NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev

Mathematics (Maths) Class 12

JEE : NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev

The document NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev is a part of the JEE Course Mathematics (Maths) Class 12.
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Continuity & Differentiability 

Question 1:
 Prove that the function 
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev


Answer
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, f is continuous at x = 5


Question 2:
 Examine the continuity of the function 
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev.

Answer

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Thus, f is continuous at x = 3


Question 3: Examine the following functions for continuity.
 (a) 
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev

(b)NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
(c) NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev

(d) NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev


Answer
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev


(c) The given function is NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
For any real number c ≠ −5, we obtain
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev


NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, f is continuous at all real numbers greater than 5.
Hence, f is continuous at every real number and therefore, it is a continuous function.


Question 4:
 Prove that the function 
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev  is continuous at x = n, where n is a positive integer.
Answer
The given function is f (x) = xn
It is evident that f is defined at all positive integers, n, and its value at n is nn.

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, f is continuous at n, where n is a positive integer.


Question 5: Is the function f defined by

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
continuous at x = 0? At x = 1? At x = 2?
Answer
The given function f is NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
At x = 0,
It is evident that f is defined at 0 and its value at 0 is 0.
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
The right hand limit of f at x = 1 is,
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, f is continuous at x = 2


Question 6: Find all points of discontinuity of f, where f is defined by

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Answer
The given function f is NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
It is evident that the given function f is defined at all the points of the real line.
Let c be a point on the real line. Then, three cases arise.
(i) c < 2
(ii) c > 2
(iii) c = 2
Case (i) c < 2

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, f is continuous at all points x, such that x < 2
Case (ii) c > 2
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
It is observed that the left and right hand limit of f at x = 2 do not coincide.
Therefore, f is not continuous at x = 2
Hence, x = 2 is the only point of discontinuity of f.


Question 7: Find all points of discontinuity of f, where f is defined by

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev

Answer
The given function f is NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
The given function f is defined at all the points of the real line.
Let c be a point on the real line.
Case I:

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, f is continuous at all points x, such that x < −3
Case II:

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
It is observed that the left and right hand limit of f at x = 3 do not coincide.
Therefore, f is not continuous at x = 3
Case V:

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, f is continuous at all points x, such that x > 3
Hence, x = 3 is the only point of discontinuity of f.


Question 8: Find all points of discontinuity of f, where f is defined by
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Answer
The given function f is NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
It is observed that the left and right hand limit of f at x = 0 do not coincide.
Therefore, f is not continuous at x = 0

Case III:

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, f is continuous at all points x, such that x > 0
Hence, x = 0 is the only point of discontinuity of f.


Question 9: Find all points of discontinuity of f, where f is defined by

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Answer
The given function f is NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, the given function is a continuous function.
Hence, the given function has no point of discontinuity.

Question 10: Find all points of discontinuity of f, where f is defined by

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Answer
The given function f is NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
The given function f is defined at all the points of the real line.
Let c be a point on the real line.
Case I:
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, f is continuous at all points x, such that x > 1
Hence, the given function f has no point of discontinuity.

Question 11: Find all points of discontinuity of f, where f is defined by

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Answer
The given function f is NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
The given function f is defined at all the points of the real line.
Let c be a point on the real line.
Case I:
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, f is continuous at all points x, such that x > 2
Thus, the given function f is continuous at every point on the real line.
Hence, f has no point of discontinuity.

Question 12: Find all points of discontinuity of f, where f is defined by

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Answer
The given function f is NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
The given function f is defined at all the points of the real line.
Let c be a point on the real line.
Case I:
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, f is continuous at all points x, such that x > 1
Thus, from the above observation, it can be concluded that x = 1 is the only point of discontinuity of f.

Question 13: Is the function defined by

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
a continuous function?
Answer
The given function is NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
The given function f is defined at all the points of the real line.
Let c be a point on the real line.
Case I:
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, f is continuous at all points x, such that x > 1
Thus, from the above observation, it can be concluded that x = 1 is the only point of
discontinuity of f.


