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# Class 12 RD Sharma Solutions – Chapter 9 Continuity – Exercise 9.1 | Set 2

• Last Updated : 26 May, 2021

### 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

### continuous at x = 2.

Solution:

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

Then, f(2) = k(2)2 = 4k ⇒ ⇒ k × 22 = 3 = 4k

⇒ 4k = 3 = 4k

⇒ 4k = 3

⇒ k = 3/4

Hence, the value of k is 3/4

### 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 is continuous at x = 2.

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 continuous at x = 4, find a, b.

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

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