NCERT Class 12 Maths · Chapter 6 Applications Of Derivatives · Exercise 6.3 Question 3

✅ Step-by-Step Solution

1. (1) f′=\(2 x\) and f″=\(2\).
2. Minimum at x=0, value 0; no local maximum. A positive-to-negative first-derivative change gives a maximum; a negative-to-positive change gives a minimum.
3. (2) f′=\(3 \left(x - 1\right) \left(x + 1\right)\) and f″=\(6 x\).
4. Maximum at x=−1, value 2; minimum at x=1, value −2. A positive-to-negative first-derivative change gives a maximum; a negative-to-positive change gives a minimum.
5. (3) f′=\(\sqrt{2} \cos{\left(x + \frac{\pi}{4} \right)}\) and f″=\(- \sqrt{2} \sin{\left(x + \frac{\pi}{4} \right)}\).
6. On (0,π/2), maximum at π/4, value √2; no local minimum. A positive-to-negative first-derivative change gives a maximum; a negative-to-positive change gives a minimum.
7. (4) f′=\(\sqrt{2} \sin{\left(x + \frac{\pi}{4} \right)}\) and f″=\(\sqrt{2} \cos{\left(x + \frac{\pi}{4} \right)}\).
8. On (0,2π), maximum at 3π/4, value √2; minimum at 7π/4, value −√2. A positive-to-negative first-derivative change gives a maximum; a negative-to-positive change gives a minimum.
9. (5) f′=\(3 \left(x - 3\right) \left(x - 1\right)\) and f″=\(6 \left(x - 2\right)\).
10. Maximum at x=1, value 19; minimum at x=3, value 15. A positive-to-negative first-derivative change gives a maximum; a negative-to-positive change gives a minimum.
11. (6) f′=\(\frac{\left(x - 2\right) \left(x + 2\right)}{2 x^{2}}\) and f″=\(\frac{4}{x^{3}}\).
12. For x>0, minimum at x=2, value 2; no local maximum. A positive-to-negative first-derivative change gives a maximum; a negative-to-positive change gives a minimum.
13. (7) f′=\(- \frac{2 x}{\left(x^{2} + 2\right)^{2}}\) and f″=\(\frac{2 \left(3 x^{2} - 2\right)}{\left(x^{2} + 2\right)^{3}}\).
14. Maximum at x=0, value 1/2; no local minimum. A positive-to-negative first-derivative change gives a maximum; a negative-to-positive change gives a minimum.
15. (8) f′=\(- \frac{3 x - 2}{2 \sqrt{1 - x}}\) and f″=\(\frac{3 x - 4}{4 \left(1 - x\right)^{\frac{3}{2}}}\).
16. On (0,1), maximum at x=2/3, value 2/(3√3); no local minimum. A positive-to-negative first-derivative change gives a maximum; a negative-to-positive change gives a minimum.