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NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced PDF Download

Evaluate :
Q.1. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Taking limit, we have
3 + 3 = 6
Hence, the answer is 6.

Q.2. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that:NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
=NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Taking limit, we have
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the answer is 2.

Q.3. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given thatNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
[Rationalizing the denominator]
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Taking the limits, we have
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the answer isNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Q.4. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Put x + 2 = y ⇒ x = y – 2
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence the answer isNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Q.5. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Ans.
Given that: NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Dividing the numerator and denominator by x, we get
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Putting 1 + x = y ⇒ x = y – 1
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer is 3.

Q.6. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that:NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer isNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Q.7. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that: NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced=NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
[Dividing the numerator and denominator of x –1]
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence the required answer is 7.

Q.8. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Ans.
Given that:NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Rationalizing the denominator, we get
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Taking limits, we have
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer is 8.

Q.9. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that : NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Taking limits we have
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer isNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Q.10. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that:
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Dividing the numerator and denominator by (x – 1) we get
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer is 1.

Q.11. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that:NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer is 0.

Q.12. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that: NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced[Dividing the Nr and Den. by x – 3]
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer isNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Q.13. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that :NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Taking limit, we have
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer isNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Q.14. Find ‘n’, ifNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that:NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
= n × (2)n-1 = 80NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
= n x 2n-1 = 5 x (2)5-1
∴ n = 5
Hence, the required answer is n = 5.

Q.15. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that:NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & AdvancedNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer isNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Q.16. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that:NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced[sin 2x = 2 sin x cos x]
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Taking limit we have
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer isNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Q.17. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that:NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced[cos 2x = 1 – 2 sin2 x]
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer is 2.

Q.18. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that:NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
= 1 × 1 × (1)2 = 1NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer is 1.

Q.19. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that:NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer isNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Q.20. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that:NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced[∴ 1 - cosθ = 2 sinθ/2]
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
= 3NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer is 3.

Q.21. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that:NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
√2 ×1 = √2
Hence, the required answer is √2 .

Q.22. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that:NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
= 2 × 1 = 2
Hence, the required answer is 2.

Q.23. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that:NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & AdvancedNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer is 1.

Q.24. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that:NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Taking limit we have
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer is 2 √a cos a.

Q.25. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that:NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Taking limit we have
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer is 4.

Q.26. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that:NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Taking limit, we get
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer isNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Q.27. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that:NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
= 1 – 6 + 5 = 0
Hence, the required answer is 0.

Q.28. If NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advancedthen find the value of k.
Ans.
Given that:NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required value of k isNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Differentiate each of the functions from Q. 29 to Q. 42
Q.29. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer isNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Q.30. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer isNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Q.31. (3x + 5) (1 + tanx)
Ans.
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
= (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.
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
[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. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
[Using quotient rule]
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer is NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Q.34. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Ans.
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
[Using quotient rule]
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer isNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Q.35. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
[Using quotient rule]
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer isNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Q.36. (ax2 + cotx) (p + q cosx)
Ans.
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
[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. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
[Using quotient rule]
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Q.38. (sin x + cosx)2 
Ans.
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
= 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.
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
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.
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
= (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.
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
[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
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer isNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Q.42. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Ans.
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
[Using quotient rule]
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer isNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

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
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
[By definitions of differentiations]
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Taking limit, we have
= – 2 sin (x2 + 1) × 1 × (x) = - 2x sin (x2 + 1)NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer is – 2x sin (x2 + 1).

Q.44. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Ans.
LetNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced     ....(i)
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced  ....(ii)
Subtracting eq. (i) from eq. (ii) we get
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Dividing both sides by Δx and take the limit, we get
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
[Using definition of differentiation]
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Taking limit, we have
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer isNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

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.
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
[By definition of differentiation]
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
[Expanding by Binomial theorem and rejecting the higher powers of Δx as Δx → 0]
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer isNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

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,
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & AdvancedTaking the limits, we have
= x[– sin x] + cos x
= – x sin x + cos x
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer is – x sin x + cos x

Evaluate each of the following limits in Q.47 to Q.53.
Q.47. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
∴ Taking the limits we have
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced 
= 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. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer isNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Q.49. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given, NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced(Taking limit)
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer is – 4.

Q.50. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given,NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Taking limits we have
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer isNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Q.51.  Show thatNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanceddoes not exists

Ans.
Given NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Since LHL ≠ RHL
Hence, the limit does not exist.

Q.52. LetNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

find the value of k.
Ans.
Given,NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
we are given thatNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the required answer is 6.

