RBSE Solutions for Class 12 Maths Chapter 6 Continuity and Differentiability Miscellaneous Exercise

Rajasthan Board RBSE Class 12 Maths Chapter 6 Continuity and Differentiability Miscellaneous Exercise

Question 1.
If function f (x) = \(\frac { { x }^{ 2 }-9 }{ x-3 } \) is continuous at x = 3, then value of (3) will be :
(a) 6
(b) 3
(c) 1
(d) 0
Solution:

Function is continuous at x = 3, so
f(3) = f(3 + 0)
f(3) = 6
Hence, option (a) is correct.

Question 2.
If function f(x)

is continuous at x = 2 then value of m will be :
(a) 3
(b) 1/3
(c) 1
(d) 0
Solution:

Question 3.

is continuous at x = 0, then the value of k will be :
(a) 0
(b) m + n
(c) m – n
(d) m.n
Solution:

Question 4.

is continuous at x = 3 then the value of λ will be :
(a) 4
(b) 3
(c) 2
(d) 1
Solution:
Left hand limit
f (3 – 0) = lim_h→0 f (3 – h)
= lim_h→0 (3 – h) + λ
= (3 – 0) + λ
= 3 + λ
∵ At x = 3, function is continuous then
f (3 – 0) = f (3)
3 + λ = 4
⇒ λ = 4 – 3
λ = 1
Hence, option (d) is correct.

Question 5.
If f(x) = cot x is discontinuous at x = \(\frac { n\pi }{ 2 } \), then :
(a) n ∈ Z
(b) n ∈ N
(c) \(\frac { n }{ 2 } \) ∈Z
(d) Only n = 0
Solution:
∵ Function is not continuous at x = \(\frac { n\pi }{ 2 } \)

Question 6.
Function f(x) = x | x | is differentiable in interval:
(a) (0, ∞)
(b) (-∞, 0)
(c) (-∞, 0)
(d) (-∞, 0) ∩ (0, ∞)
Solution:
Given function can be written as

Hence, function is differentiable at x = 0, so function is differentiable in (-∞, ∞).
Hence, option (b) correct.

Question 7.
Which of the following function is not differentiable at x = 0 :
(a) x | x |
(b) tan x
(c) e-x
(d) x + | x |
Solution:

Question 8.
Function f(x)

then value of left hand derivative of f(x) at x = 2 is :
(a) -1
(b) 1
(c) -2
(d) 2
Solution:

Question 9.
Function f(x) = [x] is not differentiable :
(a) at every integer
(b) at every rational number
(c) at x = 0
(d) everywhere
Solution:
[x] is not continuous at all integers. We know that at every intiger discontinuous function is not differentiable.
Hence, correct option is (a).

Question 10.
Function

is differentiable at x = 0, then right derivative of f(x) at x = 0 is :
(a) -1
(b) 1
(c) 0
(d) infinite
Solution:

Question 11.
Examine the function
f (x) = | sin x | + | cos x | + | x |, ∀ X ∈ R for continuity.
Solution:
Let x ∈ c be any arbitrary constant, then at x = c for continuity of function f(x).

Question 12.
If function,

is continuous at x = 0, then find m.
Solution:
At x = 0,

Question 13.
Find m and n if following function is continuous

Solution:
For continuity at x = 2,
Left hand limit
f (2 – 0) = lim_h→0 f(2 – A)
= lim_h→0 [(2 – h)² + m(2 – h) + n]
= (2 – 0)² + m(2 – 0) + n
= 4 + 2m + n

Right hand limit
f(2 + 0) = lim_h→0 f(2 + h)
= lim_h→0 [4(2 + h) – 1 ]
= 4(2 + 0) – 1
= 8 – 1
= 7
At x = 2, value of f (x),
f(2) = 4 × 2 – 1 = 8 – 1 = 7
Given function is continuous at x = 2, then
f(2 – 0) = f(2 + 0) = f(2)
4 + 2m + n = 7 = 7
⇒ 4 + 2m + n = 7
⇒ 2m + n = 7 – 4
or
2m + n = 3
For continuity at x = 4,

Left hand limit
f (4 – 0) = lim_h→0 f (4-h)
= lim_h→0 [4(4 – h) – 1]
= 4 (4 – 0) – 1
= 16 – 1
= 15

Right hand limit
f (4 + 0) = lim_h→0 f (4 – h)
= lim_h→0 [m(4 + h)² + 17n]
= m (4 + 0)² + 17n
= 16m + 17n
At x = 4, value of function
f(4) = 4 × 4 – 1 = 16 – 1 = 15
Given function is continuous at x = 4,
f (4 – 0) = f (4 + 0) = f (4)
15 = 16m + 17n = 15
⇒ 16m + 17n = 15 ……(ii)
Putting n = 3 – 2m from (i) in
16m + 17(3 – 2m) = 15
16m + 51 – 34m = 15
– 18m = 15 – 51
– 18m = – 36
m = 2
From equation (i),
2 × 2 + n = 3
n = 3 – 4
Hence, m = 2,n = – 1.

