Class 12 Maths – Chapter 5 | Continuity & Differentiability Worksheet

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⚡ Key Formulas at a Glance

d/dx(sin⁻¹x) = 1/√(1−x²) d/dx(cos⁻¹x) = −1/√(1−x²) d/dx(tan⁻¹x) = 1/(1+x²) d/dx(eˣ) = eˣ d/dx(log x) = 1/x Chain Rule: df/dx = (dv/dt)·(dt/dx)

Section A — Multiple Choice Questions

A function f(x) is said to be continuous at x = c if lim_x→c f(x) equals:

MCQ

✅ Solution

By Definition 1 from NCERT: f is continuous at c if lim_x→c f(x) = f(c). The left-hand limit, right-hand limit, and the value of the function at x = c must all be equal.

For f(x) = |x|, which of the following is true?

MCQ

✅ Solution

f(x) = |x| is continuous at x = 0 (LHL = RHL = f(0) = 0). However, the LHD = −1 and RHD = +1 at x = 0, so it is NOT differentiable at x = 0. This is a classic example showing continuity does not imply differentiability.

If f(x) = x³ + 3, what is lim_x→0 f(x)?

MCQ

✅ Solution

f(x) = x³ + 3 is a polynomial, hence continuous everywhere. lim_x→0 f(x) = 0³ + 3 = 3. Note f(0) = 3 as well, confirming continuity at x = 0.

The derivative of sin⁻¹(x) with respect to x is:

MCQ

✅ Solution

Let y = sin⁻¹x ⟹ x = sin y. Differentiating: 1 = cos y · (dy/dx) ⟹ dy/dx = 1/cos y = 1/√(1−x²). Domain: x ∈ (−1, 1).

The greatest integer function f(x) = [x] is discontinuous at:

MCQ

✅ Solution

At any integer n, lim_x→n⁻[x] = n−1 but lim_x→n⁺[x] = n. Since LHL ≠ RHL, f is discontinuous at every integer. At non-integer points, it is continuous.

Using Chain Rule, d/dx[sin(x²)] equals:

MCQ

✅ Solution

Let t = x² and f(x) = sin t. By Chain Rule: df/dx = (d sin t / dt) · (dt/dx) = cos t · 2x = 2x cos(x²).

Which statement is always TRUE?

MCQ

✅ Solution

Theorem 3 (NCERT): If f is differentiable at c, then it is continuous at c. So Differentiable ⟹ Continuous is always true. The converse is NOT true (e.g., f(x) = |x| at x = 0).

The derivative of eˣ with respect to x is:

MCQ

✅ Solution

Theorem 5 (NCERT): d/dx(eˣ) = eˣ. This is the remarkable property of the natural exponential function — it is its own derivative!

If x = at², y = 2at, then dy/dx equals:

MCQ

✅ Solution

dx/dt = 2at; dy/dt = 2a. By parametric differentiation: dy/dx = (dy/dt)/(dx/dt) = 2a / 2at = 1/t.

The second order derivative of y = 3e²ˣ + 2e³ˣ is:

MCQ

✅ Solution

y' = 6e²ˣ + 6e³ˣ. y'' = d/dx(6e²ˣ + 6e³ˣ) = 6·2·e²ˣ + 6·3·e³ˣ = 12e²ˣ + 18e³ˣ.

For logarithmic differentiation of y = xˢⁱⁿˣ, which step is taken first?

MCQ

✅ Solution

For y = [u(x)]^v(x) form, the first step in logarithmic differentiation is to take natural log on both sides: log y = sin x · log x. Then differentiate both sides w.r.t. x.

d/dx(log x) equals (for x > 0):

MCQ

✅ Solution

By Theorem 5: d/dx(log x) = 1/x for x > 0. Here log refers to natural logarithm (base e = ln).

