NCERT solutions of chapter 5-Continuity and Differentiability
Class 12 maths NCERT solutions of chapter 5- Continuity and Differentiability are published here for helping class 12 students in their preparation of forthcoming examinations and CBSE board exams of 2020-21. All the questions of chapter 5 – Continuity and Differentiability are solved by an expert teacher of CBSE maths as per the norms of the CBSE board. The NCERT solutions of chapter 5-Continuity and Differentiability of class 12 maths will help you a complete understanding of the value of the function and limits at a particular point the function is defined for. You will learn in chapter 5-Continuity and Differentiability of class 12 maths textbook that a particular function is differentiable or not means the derivatives of the function exist or not at the given point.
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability
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Class 12 maths NCERT solutions of Exercise 5.1 chapter 5-Continuity and Differentiability
Q1. Prove that the function f(x) = 5x – 3 is continuous at x= o, x = -3 and x = 5.
Ans. We are given the function f(x) = 5x – 3
The value of f(x) at x = 0
f(0) = 5 × 0 – 3 = -3
The continuety at x = 0
So, f(x) is continuous at x = 0
at x = -3, f(-3) = 5 × -3 – 3 = -15 -3 = -18
Therefore f(x) is continuous at x = -3
The value of f(x) at x = 5
f(0) = 5 × 5 – 3 = 22
The continuety at x = 5
Therefore f(x) is continuous at x = 5
Q2. Examine the continuety of the function f(x) = 2x² – 1, at x = 3.
Ans. The value of f(x) at x = 3
f(3) = 2×3² – 1 = 2 × 9 – 1 = 18 – 1 = 17
Therefore the given function is continuous at x = 3
Q3.Examine the following function for continuty.
Ans. (a) Given function f(x) = x – 5, we know f is defined at every real number k
The value of the function at k
f(k) = k – 5
The continuity of f(x) at k
Therefore f(x) is continuous at every real number, so f(x) is continuous
The value of f(x) for any real number k ≠ 5
The continuity of the function for a real number k ≠ 5
Therefore f(x) is continuous for any real number x ≠ 5 or of the domain of the function
The value of f(x) for any real number k ≠ -5
The continuity of the function for any real number k ≠ -5
The given function is
The given function f(x) is a modulus function
The given function is defined for all real numbers
Let k be a real number such that k< 5
Then f(x) = 5 – x
The value of f(x) at x = k
f(k) = 5 – k
The continuity of f(x) at x = k
Therefore f(x) is continuous at all real numbers less than 5.
In the case when k = 5
f(x) = x – 5
The value of f(x) at x =k
f(k) = k -5
The continuity of f(x) at x = k
Q4.Prove that the f(x) =x^{n }is continuous at x = n where n is a positive integer.
Ans. The given function is f(x) =x^{n }
The value of f(x) at x = n
f(n) = n^{n}
The continuty of f(x) at x = n
Therefore the given function is continuous at x = n
Q5. Is the given function f defined by
continuous at x = 0 at x = 1 at x = 2.
Ans. The given function is
The function is defined for x< 1
f (x) = x
So, its value at x = 0 is f(0) = 0
The continuty of f(x) at x= 0
Therefore f(x) is continuous at x = 0
The function is defined for x = 1
f (x) = x
The value of f(x) at x = 1
f(1) = 1
The continuty of the function f(x) at x = 1 since f(x) can be approached from LHS and RHS of 1, so calculating LHS limit and RHS limit of the f(x) at x = 1
LHS limit of f(x) at x = 1
RHS limit of f(x) at x = 1
LHS limit ≠ RHS limit
Therefore f(x) is not continuous at x = 1
For x >1 the function is defined f(x) = 5
The value of f(x) at x = 2 is f(2) = 5
Therefore f(x) is continuous at x = 2
Q6. Find all the points of discontinuty of f , where f is defined by
Ans. The given function f(x) is defined as follows
It is clear that the function is valid for x ≤ 2 and x > 2, so is defined for all values of real numbers
Let there is a real number k, so there arises 3 cases as follows
First case, when k < 2
The value of f(x) at x = k
f(k) = 2k + 3
The continuty of f(x) at x = k
So, f(x) is continuous at x = k
Second case, when x = k
The value of f(x) at x = k
f(k) = 2k + 3
Checking the LHS and RHS limit of the f(x) at k = 2
LHS limit
RHS limit
LHS limit ≠ RHS
Therefore the function f(x) is discontinuous at x = 2
The third case , when k > 2
The value of f(x) at x = k
f(k) = 2k + 3
The continuty of f(x) at x = k
So, f(x) is continuous at x = k
Therefore f(x) is discontinuous at x = 2
Q7. Find all the points of discontinuity of f, where f is defined by
Ans. The given function f is defined as follows
As it is seen that the function f(x) is defined for all values of real numbers
Let k is a point on the real number line, then, there arise following cases
First case when k < 3
The value of f(x) at x = k
For k < 3
So, f(x) is continuous at all points when x < -3.
