Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 9 Differentiation Ex 9.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 9 Differentiation 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
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (i)

(b) sin(3x)
Solution:
Let f(x) = sin 3x
f(x + h) = sin3(x + h) = sin(3x + 3h)
By first principle, we get
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (ii)
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (ii).1

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1

(c) e2x+1
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (iii)

(d) 3x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (iv)

(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
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (v)
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (v).1

(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
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (vi)
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (vi).1

(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
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (vii)
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (vii).1

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1

(h) x√x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (viii)

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) \(\sqrt{2 x+5}\) at x = 2
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q2 (i)

(ii) tan x at x = \(\frac{\pi}{4}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q2 (ii)
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q2 (ii).1

(iii) 23x+1 at x = 2
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q2 (iii)

(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
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q2 (iv)

(v) e3x-4 at x = 2
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q2 (v)

(vi) cos x at x = \(\frac{5 \pi}{4}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q2 (vi)
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q2 (vi).1
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q2 (vi).2

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1

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:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q3
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q3.1
∴ 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:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q4

Question 5.
Discuss the continuity and differentiability of
(i) f(x) = x |x| at x = 0
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q5 (i)

(ii) f(x) = (2x + 3) |2x + 3| at x = \(-\frac{3}{2}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q5 (ii)
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q5 (ii).1
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q5 (ii).2

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.
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q6
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.

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1

Question 7.
Test the continuity and differentiability of
f(x) = 3x + 2 if x > 2
= 12 – x2 if x ≤ 2 at x = 2.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q7
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q7.1

Question 8.
If f(x) = sin x – cos x if x ≤ \(\frac{\pi}{2}\)
= 2x – π + 1 if x > \(\frac{\pi}{2}\)
Test the continuity and differentiability of f at x = \(\frac{\pi}{2}\).
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q8
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q8.1
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q8.2

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1

Question 9.
Examine the function
f(x) = x2 cos(\(\frac{1}{x}\)), for x ≠ 0
= 0, for x = 0
for continuity and differentiability at x = 0.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q9