# RD Sharma Class 12 Ex 9.1 Solutions Chapter 9 Continuity

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## RD Sharma Class 12 Ex 9.1 Solutions Chapter 9 Continuity

### Solution:

Given that

Now, let us consider LHL at x = 0

Now, let us consider RHL at x = 0

So, LHL ≠ RHL

Therefore, f(x) is discontinuous at origin and the discontinuity is of 1st kind.

### Question 2. A function f(x) is defined as . Show that f(x) is continuous at x = 3.

Solution:

Given that

So, here we check the given f(x) is continuous at x = 3,

Now, let us consider LHL at x = 3

Now, let us consider RHL at x = 3

So, f(3) = 5

LHL= RHL = f(3)

Therefore, f(x) is continuous at x = 3

### Show that f(x) is continuous at x = 3.

Solution:

Given that

So, here we check the given f(x) is continuous at x = 3,

Now, let us consider LHL at x = 3

Now, let us consider RHL at x = 3

So, f(3) = 6

LHL= RHL= f(3)

Therefore, f(x) is continuous at x = 3

### Find whether f(x) is continuous at x = 1

Solution:

Given that

So, here we check the given f(x) is continuous at x = 1,

Now, let us consider LHL at x = 1

Now, let us consider RHL at x = 1

So, f(1) = 2

LHL= RHL = f(1)

Therefore, f(x) is continuous at x = 1

### Find whether f(x) is continuous at x = 0.

Solution:

Given that

So, here we check the given f(x) is continuous at x = 0,

Now, let us consider LHL at x = 0

Now, let us consider RHL at x = 0

So, f(0) = 1

LHL = RHL≠ f(0)

Therefore, f(x) is discontinuous at x = 0.

### Find whether f is continuous at x = 0.

Solution:

Given that

So, here we check the given f(x) is continuous at x = 0,

Now, let us consider LHL at x = 0

Now, let us consider RHL at x = 0

So, LHL≠ RHL

Therefore, the f(x) is discontinuous at x = 0.

### Show that f(x) is discontinuous at x = 0.

Solution:

Given that

So, here we check the given f(x) is discontinuous at x = 0,

Now, let us consider LHL at x = 0

= 2 × 1/4 = 1/2

Now, let us consider RHL at x = 0

= 2 × 1/4 = 1/2

f(0) = 1

LHL= RHL ≠ f(0)

Therefore, the f(x) is discontinuous at x = 0.

### Question 8. Show that is discontinuous at x = 0.

Solution:

Given that

So, here we check the given f(x) is discontinuous at x = 0,

Now, let us consider LHL at x = 0

Now, let us consider RHL at x = 0

f(0) = 2

Thus, LHL= RHL≠ f(0)

Therefore, f(x) is discontinuous at x = 0.

### Question 9. Show that is discontinuous at x = a.

Solution:

Given that

So, here we check the given f(x) is discontinuous at x = a,

Now, let us consider LHL at x = a

Now, let us consider RHL at x = a

Thus, LHS ≠ RHL

Therefore, the f(x) is discontinuous at x = a.

### Question 10 (i). Solution:

Given that

So, here we check the continuity of the given f(x) at x = 0,

Let us consider LHL,

Now, let us consider RHL,

f(0) = 0

Thus, LHL= RHL= f(0) = 0

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

### Question 10 (ii). at x = 0

Solution:

Given that

So, here we check the continuity of the given f(x) at x = 0,

Let us consider LHL,

Now, let us consider RHL,

f(0) = 0

Thus, LHL= RHL = f(0) = 0

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

### Question 10 (iii). at x = a

Solution:

Given that

So, here we check the continuity of the given f(x) at x = a,

Let us consider LHL,

Now, let us consider RHL,

f(a) = 0

Thus, LHL= RHL= f(a) = 0

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

### Question 10 (iv). at x = 0

Solution:

Given that

So, here we check the continuity of the given f(x) at x = 0,

= 1/2 × 1/1 = 1/2

And,

f(0) = 7 ≠ f(0)

Therefore, f(x) is discontinuous at x = 0.

### Question 10 (v). n ∈ N at x = 1

Solution:

Given that

So, here we check the continuity of the given f(x) at x = 1,

Let us consider LHL,

Now, let us consider RHL,

f(1) = n – 1

Thus, LHL = RHL ≠ f(1)

Therefore, f(x) is discontinuous at x = 1.

### Question 10 (vi). at x = 1

Solution:

Given that

So, here we check the continuity of the given f(x) at x = 1,

Let us consider LHL,

Now, let us consider RHL,

f(1) = 2

LHL= RHL = f(1) = 2

Therefore, f(x) is discontinuous at x = 1.

