📈 \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)
📈 Chapter 13: Applications of Derivatives | NEB Mathematics Notes Class 12
Based on Latest Syllabus 2080 | L’Hôpital’s Rule, Maxima-Minima, Tangents & Normals
✅ Updated according to latest syllabus of 2080 | Complete notes with 15 solved problems
📈 Applications of Derivatives — This chapter covers L’Hôpital’s Rule for indeterminate forms, increasing/decreasing functions, tangents and normals, maxima and minima, and optimization problems. The notes have been updated according to the latest syllabus of 2080. Now you don’t need to go anywhere searching for the notes of this chapter because we are here to serve you.
📚 Exercises
📖 L’Hôpital’s Rule
L’Hôpital’s Rule: For limit of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\):
\[
\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}
\]
provided the limit on the right exists or is infinite.
Conditions:
Conditions:
- Original limit must be \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\)
- f(x) and g(x) must be differentiable near c with \(g'(x) \neq 0\)
- The limit of the derivatives must exist
⚠️ Indeterminate Forms
\(\frac{0}{0}\) – Use L’Hôpital’s Rule
\(\frac{\infty}{\infty}\) – Use L’Hôpital’s Rule
\(0 \cdot \infty\) – Rewrite as \(\frac{0}{1/\infty}\) or \(\frac{\infty}{1/0}\)
\(\infty – \infty\) – Combine fractions or factor
\(0^0, \infty^0, 1^\infty\) – Use logarithmic differentiation
📌 Key Applications of Derivatives
Tangents and Normals: Slope of tangent = \(f'(x)\), slope of normal = \(-\frac{1}{f'(x)}\)
Increasing/Decreasing Functions: \(f'(x)>0\) → increasing, \(f'(x)<0\) → decreasing
Local Maxima/Minima: First derivative test or second derivative test
Second Derivative Test: \(f”(x)>0\) → local minima, \(f”(x)<0\) → local maxima
Rate of Change: \(\frac{dy}{dx}\) represents instantaneous rate of change
Optimization: Find critical points and evaluate boundaries
📝 Solved Problems (15 Important Questions)
Q1 Evaluate \(\lim_{x \to 0} \frac{\sin x}{x}\) using L’Hôpital’s Rule.
As \(x \to 0\), numerator \(\to 0\), denominator \(\to 0\) → \(\frac{0}{0}\) form.
\(\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = \cos 0 = 1\)
\(\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = \cos 0 = 1\)
✅ \(\lim_{x \to 0} \frac{\sin x}{x} = 1\)
Q2 Evaluate \(\lim_{x \to 0} \frac{1 – \cos x}{x^2}\).
\(\frac{0}{0}\) form. Apply L’Hôpital’s Rule once: \(\lim_{x \to 0} \frac{\sin x}{2x}\) → still \(\frac{0}{0}\)
Apply again: \(\lim_{x \to 0} \frac{\cos x}{2} = \frac{1}{2}\)
Apply again: \(\lim_{x \to 0} \frac{\cos x}{2} = \frac{1}{2}\)
✅ \(\frac{1}{2}\)
Q3 Evaluate \(\lim_{x \to \infty} \frac{\ln x}{x}\).
As \(x \to \infty\), numerator \(\to \infty\), denominator \(\to \infty\) → \(\frac{\infty}{\infty}\) form.
\(\lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{1/x}{1} = 0\)
\(\lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{1/x}{1} = 0\)
✅ \(0\)
Q4 Evaluate \(\lim_{x \to 0} x \ln x\).
As \(x \to 0\), \(x \to 0\), \(\ln x \to -\infty\) → \(0 \cdot \infty\) form.
Rewrite: \(\lim_{x \to 0} \frac{\ln x}{1/x}\) → \(\frac{-\infty}{\infty}\) form.
Apply L’Hôpital: \(\lim_{x \to 0} \frac{1/x}{-1/x^2} = \lim_{x \to 0} (-x) = 0\)
Rewrite: \(\lim_{x \to 0} \frac{\ln x}{1/x}\) → \(\frac{-\infty}{\infty}\) form.
Apply L’Hôpital: \(\lim_{x \to 0} \frac{1/x}{-1/x^2} = \lim_{x \to 0} (-x) = 0\)
✅ \(0\)
Q5 Evaluate \(\lim_{x \to 0} \frac{e^x – 1 – x}{x^2}\).
\(\frac{0}{0}\) form. Apply L’Hôpital: \(\lim_{x \to 0} \frac{e^x – 1}{2x}\) → still \(\frac{0}{0}\)
Apply again: \(\lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}\)
Apply again: \(\lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}\)
✅ \(\frac{1}{2}\)
Q6 Find the equation of the tangent to \(y = x^2\) at \(x = 2\).
\(f'(x) = 2x\), at \(x=2\), slope \(m = 4\)
Point: \((2,4)\). Equation: \(y – 4 = 4(x – 2)\) → \(y = 4x – 4\)
Point: \((2,4)\). Equation: \(y – 4 = 4(x – 2)\) → \(y = 4x – 4\)
✅ \(y = 4x – 4\)
Q7 Find the equation of the normal to \(y = x^3\) at \(x = 1\).
