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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,

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ -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

Question 36. In each of the following, find the value of the constant k so that the given function is continuous at the indicated point:

(i)  at x = 0

Solution:

Given that, 

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

⇒ 

⇒ 

⇒ 

⇒ 2k2 × 1 = 8

⇒ k2 = 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) at x = π

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) 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

⇒ 5k + 1 = 10

⇒ 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 ≠ 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)  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

⇒ 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

Question 37. Find the values of a and b so that the function f given by 

 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

Question 39. Discuss the continuity of the f(x) at the indicated points:

(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.

Question 42. For what value λ of is the function

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

Solution:

Given that,

Check for x = 0, 

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

Now for x = 1,

f(1) = 4x + 1 = 4 × 1 + 1 = 5

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

Now for x = -1, 

f(-1) = λ(1 + 2)= 3λ

 

Hence, for any values of λ, 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)

⇒ 5 = 5 = k

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

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

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

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 × 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, 

⇒ 1 = 1 = k

Hence, the required value of k is 1.

Question 46. Find the relationship between ‘a’ and ‘b’ so that function ‘f’ defined by

 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, 

⇒ 3a + 1 = 3b + 3

⇒ 3a – 3b = 2

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


Article Tags :
Class 12
Mathematics
RD Sharma Solutions
School Learning
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