Latest Maharashtra State Board (SSC & HSC) 2026-27 Syllabus Digest & Solutions Updated!
Class 11 (FYJC / HSC)Mathematics & Statistics2026-27 Syllabus

Chapter 9 Differentiation Ex 9.1 Solutions

Complete Maharashtra State Board Balbharati & Yuvakbharati textbook solutions for Chapter 9 Differentiation Ex 9.1. Step-by-step solved exercises, numerical problems, and digest answers.

9 Solved Questions35 Diagrams692 words

Maharashtra State Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1

Question 1 Maharashtra Board Solution
Find the derivatives of the following w.r.t. x by using the method of the first principle. (a) x2 + 3x – 1
Solution & Step-by-Step Answer:
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 Maharashtra Board Solution
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) at x = 2
Solution & Step-by-Step Answer:

(ii) tan x at x =
Solution:

(iii) 23x+1at 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-4at x = 2
Solution:

(vi) cos x at x =
Solution:


Question 3 Maharashtra Board Solution
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 & Step-by-Step Answer:
∴ L f'(-3) ≠ R f'(-3) ∴ f is not differentiable at x = -3.

Question 4 Maharashtra Board Solution
Show that f(x) = x2 is continuous and differentiable at x = 0.
Solution & Step-by-Step Answer:

Question 5 Maharashtra Board Solution
Discuss the continuity and differentiability of (i) f(x) = x |x| at x = 0
Solution & Step-by-Step Answer:

(ii) f(x) = (2x + 3) |2x + 3| at x =
Solution:


Question 6 Maharashtra Board Solution
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 & Step-by-Step Answer:
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 Maharashtra Board Solution
Test the continuity and differentiability of f(x) = 3x + 2 if x > 2 = 12 – x2 if x ≤ 2 at x = 2.
Solution & Step-by-Step Answer:

Question 8 Maharashtra Board Solution
If f(x) = sin x – cos x if x ≤ = 2x – π + 1 if x > Test the continuity and differentiability of f at x = .
Solution & Step-by-Step Answer:

Question 9 Maharashtra Board Solution
Examine the function f(x) = x2 cos(), for x ≠ 0 = 0, for x = 0 for continuity and differentiability at x = 0.
Solution & Step-by-Step Answer: