NCERT Exemplar: Limits and Derivatives

Evaluate :
Q.1.
Ans.
Given that
Taking limit, we have
3 + 3 = 6

Q.2.
Ans.
Given that:
=

Taking limit, we have

Q.3.
Ans.
Given that

[Rationalizing the denominator]

Taking the limits, we have

Q.4.
Ans.
Given that
Put x + 2 = y ⇒ x = y – 2

Q.5.

Ans.
Given that:
Dividing the numerator and denominator by x, we get

Putting 1 + x = y ⇒ x = y – 1

Hence, the required answer is 3.

Q.6.
Ans.
Given that:

Q.7.
Ans.
Given that:
=
[Dividing the numerator and denominator of x –1]

Hence the required answer is 7.

Q.8.

Ans.
Given that:
Rationalizing the denominator, we get

Taking limits, we have

Hence, the required answer is 8.

Q.9.
Ans.
Given that :

Taking limits we have

Q.10.
Ans.
Given that:

Dividing the numerator and denominator by (x – 1) we get

Hence, the required answer is 1.

Q.11.
Ans.
Given that:

Hence, the required answer is 0.

Q.12.
Ans.
Given that:
[Dividing the Nr and Den. by x – 3]

Q.13.
Ans.
Given that :

Taking limit, we have

Q.14. Find ‘n’, if
Ans.
Given that:
= n × (2)n-1 = 80
= n x 2n-1 = 5 x (2)5-1
∴ n = 5
Hence, the required answer is n = 5.

Q.15.
Ans.
Given that:

Q.16.
Ans.
Given that:

[sin 2x = 2 sin x cos x]

Taking limit we have

Q.17.
Ans.
Given that:
[cos 2x = 1 – 2 sin2 x]

Hence, the required answer is 2.

Q.18.
Ans.
Given that:

= 1 × 1 × (1)2 = 1
Hence, the required answer is 1.

Q.19.
Ans.
Given that:

Q.20.
Ans.
Given that:
[∴ 1 - cosθ = 2 sinθ/2]

= 3
Hence, the required answer is 3.

Q.21.
Ans.
Given that:

√2 ×1 = √2
Hence, the required answer is √2 .

Q.22.
Ans.
Given that:

= 2 × 1 = 2
Hence, the required answer is 2.

Q.23.
Ans.
Given that:

Hence, the required answer is 1.

Q.24.
Ans.
Given that:

Taking limit we have

Hence, the required answer is 2 √a cos a.

Q.25.
Ans.
Given that:

Taking limit we have

Hence, the required answer is 4.

Q.26.
Ans.
Given that:

Taking limit, we get

Q.27.
Ans.
Given that:

= 1 – 6 + 5 = 0
Hence, the required answer is 0.

Q.28. If then find the value of k.
Ans.
Given that:

Hence, the required value of k is

Differentiate each of the functions from Q. 29 to Q. 42
Q.29.
Ans.

Q.30.
Ans.

Q.31. (3x + 5) (1 + tanx)
Ans.

= (3x + 5) (sec2x) + (1 + tan x) (3)
= 3x sec2 x + 5 sec2 x + 3 + 3 tan x [using product rule]
Hence, the required answer is 3x sec2 x + 5 sec2 x + 3 tan x + 3

Q.32. (sec x – 1) (sec x + 1)
Ans.

[using product rule]
= (sec x – 1) (sec x tan x) + (sec x + 1) (sec x tan x)
= sec x tan x (sec x – 1 + sec x + 1)
= sec x tan x × 2 sec x
= 2 sec2 x × tan x
Hence, the required answer is 2 tan x sec2 x.

Q.33.
Ans.

[Using quotient rule]

Q.34.

Ans.

[Using quotient rule]

Q.35.
Ans.

[Using quotient rule]

Q.36. (ax2 + cotx) (p + q cosx)
Ans.

[Using Product Rule]
= (ax2 + cot x) (- q sin x) + (p + q cos x) (2ax - cosec2 x)
= (ax2 + cot x) (- q sin x) + (p + q cos x) (2ax - cosec2 x)

Q.37.
Ans.

[Using quotient rule]

Q.38. (sin x + cosx)2
Ans.