Q.53. LetNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advancedexists.
Ans.
Given,NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Since the limits exist.
∴ LHL = RHL
∴ c = 1
Hence, the required answer = 1

OBJECTIVE ANSWER  TYPE QUESTIONS

Q.54. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
(a) 1 
(b) 2 
(c) – 1 
(d) – 2
Ans. (c)
Solution.
Given,NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct option is (c).

Q.55. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
(a) 2 
(b) 3/2 
(c) – 3/2 
(d) 1
Ans. (a)
Solution.
 GivenNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
= 2 cos 0 = 2 x 1 = 2NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct option is (a).

Q.56. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
(a) n 
(b) 1 
(c) – n 
(d) 0
Ans. (a)
Solution.
Given NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct option is (a).

Q.57. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
(a) 1 
(b) m/n 
(c) – m/n 
(d) m2/n2
Ans. (b)
Solution.

Given NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & AdvancedNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct option is (b).

Q.58. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
(a) 4/9 
(b) 1/2 
(c) – 1/2 
(d) –1
Ans. (a)
Solution.

Given NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct option is (a).

Q.59. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
(a) - 1/2
(b) 1 
(c) 1/2
(d) – 1
Ans. (c)
Solution.
GivenNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
[∴ sin 2x = 2 sin x cos x]
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct option is (c).

Q.60. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

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

 Given NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Taking limit, we get
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct option is (c).

Q.61. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

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

Given,NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct option is (d).

Q.62. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
(a) 1/10
(b) - 1/10
(c) 1 
(d) None of these
Ans. (b)
Solution.

GivenNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Taking limit we have
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct option is (b).

Q.63. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced where [.] denotes the greatest integer function thenNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advancedis equal to
(a) 1
(b) 0 
(c) – 1 
(d) None of these
Ans. (d)
Solution.

Given,NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
LHL ≠ RHL
So, the limit does not exist.
Hence, the correct option is (d).

Q.64. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

(a) 1
(b) – 1 
(c) Does not exist 
(d) None of these.
Ans. (c)
Solution.
Given NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
LHL ≠ RHL,    
so the limit does not exist.
Hence, the correct option is (c).

Q.65. Let NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advancedthe quadratic equation whose roots are NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced is
(a) x2 – 6x + 9 = 0 
(b) x2 – 7x + 8 = 0 
(c) x2 – 14x + 49 = 0 
(d) x2 – 10x + 21 = 0
Ans. (d)
Solution.

GivenNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
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. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
(a) 2 
(b) 1/2
(c) - 1/2
(d) 1
Ans. (b)
Solution.

GivenNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & AdvancedNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
∴ 2x → 0
Hence, the correct option is (b).

Q.67. Let f (x) = x – [x]; ∈ R, thenNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
(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
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Since LHD = RHD
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct option is (b).

Q.68. IfNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced at x = 1 is
(a) 1
(b) 1/2
(c) 1/√2
(d) 0
Ans. (d)
Solution.

Given thatNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct option is (d).

Q.69. If NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advancedthen f ′(1) is
(a) 5/4
(b) 4/5
(c) 1 
(d) 0
Ans. (a)
Solution.

Given that NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct option is (a).

Q.70. IfNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

(a) NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
(b)NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
(c)NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
(d)NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Ans. (a)
Solution.

GivenNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct option is (a).

Q.71. IfNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advancedat x = 0 is

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

Given NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct option is (a).

Q.72. IfNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advancedat x = 0 is

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

GivenNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct option is (a).

Q.73.  IfNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advancedthen f ′(1) is equal to

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

Ans. (b)
Solution.

GivenNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
∴ f ’(1) = 1 + 1 + 1 + ..... + 1 (100 times) = 100
Hence, the correct option is (b).

Q.74. IfNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advancedfor some constant ‘a’, then f ′(a) is
(a) 1
(b) 0
(c) does not exist
(d) 1/2

Ans. (c)
Solution.

 GivenNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
SoNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced= 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
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
= 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)
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
= 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. IfNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced= _______
Ans.
GivenNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the value of the filler is 1.

Q.78. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advancedthen m = _______

Ans.
GivenNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & AdvancedNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence, the value of the filler isNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced

Q.79. ifNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced= _______
Ans.
 Given thatNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
Hence the value of the filler is y.

Q.80. NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced= _______
Ans.
GivenNCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Limits and Derivatives | Mathematics (Maths) for JEE Main & Advanced
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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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.
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