\(f'(x) = 3x^2\), at \(x=1\), slope \(m = 3\)
Normal slope = \(-\frac{1}{3}\)
Point: \((1,1)\). Equation: \(y – 1 = -\frac{1}{3}(x – 1)\) → \(y = -\frac{x}{3} + \frac{4}{3}\)
Normal slope = \(-\frac{1}{3}\)
Point: \((1,1)\). Equation: \(y – 1 = -\frac{1}{3}(x – 1)\) → \(y = -\frac{x}{3} + \frac{4}{3}\)
✅ \(y = -\frac{x}{3} + \frac{4}{3}\)
Q8 Find the intervals where \(f(x) = x^3 – 3x^2 + 4\) is increasing or decreasing.
\(f'(x) = 3x^2 – 6x = 3x(x – 2)\)
Critical points: \(x = 0, 2\)
For \(x<0\): \(f'(x)>0\) → increasing
For \(0
For \(x>2\): \(f'(x)>0\) → increasing
Critical points: \(x = 0, 2\)
For \(x<0\): \(f'(x)>0\) → increasing
For \(0
✅ Increasing on \((-\infty,0) \cup (2,\infty)\), decreasing on \((0,2)\)
Q9 Find the local maxima and minima of \(f(x) = x^3 – 3x^2 + 4\).
\(f'(x) = 3x(x-2)\), critical points \(x=0,2\)
\(f”(x) = 6x – 6\)
At \(x=0\): \(f”(0) = -6 < 0\) → local maximum, \(f(0)=4\)
At \(x=2\): \(f”(2) = 6 > 0\) → local minimum, \(f(2)=0\)
\(f”(x) = 6x – 6\)
At \(x=0\): \(f”(0) = -6 < 0\) → local maximum, \(f(0)=4\)
At \(x=2\): \(f”(2) = 6 > 0\) → local minimum, \(f(2)=0\)
✅ Local max at \((0,4)\), local min at \((2,0)\)
Q10 Find the maximum area of a rectangle with perimeter 20 cm.
Let length = l, width = w, \(2(l+w)=20\) → \(l+w=10\) → \(l=10-w\)
Area \(A = lw = w(10-w) = 10w – w^2\)
\(A'(w) = 10 – 2w = 0\) → \(w=5\), then \(l=5\)
Maximum area = \(25\) cm²
Area \(A = lw = w(10-w) = 10w – w^2\)
\(A'(w) = 10 – 2w = 0\) → \(w=5\), then \(l=5\)
Maximum area = \(25\) cm²
✅ \(25\) cm²
Q11 Evaluate \(\lim_{x \to 0} \frac{\tan x – x}{x^3}\).
\(\frac{0}{0}\) form. Apply L’Hôpital: \(\lim_{x \to 0} \frac{\sec^2 x – 1}{3x^2} = \lim_{x \to 0} \frac{\tan^2 x}{3x^2}\)
\(\lim_{x \to 0} \frac{\sin^2 x}{3x^2 \cos^2 x} = \frac{1}{3}\)
\(\lim_{x \to 0} \frac{\sin^2 x}{3x^2 \cos^2 x} = \frac{1}{3}\)
✅ \(\frac{1}{3}\)
Q12 Find the slope of the tangent to \(y = \ln x\) at \(x = e\).
\(f'(x) = \frac{1}{x}\), at \(x=e\), slope = \(\frac{1}{e}\)
✅ \(\frac{1}{e}\)
Q13 Find the radius of a cylinder of maximum volume inscribed in a sphere of radius R.
Let radius of cylinder = r, height = h. Then \(h^2 + (2r)^2 = (2R)^2\) → \(h^2 + 4r^2 = 4R^2\)
Volume \(V = \pi r^2 h = \pi r^2 \sqrt{4R^2 – 4r^2} = 2\pi r^2 \sqrt{R^2 – r^2}\)
Maximizing \(V^2\) leads to \(r = \frac{R}{\sqrt{2}}\)
Volume \(V = \pi r^2 h = \pi r^2 \sqrt{4R^2 – 4r^2} = 2\pi r^2 \sqrt{R^2 – r^2}\)
Maximizing \(V^2\) leads to \(r = \frac{R}{\sqrt{2}}\)
✅ \(r = \frac{R}{\sqrt{2}}\)
Q14 Evaluate \(\lim_{x \to 0} \frac{\sin^{-1} x}{x}\).
\(\frac{0}{0}\) form. L’Hôpital: \(\lim_{x \to 0} \frac{1/\sqrt{1-x^2}}{1} = 1\)
✅ \(1\)
Q15 Find the point on the curve \(y = x^2\) closest to the point \((0, 3)\).