= 2(sin x + cos x) (cos x - sin x) = 2(cos2 x - sin2 x) = 2 cos 2x
Hence, the required answer is 2 cos 2x.

Q.39. (2x – 7)2 (3x + 5)3
Ans.

Using product Rule]
= (2x – 7)2 × 3(3x + 5)2 × 3 + (3x + 5)3 × 2(2x - 7) × 2
= 9(2x - 7)2 (3x + 5)2 + 4(3x + 5)3 (2x - 7)
= (2x - 7) (3x + 5)2 [9(2x - 7) + 4(3x + 5)]
= (2x - 7) (3x + 5)2 (18x - 63 + 12x + 20)
= (2x - 7) (3x + 5)2 (30x - 43)
Hence, the required answer is (2x - 7) (30x - 43) (3x + 5)2

Q.40. x2 sinx + cos2x
Ans.

= (x2 cos x + sin x × 2 x) + (- 2 sin 2 x)
= x2 cos x + 2x sin x - 2 sin 2x
Hence, the required answer is x2 cos x + 2x sin x - 2 sin 2x.

Q.41. sin3x cos3x
Ans.

[Using Product Rule]
= sin3 x × 3 cos2 x (– sin x) + cos3 x × 3 sin2 x × cos x
= - 3 sin4 x cos2 x + 3 cos4 x sin2 x
= 3 sinx cos2 x (- sin2 x + cos2 x)
= 3 sin2 x cos2 x × cos 2x

Q.42.

Ans.

[Using quotient rule]

Differentiate each of the functions with respect to ‘x’ . Q.43 to Q.46 using first principle.

Q.43. cos (x2 + 1)
Ans.
Let f (x) = cos(x2 + 1)   ...(i)
⇒ f (x + Δx) = cos [(x + Δx)2 + 1]    ...(ii)
Subtracting eq. (i) from eq. (ii) we get
f (x + Δx) – f (x) = cos [(x + Δx)2 + 1] – cos(x2 + 1)
Dividing both side by Δx we get

[By definitions of differentiations]

Taking limit, we have
= – 2 sin (x2 + 1) × 1 × (x) = - 2x sin (x2 + 1)
Hence, the required answer is – 2x sin (x2 + 1).

Q.44.

Ans.
Let     ....(i)
....(ii)
Subtracting eq. (i) from eq. (ii) we get

Dividing both sides by Δx and take the limit, we get

[Using definition of differentiation]

Taking limit, we have

Q.45. x2/3
Ans.
Let f (x) = x2/3    ....(i)
f (x + Δx) = (x + Δx)2/3      ....(ii)
Subtracting eq. (i) from (ii) we get
f (x + Δx) – f (x) = (x + Δx)2/3 – x2/3
Dividing both sides by Δx and take the limit.

[By definition of differentiation]

[Expanding by Binomial theorem and rejecting the higher powers of Δx as Δx → 0]

Q.46. x cosx
Ans.
Let y = x cos x    ....(i)
y + Δy = (x + Δx) cos (x + Δx)    ....(ii)
Subtracting eq. (i) from eq. (ii) we get
y + Δy – y = (x + Δx) cos (x + Δx) – x cos x
⇒ Δy = x cos (x + Δx) + Δx cos (x + Δx) – x cos x
Dividing both sides by Δx and take the limits,

Taking the limits, we have
= x[– sin x] + cos x
= – x sin x + cos x

Hence, the required answer is – x sin x + cos x

Evaluate each of the following limits in Q.47 to Q.53.
Q.47.
Ans.

∴ Taking the limits we have

= x sec x tan x + sec x = sec x (x tan x + 1)
Hence, the required answer is sec x (x tan x + 1)

Q.48.
Ans.

Q.49.
Ans.
Given,

(Taking limit)

Hence, the required answer is – 4.

Q.50.
Ans.
Given,

Taking limits we have

Q.51.  Show thatdoes not exists

Ans.
Given

Since LHL ≠ RHL
Hence, the limit does not exist.

Q.52. Let

find the value of k.
Ans.
Given,

we are given that

Hence, the required answer is 6.