Question 14: Discuss the continuity of the function f, where f is defined by

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Answer
The given function is NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
The given function is defined at all points of the interval [0, 10].
Let c be a point in the interval [0, 10].
Case I:
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, f is continuous at all points of the interval (1, 3).
Case IV:
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, f is continuous at all points of the interval (3, 10].
Hence, f is not continuous at x = 1 and x = 3


Question 15: Discuss the continuity of the function f, where f is defined by

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Answer
The given function is NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
The given function is defined at all points of the real line.
Let c be a point on the real line.
Case I:
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, f is continuous at all points of the interval (0, 1).
Case IV:
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, f is continuous at all points x, such that x > 1
Hence, f is not continuous only at x = 1


Question 16: Discuss the continuity of the function f, where f is defined by

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Answer
The given function f is NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
The given function is defined at all points of the real line.
Let c be a point on the real line.


Case I:
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, f is continuous at x = −1


Case III:
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, f is continuous at all points x, such that x > 1
Thus, from the above observations, it can be concluded that f is continuous at all points of the real line.


Question 17: Find the relationship between a and b so that the function f defined by is continuous at x = 3.

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Answer
The given function f is NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
If f is continuous at x = 3, then
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev


Question 18: For what value of is the function defined by

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
continuous at x = 0? What about continuity at x = 1?
Answer
The given function f is NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
If f is continuous at x = 0, then

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, there is no value of λ for which f is continuous at x = 0 At x = 1,
f (1) = 4x + 1 = 4 × 1 + 1 = 5

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, for any values of λ, f is continuous at x = 1


Question 19: Show that the function defined by g(x)= x-[x] is discontinuous at all integral point.
 Here [x] denotes the greatest integer less than or equal to x.

Answer
The given function is g(x)= x-[x]
It is evident that g is defined at all integral points.
Let n be an integer.
Then,

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
It is observed that the left and right hand limits of f at x = n do not coincide.
Therefore, f is not continuous at x = n
Hence, g is discontinuous at all integral points.


Question 20:
 Is the function defined by 
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev  continuous at x = p?
Answer
The given function is NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
It is evident that f is defined at x = p

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, the given function f is continuous at x = π


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

Answer
It is known that if g and h are two continuous functions, then
g+h, g-h, and g.h are also continuous.
It has to proved first that g (x) = sin x and h (x) = cos x are continuous functions.
Let g (x) = sin x
It is evident that g (x) = sin x is defined for every real number.
Let c be a real number. Put x = c + h
If x → c, then h → 0
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev

Therefore, g is a continuous function.
Let h (x) = cos x
It is evident that h (x) = cos x is defined for every real number.
Let c be a real number. Put x = c + h

If x → c, then h → 0

h (c) = cos c

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, h is a continuous function.
Therefore, it can be concluded that
(a) f (x) = g (x) + h (x) = sin x + cos x is a continuous function
(b) f (x) = g (x) − h (x) = sin x − cos x is a continuous function
(c) f (x) = g (x) × h (x) = sin x × cos x is a continuous function.


Question 22: Discuss the continuity of the cosine, cosecant, secant and cotangent functions,
Answer:

It is known that if g and h are two continuous functions, then

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev

It has to be proved first that g (x) = sin x and h (x) = cos x are continuous functions. Let g (x) = sin x 

It is evident that g (x) = sin x is defined for every real number. 

Let c be a real number. Put x = c + h  

Therefore, g is a continuous function.
Let h (x) = cos x
It is evident that h (x) = cos x is defined for every real number.
Let c be a real number.

Put x = c + h

If x → c, then h → 0
Therefore, h (x) = cos x is continuous function.

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, cotangent is continuous except at x = np, n Î Z


Question 23: Find the points of discontinuity of f, where 

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Answer

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, f is continuous at x = 0
From the above observations, it can be concluded that f is continuous at all points of
the real line.
Thus, f has no point of discontinuity.

Question 24:Determine if f defined by

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
is a continuous function?
Answer
The given function f is NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
It is evident that f is defined at all points of the real line.
Let c be a real number.
Case I:
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev

Therefore, f is continuous at x = 0
From the above observations, it can be concluded that f is continuous at every point of the real line.
Thus, f is a continuous function.


