RBSE Class 12 Maths Chapter 6 – Continuity and Differentiability Important questions and solutions are given here. All these important questions on functions contain detailed explanations. The RBSE Class 12 important questions and solutions available at BYJU’S cover all the concepts of Class 12, so that students can get a complete idea on the syllabus and important concepts for exams.

Chapter 6 of RBSE Class 12 involves the concepts of continuity of the function and differentiability of the function with two exercises of practice questions. Students will be introduced with the various strategies for testing the continuity of different types of functions as well as differentiability in this chapter.

### RBSE Maths Chapter 6: Exercise 6.1 Textbook Important Questions and Solutions

**Question 1: Examine the continuity of the following function.**

**\(f(x)=\left\{\begin{matrix} 1+x; &x\le 3 \\ 7-x; & x>3 \end{matrix}\right.\)**

**at x = 3.**

**Solution:**

Given function is:

\(f(x)=\left\{\begin{matrix} 1+x; &x\le 3 \\ 7-x; & x>3 \end{matrix}\right.\)Left hand limit:

f(3 – 0) = lim_{h→0} f(3 – h)

= lim_{h→0} 1 + (3 – h)

= 1 + (3 – 0)

= 1 + 3

= 4

Right hand limit

f(3 + 0) = lim_{h→0} f(3 + h)

= lim_{h→0} 7 – (3 + h)

= 7 – (3 + 0)

= 7 – 3

= 4

At x = 3,

f(3) = 1 + x = 1 + 3 = 4

Thus, f(3 – 0) = f(3 + 0) = f(3)

Hence, the given function is continuous at x = 3.

**Question 2: Examine the continuity of f(x) = x – [x] at x = 3.**

**Solution:**

Given function is:

f(x) = x – [x], at x = 3

Left hand limit

f(3 – 0) = lim_{h→0} f(3 – h)

= lim_{h→0} (3 – h) – [3 – h]

= 3 – 2 (since 2 is just before the greatest integer 3 here)

= 1

Right hand limit

f(3 + 0) = lim_{h→0} f(3 + h)

= lim_{h→0} (3 + h) – [3 + h]

= 3 – 3

= 0

∴ f(3 – 0) ≠ f(3 + 0)

Hence, the given function is not continuous at x = 3.

**Question 3: Examine the continuity of the following function in [-1, 2].**

**\(f(x)=\left\{\begin{matrix} -x^2; &-1\le x<0 \\ 4x-3;& 0<x\le 1\\ 5x^2-4x; & 1<x\le 2 \end{matrix}\right.\)**

**Solution:**

Given function is:

\(f(x)=\left\{\begin{matrix} -x^2; &-1\le x<0 \\ 4x-3;& 0<x\le 1\\ 5x^2-4x; & 1<x\le 2 \end{matrix}\right.\)Now, we will test the continuity of the given function at x = 0 and 0 ∈ [-1,2].

Left hand limit

f(0 – 0) = lim_{h→0} f(0 – h)

= lim_{h→0} (0 – h)^{2}

= lim_{h→0} h^{2}

= 0

Right hand limit

f (0 + 0) = lim_{h→0} f(0 + h)

= lim_{h→0} 4(0 + h) – 3

= lim_{h→0} 4h – 3

= 0 – 3

= – 3

∴ f(0 – 0) ≠ f(0 + 0)

LHL ≠ RHL

Hence, the function is not continuous at x = 0 and x ∈ [- 1, 2]

At x = 1

Left hand limit

f(1 – h) = lim_{h→0} 4(1 – h) – 3

= lim_{h→0} 4 – 3 – 4h

= 4 – 3 – 0

= 1

Right hand limit

f(1 + h) = lim_{h→0} 5(1 + h)^{2} – 4(1 + h)

= lim_{h→0} 5(1 + h^{2} + 2h) – (4 + Ah)

= lim_{h→0} 5h^{2} +10h + 5 – 4 – 4h

= 5 × 0 + 10 × 0 + 1 – 4(0)

= 1

Value of function at x = 1

f(1) = 4 × 1 – 3 = 4 – 3 = 1

∵ lim_{h→0} f(1 – h) = f(1) = lim_{h→0} f(1 + h)

Thus, the function is continuous at x = 1.

