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NCERT Solutions for Maths Class 12 Exercise 5.2
NCERT Solutions for Maths Class 12 Exercise 5.3
NCERT Solutions for Maths Class 12 Exercise 5.4
NCERT Solutions for Maths Class 12 Exercise 5.5
NCERT Solutions for Maths Class 12 Exercise 5.6
NCERT Solutions for Maths Class 12 Exercise 5.7
NCERT Solutions for Maths Class 12 Exercise 5.1
Maths Class 12 Ex 5.1 Question 1.
Prove that the function f(x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5.Solution:
The given function is f(x) = 5x – 3.
(i) At x = 0, limx→0 f(x) = limx→0 (5x – 3) = 5 × 0 – 3 = – 3 and f(0) = 5 × 0 – 3 = – 3
Since limx→0 f(x) = f(x), therefore, f is continuous at x = 0.
(ii) At x = – 3, limx→3 f(x)
= limx→–3 (5x – 3) = 5 × (–3) – 3 = –18 and f(– 3) = –18
Since
limx→–3 f(x)
= f(x), therefore,
f is continuous
at x = –3.
(iii) At x = 5, limx→5 f(x)
= limx→5 (5x – 3) = 5 × 5 – 3 = 22 and f(5) = 5 × 5 – 3 = 22
Since
limx→5 f(x)
= f(x), therefore, f is continuous at x = 5.
Maths Class 12 Ex 5.1 Question 2.
Examine the continuity of the function f(x) = 2x² – 1 at x = 3.Solution:
The given function is f(x)
= 2x² – 1.
At x = 3, limx→3 f(x) = limx→3 (2x² – 1) = 2 ×
(3)2 – 1 = 17 and f(3) = 2 × (3)2 – 1 = 17
Since
limx→3 f(x)
= f(x), therefore, f is continuous at x = 3.
Maths Class 12 Ex 5.1 Question 3.
Examine the following functions for continuity.(a) f(x) = x – 5
(b) f(x) = 1/(x – 5), x ≠ 5
(c) f(x) = (x2 – 25)/(x + 5), x ≠ 5
(d) f(x) = |x – 5|
Solution:
(a) f(x) = x – 5 => x – 5 is a polynomial.
At x = c ∈ R, limx→c f(x) = limx→c (x
– 5) = c – 5 and f(c) = c – 5
Since
limx→c f(x)
= f(c), therefore, f is continuous at every c ∈ R.
Maths Class 12 Ex 5.1 Question 4.
Prove that the function f(x) = xn is continuous at x = n, where n is a positive integer.Solution:
The given
function is f(x) = xn which is a polynomial.
At x = n ∈ N, limx→n f(x) = limx→n xn = xn and f(x) = xn
Since
limx→n f(x)
= f(x), therefore, f is continuous at x = n ∈ N.
Maths Class 12 Ex 5.1 Question 5.
Is the function f defined by
continuous at x =
0? At x = 1? At x = 2?
Solution:
(i) At x = 0, limx→0- f(x) = limx→0- x
= 0
And limx→0+ f(x)
= limx→0+ x = 0 ⇒ f(0) = 0
∴ f is continuous at x = 0.
(ii) At x = 1, limx→1- f(x) = limx→1- (x)
= 1
And limx→1+ f(x)
= limx→1+ (x) = 5
∴ limx→1- f(x) ≠ limx→1+ f(x)
∴ f is discontinuous at x = 1.
(iii) At x = 2, limx→2 f(x) = 5,
f(2) = 5
∴ f is continuous at x = 2.
Find all points
of discontinuity of f, where f is defined by
Maths Class 12 Ex 5.1 Question 6.
Solution:
Maths Class 12 Ex 5.1 Question 7.
Solution:
Maths Class 12 Ex 5.1 Question 8.
Test the
continuity of the function f (x) at x = 0
Solution:
Maths Class 12 Ex 5.1 Question 9.
Solution:
Maths Class 12 Ex 5.1 Question 10.
Solution:
Maths Class 12 Ex 5.1 Question 11.