### Question 10 (vii). at x = 0

Solution:

Given that

So, here we check the continuity of the given f(x) at x = 0,

Let us consider LHL,

Let us consider RHL,

Thus, LHL ≠ RHL

Therefore, f(x) is discontinuous at x = 0.

### Question 10 (viii). at x = a

Solution:

Given that,

f(x) = (x – a)sin{1/(x – a)}, x > 0

= (x – a)sin{1/(x – a)}, x < 0

= 0, x = a

Let us consider LHL,

Now, let us consider RHL,

⇒ Therefore, f(x) is continuous at x = a.

### Question 11. Show that Solution:

Given that,

So, here we check the given f(x) is discontinuous at x = 1,

Let us consider LHL,

Now, let us consider RHL,

LHL ≠ RHL

Therefore, f(x) is discontinuous at x = 1.

### Question 12. Show that Solution:

Given that,

So, here we check the given f(x) is continuous at x = 0,

Let us consider LHL,

Let us consider RHL,

f(0) = 3/2

Thus, LHL = RHL = f(0) = 3/2

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

### Also sketch the graph of this function.

Solution:

Given that,

So, here we check the continuity of the given f(x) at x = 0,

Let us consider LHL,

Now, let us consider RHL,

LhL ≠ RHL

So, the f(x) is discontinuous.

### at the point x = 0.

Solution:

Given that,

So, here we check the continuity of the given f(x) at x = 0,

Let us consider LHL,

Now, let us consider RHL,

f(0) = 1

LHL = RHL ≠ f(0)

Hence, the f(x) is discontinuous at x = 0.

### at the point x = 1/2.

Solution:

Given that,

So, here we check the continuity of the given f(x) at x = 1/2,

Let us consider LHL,

Now, let us consider RHL,

f(1/2) = 1/2

Thus, LHL= RHL = f(1/2) = 1/2

Hence, the f(x) is continuous at x = 1/2.

### Question 17. Discuss the continuity of at the point x = 0.

Solution:

Given that,

So, here we check the continuity of the given f(x) at x = 10,

Let us consider LHL,

Now, let us consider RHL,

Thus, LHL ≠ RHL

Hence, the f(x) is discontinuous at x = 0.

### Question 18. For what value of k is the function continuous at x = 1 ?

Solution:

Given that,

Also, f(x) is continuous at x = 1

So,

LHL = RHL = f(1)        ……(i)

Let us consider LHL,

f(1) = k

From eq(i), we get

LHL = F(1)

Therefore, k = 2

### continuous at x = 1.

Solution:

Given that,

Also, f(x) is continuous at x = 1

So, LHL = RHL = f(1)        …..(i)

Let us consider LHL,

f(1) = k

From eq(i), we get

LHL = F(1)

Therefore, k = -1

### Question 20. For what value of k is the function continuous at x = 0 ?

Solution:

Given that,

Also, f(x) is continuous at x = 0

So, LHL = RHL = f(0)        …..(i)

Let us consider LHL,

f(0) = k

Thus, from eq(i), we get

k = 5/3

Therefore, k = 5/3

### is continuous at x = 0.

Solution:

Given that,

Also, f(x) is continuous at x = 0

So, LHL = RHL = f(0)        ….(i)

Let us consider LHL,

f(0) = k

From eq(i), we get

k = 2/5

### Question 23. Find the values of a so that the function Solution:

Given that,

Also, f(x) is continuous at x = 2

So, LHL = RHL = f(2)        …….(i)

Let us consider LHL,

= 2a + 5

Now, let us consider RHL,

From eq(i), we get

2a + 5 = 1

⇒ a = -2

### remains discontinuous at x = 0, regardless the choice of k.

Solution:

Given that,

We have, at x = 0

Let us consider LHL,

f(0) = k

Now, let us consider RHL,

Since, LHL ≠ RHL,

Therefore, f(x) will remain discontinuous at x = 0, regardless the value of k.

### Solution:

Given that,

Also, f(x) is continuous at x = π/2

LHL = RHL

⇒ ⇒ ⇒ ⇒ ⇒ k/2 = 3

⇒ k = 6

### is continuous at x = 0.

Solution:

Given that,

Also, f(x) is continuous at x = 0

So, LHL = RHL = f(0)        …..(i)

f(0) = 0

Let us consider LHL,   = a + 1 + 1 = a + 2

Now, let us consider RHL,    From eq(i), we get

a + 2 = 1/2 ⇒ a = -3/2

c = 1/2 and b ∈ R -{0}

Hence, a = -3/2, b ∈ R -{0}, c =1/2

### Question 27. If is continuous at x = 0, find k.

Solution:

