ilovepdf_merged (6).pdf). Content covers sections 29.1–29.x.Derivatives solve geometric and optimisation problems: increasing/decreasing behaviour, maxima/minima, rate problems, and approximations.
If f′(x) > 0 on an interval, f is strictly increasing there; if f′ < 0, strictly decreasing.
Related rates: differentiate both sides of a relation with respect to t. Approximation: Δy ≈ f′(x) Δx for small Δx.
Most exam-important points from this chapter:
Start every answer with the key definition or standard form from Application of Derivatives.
Copy the formula, then substitute values; never jump to the number alone.
Check domain, quadrant, non-zero denominators, and applicability of the theorem.
Memorise and apply: Average change in y per unit change in x =
Memorise and apply: As x 0, the limiting value of the average rate of change of y with respect to x.
English questions from NIOS Mathematics (311) public / sample papers (Sample 2024, Apr 2024, Oct 2024, Apr 2025; board-style where scanned papers had no extractable text). Mapped exclusively to this chapter — no cross-chapter duplicates. Use Model Answer for the marking key and Explanation for working.
4 question(s) · Sources: Board-style approximation, Board-style increasing, Board-style max/min, Board-style rate
PYQ1. Find the local maximum and local minimum values of f(x)=x³−3x.
Model Answer
1) Local max f(−1)=2; local min f(1)=−2
Explanation
1) f′=3x²−3=0 ⇒ x=±
1. f″=6x: f″(−1)<0 max, f″(1)>0 min. Values 2 and −2.
PYQ2. Show that f(x)=x³+x is strictly increasing on ℝ.
Model Answer
1) f′(x)=3x²+1>0 for all x ⇒ strictly increasing
Explanation
1) 3x²+1 ≥ 1 > 0 everywhere.
PYQ3. Using differentials, approximate √26.
Model Answer
1) ≈ 5.1
Explanation
1) f(x)=√x at 25: f=5, f′=1/(2√x)=1/10, Δx=1
2) Δy≈0.1
3) √26≈5.1.
PYQ4. Side of a square increases at 0.1 cm/s. Rate of increase of area when side is 5 cm?
Model Answer
1) 1 cm²/s
Explanation
1) A=s² ⇒ dA/dt=2s ds/dt=2·5·0.1=1.
10 English exam-style problems for this chapter (NIOS Mathematics 311). Each question states the formula, gives full step-by-step working with the final answer, plus formal/easy explanations. English only.
Show that f(x)=x³ + x is strictly increasing on ℝ.
1) f′(x)=3x²+1 ≥ 1 > 0 for all real x.
2) Hence f is strictly increasing on ℝ.
Answer: f′(x)=3x²+1>0 ⇒ strictly increasing on ℝ
Positive derivative everywhere implies strict increase.
3x²+1 never zero or negative.
3x²+1 has minimum 1.
Do not stop at f′≥0 without checking zeros if needed.
Find the local maximum/minimum of f(x)=x³ − 3x.
1) f′=3x²−3=3(x−1)(x+1); critical points x=±1.
2) f″=6x; f″(1)=6>0 local min; f″(−1)=−6<0 local max.
3) f(1)=−2 (local min); f(−1)=2 (local max).
Answer: Local max 2 at x=−1; local min −2 at x=1
Second derivative test classifies the stationary points.
Critical ±1; second derivative sign decides max/min.
Values f(±1) are the local extreme values.
State both x and f(x).
Approximate √26 using f(x)=√x near 25.
1) f(25)=5; f′(x)=1/(2√x); f′(25)=1/10.
2) Δx=1 ⇒ Δy≈0.1 ⇒ √26≈5.1.
Answer: ≈ 5.1
Linear approximation at a perfect square.
From 25, go up by about 0.1.
True √26≈5.099, so good.
Use x=25 not 26 in f′.
A square’s side increases at 0.1 cm/s. Find the rate of increase of area when side is 5 cm.
1) A=s² ⇒ dA/dt=2s ds/dt.
2) s=5, ds/dt=0.1 ⇒ dA/dt=2·5·0.1=1 cm²/s.
Answer: 1 cm²/s
Related rates via chain rule on A=s².
2 times side times side-speed.
Units area per time.
Do not use 2s only.
Equation of tangent to y=x²−2x at x=3.
1) y(3)=3; f′=2x−2; f′(3)=4.
2) y−3=4(x−3) ⇒ y=4x−9.
Answer: y = 4x − 9
Point-slope with derivative slope.
Point (3,3), slope 4.
Check: at x=3, 12−9=3.
y value is f(3)=9−6=3.
Find absolute maximum of f(x)=x² on [−1,2].
1) f′=2x=0 at x=0; f(0)=0.
2) f(−1)=1; f(2)=4.
3) Absolute max is 4 at x=2.
Answer: Absolute maximum 4 at x=2
On a closed interval extrema occur at critical points or endpoints.
Compare 0,1,4 — largest is 4.
Absolute min is 0 at x=0 here.
Never skip endpoints.
Find the critical points of f(x) = x³ − 12x.
1) f′(x) = 3x² − 12 = 3(x² − 4) = 3(x−2)(x+2).
2) Critical points x = 2 and x = −2.
Answer: x = −2, x = 2
Set the first derivative equal to zero.
Factor 3(x²−4).
Also check where f′ undefined (none here).
Check each algebraic step carefully.
Find the equation of the tangent to y = x² at (2, 4).
1) f′(x)=2x; slope at 2 is 4.
2) Y − 4 = 4(X − 2) ⇒ Y = 4X − 4.
Answer: Y = 4X − 4
Point–slope with derivative as slope.
Slope 4 through (2,4).
Check the point lies on the curve.
Check each algebraic step carefully.
Classify the critical points of f(x)=x² − 4x + 1.
1) f′=2x−4=0 ⇒ x=2.
2) f″=2>0 ⇒ local (and global) minimum at x=2.
3) f(2)=4−8+1=−3.
Answer: Local minimum value −3 at x=2
Second derivative positive means concave up / min.
Parabola opens upward.
State both x and f(x).
Check each algebraic step carefully.
If A = πr² and dr/dt = 2 cm/s, find dA/dt when r = 5 cm.
1) dA/dt = 2πr dr/dt = 2π·5·2 = 20π cm²/s.
Answer: 20π cm²/s
Related rates via chain rule.
Area rate = circumference times radius rate.
Include units if asked.
Check each algebraic step carefully.