Q5. Find the intervals in which (i) f(x) = 10 − 6x − 2x2 and (ii) f(x) = −2x3 − 9x2 − 12x + 1 are strictly increasing or decreasing (NCERT Exercise 6.2, Q6 (b) and (c), page 158).
(i) f′(x) = −6 − 4x, zero at x = −3/2: increasing for x < −3/2, decreasing for x > −3/2. (ii) f′(x) = −6x2 − 18x − 12 = −6(x + 1)(x + 2): increasing for −2 < x < −1, decreasing for x < −2 and for x > −1.
Q6. Find the values of x for which y = [x(x − 2)]2 is increasing (NCERT Exercise 6.2, Q8, page 158).
y′ = 2x(x − 2) × (2x − 2) = 4x(x − 1)(x − 2). Its sign is negative for x < 0, positive on (0, 1), negative on (1, 2) and positive for x > 2. So y is increasing for 0 < x < 1 and x > 2.
Q7. Prove that the logarithmic function is increasing on (0, ∞), and that log |cos x| is decreasing on (0, π/2) (NCERT Exercise 6.2, Q10 and Q17, page 159).
d/dx (log x) = 1/x > 0 for x > 0. On (0, π/2), log |cos x| = log cos x, whose derivative is −sin x/cos x = −tan x < 0.
Q8. Prove that f(x) = x2 − x + 1 is neither strictly increasing nor strictly decreasing on (−1, 1) (NCERT Exercise 6.2, Q11, page 159).
f′(x) = 2x − 1 is negative on (−1, ½) and positive on (½, 1). So f decreases and then increases: for example f(0) = 1 > f(½) = ¾, but f(½) = ¾ < f(¾) = 13/16, so neither holds on the whole interval.
Q9. Let I be an interval disjoint from [−1, 1]. Prove that f(x) = x + 1/x is increasing on I (NCERT Exercise 6.2, Q15, page 159).
f′(x) = 1 − 1/x2 = (x2 − 1)/x2. For |x| > 1, x2 > 1, so f′(x) > 0 throughout I.
Q10. On which interval is f(x) = x100 + sin x − 1 decreasing? (A) (0, 1); (B) (π/2, π); (C) (0, π/2); (D) none of these (NCERT Exercise 6.2, Q13, page 159).
f′(x) = 100x99 + cos x. On (0, 1) and (0, π/2), cos x > 0 and x99 > 0, so f′ > 0. On (π/2, π), 100x99 > 100 × (1.57)99, far more than |cos x| ≤ 1, so f′ > 0 again. The answer is (D).
Q11. Find all a for which f(x) = x3 + ax2 + 3x + 5 is increasing on R.
f′(x) = 3x2 + 2ax + 3 must be ≥ 0 for every x, so its discriminant 4a2 − 36 must be ≤ 0: −3 ≤ a ≤ 3. Keep the end values: at a = 3, f′ = 3(x + 1)2 is 0 only at x = −1, and f still increases.