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.
Please Login to comment...