8$; f($0$)=$0$; f($2$)=$8$. Step 4: Comparing these values gives the absolute maximum and minimum; endpoints must be included. Step 5: Part 2: Step 6: Differentiate: f′=$\sqrt{2} \cos{\left(x + \frac{\pi}{4} \right)}$. The interior stationary points in this interval are $\frac{\pi}{4}$. Step 7: Evaluate every candidate: f($0$)=$1$; f($\frac{\pi}{4}$)=$\sqrt{2}$; f($\pi$)=$-1$. Step 8: Comparing these values gives the absolute maximum and minimum; endpoints must be included. Step 9: Part 3: Step 10: Differentiate: f′=$4 - x$. The interior stationary points in this interval are $4$. Step 11: Evaluate every candidate: f($-2$)=$-10$; f($4$)=$8$; f($\frac{9}{2}$)=$\frac{63}{8}$. Step 12: Comparing these values gives the absolute maximum and minimum; endpoints must be included. Step 13: Part 4: Step 14: Differentiate: f′=$2 \left(x - 1\right)$. The interior stationary points in this interval are none. Step 15: Evaluate every candidate: f($-3$)=$19$; f($1$)=$3$. Step 16: Comparing these values gives the absolute maximum and minimum; endpoints must be included.

NCERT Class 12 Maths · Chapter 6 Applications Of Derivatives · Exercise 6.3 Question 5

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