## 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} 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} 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 + 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 = 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],

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