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CBSE Important Questions Class 12 Maths Chapter 5 – Continuity and Differentiability

Class 12 Maths Chapter 5 Continuity and Differentiability is part of the CBSE Class 12 Maths syllabus. It builds on the limits and derivatives studied in Class 11. Students learn what it means for a function to be continuous at a point, the relation between continuity and differentiability, the chain rule, derivatives of implicit, inverse trigonometric, exponential and logarithmic functions, logarithmic differentiation, parametric differentiation and second order derivatives.

These CBSE important questions for Class 12 Maths Chapter 5 cover checking continuity and differentiability of a modulus function at specific points, logarithmic differentiation of x raised to a power, applying the chain rule, implicit differentiation, derivatives in parametric form, second order derivatives, and the conditions in Rolle's theorem and the Mean Value Theorem. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. Define continuity of a function at a point. What is the relation between continuity and differentiability?

Answer: A function f is continuous at x = a if all three conditions hold:
1. f(a) exists
2. lim_x→a f(x) exists, that is the left-hand and right-hand limits are equal
3. lim_x→a f(x) = f(a)
Relation with differentiability: If a function is differentiable at a point, then it must be continuous there.
The converse is not true: a function can be continuous but not differentiable. The standard example is f(x) = |x| at x = 0, which is continuous there but has a sharp corner, so it is not differentiable.

Q2. Let f(x) = |x − 2| + |x + 1|. Examine the continuity and differentiability of f(x) at x = −1 and x = 2.

Answer: The critical points are x = −1 and x = 2. Write f piecewise:
f(x) = −2x + 1 for x < −1
f(x) = 3 for −1 ≤ x ≤ 2
f(x) = 2x − 1 for x > 2
At x = −1: Left limit = −2(−1) + 1 = 3, right limit = 3, and f(−1) = 3.
All three agree, so f is continuous at x = −1. But the left derivative is −2 and the right derivative is 0, so f is not differentiable there.
At x = 2: Left limit = 3, right limit = 2(2) − 1 = 3, and f(2) = 3.
So f is continuous at x = 2. But the left derivative is 0 and the right derivative is 2, so f is not differentiable there.

Q3. Differentiate x^√x with respect to x.

Answer: Here both the base and the exponent contain x, so we use logarithmic differentiation.
Let y = x^√x. Taking the logarithm of both sides:
log y = √x · log x
Differentiating both sides with respect to x:
(1/y) dy/dx = √x · (1/x) + log x · (1/(2√x))
= 1/√x + log x/(2√x)
Multiplying by y:
dy/dx = x^√x [ 1/√x + log x/(2√x) ]
= x^√x · (2 + log x) / (2√x)

Q4. Write the chain rule and use it to differentiate y = sin(x²).

Answer: Chain rule: If y = f(u) and u = g(x), then
dy/dx = (dy/du) × (du/dx)
For y = sin(x²): Let u = x², so y = sin u.
dy/du = cos u = cos(x²)
du/dx = 2x
dy/dx = cos(x²) × 2x
= 2x cos(x²)

Q5. If x² + y² = 25, find dy/dx using implicit differentiation.

Answer: Differentiate both sides with respect to x, remembering that y is a function of x:
2x + 2y (dy/dx) = 0
2y (dy/dx) = −2x
dy/dx = −x/y
(Here we did not need to solve for y first — that is the advantage of implicit differentiation.)

Q6. Find dy/dx if x = a cosθ and y = a sinθ.

Answer: This is a parametric form, so we use dy/dx = (dy/dθ) ÷ (dx/dθ).
dx/dθ = −a sinθ
dy/dθ = a cosθ
dy/dx = (a cosθ) / (−a sinθ)
= −cosθ/sinθ
= −cotθ

Q7. Write the derivatives of the inverse trigonometric functions sin⁻¹x, cos⁻¹x and tan⁻¹x.

Answer: • d/dx (sin⁻¹x) = 1/√(1 − x²), for −1 < x < 1
• d/dx (cos⁻¹x) = −1/√(1 − x²), for −1 < x < 1
• d/dx (tan⁻¹x) = 1/(1 + x²), for all real x
Also useful: d/dx (cot⁻¹x) = −1/(1 + x²), d/dx (sec⁻¹x) = 1/(|x|√(x² − 1)) and d/dx (cosec⁻¹x) = −1/(|x|√(x² − 1)).

Q8. Find the second order derivative of y = x³ + 2x² − 5x + 3.

Answer: First derivative:
dy/dx = 3x² + 4x − 5
Second derivative: Differentiate again.
d²y/dx² = 6x + 4
So the second order derivative is 6x + 4.

Q9. Find the value of k for which f(x) = kx + 1 for x ≤ 5 and f(x) = 3x − 5 for x > 5 is continuous at x = 5.

Answer: For continuity at x = 5, the left limit, right limit and f(5) must all be equal.
Left limit and f(5): k(5) + 1 = 5k + 1
Right limit: 3(5) − 5 = 10
Setting them equal:
5k + 1 = 10
5k = 9
k = 9/5

Q10. State Rolle's theorem and the Lagrange Mean Value Theorem.