If f(x) = x³ − 3 for x ≠ 0, and f(0) = 1, then f is discontinuous at x = 0 because:

MCQ

✅ Solution

lim_x→0(x³ − 3) = 0 − 3 = −3, but f(0) = 1. Since the limit ≠ function value, f is discontinuous at x = 0 (Example 4 type from NCERT).

For x = a cosθ, y = a sinθ, dy/dx equals:

MCQ

✅ Solution

dx/dθ = −a sinθ; dy/dθ = a cosθ. So dy/dx = (a cosθ)/(−a sinθ) = −cot θ. (Example 31 in NCERT)

d/dx(aˣ), where a > 0 is a constant, equals:

MCQ

✅ Solution

Let y = aˣ ⟹ log y = x log a. Differentiating: (1/y)·(dy/dx) = log a ⟹ dy/dx = y log a = aˣ log a. (Example 28 in NCERT)

Section B — One Word / Numerical Answers

Find d/dx[sin(x²)] at x = √(π/2). (Enter the numerical value)

Numerical

✅ Solution

d/dx[sin(x²)] = 2x cos(x²). At x = √(π/2): = 2√(π/2) · cos(π/2) = 2√(π/2) · 0 = 0.

If y = e⁻ˣ, find dy/dx. (Enter in terms of the function, e.g. -e^-x)

Numerical

✅ Solution

y = e⁻ˣ. Using chain rule: dy/dx = e⁻ˣ · d/dx(−x) = e⁻ˣ · (−1) = −e⁻ˣ. (Example 26(i) NCERT)

How many points of discontinuity does f(x) = 1/x have? (Answer: 1 or 0 or infinite)

Numerical

✅ Solution

f(x) = 1/x is continuous at every point in its domain (x ≠ 0). The only point NOT in domain is x = 0, which is the 1 point of discontinuity.

Find d²y/dx² if y = x³ + tan x. What is the coefficient of sec²x·tan x in the answer?

Numerical

✅ Solution

y = x³ + tan x. y' = 3x² + sec²x. y'' = 6x + 2 sec x · sec x · tan x = 6x + 2 sec²x tan x. Coefficient = 2. (Example 35 NCERT)

For f(x) = 2x + 3, find lim_x→1 f(x). (Numerical answer)

Numerical

✅ Solution

lim_x→1(2x + 3) = 2(1) + 3 = 5. Also f(1) = 5, so f is continuous at x = 1. (Example 1, NCERT)

If y = A sin x + B cos x, what is d²y/dx² + y? (Numerical)

Numerical

✅ Solution

y' = A cos x − B sin x. y'' = −A sin x − B cos x = −y. Therefore d²y/dx² + y = −y + y = 0. (Example 36, NCERT)

Find f'(x) for f(x) = 5x − 3 at x = 0. (Numerical value)

Numerical

✅ Solution

f(x) = 5x − 3 is a polynomial. f'(x) = 5 (constant). This value is the same at every x, including x = 0.

If y = sin(log x), find dy/dx as a simplified expression. The numerator is: cos(log x) divided by what? (Enter denominator)

Numerical

✅ Solution

y = sin(log x). By chain rule: dy/dx = cos(log x) · d/dx(log x) = cos(log x) · (1/x) = cos(log x) / x. Denominator = x. (Example 26(ii) NCERT)

The domain of the function f(x) = sin⁻¹(x) is (−a, a). What is the value of a?

Numerical

✅ Solution

d/dx(sin⁻¹x) = 1/√(1−x²) is defined only when 1−x² > 0 ⟹ x² < 1 ⟹ −1 < x < 1. So domain = (−1, 1) and a = 1.

If y + sin y = cos x, find dy/dx. The answer is −sin x divided by (1 + cos y). What is the sign in numerator? (Enter +1 or -1)

Numerical

✅ Solution

Differentiating y + sin y = cos x w.r.t. x: dy/dx + cos y · dy/dx = −sin x ⟹ dy/dx(1 + cos y) = −sin x ⟹ dy/dx = −sin x/(1 + cos y). Sign = −1. (Example 23, NCERT)

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