Therefore, the given function is not continuous in interval [-1, 2]

### RBSE Maths Chapter 6: Exercise 6.2 Textbook Important Questions and Solutions

**Question 4: Prove that the function f(x) = |x| is not differentiable at x = 0.**

**Solution:**

Given function is:

f(x) = |x|

Left hand derivative

f'(0 – 0) = lim_{h → 0} [f(0 – h) – f(0)]/(-h)]

= lim_{h → 0} [|0 – h| – |0|]/(-h)

= lim_{h → 0} [|-h| – 0]/(-h)

= lim_{h → 0} [h/(-h)]

= lim_{h → 0} (-1)

= -1

Right hand derivative

f'(0 + 0) = lim_{h → 0} [f(0 + h) – f(0)]/(h)]

= lim_{h → 0} [|0 + h| – |0|]/h

= lim_{h → 0} [|h| – 0]/(h)

= lim_{h → 0} (h/h)

= lim_{h → 0} (1)

= 1

Thus, f'(0 – 0) ≠ f'(0 + 0)

Therefore, f(x) is not differentiable at x = 0.

**Question 5: Examine the differentiability of the function f(x) = |x – 1| + |x – 2| in [0, 2].**

**Solution:**

Given function is:

f(x) = |x – 1| + |x – 2|

This can be written as \(f(x)=\left\{\begin{matrix} 3-2x; & 0\le x<1\\ 1; & 1\le x < 2\\ 2x-3; & x\ge 2 \end{matrix}\right.\)

Now, we will test the differentiability of the function at x = 1, 1 ∈ [0, 2]

For differentiability of f(x) at x = 1

Left hand derivative

f'(1 – 0) = lim_{h→0} [F(1 – h) – f(1)]/(-h)

= lim_{h→0} [3 – 2(1 – h) – 1]/(-h)

= lim_{h→0} [3 – 2 + 2h – 1]/(-h)

= lim_{h→0} [2h/(-h)]

= lim_{h→0} (-2)

= -2

Right hand derivative

f'(1 + 0) = lim_{h→0} [f(1 + h) – f(1)]/h

= lim_{h→0} (1 – 1)/h

= lim_{h→0} = (0/h)

= lim_{h→0} (0)

= 0

f'(1 – 0) ≠ f'(1 + 0)

Thus, f(x) is not differentiable at x = 1.

Hence, the given function is not differentiable in the interval [0, 2].

**Question 6: Examine the differentiability of the function \(f(x)=\left\{\begin{matrix} xtan^{-1}x; & x\ne 0\\ 0; & x=0 \end{matrix}\right.\)**

**Solution:**

Given function is:

\(f(x)=\left\{\begin{matrix} xtan^{-1}x; & x\ne 0\\ 0; & x=0 \end{matrix}\right.\)Left hand derivative

f'(0 – 0) = lim_{h→0} [f(0 – h) – f(0)]/(-h)

= lim_{h→0} [(0 – h)tan^{-1}(0 – h) – 0]/(-h)

= lim_{h→0} [(-h)tan^{-1}(-h)]/(-h)

= lim_{h→0} tan^{-1}(-h)

= 0

Right hand derivative

f'(0 + 0) = lim_{h→0} [f(0 + h) – f(0)]/(h)

= lim_{h→0} [(0 + h)tan^{-1}(0 + h) – 0]/(h)

= lim_{h→0} [(h)tan^{-1}(h)]/(h)

= lim_{h→0} tan^{-1}(h)

= 0

Thus, f'(0 – 0) = f'(0 + 0)

Hence, the given function is differentiable at x = 0.

**Question 7: Find the value of m and n if \(f(x)=\left\{\begin{matrix} x^2+3x+m; & x\le 1\\ nx+2; & x>1 \end{matrix}\right.\) is differentiable at every point.**

**Solution:**

Given function is:

\(f(x)=\left\{\begin{matrix} x^2+3x+m; & x\le 1\\ nx+2; & x>1 \end{matrix}\right.\)Given that f(x) is differentiable at x = 1.

We know that every differentiable function is also continuous.

Thus, the given function is continuous at x = 1.

Left hand limit

f(1 – 0) = lim_{h→0} f(1 – h)

= lim_{h→0} (1 + h)^{2} + 3(1 – h) + m

= (1 – 0)^{2} + 3(1 – 0) + m

= 1 + 3 + m

= 4 + m

Right hand limit

f(1 + 0) = lim_{h→0} f(1 + h)

= = lim_{h→0} n(1 + h) + 2

= n(1 + 0) + 2

= n + 2

Therefore, The given function is continuous at x = 1.

Thus, f(1 – 0) = f(1 + 0)

4 + m = n + 2

m – n = 2 – 4

m – n = -2 ….(i)

Again, at x = 1, f(x) is differentiable.

