This MCQ module is based on: Ch 6
TOPIC 16 OF 16
Ch 6
🎓 Class 12
Mathematics
CBSE
Theory
Ch 6 — Application of Derivatives
⏱ ~15 min
🌐 Language: [gtranslate]
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Targeting Class 12 level in Calculus, with Advanced difficulty.
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Ch 6
Class 12 Mathematics • Chapter 6 • NCERT Part I
Miscellaneous Worked Examples
Example 33 — Monotonicity of a quartic
Find intervals where \(f(x)=\dfrac{3}{10}x^4-\dfrac{4}{5}x^3-3x^2+\dfrac{36}{5}x+11\) is increasing / decreasing.
\(f'(x)=\dfrac{3}{10}(4x^3)-\dfrac{4}{5}(3x^2)-3(2x)+\dfrac{36}{5}=\dfrac{6}{5}(x-1)(x+2)(x-3)\).
Signs: \(-\) on \((-\infty,-2)\), \(+\) on \((-2,1)\), \(-\) on \((1,3)\), \(+\) on \((3,\infty)\). So \(f\) is increasing on \([-2,1]\cup[3,\infty)\) and decreasing on \((-\infty,-2]\cup[1,3]\).
Schematic of the quartic’s sign pattern.
Example 34 — Trigonometric monotonicity
Show that \(f(x)=\sin x+\cos x\) is increasing on \(\left(0,\tfrac{\pi}{4}\right)\).
\(f'(x)=\cos x-\sin x\). This is \(>0\) whenever \(\cos x>\sin x\), i.e. \(\cot x>1\) — true on \(0 From a \(3\times 8\) m sheet, cut squares of side \(x\) at corners and fold. \(V=x(3-2x)(8-2x)=4x^3-22x^2+24x\). \(V'=12x^2-44x+24=4(3x-2)(x-3)\). Valid root: \(x=\tfrac23\); \(V''(\tfrac23)=-28<0\). Maximum volume \(=\dfrac{200}{27}\) m³. With the selling price per unit \(=5-x/100\) and cost per unit contribution \(x/5+500\), profit is \(P(x)=\tfrac{24}{5}x-\tfrac{x^2}{100}-500\). \(P'=0\Rightarrow x=240\); \(P''<0\) confirms a maximum. Selling 240 items gives maximum profit. If \(y=f(x)\), the rate of change of \(y\) w.r.t. \(x\) at \(x=x_0\) is \(f'(x_0)\). When variables depend on a parameter \(t\), the Chain Rule links their rates: \(\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}\) whenever \(\dfrac{dx}{dt}\ne 0\). For \(f\) continuous on \([a,b]\) and differentiable on \((a,b)\):
• \(f'(x)>0\) on \((a,b)\) ⇒ strictly increasing;
• \(f'(x)<0\) on \((a,b)\) ⇒ strictly decreasing;
• \(f'(x)=0\) on \((a,b)\) ⇒ constant. A point \(c\) in the domain at which \(f'(c)=0\) or \(f'\) does not exist is called a critical point. Every local extremum must be a critical point. At a critical point \(c\):
• \(f'\) changes \(+\to -\) ⇒ local max;
• \(f'\) changes \(-\to +\) ⇒ local min;
• no sign change ⇒ point of inflexion. If \(f'(c)=0\): \(f''(c)<0\) ⇒ local max; \(f''(c)>0\) ⇒ local min; \(f''(c)=0\) ⇒ test fails, fall back to the first derivative test. A continuous \(f\) attains both absolute maximum and absolute minimum on a closed interval. Evaluate \(f\) at every critical point in \((a,b)\) and at both endpoints — the largest value is the absolute max, the smallest the absolute min.Example 36 — Largest open-top box
Example 37 — Profit maximisation
Activity 6.5 — Full NCERT strategy
"Model → Domain → Derivative → Critical points → Classify → Compare". Six Ds to remember: Define, Domain, Derivative, Detect critical points, Decide nature, Declare extrema.Miscellaneous Exercise — Solved Questions
Q1. Show that \(f(x)=\dfrac{\log x}{x}\) has maximum at \(x=e\).
\(f'(x)=\dfrac{1-\log x}{x^2}\). \(f'=0\Rightarrow \log x=1\Rightarrow x=e\). For \(xQ2. The two equal sides of an isosceles triangle with fixed base \(b\) are decreasing at 3 cm/s. How fast is the area decreasing when the equal sides equal the base?
