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NCERT Exemplar: Limits and Derivatives
Evaluate :
Q.1.
Ans.
Given that
Taking the limit, the expression simplifies to
3 + 3 = 6
Hence, the answer is 6.
Q.2.
Ans.
Given that:
=
Taking the limit, the terms simplify and we obtain
Hence, the answer is 2.
Q.3.
Ans.
Given that
[Rationalizing the denominator]
Taking the limits, the simplified form is
Hence, the answer is
Q.4.
Ans.
Given that
Put x + 2 = y ⇒ x = y - 2
Hence the answer is
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:
Therefore, on taking the limit the result is
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 the limits, the expression simplifies to
Hence, the required answer is 8.
Q.9.
Ans.
Given that :
Taking limits we obtain
Hence, the required answer is
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:
Therefore, on taking the limit we obtain 0.
Hence, the required answer is 0.
Q.12.
Ans.
Given that:
Hence, the required answer is
Q.13.
Ans.
Given that :
Taking limit, the expression reduces to
Hence, the required answer is
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:
Therefore, simplifying the terms gives
Hence, the required answer is
Q.16.
Ans.
Given that:
Taking the limit we obtain
Hence, the required answer is
Q.17.
Ans.
Given that:
Therefore, evaluating the limit gives 2.
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:
On simplification we get
Hence, the required answer is
Q.20.
Ans.
Given that:
= 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:
Therefore, the simplified limit equals 1.
Hence, the required answer is 1.
Q.24.
Ans.
Given that:
Taking the limit we have
Hence, the required answer is 2 √a cos a.
Q.25.
Ans.
Given that:
Taking the limit we obtain
Hence, the required answer is 4.
Q.26.
Ans.
Given that:
Taking the limit, we get
Hence, the required answer is
Q.27.
Ans.
Given that:
= 1 - 6 + 5 = 0
Hence, the required answer is 0.
Q.28. If
Ans.
Given that:
Therefore, on simplifying we obtain the value of k as
Differentiate each of the functions from Q. 29 to Q. 42
Q.29.
Ans.
Hence, the required answer is
Q.30.
Ans.
Hence, the required answer is
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]
Hence, the required answer is
Q.34.
Ans.
[Using quotient rule]
Hence, the required answer is
Q.35.
Ans.
[Using quotient rule]
Hence, the required answer is
Q.36. (ax2 + cotx) (p + q cosx)
Ans.
[Using Product Rule]
= (ax2 + cot x) (- q sin x) + (p + q cos x) (2ax - cosec2 x)
Hence, the required answer is
= (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 sin2 x cos2 x (- sin2 x + cos2 x)
= 3 sin2 x cos2 x × cos 2x
Hence, the required answer is
Q.42.
Ans.
[Using quotient rule]
Hence, the required answer is
LONG ANSWER TYPE QUESTIONS
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
⇒
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
Hence, the required answer is
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]
Hence, the required answer is
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,
= 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.
Hence, the required answer is
Q.49.
Ans.
Given, 
Hence, the required answer is - 4.
Q.50.
Ans.
Given,
Taking limits we have
Hence, the required answer is
Q.51. Show that
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. Let
Ans.
Given,
Since the limits exist.
∴ LHL = RHL
∴ c = 1
Hence, the required answer = 1
OBJECTIVE ANSWER TYPE QUESTIONS
Q.54.
(a) 1
(b) 2
(c) - 1
(d) - 2
Ans. (c)
Solution.
Given,
Therefore, on evaluating the limit we get option (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.
(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
(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
(a) 1
(b) 1/2
(c) 1/√2
(d) 0
Ans. (d)
Solution.
Given that
Hence, the correct option is (d).
Q.69. If
(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. If
(a) - 2
(b) 0
(c) 1/2
(d) Does not exist
Ans. (a)
Solution.
Given
Hence, the correct option is (a).
Q.72. If
(a) cos 9
(b) sin 9
(c) 0
(d) 1
Ans. (a)
Solution.
Given
Hence, the correct option is (a).
Q.73. If
(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. If
(a) 1
(b) 0
(c) does not exist
(d) 1/2
Ans. (c)
Solution.
Given
∴
So
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.
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 is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
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FAQs on NCERT Exemplar: Limits and Derivatives
| 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.
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