Class 11: Maths Chapter 9 solutions. Complete Class 11 Maths Chapter 9 Notes.
Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 2): Chapter 9- Differentiation
Maharashtra Board 11th Maths Chapter 9, Class 11 Maths Chapter 9 solutions
Ex 9.1
Question 1.
Find the derivatives of the following w.r.t. x by using the method of the first principle.
(a) x2 + 3x – 1
Solution:
Let f(x) = x2 + 3x – 1
∴ f(x + h) = (x + h)2 + 3(x + h) – 1
= x2 + 2xh + h2 + 3x + 3h – 1
By first principle, we get
(b) sin(3x)
Solution:
Let f(x) = sin 3x
f(x + h) = sin3(x + h) = sin(3x + 3h)
By first principle, we get
(c) e2x+1
Solution:
(d) 3x
Solution:
(e) log(2x + 5)
Solution:
Let f(x) = log(2x + 5)
∴ f(x + h) = log[2(x + h) + 5] = log (2x + 2h + 5)
By first principle, we get
(f) tan(2x + 3)
Solution:
Let f(x) = tan(2x + 3)
∴ f(x + h) = tan[2(x + h) + 3] = tan(2x + 2h + 3)
By first principle, we get
(g) sec(5x – 2)
Solution:
Let f(x) = sec(5x – 2)
f(x + h) = sec[5(x + h) – 2] = sec(5x + 5h – 2)
By first principle, we get
(h) x√x
Solution:
Question 2.
Find the derivatives of the following w.r.t. x. at the points indicated against them by using the method of the first principle.
(i) 2x+5−−−−−√ at x = 2
Solution:
(ii) tan x at x = π4
Solution:
(iii) 23x+1 at x = 2
Solution:
(iv) log(2x + 1) at x = 2
Solution:
Let f(x) = log(2x + 1)
∴ f(2) = log [2(2) + 1] = log 5 and
f(2 + h) = log [2(2 + h) + 1] = log(2h + 5)
By first principle, we get
(v) e3x-4 at x = 2
Solution:
(vi) cos x at x = 5π4
Solution:
Question 3.
Show that the function f is not differentiable at x = -3,
where f(x) = x2 + 2 for x < -3
= 2 – 3x for x ≥ -3
Solution:
∴ L f'(-3) ≠ R f'(-3)
∴ f is not differentiable at x = -3.
Question 4.
Show that f(x) = x2 is continuous and differentiable at x = 0.
Solution:
Question 5.
Discuss the continuity and differentiability of
(i) f(x) = x |x| at x = 0
Solution:
(ii) f(x) = (2x + 3) |2x + 3| at x = −32
Solution:
Question 6.
Discuss the continuity and differentiability of f(x) at x = 2.
f(x) = [x] if x ∈ [0, 4). [where [ ] is a greatest integer (floor) function]
Solution:
Explanation:
x ∈ [0, 4)
∴ 0 ≤ x < 4
We will plot graph for 0 ≤ x < 4
not for x < 0 and upto x = 4 making on X-axis.
f(x) = [x]
∴ Greatest integer function is discontinuous at all integer values of x and hence not differentiable at all integers.
∴ f is not continuous at x = 2.
∵ f(x) = 1, x < 2
= 2, x ≥ 2
x ∈ neighbourhood of x = 2.
∴ L.H.L. = 1, R.H.L. = 2
∴ f is not continuous at x = 2.
∴ f is not differentiable at x = 2.
Question 7.
Test the continuity and differentiability of
f(x) = 3x + 2 if x > 2
= 12 – x2 if x ≤ 2 at x = 2.
Solution:
Question 8.
If f(x) = sin x – cos x if x ≤ π2
= 2x – π + 1 if x > π2
Test the continuity and differentiability of f at x = π2.
Solution:
Question 9.
Examine the function
f(x) = x2 cos(1x), for x ≠ 0
= 0, for x = 0
for continuity and differentiability at x = 0.
Solution:
Ex 9.2
(I) Differentiate the following w.r.t. x
Solution:
∴ 4a + b = 2 …..(iii)
Also, f'(3) = 2a(3) + b
∴ f'(3) = 6a + b
But, f'(3) = 12 …..(given)
∴ 6a + b = 12 …..(iv)
equation (iv) – equation (iii), we get
2a = 10
∴ a = 5
Substituting a = 5 in (iii), we get
4(5) + b = 2
∴ b = -18
∴ a = 5, b = -18, c = 3
∴ f(x) = 5x2 – 18x + 3
Check:
If f(0) = 3, f'(2) = 2 and f'(3) = 12, then our answer is correct.
f(x) = 5x2 – 18x + 3 and f'(x) = 10x – 18
f(0) = 5(0)2 – 18(0) + 3 = 3
f'(2) = 10(2) – 18 = 2
f'(3) = 10(3) – 18 = 12
Thus, our answer is correct.
Solution:
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Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 2): Chapter 9- Differentiation
Chapterwise Maharashtra Board Solutions Class 11 Arts & Science Maths (Part 2) :
- Chapter 1- Complex Numbers
- Chapter 2- Sequences and Series
- Chapter 3- Permutations and Combination
- Chapter 4- Methods of Induction and Binomial Theorem
- Chapter 5- Sets and Relations
- Chapter 6- Functions
- Chapter 7- Limits
- Chapter 8- Continuity
- Chapter 9- Differentiation