RBSE Solutions for Class 12 Maths Chapter 6 Continuity and Differentiability Ex 6.1

Rajasthan Board RBSE Class 12 Maths Chapter 6 Continuity and Differentiability Ex 6.1

Question 1.
Examine the following functions for continuity :

Solution:
(a) Given function.

Left hand limit

Right hand limit

Left hand limit

Right hand limit

Hence, given function is not continuous at x = 0.

and at x = 3
f (3) = 1 + x ⇒ 1 + 3 = 4
∴ f (3 – 0) = f(3 + 0) = f(3) = 4
Hence, given function is continuous at x = 3.

(d) Given function,

Left hand limit
f(0 – 0) = limh→0 f(0 – h)
= lim_h→0 sin (0 – h)
= lim_h→0 sin (- h)
= lim_h→0 (-sin h)
[∵ sin (-θ) = -sin θ]
= 0
Right hand limit

Hence, given function is continuous at x = 0.

(e) Given function,

Left hand limit

Right hand limit


Hence, given function is not continuous at x = 0.

(f) Given function,

Left hand limit

Right hand limit

Hence, given function is not continuous at x = a.

(g) Given function,

Left hand limit

Right hand limit

Hence, given function is continuous at x = a.

Question 2.
Test the continuity of f(x) at x = 3 if f(x) = x – [x].
Solution:
Given function 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
= 1
[∵ 2 is just before greatest integer 3]

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, given function at not continuous at x = 3.

Question 3.

continuous at x = 2, then find k.
Solution :
Left hand limit

Right hand limit

Question 4.

the continuity of f (x) is closed interval [- 1, 2].
Solution:
Here, we will test the continuity of 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 h²
= 0

Right hand limit
f (0 + 0) = lim_h→0 f(0 + h)
= lim_h→0 4(0 + h) – 3
= lim_h→0 Ah – 3
= 0 – 3
= – 3
∴ f (0 – 0) ≠ f(0 + 0)
LHL ≠ RHL
So, function is not continuous at x = 0 and x ∈ [- 1, 2]
At x = 1
Left hand limit
f(1 – h) = limh→0 4(1 – h) – 3
= lim_h→0 4 – 3 – Ah
=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 + h² + 2h) – (4 + Ah)
= lim_h→0 5h² +10h + 5 – 4 – 4h
= 5 × 0 + 10 × 0 + 1 – 4(0)
= 1
Value of function at x = 1
f(1) = 4 × 1 – 3 = 1
∵ lim_h→0 f(1 – A) = f(1) = lim_h→0 f(1 + h)
∴Function is continuous at x = 1.
Hence, function is not continuous in interval [-1, 2],

RBSE Solutions for Class 12 Maths