Question 25: Examine the continuity of f, where f is defined by

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Answer
The given function f is NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
It is evident that f is defined at all points of the real line.
Let c be a real number.
Case I:
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, f is continuous at x = 0
From the above observations, it can be concluded that f is continuous at every point of
the real line.
Thus, f is a continuous function.


Question 26: Find the values of k so that the function f is continuous at the indicated point.

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Answer
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRevNCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, the required value of k is 6.


Question 27: Find the values of k so that the function f is continuous at the indicated point.

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev

The given function is  NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev

The given function f is continuous at x = 2, if f is defined at x = 2 and if the value of f at x = 2 equals the limit of f at x = 2

It is evident that f is defined at x = 2 and f(2) = k(2)2 = 4k

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev


Question 28: Find the values of k so that the function f is continuous at the indicated point.

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Answer
The given function is NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
The given function f is continuous at x = p, if f is defined at x = p and if the value of f at
x = p equals the limit of f at x = p
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev


Question 29: Find the values of k so that the function f is continuous at the indicated point.

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Answer
The given function f is NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
The given function f is continuous at x = 5, if f is defined at x = 5 and if the value of f at
x = 5 equals the limit of f at x = 5

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev


Question 30: Find the values of a and b such that the function defined by

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev, is a continuous function.
Answer
The given function f is NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
It is evident that the given function f is defined at all points of the real line.
If f is a continuous function, then f is continuous at all real numbers.
In particular, f is continuous at x = 2 and x = 10
Since f is continuous at x = 2, we obtain
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, the values of a and b for which f is a continuous function are 2 and 1 respectively.


Question 31: Show that the function defined by f(x) = cos (x2) is a continuous function.
Answer
The given function is f (x) = cos (x2)
This function f is defined for every real number and f can be written as the composition
of two functions as,
f = g o h, where g (x) = cos x and h (x) = x2
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, g (x) = cos x is continuous function.
h (x) = x2
Clearly, h is defined for every real number.
Let k be a real number, then h (k) = k2

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, h is a continuous function.
It is known that for real valued functions g and h, such that (g o h) is defined at c, if g is continuous at c and if f is continuous at g (c), then (f o g) is continuous at c.
Therefore, NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev  is a continuous function.


Question 32: Show that the function defined by NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev is a continuous function.
Answer
The given function is  NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
This function f is defined for every real number and f can be written as the composition of two functions as,

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Clearly, g is defined for all real numbers.
Let c be a real number.
Case I:
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, g is continuous at all points x, such that x > 0.


Question 33: Examine that sin|x| is a continuous function.

Answer
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
This function f is defined for every real number and f can be written as the composition
of two functions as,

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Clearly, g is defined for all real numbers.
Let c be a real number.
Case I:

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, g is continuous at x = 0
From the above three observations, it can be concluded that g is continuous at all
points. h (x) = sin x
It is evident that h (x) = sin x is defined for every real number.
Let c be a real number. Put x = c +k

If x → c, then k → 0 h (c) = sin c

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev

Therefore, h is a continuous function.  

It is known that for real valued functions g and h,such that (g o h) is defined at c, if g is continuous at c and if f is continuous at g (c), then (f o g) is continuous at c.  

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev is a continuous function. 


Question 34:
Find all the points of discontinuity of f defined by 
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev.

Answer
The given function is NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
The two functions, g and h, are defined as

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Clearly, g is defined for all real numbers.
Let c be a real number.

Case I:

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev

From the above three observations, it can be concluded that g is continuous at all 

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev

Case II:  

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev

Clearly, h is defined for every real number.
Let c be a real number.
Case I:

NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, h is continuous at all points x, such that x < −1
Case II:
NCERT Solutions - Continuity & Differentiability, Exercise 5.1 JEE Notes | EduRev
Therefore, h is continuous at x = −1
From the above three observations, it can be concluded that h is continuous at all points of the real line. g and h are continuous functions. Therefore, f = g − h is also acontinuous function.
Therefore, f has no point of discontinuity.

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