Q.53. Letexists.
Ans.
Given,

Since the limits exist.
∴ LHL = RHL
∴ c = 1
Hence, the required answer = 1

Q.54.
(a) 1
(b) 2
(c) – 1
(d) – 2
Ans. (c)
Solution.
Given,

Hence, the correct option is (c).

Q.55.
(a) 2
(b) 3/2
(c) – 3/2
(d) 1
Ans. (a)
Solution.
Given

= 2 cos 0 = 2 x 1 = 2
Hence, the correct option is (a).

Q.56.
(a) n
(b) 1
(c) – n
(d) 0
Ans. (a)
Solution.
Given

Hence, the correct option is (a).

Q.57.
(a) 1
(b) m/n
(c) – m/n
(d) m2/n2
Ans. (b)
Solution.

Given

Hence, the correct option is (b).

Q.58.
(a) 4/9
(b) 1/2
(c) – 1/2
(d) –1
Ans. (a)
Solution.

Given

Hence, the correct option is (a).

Q.59.
(a) - 1/2
(b) 1
(c) 1/2
(d) – 1
Ans. (c)
Solution.
Given

[∴ sin 2x = 2 sin x cos x]

Hence, the correct option is (c).

Q.60.

(a) 2
(b) 0
(c) 1
(d) – 1
Ans. (c)
Solution.

Given

Taking limit, we get

Hence, the correct option is (c).

Q.61.

(a) 3
(b) 1
(c) 0
(d) 2
Ans. (d)
Solution.

Given,

Hence, the correct option is (d).

Q.62.
(a) 1/10
(b) - 1/10
(c) 1
(d) None of these
Ans. (b)
Solution.

Given

Taking limit we have

Hence, the correct option is (b).

Q.63.  where [.] denotes the greatest integer function thenis equal to
(a) 1
(b) 0
(c) – 1
(d) None of these
Ans. (d)
Solution.

Given,

LHL ≠ RHL
So, the limit does not exist.
Hence, the correct option is (d).

Q.64.

(a) 1
(b) – 1
(c) Does not exist
(d) None of these.
Ans. (c)
Solution.
Given

LHL ≠ RHL,
so the limit does not exist.
Hence, the correct option is (c).

Q.65. Let the quadratic equation whose roots are  is
(a) x2 – 6x + 9 = 0
(b) x2 – 7x + 8 = 0
(c) x2 – 14x + 49 = 0
(d) x2 – 10x + 21 = 0
Ans. (d)
Solution.

Given

Therefore, the quadratic equation whose roots are 3 and 7 is
x2 – (3 + 7)x + 3 x 7 = 0 i.e., x2 - 10x + 21 = 0.
Hence, the correct option is (d).

Q.66.
(a) 2
(b) 1/2
(c) - 1/2
(d) 1
Ans. (b)
Solution.

Given

∴ 2x → 0
Hence, the correct option is (b).

Q.67. Let f (x) = x – [x]; ∈ R, then
(a) 3/2
(b) 1
(c) 0
(d) – 1

Ans. (b)
Solution.

Given f (x) = x – [x]
we have to first check for differentiability of f (x) at x = 1/2

Since LHD = RHD

Hence, the correct option is (b).

Q.68. If at x = 1 is
(a) 1
(b) 1/2
(c) 1/√2
(d) 0
Ans. (d)
Solution.

Given that

Hence, the correct option is (d).

Q.69. If then f ′(1) is
(a) 5/4
(b) 4/5
(c) 1
(d) 0
Ans. (a)
Solution.

Given that

Hence, the correct option is (a).

Q.70. If

(a)
(b)
(c)
(d)
Ans. (a)
Solution.

Given

Hence, the correct option is (a).

Q.71. Ifat x = 0 is

(a) – 2
(b) 0
(c) 1/2
(d) Does not exist
Ans. (a)
Solution.

Given

Hence, the correct option is (a).

Q.72. Ifat x = 0 is

(a) cos 9
(b) sin 9
(c) 0
(d) 1
Ans. (a)
Solution.

Given

Hence, the correct option is (a).

Q.73.  Ifthen f ′(1) is equal to

(a) 1/100
(b) 100
(c) does not exist
(d) 0

Ans. (b)
Solution.

Given

∴ f ’(1) = 1 + 1 + 1 + ..... + 1 (100 times) = 100
Hence, the correct option is (b).