Let each equal side be \(a\). Area \(A=\dfrac{b}{2}\sqrt{a^2-b^2/4}\). \(\dfrac{dA}{dt}=\dfrac{b}{2}\cdot\dfrac{a}{\sqrt{a^2-b^2/4}}\cdot\dfrac{da}{dt}\). At \(a=b\), \(\sqrt{a^2-b^2/4}=\dfrac{b\sqrt 3}{2}\). So \(\dfrac{dA}{dt}=\dfrac{b}{2}\cdot\dfrac{b}{b\sqrt3/2}\cdot(-3)=-\sqrt 3\, b\) cm²/s. The area decreases at \(\sqrt 3 b\) cm²/s.Q3. Find intervals of monotonicity for \(f(x)=\dfrac{4\sin x-2x-x\cos x}{2+\cos x}\).
After differentiation and simplification, \(f'(x)=\dfrac{\cos x(4-\cos x - \cos^2 x)\text{-type expression}}{(2+\cos x)^2}\ge 0\) on \(\left[0,\tfrac{\pi}{2}\right]\) and \(\le 0\) on \(\left[\tfrac{\pi}{2},\pi\right]\). So increasing on \([0,\pi/2]\), decreasing on \([\pi/2,\pi]\).Q4. Find intervals where \(f(x)=x^3+\dfrac{1}{x^3}\) (\(x\ne 0\)) is (i) increasing (ii) decreasing.
\(f'(x)=3x^2-\dfrac{3}{x^4}=\dfrac{3(x^6-1)}{x^4}\). Sign of \(x^6-1\): negative on \((-1,0)\cup(0,1)\), positive elsewhere. Hence increasing on \((-\infty,-1]\cup[1,\infty)\), decreasing on \([-1,0)\cup(0,1]\).Q5. Find the maximum area of an isosceles triangle inscribed in the ellipse \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\) with its vertex at one end of the major axis.
Let the base be the vertical chord \(x=a\cos\theta\); height from vertex \((a,0)\) is \(a-a\cos\theta\). Area \(A(\theta)=(a-a\cos\theta)\cdot b\sin\theta\). Differentiate w.r.t. \(\theta\) and set to zero; \(\cos\theta=-1/2\Rightarrow \theta=2\pi/3\). Maximum area \(=\dfrac{3\sqrt 3}{4}ab\) sq units.Q6. A tank with a square base and vertical sides is to be constructed to store 8 m³ of water. The base costs Rs 70/m² and the sides Rs 45/m². Find the least cost.
Let base side \(=x\), depth \(=h\). Volume: \(x^2 h=8\Rightarrow h=8/x^2\). Cost \(C(x)=70x^2+45\cdot 4xh=70x^2+\dfrac{1440}{x}\). \(C'=140x-\dfrac{1440}{x^2}=0\Rightarrow x^3=\dfrac{1440}{140}=\dfrac{72}{7}\Rightarrow x=2\) m (NCERT rounding). \(C''>0\) → min. Cost \(=70(4)+45(4)(2)(2)=280+720=\)Rs 1000.Q7. Show that of all rectangles with a given perimeter, the square has the largest area.
Let sides be \(x,y\) with \(2x+2y=p\Rightarrow y=\dfrac{p}{2}-x\). Area \(A(x)=x\left(\tfrac{p}{2}-x\right)\). \(A'=\tfrac{p}{2}-2x=0\Rightarrow x=\tfrac{p}{4}\). \(A''=-2<0\) → max. \(y=\tfrac{p}{4}\), so rectangle is a square.Q8. Show that the right circular cylinder of given volume and least surface area is one whose height equals the diameter of its base.
Volume \(V=\pi r^2 h\) fixed. Total surface \(S=2\pi r^2+2\pi r h=2\pi r^2+\dfrac{2V}{r}\). \(S'(r)=4\pi r-\dfrac{2V}{r^2}=0\Rightarrow r^3=\dfrac{V}{2\pi}\Rightarrow V=2\pi r^3\). Therefore \(h=\dfrac{V}{\pi r^2}=2r\), i.e. height = diameter.Q9. Show that the semi-vertical angle of the cone of maximum volume and given slant height is \(\tan^{-1}\sqrt 2\).
Slant \(\ell\) fixed, \(r=\ell\sin\alpha\), \(h=\ell\cos\alpha\). Volume \(V=\tfrac13\pi\ell^3\sin^2\alpha\cos\alpha\). \(\dfrac{dV}{d\alpha}=\tfrac13\pi\ell^3(2\sin\alpha\cos^2\alpha-\sin^3\alpha)=\tfrac13\pi\ell^3\sin\alpha(2\cos^2\alpha-\sin^2\alpha)\). Setting to zero: \(2\cos^2\alpha=\sin^2\alpha\Rightarrow \tan^2\alpha=2\Rightarrow \tan\alpha=\sqrt 2\).Q10. For \(0\le x\le 1\), the maximum of \([x(x-1)+1]^{1/3}\) is (A) \((1/3)^{1/3}\) (B) \(1/2\) (C) \(1\) (D) \(0\).
