RowQ
The Vault
RowQ
The Vault
CBSE Class 12 Maths · 11 questions · 26 marks
This chapter tightens the informal idea of an unbroken graph into the precise condition that the left-hand limit, right-hand limit and function value all agree at a point. Building on that, differentiability is introduced as a stronger requirement — every differentiable function is continuous, but not every continuous function is differentiable, as the sharp corner of |x| at the origin shows. The chain rule, implicit differentiation and logarithmic differentiation then extend derivative techniques to far more complicated functions.
For f(x) = (x²-4)/(x-2) when x≠2 and f(2)=k, find the value of k that makes f continuous at x=2.
Answer
k = 4. For x≠2, (x²-4)/(x-2) = (x-2)(x+2)/(x-2) = x+2. So lim(x→2) f(x) = 2+2 = 4. For continuity at x=2, f(2) must equal this limit, so k = 4.
The derivative d/dx(sin⁻¹x), for x ∈ (-1, 1), equals:
Answer
The correct derivative is 1/√(1-x²). This is a standard derivative result for the inverse sine function, valid for x∈(-1,1). The negative version, -1/√(1-x²), is instead the derivative of cos⁻¹x.
If y = log(sin x), then dy/dx equals:
Answer
dy/dx = cot x. By the chain rule, d/dx[log(sin x)] = (1/sin x) × d/dx(sin x) = (1/sin x) × cos x. cos x / sin x = cot x.
The function f(x) = |x-2| is:
Answer
f is continuous everywhere but not differentiable at x=2. |x-2| is built from continuous pieces (x-2 for x≥2, 2-x for x<2) that agree at x=2, so it is continuous everywhere. However, the left-hand derivative at x=2 is -1 and the right-hand derivative is +1; since these differ, f is not differentiable at x=2, just like |x| at the origin.
Assertion (A): Every differentiable function is continuous. Reason (R): Every continuous function is differentiable.
Answer
A is true but R is false. Assertion A is a genuine theorem: differentiability at a point forces continuity there, because the existence of the limit defining the derivative requires the function to have no jump. Reason R is false in general — f(x)=|x| is continuous everywhere but fails to be differentiable at x=0, so continuity does not guarantee differentiability.
Find dy/dx if y = sin(x² + 1).
Answer
This is a composite function, so apply the chain rule with u = x²+1 and y = sin(u). dy/du = cos(u) = cos(x²+1), and du/dx = 2x. By the chain rule, dy/dx = cos(x²+1) × 2x = 2x cos(x²+1).
Find dy/dx if x² + y² = 25.
Answer
Differentiate both sides with respect to x, treating y as a function of x. d/dx(x²) + d/dx(y²) = d/dx(25) gives 2x + 2y(dy/dx) = 0. Solving for dy/dx: 2y(dy/dx) = -2x, so dy/dx = -x/y.
Show that f(x) = |x| is continuous at x=0 but not differentiable at x=0.
Answer
Continuity: lim(x→0⁻)|x| = lim(x→0⁻)(-x) = 0, lim(x→0⁺)|x| = lim(x→0⁺)(x) = 0, and f(0) = 0. Since all three agree, f is continuous at x=0. Differentiability: the left-hand derivative is lim(h→0⁻)[|h|-0]/h = lim(h→0⁻)(-h)/h = -1. The right-hand derivative is lim(h→0⁺)[|h|-0]/h = lim(h→0⁺)(h)/h = 1. Since the left-hand derivative (-1) and right-hand derivative (1) are unequal, f is not differentiable at x=0.
If y = xˣ, find dy/dx using logarithmic differentiation.
Answer
Since the variable appears in both the base and the exponent, take the natural log of both sides first. ln y = ln(xˣ) = x ln x. Differentiate both sides with respect to x, using the product rule on the right: (1/y)(dy/dx) = ln x + x×(1/x) = ln x + 1. Multiply both sides by y: dy/dx = y(ln x + 1). Substituting back y = xˣ: dy/dx = xˣ(ln x + 1).
Verify Rolle's theorem for f(x) = x² - 4x + 3 on the interval [1,3].
Answer
f is a polynomial, so it is continuous on [1,3] and differentiable on (1,3) — the first two conditions of Rolle's theorem hold automatically. Check the endpoint condition: f(1) = 1 - 4 + 3 = 0, and f(3) = 9 - 12 + 3 = 0. Since f(1) = f(3), the third condition holds. By Rolle's theorem, there must exist c in (1,3) with f'(c) = 0. f'(x) = 2x - 4. Setting f'(c) = 0 gives 2c - 4 = 0, so c = 2, which does lie in (1,3). Hence Rolle's theorem is verified, with c = 2.
Read the following and answer the questions that follow: A firm's cost (in thousands of rupees) to produce x hundred units is modelled piecewise as C(x) = 3x + 2 for 0 ≤ x < 4, and C(x) = ax + b for x ≥ 4. Management wants no sudden jump or kink in cost at x = 4, so C must be both continuous and differentiable there. (a) Find the value the first piece gives as x approaches 4, i.e. C(4⁻). (b) Using continuity at x = 4, write a relation between a and b. (c) Using differentiability at x = 4, find the value of a. (d) Hence find the value of b.
Answer
(a) Using the first piece, C(4⁻) = 3(4) + 2 = 14. (b) For continuity at x=4, the second piece must give the same value: a(4) + b = 14, i.e. 4a + b = 14. (c) The derivative of the first piece is 3 (constant slope), and the derivative of the second piece is a. For differentiability at x=4, these slopes must match, so a = 3. (d) Substituting a = 3 into 4a + b = 14 gives 12 + b = 14, so b = 2.
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