Solution:
Maths Class 12 Ex 5.1 Question 12.
Solution:
Maths Class 12 Ex 5.1 Question 13.
Is the function defined by 
a continuous
function?
Solution:
At x = 1, L.H.L. =
limx→1- f(x) = limx→1- (x + 5) = 6,
R.H.L. = limx→1+ f(x) = limx→1+ (x – 5) = –4
f(1) = 1 + 5 = 6,
f(1) = L.H.L. ≠ R.H.L.
Therefore, f is not continuous at x = 1.
At x = c < 1, limx→c (x + 5) = c + 5 = f(c)
At x = c > 1, limx→c (x – 5) = c – 5 = f(c)
∴ f is continuous at all points x ∈ R except x = 1.
Discuss the
continuity of the function f, where f is defined by
Maths Class 12 Ex 5.1 Question 14.
Solution:
In the interval 0 ≤ x ≤ 1, f(x) = 3; therefore, f is continuous in this
interval.
At x = 1, L.H.L. = lim x→1- f(x) = 3,
R.H.L. = limx→1+ f(x) = 4, therefore, f is discontinuous at x =
1.
At x = 3, L.H.L. = limx→3- f(x) = 4,
R.H.L. = limx→3+ f(x) = 5, therefore, f is discontinuous at x =
3.
Hence, f is not continuous at x = 1 and x = 3.
Maths Class 12 Ex 5.1 Question 15.
Solution:
At x = 0, L.H.L. = limx→0- (2x) = 0,
R.H.L. = limx→0+ (0) = 0, f(0) = 0
Therefore, f is continuous at x = 0.
At x = 1, L.H.L. = limx→1- (0) = 0,
R.H.L. = limx→1+ (4x) = 4 and f(1) = 0.
f(1) = L.H.L. ≠ R.H.L.
∴ f is not continuous at x = 1.
When x < 0,
f(x) = 2x is a polynomial.
It is continuous
at all points x < 0.
When x > 1, f(x)
= 4x is a polynomial.
It is continuous
at all points x > 1.
When 0 ≤ x ≤ 1, f(x) = 0 is a continuous function.
Hence, the point
of discontinuity is x = 1.
Maths Class 12 Ex 5.1 Question 16.
Solution:
At x = –1, L.H.L. = limx→1- f(x) = –2, f(–1) = –2,
R.H.L. = limx→1+ f(x) = –2
Therefore, f is continuous at x = –1.
At x = 1, L.H.L. = limx→1- f(x) = 2, R.H.L. = limx→1+ f(x)
= 2 and f(1) = 2
Therefore, f is continuous at x = 1.
Hence, f is a continuous function.
Maths Class 12 Ex 5.1 Question 17.
Find the relationship between a and b so that the function f defined by
is continuous at x = 3.
Solution:
At x = 3, L.H.L.
= limx→3- (ax + 1) = 3a + 1,
R.H.L. = limx→3+ (bx + 3) = 3b + 3 and f(3) = 3a + 1
f is continuous if L.H.L. = R.H.L. = f(3)
3a + 1 = 3b + 3 or 3(a – b) = 2
a – b = 2/3 or a = b + 2/3, for any arbitrary value of b.
Therefore, the value of a is corresponding to the value of b.
Maths Class 12 Ex 5.1 Question 18.
For what value of λ is the function defined by
continuous at x = 0? What about continuity at x = 1?
At x = 0, L.H.L.
= limx→0- λ(x² – 2x) = 0,
R.H.L. = limx→0+ (4x + 1) = 1 and f(0) = 0
f(0) = L.H.L. ≠ R.H.L.
Therefore, f is not continuous at x = 0, for every value of λ ∈ R.
At x = 1, limx→1 f(x) = limx→1 (4x + l) = f(1),
therefore, f is continuous at x = 1.
Hence, f is not continuous at x = 0 for any value of λ but f is continuous at x
= 1 for all values of λ.
Maths Class 12 Ex 5.1 Question 19.
