Q5. Find the maximum and minimum values, if any, of (i) (2x − 1)2 + 3; (ii) 9x2 + 12x + 2; (iii) −(x − 1)2 + 10; (iv) x3 + 1 (NCERT Exercise 6.3, Q1, page 174).
(i) minimum 3, at x = ½, no maximum. (ii) 9x2 + 12x + 2 = (3x + 2)2 − 2: minimum −2, no maximum. (iii) maximum 10, no minimum. (iv) x3 + 1 takes every real value: neither.
Q6. Find the maximum profit if the profit function is p(x) = 41 − 72x − 18x2 (NCERT Exercise 6.3, Q6, page 175).
p′(x) = −72 − 36x = 0 at x = −2, and p″ = −36 < 0, so the largest value over all real x is p(−2) = 41 + 144 − 72 = 113, NCERT’s answer. Note that x = −2 is a negative output: if x must be at least 0, p decreases for x ≥ 0 and the largest profit is p(0) = 41.
Q7. Find the maximum and minimum values of x + sin 2x on [0, 2π] (NCERT Exercise 6.3, Q12, page 176).
f′(x) = 1 + 2 cos 2x = 0 when cos 2x = −½, that is x = π/3, 2π/3, 4π/3, 5π/3. The values there are π/3 + √3/2, 2π/3 − √3/2, 4π/3 + √3/2 and 5π/3 − √3/2, and at the ends f(0) = 0 and f(2π) = 2π. The largest is 2π (at x = 2π) and the smallest 0 (at x = 0).
Q8. At x = 1 the function x4 − 62x2 + ax + 9 attains its maximum value on [0, 2]. Find a (NCERT Exercise 6.3, Q11, page 176).
An interior maximum needs f′(1) = 0: 4 − 124 + a = 0, so a = 120. For sufficiency on the whole interval, f″(x) = 12x − 124 < 0 throughout [0, 2], so f′ strictly decreases through zero at 1: the function rises up to 1 and falls afterwards.
Q9. A rectangular sheet 45 cm by 24 cm is made into an open box by cutting squares from the corners. Find the side of the square for the largest volume (NCERT Exercise 6.3, Q18, page 176).
V = x(45 − 2x)(24 − 2x) = 4x3 − 138x2 + 1080x, so V′ = 12x2 − 276x + 1080 = 12(x − 5)(x − 18). Only x = 5 gives a box (x < 12), and V″(5) = 120 − 276 < 0. So x = 5 cm.
Q10. Of all closed cylindrical cans of volume 100 cm3, find the dimensions of the one with least surface area (NCERT Exercise 6.3, Q21, page 176).
With πr2h = 100, S = 2πr2 + 200/r, so S′ = 4πr − 200/r2 = 0 at r3 = 50/π, and S″ > 0. Then h = 100/(πr2) = 2r: radius (50/π)1/3 cm and height 2(50/π)1/3 cm.
Q11. For all real x, the minimum value of (1 − x + x2)/(1 + x + x2) is (A) 0; (B) 1; (C) 3; (D) 1/3 (NCERT Exercise 6.3, Q28, page 177).
The derivative of the quotient has top (2x − 1)(1 + x + x2) − (1 − x + x2)(2x + 1) = 2x2 − 2, zero at x = ±1. The value is 1/3 at x = 1 and 3 at x = −1, and it tends to 1 as x → ±∞. So the minimum is 1/3: (D).
Q12. Find the absolute maximum and minimum values of f(x) = (x − 1)2 + 3 on [−3, 1] (NCERT Exercise 6.3, Q5 (iv), page 175).
f′(x) = 2(x − 1) is 0 only at x = 1, an end point. Compare f(−3) = 16 + 3 = 19 and f(1) = 3: the absolute maximum is 19, at x = −3, and the absolute minimum is 3, at x = 1. NCERT slip (answers, page 221): the key prints “Absolute minimum value = 19, absolute maximum value = 3”, with the two values swapped. In an exam, check that your maximum is not less than your minimum.