# Class 12 RD Sharma Solutions – Chapter 9 Continuity – Exercise 9.1 | Set 3

### 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,

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’ -2 Ã— 1 Ã— (1 + 1) = k

â‡’ k = -4

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

Solution:

Given that,

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

LHL = RHL = f(Ï€)  ……(i)

Let us consider LHL,

= (2/5) Ã— (49/4) = 49/10

Thus, from eq(i) we get,

f(Ï€) = 49/10

Hence, f(x) is continuous at x = Ï€

### Question 34. If f(x) = , x â‰  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

â‡’

â‡’

â‡’

â‡’ 2k2 Ã— 1 = 8

â‡’ k2 = 4

â‡’ k = Â±2

### (ii)  at x = 1

Solution:

Given that,

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

â‡’

Now, on putting x – 1 = y, we get

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’ (-2/Ï€) Ã— (1/1) = k

â‡’ k = (-2/Ï€)

### (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) at x = Ï€

Solution:

Given that,

Also, f(x) is continuous at x = Ï€

Let us consider LHL

Let us consider RHL

cosÏ€ = -1

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

â‡’ kÏ€ + 1 = -1

â‡’ k = (-2/Ï€)

### (v) at x = 5

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

â‡’ 5k + 1 = 10

â‡’ k = 9/5

### (vi)  at x = 5

Solution:

Given that,

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

So,

f(x) = (x2 – 25)/(x – 5), if x â‰  5 & f(x) = k, if x = 5

â‡’ f(x)= {(x – 5)(x+5)/(x-5)}, if x â‰  5 & f(x) = k, if x = 5

â‡’ f(x)= (x + 5), if x â‰  5 & f(x) = k, if x = 5

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

â‡’

â‡’ 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

â‡’ k = 4

### (viii)  at x = 0

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

â‡’ 2k = 1

â‡’ k = 1/2

### (ix)  at x = 2

Solution:

Given that,

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

f(x)= , if x â‰  2 & f(x) = k, if x = 2

â‡’ f(x)=  , if x â‰  2 & f(x) = k, if x = 2

â‡’ f(x)=  , if x â‰  2 & f(x) = k, if x = 2

â‡’ f(x)= (x + 5), if x â‰  2 & f(x) = k, if x = 2

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

â‡’

â‡’ 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

â‡’ 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 â‰  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)

â‡’

k can be any real number.

### continuous at x = 0 ? What about continuity at x = Â±1?

Solution:

Given that,

Check for x = 0,

Hence, there is no value of Î» for which f(x) is continuous at x = 0.

Now for x = 1,

f(1) = 4x + 1 = 4 Ã— 1 + 1 = 5

Hence, for any values of Î», f is continuous at x = 1.

Now for x = -1,

f(-1) = Î»(1 + 2)= 3Î»

Hence, for any values of Î», f is continuous at x=-1.

### Question 43. For what values of k is the following function continuous at x = 2?

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)

â‡’ 5 = 5 = k

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

### Question 44. Let  If f(x) is continuous at x = (Ï€/2), find a and b.

Solution:

Given that,

Let us consider LHL at x = Ï€/2

= 1/2

Let us consider RHL at x = Ï€/2

= b/8 Ã— 1

= b/8

Also,

f(Ï€/2) = a

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

LHL = RHL = f(Ï€/2)

So,

â‡’ 1/2 = b/8 = a

â‡’ a = 1/2 and b = 4

### Question 45. If the functions f(x), defined below is continuous at x = 0, find the value of k,

Solution:

Given that,

Let us consider LHL at x = 0,

= 1 Ã— 1

Let us consider RHL at x = 0,

Also,

f(0) = k

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

LHL = RHL = f(0)

So,

â‡’ 1 = 1 = k

Hence, the required value of k is 1.

### is continuous at x = 3.

Solution:

Given that,

Let us consider LHL at x = 3,

= 3a + 1

Let us consider RHL at x = 3,

= 3b + 3

It is given that f(x) is continuous at x = 3,

LHL = RHL = f(3)

So,

â‡’ 3a + 1 = 3b + 3

â‡’ 3a – 3b = 2

Hence, the required relationship between a and b is 3a – 3b = 2.

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