Given that,

Also, f(x) is continuous at x = 0

So, LHL = RHL = f(0)       …….(i)

f(0) = 1/2

Let us consider LHL,

= k2/2

Using eq(i) we get,

k2/2 = 1/2 ⇒ k = ±1

### Question 28. If Solution:

Given that,

Also, f(x) is continuous at x = 4

So, LHL = RHL = f(4)        ……(i)

f(4) = a + b     …..(ii)

Let us consider LHL,

= a – 1        ……(iii)

Now, let us consider RHL,

= b + 1           ……(iv)

From eq(i), we get

a – 1 = b + 1 ⇒ a – b = 2         …..(v)

From eq(ii) and eq(iii), we get

a + b = a – 1 ⇒ a – b = -1

From eq(ii) and (iv), we get

a + b = b + 1 ⇒ a = 1

Thus, a = 1 and b = -1

### continuous at x = 0 ?

Solution:

Given that,

Also, f(x) is continuous at x = 0

So, LHL = RHL = f(0)        …..(i)

f(0) = k

Let us consider LHL,

Using eq(i), we get

k = 2

### Question 30. Let f(x) = , x ≠ 0. Find the value of f at x = 0 so that f becomes continuous at x = 0.

Solution:

Given that,

f(x) = Also, f(x) is continuous at x = 0

So, LHL=RHL=f(0)        ….(i)

Let us consider LHL,

= 1/a + 1/b = (a + b)/ab

From eq(i), we get

f(0) = (a + b)/ab

### Question 31. If is continuous at x = 2, find k.

Solution:

Given that,

Also, f(x) is continuous at x = 2

So, LHL = RHL = f(2)        …..(i)

Now,

f(2) = k  ……(ii)

Let us consider LHL, ……(iii)

Using eq(i), (ii) and (iii), we get

k = 1/2

### Question 32. If is continuous at x = 0, find k.

Solution:

Given that,

Also, f(x) is continuous at x = 2

So, LHL = RHL

Now,

⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ -2 × 1 × (1 + 1) = k

⇒ k = -4

### Question 33. Extend the definition of the following by continuity f(x) = at the point x = π.

Solution:

Given that,

As we know that a f(x) is continuous at x = π if,

LHL = RHL = f(π)  ……(i)

Let us consider LHL,

= (2/5) × (49/4) = 49/10

Thus, from eq(i) we get,

f(π) = 49/10

Hence, f(x) is continuous at x = π

### Question 34. If f(x) = , x ≠ 0 is continuous at x = 0, then find f(0).

Solution:

Given that,

f(x) = Also, f(x) is continuous at x = 0

So, LHL = RHL = f(0)     ……(i)

Let us consider LHL,

From eq(i) we get,

f(0) = 1

### Question 35. Find the value of k for which is continuous at x = 0

Solution:

Given that,

Also, f(x) is continuous at x = 0

LHL = RHL = f(0)     …..(i)

f(0) = k

Let us consider LHL,

Thus, from eq(i) we get,

k = 1

### (i) at x = 0

Solution:

Given that,

Also, f(x) is continuous at x = 0

⇒ ⇒ ⇒ ⇒ 2k× 1 = 8

⇒ k= 4

⇒ k = ±2

### (ii) at x = 1

Solution:

Given that,

Also, f(x) is continuous at x = 1

⇒ Now, on putting x – 1 = y, we get

⇒ ⇒ ⇒ ⇒ ⇒ ⇒ (-2/π) × (1/1) = k

⇒ k = (-2/π)

### (iii) at x = 0

Solution:

Given that,

Also, f(x) is continuous at x = 0

Let us consider LHL, at x = 0

Let us consider RHL at x = 0

Hence, no value of k exists for which function is continuous at x = 0.

### (iv) Solution:

Given that,

Also, f(x) is continuous at x = π

Let us consider LHL

Let us consider RHL

cosπ = -1

As we know that f(x) is continuous at x = π, so

⇒ kπ + 1 = -1

⇒ k = (-2/π)

### (v) Solution:

Given that,

Also, f(x) is continuous at x = 5

Let us consider LHL

= 5k + 1

Let us consider RHL

= 10

As we know that f(x) is continuous at x = 5, so

⇒ 5k + 1 = 10

⇒ k = 9/5

### (vi) at x = 5

Solution:

Given that,

Also, f(x) is continuous at x = 5

So,

f(x) = (x– 25)/(x – 5), if x ≠ 5 & f(x) = k, if x = 5

⇒ f(x)= {(x – 5)(x+5)/(x-5)}, if x ≠ 5 & f(x) = k, if x = 5

⇒ f(x)= (x + 5), if x ≠ 5 & f(x) = k, if x = 5

As we know that f(x) is continuous at x = 5, so

⇒ ⇒ k = 5 + 5 = 10

### (vii) at x = 1

Solution:

Given that,

Also, f(x) is continuous at x = 1

Let us consider LHL

Let us consider RHL

= k

As we know that f(x) is continuous at x = 1, so

⇒ k = 4

### (viii) Solution:

Given that,

Also, f(x) is continuous at x = 0

Let us consider LHL

= 2k

Let us consider RHL

= 1

As we know that f(x) is continuous at x = 0, so

⇒ 2k = 1

⇒ k = 1/2

### (ix) at x = 2

Solution:

Given that,

Also, f(x) is continuous at x = 2

f(x)= , if x ≠ 2 & f(x) = k, if x = 2

⇒ f(x)= , if x ≠ 2 & f(x) = k, if x = 2

⇒ f(x)= , if x ≠ 2 & f(x) = k, if x = 2

⇒ f(x)= (x + 5), if x ≠ 2 & f(x) = k, if x = 2

As we know that f(x) is continuous at x = 2, so

⇒ ⇒ k = 2 + 5 = 7

### is continuous at x = 3 and x = 5.

Solution:

Given that,

Let us consider LHL at x = 3,

= 1

Let us consider RHL at x = 3,

= 3a + b

Let us consider LHL at x = 5,

= 5a + b

Let us consider RHL at x = 5,

= 7

It is given that f(x) is continuous at x = 3 and x = 5, then and ⇒ 1 = 3a + b …..(i)

and 5a + b = 7 …….(ii)

On solving eq(i) and (ii), we get

a = 3 and b = -8

### Question 38. If . Show that f is continuous at x = 1.

Solution:

Given that,

So,

Let us consider LHL at x = 1,

= 1/2

Let us consider RHL at x = 1,

= 2 – 3 + 3/2 = 1/2

Also,

f(1) = (1)2/2 = 1/2

LHL = RHL = f(1)

Hence, the f(x) is continuous at x = 1

### (i) f(x) = |x| + |x – 1| at x = 0, 1.

Solution:

Given that,

f(x) = |x| + |x – 1|

So, here we check the continuity of the given f(x) at x = 0,

Let us consider LHL at x = 0,

Let us consider RHL at x = 0,

Also,

f(0) = |0| + |0 – 1| = 0 + 1 = 1

LHL = RHL = f(0)

Now, we check the continuity of the given f(x) at x = 1,

Let us consider LHL at x = 1,

= 1

Let us consider RHL at x = 1

= 1

Also,

f(1) = |1| + |1 – 1| = 1 + 0 = 1

LHL = RHL = f(1)

Hence, f(x) is continuous at x = 0, 1.

### (ii) f(x) = |x – 1| + |x + 1| at x = -1, 1.

Solution:

Given that,

f(x) = |x – 1| + |x + 1| at x = -1, 1.

So, here we check the continuity of the given f(x) at x = -1,

Let us consider LHL at x = -1,

Let us consider RHL at x = -1,

Also,

f(-1) = |-1 – 1| + |-1 + 1| = |-2| = 2

LHL = RHL = f(-1)

Now, we check the continuity of the given f(x) at x = 1,

Let us consider LHL at x = 1,

= 2

= 2

Also,

f(1) = |1 + 1| + |1 – 1| = 2

LHL = RHL = f(1)

Hence, f(x) is continuous at x = -1, 1.

### Question 40. Prove that is discontinuous at x = 0.

Solution:

Prove that is discontinuous at x = 0.

Proof:

Let us consider LHL at x = 0,

Let us consider RHL at x = 0,

LHL ≠ RHL

Hence, f(x) is discontinuous at x = 0.

### Question 41. If then what should be the value of k so that f(x) is continuous at x = 0.

Solution:

Given that,

Let us consider LHL at x = 0,

= k

Let us consider RHL at x = 0,

= k

It is given that f(x) is continuous at x = 0.

LHL = RHL = f(0)

⇒ k can be any real number.

### Solution:

Given that,

We have,

Let us consider LHL at x = 2,

= 5

Let us consider RHL at x = 2,

= 5

Also,

f(2) = k

It is given that f(x) is continuous at x = 2.

LHL = RHL = f(2)

⇒ 5 = 5 = k

Hence, for k = 5, f(x) is continuous at x = 2.

### Question 44. Let Solution:

Given that,

Let us consider LHL at x = π/2

= 1/2

Let us consider RHL at x = π/2

= b/8 × 1

= b/8

Also,

f(π/2) = a

It is given that f(x) is continuous at x = π/2.

LHL = RHL = f(π/2)

So,

⇒ 1/2 = b/8 = a

⇒ a = 1/2 and b = 4