Answer: Rolle's theorem: If a function f is
(i) continuous on the closed interval [a, b],
(ii) differentiable on the open interval (a, b), and
(iii) f(a) = f(b),
then there exists at least one point c in (a, b) such that f′(c) = 0.
Mean Value Theorem: If f satisfies conditions (i) and (ii) above, then there exists at least one c in (a, b) such that
f′(c) = [f(b) − f(a)] / (b − a)
Rolle's theorem is the special case of the Mean Value Theorem when f(a) = f(b).

Q11. Differentiate y = e^x log x with respect to x.

Answer: This is a product of two functions, so we use the product rule.
Let u = e^x and v = log x.
du/dx = e^x and dv/dx = 1/x
dy/dx = u (dv/dx) + v (du/dx)
= e^x (1/x) + log x (e^x)
= e^x (1/x + log x)

Q12. If y = tan⁻¹[(1 − x)/(1 + x)], find dy/dx.

Answer: Use the standard result tan⁻¹[(1 − x)/(1 + x)] = π/4 − tan⁻¹x.
This comes from the formula tan⁻¹A − tan⁻¹B = tan⁻¹[(A − B)/(1 + AB)] with A = 1.
So y = π/4 − tan⁻¹x
Differentiating, and remembering that π/4 is a constant:
dy/dx = 0 − 1/(1 + x²)
= −1/(1 + x²)
(Simplifying the expression first made the differentiation far easier than using the quotient rule directly.)

Related Study Material on Extramarks

Q1.

$If y = \log_{cosx} sinx, then \frac{dy}{dx} is equal to$

Opt.

$\frac{cotx . logcosx ? tanx . logxsinx}{{(logcosx)}^{2}}$
$\frac{? tanxlogsinx ? logcosxcotx}{{(logsinx)}^{2}}$
$\frac{tanxlogsinx ? logcosxcotx}{{(logsinx)}^{2}}$
$\frac{tanxlogsinx + logcosxcotx}{{(logsinx)}^{2}}$

Ans.

$\frac{cotx . logcosx tanx . logxsinx}{{(logcosx)}^{2}}$

Q2.

$I f (x) = (2 x ? 1) \tan^{1} (e^{2 x}), then f ‘ (0) is equal to$

Opt.

$\frac{?}{2}$
$\frac{?}{2} + 1$
$\frac{?}{2} - 1$
$\frac{?}{6} + 5$

Ans.

Q3.

$Find \frac{dy}{dx} if y = \sin (\tan^{- 1} e^{- x}) .$

Opt.

$\begin{array}{l} Here y = \sin (\tan^{1} e^{1}) \\ differentiating y w . r . t x \\ \frac{dy}{dx} = \frac{d}{dx} [\sin (\tan^{1} e^{1})] \end{array}$

Ans.

$\begin{array}{l} Here y = \sin (\tan^{1} e^{1}) \\ differentiating y w . r . t x \\ \frac{dy}{dx} = \frac{d}{dx} [\sin (\tan^{1} e^{1})] \\ = \cos (\tan^{1} e^{1}) \frac{d}{dx} (\tan^{1} e^{x}) \\ = \cos (\tan^{1} e^{1}) . \frac{1}{1 + {(e^{x})}^{2}} \frac{d}{dx} (e^{x}) \\ = \cos (\tan^{1} e^{x}) . \frac{1}{1 + {(e^{x})}^{2}} . e^{x} \frac{d}{dx} (x) \\ = \cos (\tan^{1} e^{x}) . \frac{1}{1 + {(e^{x})}^{2}} . e^{x} . (1) \\ = \frac{e^{x} \cos (\tan^{1} e^{x})}{1 + e^{2 x}} \end{array}$

Q4.

$If y = {(\tan^{1} x)}^{2}, ? show that {(x^{2} + 1)}^{2} y_{2} + 2 x (x^{2} + 1) y_{1} ? = ? 2$

Opt.

$\begin{array}{l} Here, y = {(\tan^{1} x)}^{2} \\ differentiating y w . r . t . x \\ y_{1} = \frac{dy}{dx} = \frac{d}{dx} {(\tan^{1} x)}^{2} \end{array}$

Ans.

$\begin{array}{l} Here, y = {(\tan^{1} x)}^{2} \\ differentiating y w . r . t . x \\ y_{1} = \frac{dy}{dx} = \frac{d}{dx} {(\tan^{1} x)}^{2} \\ = 2 (\tan^{1} x) \frac{d}{dx} (\tan^{1} x) \\ = \frac{2 \tan^{1} x}{x^{2} + 1} \\ (x^{2} + 1) y_{1} = 2 \tan^{1} x \\ again differentiating w . r . t . x \\ \frac{d}{dx} [(x^{2} + 1) y_{1}] = \frac{d}{dx} (2 \tan^{1} x) \\ (x^{2} + 1) \frac{d}{dx} y_{1} + y_{1} \frac{d}{dx} (x^{2} + 1) = \frac{d}{dx} (2 \tan^{1} x) \\ (x^{2} + 1) y_{2} + y_{1} (2 x) = \frac{2}{x^{2} + 1} [as y_{2} = \frac{d^{2 y}}{{dx}^{2}}] \\ {(x^{2} + 1)}^{2} y_{2} + y_{1} (x^{2} + 1) (2 x) = 2 \\ {(x^{2} + 1)}^{2} y_{2} + 2 x (x^{2} + 1) y_{1} = 2 . \end{array}$

CBSE Important Questions Class 12 Maths Chapter 5 – Continuity and Differentiability

Related Study Material on Extramarks

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