NCERT Solutions Class 11 Maths Chapter 12 Limits and Derivatives
Understanding the ins and outs of class 11 Maths chapter 12 in Orchids International School lays the foundation for the basics of Calculus. The chapter is called "Limits and Derivatives," and it elaborates on the fundamental principles of limits and the derivative, which are very important in advanced mathematical studies. The class 11 Maths chapter 12 PDF is the umbrella under which all these concepts are generously explained in the form of comprehensive explanations coupled with numerous practice problems to aid learning. It will enable students to work their way through the material and have a proper realization of exactly how limits are used to define derivatives and therefore approach problems in calculus. Ample resources available in Orchids International School ensure that learners are well-equipped to excel in studies and apply these mathematical principles effectively in their academic journey.
Access Answers to NCERT Solutions Class 11 Maths Chapter 12 Limits and Derivatives
Students can access the NCERT Solutions Class 11 Maths Chapter 12 Limits and Derivatives. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Limits and Derivatives
Evaluate the given limit:
Given,
Substituting x = 3, we get
= 3 + 3
= 6
Evaluate the given limit:
Given limit,
Substituting x = π, we get
Evaluate the given limit:
Given limit,
Substituting r = 1, we get
= π
Evaluate the given limit:
Given limit,
Substituting x = 4, we get
= (16 + 3) / 2
= 19 / 2
Evaluate the given limit:
Given limit,
Substituting x = -1, we get
= [(-1)10 + (-1)5 + 1] / (-1 – 1)
= (1 – 1 + 1) / – 2
= – 1 / 2
Evaluate the given limit:
Given limit,
= [(0 + 1)5 – 1] / 0
=0
Since this limit is undefined,
Substitute x + 1 = y, then x = y – 1
Evaluate the given limit:
= [a (0) + b] / c (0) + 1
= b / 1
= b
Evaluate the given limit:
Evaluate the given limit:
Evaluate the given limit:
Evaluate the given limit:
Given limit,
Substituting x = 1,
= [a (1)2 + b (1) + c] / [c (1)2 + b (1) + a]
= (a + b + c) / (a + b + c)
Given,
= 1
Evaluate the given limit:
By substituting x = – 2, we get
Evaluate the given limit:
Given
Evaluate the given limit:
Evaluate the given limit:
Evaluate the given limit:
Evaluate the given limit:
Evaluate the given limit:
Evaluate the given limit:
Evaluate the given limit:
= b
10. Evaluate the given limit:
Solution:
11. Evaluate the given limit:
Solution:
Given limit,
Substituting x = 1,
= [a (1)2 + b (1) + c] / [c (1)2 + b (1) + a]
= (a + b + c) / (a + b + c)
Given,
= 1
12. Evaluate the given limit:
Solution:
By substituting x = – 2, we get
13. Evaluate the given limit:
Solution:
Given
14. Evaluate the given limit:
Solution:
15. Evaluate the given limit:
Solution:
16. Evaluate the given limit:
Solution:
17. Evaluate the given limit:
Solution:
18. Evaluate the given limit:
Solution:
19. Evaluate the given limit:
Solution:
20. Evaluate the given limit:
Solution:
Evaluate the given limit:
Evaluate the given limit:
Find
Evaluate
Find
Find
Suppose
Let a1, a2,………an be fixed real numbers and define a function
f (x) = (x – a1) (x – a2) ……. (x – an).
What is
If 
If the function f(x) satisfies 
If 
Exercise 12.2
Find the derivative of x2– 2 at x = 10.
Let f (x) = x2 – 2
Find the derivative of
(i) 2x – 3 / 4
(ii) (5x3 + 3x – 1) (x – 1)
(iii) x-3 (5 + 3x)
(iv) x5 (3 – 6x-9)
(v) x-4 (3 – 4x-5)
(vi) (2 / x + 1) – x2 / 3x – 1
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Find the derivative of x at x = 1.
Let f (x) = x
then,
Find the derivative of 99x at x = l00.
Let f (x) = 99x,
From the first principle,
= 99
Find the derivative of the following functions from the first principle.
(i) x3 – 27
(ii) (x – 1) (x – 2)
(iii) 1 / x2
(iv) x + 1 / x – 1
(i) Let f (x) = x3 – 27
From the first principle,
(ii) Let f (x) = (x – 1) (x – 2)
From the first principle,
(iii) Let f (x) = 1 / x2
From the first principle, we get
(iv) Let f (x) = x + 1 / x – 1
From the first principle, we get
For the function 
Find the derivative of 
For some constants a and b, find the derivative of
(i) (x − a) (x − b)
(ii) (ax2 + b)2
(iii) x – a / x – b
(i) (x – a) (x – b)
(ii) (ax2 + b)2
= 4ax (ax2 + b)
(iii) x – a / x – b
Find the derivative of 
Find the derivative of cos x from the first principle.
Find the derivative of the following functions.
(i) sin x cos x
(ii) sec x
(iii) 5 sec x + 4 cos x
(iv) cosec x
(v) 3 cot x + 5 cosec x
(vi) 5 sin x – 6 cos x + 7
(vii) 2 tan x – 7 sec x
(i) sin x cos x
(ii) sec x
(iii) 5 sec x + 4 cos x
(iv) cosec x
(v) 3 cot x + 5 cosec x
(vi)5 sin x – 6 cos x + 7
(vi)5 sin x – 6 cos x + 7
(vii) 2 tan x – 7 sec x
Miscellaneous exercise
Find the derivative of the following functions from the first principle.
(i) –x
(ii) (–x)–1
(iii) sin (x + 1)
(iv)
(ii) (-x)-1
= 1 / x2
(iii) sin (x + 1)
(iv)
We get,
Find the derivative of the following functions. (It is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants, and m and n are integers.)
2. (x + a)
(px + q) (r / x + s)
(ax + b) (cx + d)2
(ax + b) / (cx + d)
(1 + 1 / x) / (1 – 1 / x)
1 / (ax2 + bx + c)
(ax + b) / px2 + qx + r
(px2 + qx + r) / ax + b
(a / x4) – (b / x2) + cos x
(ax + b)n
(ax + b)n (cx + d)m
sin (x + a)
cosec x cot x
So, we get
sinn x
(x + sec x) (x – tan x)
x4 (5 sin x – 3 cos x)
(x2 + 1) cos x
(ax2 + sin x) (p + q cos x)
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