Show that the function defined by g(x) = x – [x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.Solution:
Let c be an
integer, [c – h] = c – 1, [c + h] = c, [c] = c, g(x) = x – [x].
At x = c, L.H.L. = limx→c- (x – [x]) = limh→0 [(c
– h) – (c – 1)]
= limh→0 (c – h – c + 1)) = 1 [∵ [c – h] = c – 1]
R.H.L. = limx→c+ (x – [x]) = limh→0 (c + h – [c
+ h])
= limh→0 (c + h – c) = 0 and f(c) = c – [c] = 0.
Thus, L.H.L. ≠ R.H.L. = f (c), therefore, f is not continuous at integral
points.
Maths Class 12 Ex 5.1 Question 20.
Is the function defined by f(x) = x² – sin x + 5 continuous at x = Ï€?Solution:
Maths Class 12 Ex 5.1 Question 21.
Discuss the continuity of the following functions:(a) f(x) = sin x + cos x
(b) f(x) = sin x – cos x
(c) f(x) = sin x · cos x
Solution:
Maths Class 12 Ex 5.1 Question 22.
Discuss the
continuity of the cosine, cosecant, secant and cotangent functions.
Solution:
Maths Class 12 Ex 5.1 Question 23.
Find all points
of discontinuity of f, where
Solution:
Maths Class 12 Ex 5.1 Question 24.
Determine if f defined by
is a continuous
function?
Solution:
Maths Class 12 Ex 5.1 Question 25.
Examine the continuity of f, where f is defined by 
Solution:
Maths Class 12 Ex 5.1 Question 26.
Solution:
Maths Class 12 Ex 5.1 Question 27.
Solution:
Maths Class 12 Ex 5.1 Question 28.
Solution:
Maths Class 12 Ex 5.1 Question 29.
Solution:
Maths Class 12 Ex 5.1 Question 30.
Find the values
of a and b such that the function defined by
is a continuous function.
Solution:
Maths Class 12 Ex 5.1 Question 31.
Show that the
function defined by f(x) = cos (x²) is a continuous function.
Solution:
We have, f(x) =
cos x²,
Let g(x) = cos x
and h(x) = x²
∴ goh(x) = g(h(x)) = cos x²
Now, g and h both are continuous ∀ x ∈ R.
Hence, f(x) = goh(x) = cos x² is also a continuous function at all x ∈ R.
Maths Class 12 Ex 5.1 Question 32.
Show that the function defined by f(x) = |cos x| is a continuous function.Solution:
We have, f(x) =
|cos x|,
Let g(x) =|x| and
h(x) = cos x,
∴ goh(x) = g(h(x)) = g(cos x) = |cos x|
Now, g(x) = |x| and h(x) = cos x both are continuous for all values of x ∈ R.
∴ (goh)(x) is also continuous.
Hence, f(x) = goh(x) = |cos x| is continuous for all values of x ∈ R.
Maths Class 12 Ex 5.1 Question 33.
Examine that sin |x| is a continuous function.Solution:
We have, f(x) = sin
|x|,
Let g(x) = sin x,
h(x) = |x|, goh(x) = g(h(x)) = g(|x|) = sin |x| = f(x)
Now, g(x) = sin x and h(x) = |x| both are continuous for all x ∈ R.
Hence,
f(x) = goh(x) =
sin |x| is continuous at all x ∈
R.
Maths Class 12 Ex 5.1 Question 34.
Find all the points of discontinuity of f defined by f(x) = |x| – |x + 1|.Solution:
We have, f(x) =
|x| – |x + 1|, when x < –1,
f(x) = –x – [– (x + 1)] = –x + x + 1 = 1
When –1 ≤ x < 0, f(x) = –x – (x + 1) = –2x – 1,
When x ≥ 0, f(x) = x – (x + 1) = –1
Related Links:
NCERT Solutions for Maths Class 12 Exercise 5.2
NCERT Solutions for Maths Class 12 Exercise 5.3
NCERT Solutions for Maths Class 12 Exercise 5.4
NCERT Solutions for Maths Class 12 Exercise 5.5