Continuity and Differentiability: Miscellaneous Exercises (Class 12 Maths NCERT)

Understanding Miscellaneous Exercises

Miscellaneous exercises in the NCERT textbook serve as a comprehensive review and a formidable challenge, consolidating all the concepts covered in a chapter. For "Continuity and Differentiability," this means problems that often require you to apply multiple differentiation rules (like product, quotient, chain rule) in a single question, deal with piecewise functions for continuity and differentiability checks, or differentiate complex implicit, parametric, and inverse trigonometric functions. These problems are typically more intricate and require a deeper analytical approach compared to the standard exercises. They are excellent for identifying any gaps in your understanding and pushing your mathematical reasoning. Mastering them is key to truly owning the chapter and performing well in board examinations, where similar challenging questions often appear. It's not just about finding the derivative; it's about choosing the right technique and executing it flawlessly while also rigorously checking conditions for continuity. Expect to use logarithmic differentiation for complex power functions, implicit differentiation where variables aren't easily separable, and parametric differentiation when both x and y are functions of a third variable.

Key Concepts to Master for Miscellaneous Problems

Before diving into complex problems, a quick revisit of the core concepts is vital. Continuity demands that the left-hand limit, right-hand limit, and the function's value at a point must all be equal. Remember to check for conditions like f(c) = lim (x->c-) f(x) = lim (x->c+) f(x). For Differentiability, a function is differentiable at a point if its left-hand derivative (LHD) equals its right-hand derivative (RHD) at that point. An important theorem states that if a function is differentiable at a point, it must also be continuous at that point (though the converse is not always true). You'll frequently encounter the Chain Rule (for composite functions), the Product Rule ((uv)' = u'v + uv'), and the Quotient Rule ((u/v)' = (u'v - uv')/v^2). Be proficient in differentiating implicit functions (where y cannot be explicitly expressed in terms of x), parametric functions (where x and y are functions of a third parameter t or θ), and inverse trigonometric functions (like sin^-1(x), tan^-1(x)). Logarithmic Differentiation is a powerful tool for functions involving products, quotients, or variables in the exponent, simplifying complex expressions before differentiation. Always remember the properties of logarithms to simplify expressions effectively.

Worked Examples: Mastering Complex Problems

• Example 1: Continuity of a Piecewise Function Question: Discuss the continuity of the function f(x) given by: `` f(x) = { (sin x) / x, if x < 0 ax + 1, if x = 0 cos x, if x > 0 } ` Solution: Step 1: Identify critical points. The function definition changes at x = 0. We need to check continuity at x = 0. Step 2: Calculate the Left Hand Limit (LHL) at x = 0. LHL = lim (x->0-) f(x) = lim (x->0-) (sin x) / x Using the standard limit lim (x->0) (sin x) / x = 1, LHL = 1. Step 3: Calculate the Right Hand Limit (RHL) at x = 0. RHL = lim (x->0+) f(x) = lim (x->0+) cos x Substitute x = 0: cos(0) = 1. RHL = 1. Step 4: Calculate the function value at x = 0. f(0) = a(0) + 1 = 1` (from the second part of the definition). Step 5: Compare LHL, RHL, and f(0). Since LHL = RHL = f(0) = 1, the function f(x) is continuous at x = 0 for any value of 'a'. The value 'a' does not affect the continuity at x=0 in this specific function. This function is continuous for all x. Final answer: The function f(x) is continuous for all real values of x, including x=0, irrespective of the value of 'a'.
• Example 2: Logarithmic Differentiation Question: Differentiate y = (x^x) + (x^sin x) with respect to x. Solution: Step 1: Recognize the form. This is a sum of two functions, each with variable in the exponent. Direct differentiation is not possible; we need logarithmic differentiation for each term. Let u = x^x and v = x^sin x, so y = u + v. Then dy/dx = du/dx + dv/dx. Step 2: Differentiate u = x^x. Take logarithm on both sides: log u = log (x^x) = x log x. Differentiate implicitly with respect to x: (1/u) (du/dx) = (1 log x) + (x (1/x)) (using product rule) (1/u) (du/dx) = log x + 1 du/dx = u (log x + 1) = x^x (1 + log x). Step 3: Differentiate v = x^sin x. Take logarithm on both sides: log v = log (x^sin x) = (sin x) log x. Differentiate implicitly with respect to x: (1/v) (dv/dx) = (cos x log x) + (sin x (1/x)) (using product rule) (1/v) (dv/dx) = cos x log x + (sin x) / x dv/dx = v (cos x log x + (sin x) / x) = x^sin x (cos x log x + (sin x) / x). Step 4: Combine the derivatives. dy/dx = du/dx + dv/dx = x^x (1 + log x) + x^sin x (cos x log x + (sin x) / x). Final answer: dy/dx = x^x (1 + log x) + x^sin x (cos x log x + (sin x) / x).

