## 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) = x^{3} 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:

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