Left hand derivative

f'(1 – 0) = lim_{h→0} [f(1 – h) – f(1)]/(-h)

= lim_{h→0} {[(1 – h)^{2} + 3(1 – h) + m] – [(1)^{2} + 3(1) + m]}/(-h)

= lim_{h→0} [1 + h^{2} – 2h + 3 – 3h + m – 1 – 3 – m]/(-h)

= lim_{h→0} [(h^{2} – 5h + 4 + m – 4 – m)/(-h)]

= lim_{h→0} [-h(5 – h)/(-h)]

= lim_{h→0} (5 – h)

= 5 – 0

= 5

Right hand derivative

f'(1 + 0) = lim_{h→0} [f(1 + h) – f(1)]/h

= lim_{h→0} {[n(1 + h) + 2] – [(1)^{2} + 3(1) + m]}/h

= lim_{h→0} [(n + nh + 2) – (1 + 3 + m)]/h

= lim_{h→0} (n + nh + 2 – 4 – m)/h

= lim_{h→0} (2 + nh + 2 – 4)/h

= n [From (i)]

Therefore, the function is differentiable at x = 1.

Thus, f'(1 – 0) = f'(1 + 0)

5 = n

⇒ n = 5

Substituting n = 5 in (i),

m – 5 = – 2

⇒ m = – 2 + 5

⇒ m = 3

Hence, m = 3 and n = 5.

### RBSE Maths Chapter 6: Additional Important Questions and Solutions

**Question 1: If f(x) = (x ^{2} – 9)/(x – 3) is continuous at x = 3, then the value of f(3) will be**

**(a) 6**

**(b) 3**

**(c) 1**

**(d) 0**

**Solution:**

Correct answer: (a)

Given function is:

f(x) = (x^{2} – 9)/(x – 3)

Right hand limit

f(3 + 0) = lim_{h→0} f(3 + h)

= lim_{h→0} [(3 + h)^{2} – 9]/(3 + h – 3)

= lim_{h→0} [(9 + 6h + h^{2} – 9)/h]

= lim_{h→0} [h(6 + h)/h]

= lim_{h→0} (6 + h)

= 6

The function is continuous at x = 3, then f(3) = f(3 + 0)

f(3) = 6

**Question 2: If f(x) = cot x is not continuous at x = nπ/2 when**

**(a) n ∈ Z**

**(b) n ∈ N**

**(c) n/2 ∈ Z**

**(d) only n = 0**

**Solution:**

Correct answer: (c)

We know that the function is not continuous at x = nπ/2.

lim_{x→nπ/2} cot x will not exist.

⇒ cot(nπ/2) = cot[(n/2)π]

Thus, n/2 ∈ Z.

**Question 3: The set of those points on f(x) = x|x|, where the function is differentiable**

**(a) (0, ∞)**

**(b) (-∞, ∞)**

**(c) (-∞, 0)**

**(d) (-∞, 0) U (0, ∞)**

**Solution:**

Correct answer: (b)

Given function is:

f(x) = x|x|

This can be written as:

\(f(x)=\left\{\begin{matrix} x^2; & x\ge 0\\ -x^2; & x<0 \end{matrix}\right.\)At x = 0,

Left hand derivative

f'(0 – 0) = lim_{h→0} [f(0 – h) – f(0)]/(-h)

= lim_{h→0} [(0 – h)^{2} – (0)^{2}]/(-h)

= lim_{h→0} [h^{2}/(-h)]

= lim_{h→0} (-h)

= 0

Right hand derivative

f'(0 + 0) = lim_{h→0} [f(0 0 h) – f(0)]/(h)

= lim_{h→0} [(0 + h)^{2} – (0)^{2}]/(h)

= lim_{h→0} [h^{2}/h)]

= lim_{h→0} (h)

= 0

f'(0 – 0) = f'(0 + 0)

Hence, the given function is differentiable at x = 0.

Therefore, the function is differentiable in (-∞, ∞).

**Question 4: Examine the continuity of the following function.**

**f(x) = |sin x| + |cos x| + |x|, ∀ x ∈ R**

**Solution:**

Given function is:

f(x) = |sin x| + |cos x| + |x|, ∀ x ∈ R

Let x ∈ c be any arbitrary constant, such that, at x = c check for the continuity of f(x).

Left hand limit

f(c – 0) = lim_{h→0} f(c – h)

= lim_{h→0} |sin (c – h)| + |cos (c – h)| + |(c – h)|

= |sin (c – 0)| + |cos (c – 0)| + |(c – 0)|

= |sin c| + |cos c| + |c|

= sin c + cos c + c

Right hand limit

f(c + 0) = lim_{h→0} f(c + h)

= lim_{h→0} |sin (c + h)| + |cos (c + h)| + |(c + h)|

= |sin (c + 0)| + |cos (c + 0)| + |(c + 0)|

= |sin c| + |cos c| + |c|

= sin c + cos c + c

At x = c, the value of the given function is

f(c) = |sin c| + |cos c| + |c|

= sin c + cos c + c

Therefore, f(c – 0) = f(c + 0) + f(c)

Thus, f(x) is continuous at x = 0.

Hence, the function f(x) is continuous everywhere in R.