Question 14.
Examine the function

for continuity at x = 0.
Solution :
Continuity at x = 0,

Question 15.
Examine the function

for continuity at x = 1 and 3.
Solution:
For continuity at x = 1

= | 1 + 0 – 3 |
= | – 2 |
= 2
Value of function f(x) at x = 1,
f(1) = | 1 – 3 |
= | -2 |
= 2
∴ f (1 – 0) = f(1 + 0) = f (1) = 2
So, f(x) is continue at f (x), x = 1
For continuity at x = 3

Left hand limit
f (3 – 0) = lim_h→0 f (3 – h)
= lim_h→0 – {(3-A)-3}
= lim_h→0 (3 – 3 + h)
= 0

Right hand limit
f(3 + 0) = lim_h→0 f(3 + h)
= lim_h→0 (3 + h – 3)
= 0
f(3) = 3 – 3 = 0
∴ f (3 – 0) = f(3 + 0)
= f(0) = o
Hence, function is continue at x = 3.

Question 16.
Find a, b, c if the function

is continuous at x = 0.
Solution:
At x = 0, given function is continuous then,

Question 17.
Test the continuity of the function

Solution:
At x = \(\frac { 4 }{ 3 } \), for continuity

Question 18.
Test the continuity of the function f(x) = | x | + | x – 1 | in interval [- 1, 2].
Solution:
Given, f(x) = | x | + | x – 1 |
It can be written as

Here, we will test of continuity at x = 0 and x = 1 only. Test of continuity at x = 0
Here, f(0) = 1

Left hand limit
f(0 – 0) = lim_h→0 f(0 – h)
= lim_h→0 1 – 2(0 – h)
= 1

Right hand limit
f(0 + 0) = lim_h→0 f(0 + h)
= 1
∴ f(0) = f (0 – c) = f (0 + 0)
So, at x = 0,f(x) is continuous
Now, at x = 1 test of continuity
Here, f(1) = 2 × 1 – 1 = 1

Left hand limit
f (1 – 0) = lim_h→0 f(1 – h)
= lim_h→0 1

Right hand limit
f (1 + 0) = lim_h→0 f(1 + h)
= lim_h→0 2(1 + h) – 1
= 2(1 + 0) – 1
= 2 – 1 = 1
∴ f (1) = f(1 – 0) = f(1 + 0)
So, at x = 1, f(x) is continuous.
∴ Function is continuous at x = 0 and x = 1
Hence, function is continuous in interval [-1,2].

Question 19.
Find the value of f(0) if function

is continuous at x = 0.
Solution:
Function is continuous at x = 0

Question 20.
Examine the function fix) for continuity at

Solution:

Question 21.
For which value of x, f(x) = sin x is differentiable.
Solution:
Given function f(x) = sin x
Domain (f) = R
Let a ∈ R be any arbitrary real number.
Then at x = a,

Question 22.
Examine the function f(x) for differentiablity

Solution:

∵ f’ (0 – 0) = f'(0 + 0)
Hence, At x = 0 function is differentiable,
∵ x ∈ R, then for every x ∈ R function will be differentiable and f'(0) = 0.

Question 23.
Examine the function for differentiability at x = a if

Solution:

Question 24.
Prove that the function

is not differentiable at x = 1.
Solution:

Question 25.
Examine the function for differentiability at x = 0, if

Solution:

Question 26.
Show that the function

is differentiable at x = 0
Solution:

Question 27.
Check the differentiablity of function f(JC) = |x – 2| + 2 | x – 3 | in the interval [1, 3].
Solution:
Given function can be written as :


Hence, at x = 2, function is not differentiable. So, we can
say that this function is not differentiable in interval [1, 3],

Question 28.
If flnction f(x) = x³ is differentiable at x = 2, then find f’ (2).
Solution:
At, x = 2,

Hence, at x = 2, function is differentiable,
and f’ (2) = 12.

Question 29.
Show that the greatest integer function f(x) = [x] is not differentiable at x = 2
Solution:
At x = 2,

Question 30.

Solution:

RBSE Solutions for Class 12 Maths