Q.74. Iffor some constant ‘a’, then f ′(a) is
(a) 1
(b) 0
(c) does not exist
(d) 1/2

Ans. (c)
Solution.

Given

So= does not exist
Hence, the correct option is (c).

Q.75. If f (x) = x100 + x99 + ... + x + 1, then f′(1) is equal to
(a) 5050
(b) 5049
(c) 5051
(d) 50051
Ans. (a)
Solution.

Given, f(x) = x100 + x99 + … + x + 1
∴ f(x) = 100x99 + 99.x98 + ... + 1
So, f (1) = 100 + 99 + 98 + ... + 1

= 50[200 – 99] = 50 x 101 = 5050
Hence, the correct option is (a).

Q.76. If f (x) = 1 – x + x2 – x3 ... – x99 + x100, then f ′(1) is equal to

(a) 150
(b) – 50
(c) – 150
(d) 50
Ans. (d)
Solution.

Given that f (x) = 1 – x + x2 – x3 + … – x99 + x100
f(x) = - 1 + 2x - 3x2 + ... - 99x98 + 100 x99
∴ f(1) = - 1 + 2 - 3 + ... - 99 + 100
= (- 1 - 3 - 5 ... - 99) + (2 + 4 + 6 + ... + 100)

= 25[–2 – 98] + 25 [4 + 98]
= 25 x - 100 + 25 x 102
= 25[–100 + 102] = 25 x 2 = 50
Hence, the correct option is (d).

FILL IN THE BLANKS

Q.77. If= _______
Ans.
Given
Hence, the value of the filler is 1.

Q.78. then m = _______

Ans.
Given

Hence, the value of the filler is

Q.79. if= _______
Ans.
Given that

Hence the value of the filler is y.

Q.80. = _______
Ans.
Given

Hence, the value of the filler is 1.

The document NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
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## Mathematics (Maths) for JEE Main & Advanced

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## FAQs on NCERT Exemplar: Limits and Derivatives - Mathematics (Maths) for JEE Main & Advanced

 1. What is the concept of a limit in calculus?
Ans. In calculus, a limit is a fundamental concept that describes the behavior of a function as the input approaches a certain value. It is used to find the values that a function approaches as its input gets arbitrarily close to a particular value. Limits play a crucial role in defining derivatives, integrals, and other concepts in calculus.
 2. How do you calculate the limit of a function?
Ans. To calculate the limit of a function, you need to evaluate the behavior of the function as the input approaches a specific value. There are various methods to determine limits, such as direct substitution, factoring, rationalization, and using special limit theorems like the Squeeze theorem or L'Hôpital's rule. These methods allow you to analyze the function's behavior and find the value it approaches or determine if the limit does not exist.
 3. What are the properties of limits in calculus?
Ans. Limits in calculus have several important properties. Some of them include: - The limit of a sum or difference of functions is equal to the sum or difference of their limits. - The limit of a product of functions is equal to the product of their limits. - The limit of a constant times a function is equal to the constant times the limit of the function. - The limit of a quotient of functions is equal to the quotient of their limits, provided the denominator's limit is not zero. - The limit of a composite function is equal to the composition of their limits. These properties help simplify limit calculations and provide a better understanding of the behavior of functions.
 4. How are limits used to find derivatives?
Ans. Limits are the foundation of finding derivatives in calculus. The derivative of a function at a certain point represents the rate of change of the function at that point. To find the derivative of a function, we often start by finding the slope of the tangent line to the function's graph at a given point. This slope is determined by taking the limit of the difference quotient as the change in x approaches zero. By using limits, we can express the instantaneous rate of change and define the derivative as a limit.
 5. What are the applications of limits and derivatives in real-life situations?
Ans. Limits and derivatives have numerous applications in real-life situations. Some examples include: - Calculating velocity and acceleration of moving objects. - Analyzing population growth and decay. - Determining maximum and minimum values of functions to optimize resources. - Modeling and predicting economic trends. - Understanding the behavior of electrical circuits. - Analyzing the spread of diseases or epidemics. These applications demonstrate the relevance of limits and derivatives in various fields, making them essential tools for solving real-world problems.

## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests

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