Exam Tips for Solving Miscellaneous Exercises

Miscellaneous exercises are often the trickiest part of the chapter, but with the right strategy, you can conquer them. Here are some key tips:

1. Break Down Complex Problems: For questions like y = u + v where u and v are complex, differentiate u and v separately first, then add their derivatives. This compartmentalizes the problem and reduces errors.
2. Continuity First, Then Differentiability: When discussing differentiability of piecewise functions, always check for continuity at the critical point first. If the function is not continuous, it cannot be differentiable. This can save you time.
3. Choose the Right Differentiation Technique: For f(x)^g(x) form, logarithmic differentiation is essential. For products and quotients, apply the respective rules. For sin^-1(u) or tan^-1(u), remember the derivative formulas and apply the chain rule for u.
4. Simplify Before Differentiating (where possible): Especially with inverse trigonometric functions, try to simplify the expression using trigonometric identities before applying the differentiation formula. This often makes the differentiation much simpler.
5. Be Meticulous with Algebra: Complex problems involve many steps and terms. Small algebraic errors can lead to completely wrong answers. Double-check your calculations, especially signs and multiplication/division of terms.
6. Practice, Practice, Practice: The more miscellaneous problems you solve, the more familiar you'll become with various problem types and the techniques required to solve them efficiently and accurately. Don't just read solutions; try solving them yourself first.

Practice Questions with Solutions

• Q: If y = (log x)^cos x, find dy/dx. A: Step 1: Take logarithm on both sides: log y = log((log x)^cos x) = cos x log(log x). Step 2: Differentiate both sides with respect to x using the product rule. (1/y) (dy/dx) = (-sin x log(log x)) + (cos x (1/(log x)) (1/x)) Step 3: Isolate dy/dx. dy/dx = y [ -sin x log(log x) + (cos x) / (x log x) ] dy/dx = (log x)^cos x [ -sin x log(log x) + (cos x) / (x log x) ] Final answer: dy/dx = (log x)^cos x [ (cos x) / (x log x) - sin x log(log x) ]
• Q: Find dy/dx if x = a(cos t + t sin t) and y = a(sin t - t cos t). A: Step 1: Find dx/dt. dx/dt = a(-sin t + 1 sin t + t cos t) = a(t cos t). Step 2: Find dy/dt. dy/dt = a(cos t - (1 cos t + t (-sin t))) = a(cos t - cos t + t sin t) = a(t sin t). Step 3: Use the formula dy/dx = (dy/dt) / (dx/dt). dy/dx = (a(t sin t)) / (a(t cos t)) dy/dx = sin t / cos t = tan t. Final answer: dy/dx = tan t
• Q: Is the function f(x) = |x-1| + |x-2| differentiable at x = 1 and x = 2? Justify. A: Step 1: Rewrite the function for intervals around x=1 and x=2. For x < 1: f(x) = -(x-1) - (x-2) = -x + 1 - x + 2 = -2x + 3 For 1 <= x < 2: f(x) = (x-1) - (x-2) = x - 1 - x + 2 = 1 For x >= 2: f(x) = (x-1) + (x-2) = x - 1 + x - 2 = 2x - 3 Step 2: Check differentiability at x = 1. LHD at x=1: f'(x) for x < 1 is -2. So LHD = -2. RHD at x=1: f'(x) for 1 < x < 2 is 0. So RHD = 0. Since LHD != RHD at x=1, f(x) is not differentiable at x = 1. Step 3: Check differentiability at x = 2. LHD at x=2: f'(x) for 1 < x < 2 is 0. So LHD = 0. RHD at x=2: f'(x) for x > 2 is 2. So RHD = 2. Since LHD != RHD at x=2, f(x) is not differentiable at x = 2. Final answer: The function f(x) is not differentiable at x = 1 and not differentiable at x = 2 because the left-hand derivative and right-hand derivative are not equal at these points.
• Q: If y = sin^-1(x sqrt(1-x) - sqrt(x) sqrt(1-x^2)), find dy/dx. A: Step 1: Recognize the form sin(A-B) = sin A cos B - cos A sin B. Let x = sin A and sqrt(x) = sin B. Then A = sin^-1(x) and B = sin^-1(sqrt(x)). Also, sqrt(1-x^2) = cos A and sqrt(1-x) = cos B (assuming x in suitable domain). Step 2: Substitute into the expression for y. y = sin^-1(sin A cos B - cos A sin B) y = sin^-1(sin(A-B)) y = A - B y = sin^-1(x) - sin^-1(sqrt(x)) Step 3: Differentiate y with respect to x. dy/dx = d/dx(sin^-1(x)) - d/dx(sin^-1(sqrt(x))) dy/dx = 1/sqrt(1-x^2) - (1/sqrt(1-(sqrt(x))^2)) d/dx(sqrt(x)) (using chain rule) dy/dx = 1/sqrt(1-x^2) - (1/sqrt(1-x)) (1/(2sqrt(x))) dy/dx = 1/sqrt(1-x^2) - 1/(2sqrt(x(1-x))) Final answer: dy/dx = 1/sqrt(1-x^2) - 1/(2sqrt(x-x^2))

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