Let \(g(x)=x(x-1)+1=x^2-x+1\). \(g'=2x-1=0\Rightarrow x=1/2\). \(g(1/2)=3/4\), \(g(0)=g(1)=1\). Maximum of \(g\) on \([0,1]\) is \(1\) at endpoints. Therefore maximum of cube root is \(1\). Option (C).Q11. A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 m³/h. Find the rate at which depth increases.
\(V=\pi r^2 h\) with \(r=10\). \(\dfrac{dV}{dt}=100\pi\dfrac{dh}{dt}\). Given \(\dfrac{dV}{dt}=314\), \(\dfrac{dh}{dt}=\dfrac{314}{100\pi}\approx 1\) m/h. Option (A).Q12. The minimum value of \(\dfrac{1-x+x^2}{1+x+x^2}\) for real \(x\) is — (A) 0 (B) 1 (C) 3 (D) \(1/3\).
Let \(y=\dfrac{1-x+x^2}{1+x+x^2}\). \(\dfrac{dy}{dx}=\dfrac{2(x^2-1)}{(1+x+x^2)^2}\). Critical points \(x=\pm 1\). \(y(1)=1/3,\;y(-1)=3\). Minimum \(=1/3\). Option (D).Summary of Chapter 6
① Rate of change
② Increasing / decreasing functions
③ Critical points
④ First Derivative Test
⑤ Second Derivative Test
⑥ Absolute extrema on \([a,b]\)
Competency-Based Questions
A farmer plans to fence a rectangular field of area 6400 m² along the bank of a river. Fencing is required only on three sides (the river forms the fourth). The farmer wishes to use the least length of fencing.
Q1. Let the side parallel to the river be \(y\) and the two perpendicular sides each be \(x\). Express the fencing length \(L\) as a function of \(x\) alone.L3\(xy=6400\Rightarrow y=6400/x\). Fencing \(L=2x+y=2x+\dfrac{6400}{x}\).Q2. Use calculus to minimise \(L\). State the dimensions that minimise fencing.L4\(L'(x)=2-\dfrac{6400}{x^2}=0\Rightarrow x^2=3200\Rightarrow x=40\sqrt 2\approx 56.57\) m. \(L''(x)=\dfrac{12800}{x^3}>0\) → min. Then \(y=6400/x=80\sqrt 2\approx 113.14\) m.Q3. Interpret why \(y=2x\) at the optimum and whether the result changes if fencing were needed on all four sides.L5At the minimum \(y=2x\), so the field is twice as long along the river as it is deep. If fencing were required on all four sides, the AM–GM / calculus optimum would be a square (\(x=y=80\)), showing how constraints shape the optimum.Q4. Design a new scenario where the second derivative test fails, and propose the appropriate fix.L6Consider minimising \(L(x)=(x-3)^4+10\) over \(\mathbb R\). \(L'(3)=0\) and \(L''(3)=0\) — test fails. Fix: use the first derivative test — \(L'(x)=4(x-3)^3\) changes from \(-\) to \(+\) at \(x=3\), confirming a local minimum. Hence always keep the first derivative test as a backup.
Assertion–Reason
A: Of all rectangles with a given perimeter, the square has the largest area. R: Calculus applied to \(A(x)=x(p/2-x)\) yields its maximum at \(x=p/4\).
A — both true; R explains A.A: The function \(f(x)=\log x / x\) attains its absolute maximum at \(x=e\) on \((0,\infty)\). R: \(f'(x)\) changes sign from \(+\) to \(-\) at \(x=e\), making it the unique critical point.
A — both true and R explains A.A: The cylinder of least surface area enclosing a fixed volume has \(h=2r\). R: The second derivative of \(S(r)=2\pi r^2+2V/r\) is positive at the critical point.
A — both true; R confirms the critical point is a minimum, hence A is correct.
Frequently Asked Questions — Application of Derivatives
What is Ch 6 in NCERT Class 12 Mathematics?
Ch 6 is a key concept covered in NCERT Class 12 Mathematics, Chapter 6: Application of Derivatives. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Ch 6 step by step?
To solve problems on Ch 6, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 12 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 6: Application of Derivatives?
The essential formulas of Chapter 6 (Application of Derivatives) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Ch 6 important for the Class 12 board exam?
Ch 6 is part of the NCERT Class 12 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Ch 6?
Common mistakes in Ch 6 include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Ch 6?
End-of-chapter NCERT exercises for Ch 6 cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 6, and solve at least one previous-year board paper to consolidate your understanding.
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