Second case when k = -3
The value of f(x) at k = -3
f(-3) = -(-3) + 3 = 6
LHS limit of f(x) at x = -3
RHS limit of f(x) at x = -3
f(x) = LHS limit = RHS limit
So, the f(x) is continuous at x = -3
Third case when -3 < x < 3
Let – 3 < k < 3
The value of f(x) at x = k
f(k) = -2k
Thus
So, f(x) is continuous at -3 < x < 3
Fourth case when k > 3
f(x) = 6x + 2
The value of f(x) at x = k
f(k) = 6 × k + 2 = 6k + 2
The limit of the f(x) at x = k
So, f(x) is continuous at k > 3
Fifth case when k = 3
LHS limit at k = 3
RHS limit at k = 3
RHS limit ≠ LHS limit
So, f(x) is not continuous at k = 3
Q8. Find all the points of discontinuity of f, where f is defined by
Ans. The given function is as follows
There arises the following cases
For x ≠ 0
When x < 0, then
When x > 0
First case, let there is a point on real number line k <0
The value of f(x) at x = k
f(k) = -1
Limit of f(x) at x = k
So, f(x) is continuous for all values x < 0
Second case, when k > o
Value of f(x) at x = k
f(k) = 1
The limit of f(x) at x = k
So, f(x) is continuous for all values x > 0
Third case when x = 0
LHS limit at x = 0
RHS limit at x = 0
LHS limit ≠ RHS limit
So, f(x) is not continuous at x = 0
Q9.Find all the points of discontinuity of f, where f is defined by
Ans. The given function is f(x) is defined as follows
If x < 0
Let there is a point k on the real number line such that k < 0
The value of f(x) at x = k
f(k) = -1
The limit of f(x) at x = k
So, f(x) is continuous at x = k
If x ≥ 0, then f(x) = -1
Let there is a point k on real number line such that k > 0
f(k) = -1
The limit at x = k, will be = -1
So, the f(x) is continuous for all values x > 0
If x = 0, then f(x) = -1
Let there is a point k on real number line such that k = 0
f(k) =f(0) =-1
LHS limit
RHS limit
Since LHS limit = RHS limit = f(k) = -1
So, the function f(x) is continuous at x = 0
Hence the f(x) is continuous at all points ,thus f(x) is not discontinuous at any point.
Q10.Find all the points of discontinuity of f, where f is defined by
Ans. The given function f(x) is defined as follows
First case when x > 1
f(x) = x + 1
Let there is point on real number line such that k >1
The value of f(x) at x = k
f(k) = k + 1
The limit of f(x) at x = k
So, f(x) is continuous at all values of x > 1
When x = 1, f(x) = x + 1
The value of f(x) at x = 1
f(1) = 1 + 1 = 2
RHS limit of f(x)
LHS limit of f(x) at x = 1
So, f(x) = LHS limit = RHS limit
So, f(x) is continuous at x = 1
If x <1, then f(x) = x² + 1
Let there is a point k on real number line such that k < 1
The value of f(x) at x = k
f(k) = k² + 1
The limit of f(x) at x = k
So, the f(x) is continuous at all values of x> 1
Hence f(x) is continuous at all real values of x, so f(x) is discontinuous at any point.
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