Distance²: \(D = (x-0)^2 + (x^2-3)^2 = x^2 + x^4 – 6x^2 + 9 = x^4 – 5x^2 + 9\)
\(D’ = 4x^3 – 10x = 2x(2x^2 – 5) = 0\) → \(x=0\) or \(x = \pm \sqrt{\frac{5}{2}}\)
Check distances: \(x=0\): \(D=9\), \(x=\pm\sqrt{2.5}\): \(D= (\frac{5}{2})^2 – 5(\frac{5}{2}) + 9 = \frac{25}{4} – \frac{25}{2} + 9 = \frac{25}{4} – \frac{50}{4} + \frac{36}{4} = \frac{11}{4}\)
Minimum at \(x = \pm\sqrt{2.5}\), points \((\pm\sqrt{2.5}, 2.5)\)
\(D’ = 4x^3 – 10x = 2x(2x^2 – 5) = 0\) → \(x=0\) or \(x = \pm \sqrt{\frac{5}{2}}\)
Check distances: \(x=0\): \(D=9\), \(x=\pm\sqrt{2.5}\): \(D= (\frac{5}{2})^2 – 5(\frac{5}{2}) + 9 = \frac{25}{4} – \frac{25}{2} + 9 = \frac{25}{4} – \frac{50}{4} + \frac{36}{4} = \frac{11}{4}\)
Minimum at \(x = \pm\sqrt{2.5}\), points \((\pm\sqrt{2.5}, 2.5)\)
✅ \((\pm\sqrt{2.5}, 2.5)\)
✏️ Practice Questions
P1 Evaluate \(\lim_{x \to 0} \frac{e^x – 1}{x}\)
✅ Answer: \(1\)
P2 Find the tangent line to \(y = \sqrt{x}\) at \(x = 4\)
✅ Answer: \(y = \frac{x}{4} + 1\)
P3 Find the critical points of \(f(x) = x^4 – 4x^3\)
✅ Answer: \(x = 0, 3\)
P4 Evaluate \(\lim_{x \to 1} \frac{\ln x}{x-1}\)
✅ Answer: \(1\)
P5 Find two positive numbers whose sum is 20 and product is maximum
✅ Answer: \(10\) and \(10\)
📋 Multiple Choice Questions (HSEB Pattern)
MCQ 1 The value of \(\lim_{x \to 0} \frac{\sin x}{x}\) is:
A) 0 B) 1 C) \(\infty\) D) Does not exist
✅ Answer: B) 1
MCQ 2 The slope of the tangent to \(y = x^2\) at \(x = 3\) is:
A) 3 B) 6 C) 9 D) 12
✅ Answer: B) 6
MCQ 3 L’Hôpital’s Rule is applicable to which indeterminate form?
A) \(\frac{0}{\infty}\) B) \(\frac{0}{0}\) C) \(\infty – \infty\) D) \(0 \cdot 1\)
✅ Answer: B) \(\frac{0}{0}\)
MCQ 4 A function is increasing if:
A) \(f'(x) < 0\) B) \(f'(x) = 0\) C) \(f'(x) > 0\) D) \(f”(x) = 0\)
✅ Answer: C) \(f'(x) > 0\)
MCQ 5 At a local maximum, \(f'(x) = 0\) and \(f”(x)\) is:
A) Positive B) Negative C) Zero D) None
✅ Answer: B) Negative
MCQ 6 \(\lim_{x \to 0} \frac{\tan x}{x} =\)
A) 0 B) 1 C) \(\infty\) D) \(\frac{1}{2}\)
✅ Answer: B) 1
MCQ 7 The normal line is perpendicular to the:
A) Secant B) Tangent C) Chord D) Asymptote
✅ Answer: B) Tangent
MCQ 8 \(\lim_{x \to 0} \frac{1 – \cos x}{x} =\)
A) 0 B) 1 C) \(\frac{1}{2}\) D) Does not exist
✅ Answer: A) 0
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This PDF provides the solutions of every question from the exercises of class 12 applications of derivatives chapter. If you want the notes of other exercises then you can choose the exercise from the button given above.
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📌 Key Formulas Summary:
• L’Hôpital’s Rule: \(\lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)}\) for \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\)
• Tangent Line: \(y – y_0 = f'(x_0)(x – x_0)\)
• Normal Line: \(y – y_0 = -\frac{1}{f'(x_0)}(x – x_0)\)
• Increasing/Decreasing: \(f'(x) > 0\) → increasing, \(f'(x) < 0\) → decreasing
• First Derivative Test: Critical points where \(f'(x) = 0\)
• Second Derivative Test: \(f”(x_0) > 0\) → min, \(f”(x_0) < 0\) → max
• Optimization: Maximize/minimize a function subject to constraints
• L’Hôpital’s Rule: \(\lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)}\) for \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\)
• Tangent Line: \(y – y_0 = f'(x_0)(x – x_0)\)
• Normal Line: \(y – y_0 = -\frac{1}{f'(x_0)}(x – x_0)\)
• Increasing/Decreasing: \(f'(x) > 0\) → increasing, \(f'(x) < 0\) → decreasing
• First Derivative Test: Critical points where \(f'(x) = 0\)
• Second Derivative Test: \(f”(x_0) > 0\) → min, \(f”(x_0) < 0\) → max
• Optimization: Maximize/minimize a function subject to constraints