{"id":564234,"date":"2021-12-23T11:43:02","date_gmt":"2021-12-23T11:43:02","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=564234"},"modified":"2021-12-28T06:45:25","modified_gmt":"2021-12-28T06:45:25","slug":"maharashtra-board-solutions-class-11-arts-science-maths-part-2-chapter-9-differentiation","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-11-arts-science-maths-part-2-chapter-9-differentiation\/","title":{"rendered":"Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 2): Chapter 9- Differentiation"},"content":{"rendered":"\n

Class 11: Maths Chapter 9 solutions. Complete Class 11 Maths Chapter 9 Notes.<\/p>\n\n\n\n

Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 2): Chapter 9- Differentiation<\/h2>\n\n\n\n

Maharashtra Board 11th Maths Chapter 9, Class 11 Maths Chapter 9 solutions<\/p>\n\n\n\n

Ex 9.1<\/h4>\n\n\n\n

Question 1.
Find the derivatives of the following w.r.t. x by using the method of the first principle.
(a) x2<\/sup> + 3x \u2013 1
Solution:<\/strong>
Let f(x) = x2<\/sup> + 3x \u2013 1
\u2234 f(x + h) = (x + h)2<\/sup> + 3(x + h) \u2013 1
= x2<\/sup> + 2xh + h2<\/sup> + 3x + 3h \u2013 1
By first principle, we get<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

(b) sin(3x)
Solution:<\/strong>
Let f(x) = sin 3x
f(x + h) = sin3(x + h) = sin(3x + 3h)
By first principle, we get
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

(c) e2x+1<\/sup>
Solution:<\/strong><\/p>\n\n\n\n

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(d) 3x<\/sup>
Solution:<\/strong><\/p>\n\n\n\n

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(e) log(2x + 5)
Solution:<\/strong>
Let f(x) = log(2x + 5)
\u2234 f(x + h) = log[2(x + h) + 5] = log (2x + 2h + 5)
By first principle, we get<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

(f) tan(2x + 3)
Solution:<\/strong>
Let f(x) = tan(2x + 3)
\u2234 f(x + h) = tan[2(x + h) + 3] = tan(2x + 2h + 3)
By first principle, we get<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

(g) sec(5x \u2013 2)
Solution:<\/strong>
Let f(x) = sec(5x \u2013 2)
f(x + h) = sec[5(x + h) \u2013 2] = sec(5x + 5h \u2013 2)
By first principle, we get<\/p>\n\n\n\n

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\"Maharashtra<\/figure>\n\n\n\n

(h) x\u221ax
Solution:<\/strong><\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

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\u2212\u2212\u2212\u2212\u2212\u221a at x = 2
Solution:<\/strong><\/p>\n\n\n\n

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(ii) tan x at x = \u03c04
Solution:<\/strong><\/p>\n\n\n\n

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\"Maharashtra<\/figure>\n\n\n\n

(iii) 23x+1<\/sup> at x = 2
Solution:<\/strong><\/p>\n\n\n\n

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(iv) log(2x + 1) at x = 2
Solution:<\/strong>
Let f(x) = log(2x + 1)
\u2234 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<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

(v) e3x-4<\/sup> at x = 2
Solution:<\/strong><\/p>\n\n\n\n

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(vi) cos x at x = 5\u03c04
Solution:<\/strong><\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

Question 3.
Show that the function f is not differentiable at x = -3,
where f(x) = x2<\/sup> + 2 for x < -3
= 2 \u2013 3x for x \u2265 -3
Solution:<\/strong><\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

\u2234 L f'(-3) \u2260 R f'(-3)
\u2234 f is not differentiable at x = -3.<\/p>\n\n\n\n

Question 4.
Show that f(x) = x2<\/sup> is continuous and differentiable at x = 0.
Solution:<\/strong><\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

Question 5.
Discuss the continuity and differentiability of
(i) f(x) = x |x| at x = 0
Solution:<\/strong><\/p>\n\n\n\n

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(ii) f(x) = (2x + 3) |2x + 3| at x = \u221232
Solution:<\/strong><\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

Question 6.
Discuss the continuity and differentiability of f(x) at x = 2.
f(x) = [x] if x \u2208 [0, 4). [where [ ] is a greatest integer (floor) function]
Solution:<\/strong>
Explanation:
x \u2208 [0, 4)
\u2234 0 \u2264 x < 4
We will plot graph for 0 \u2264 x < 4
not for x < 0 and upto x = 4 making on X-axis.<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

f(x) = [x]
\u2234 Greatest integer function is discontinuous at all integer values of x and hence not differentiable at all integers.
\u2234 f is not continuous at x = 2.
\u2235 f(x) = 1, x < 2
= 2, x \u2265 2
x \u2208 neighbourhood of x = 2.
\u2234 L.H.L. = 1, R.H.L. = 2
\u2234 f is not continuous at x = 2.
\u2234 f is not differentiable at x = 2.<\/p>\n\n\n\n

Question 7.
Test the continuity and differentiability of
f(x) = 3x + 2 if x > 2
= 12 \u2013 x2<\/sup> if x \u2264 2 at x = 2.
Solution:<\/strong><\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

Question 8.
If f(x) = sin x \u2013 cos x if x \u2264 \u03c02
= 2x \u2013 \u03c0 + 1 if x > \u03c02
Test the continuity and differentiability of f at x = \u03c02.
Solution:<\/strong><\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

Question 9.
Examine the function
f(x) = x2<\/sup> cos(1x), for x \u2260 0
= 0, for x = 0
for continuity and differentiability at x = 0.
Solution:<\/strong><\/p>\n\n\n\n

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Ex 9.2<\/h4>\n\n\n\n

(I) Differentiate the following w.r.t. x<\/strong><\/p>\n\n\n\n

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\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
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\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

Solution:<\/p>\n\n\n\n

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\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

\u2234 4a + b = 2 \u2026..(iii)
Also, f'(3) = 2a(3) + b
\u2234 f'(3) = 6a + b
But, f'(3) = 12 \u2026..(given)
\u2234 6a + b = 12 \u2026..(iv)
equation (iv) \u2013 equation (iii), we get
2a = 10
\u2234 a = 5
Substituting a = 5 in (iii), we get
4(5) + b = 2
\u2234 b = -18
\u2234 a = 5, b = -18, c = 3
\u2234 f(x) = 5x2<\/sup> \u2013 18x + 3<\/p>\n\n\n\n

Check:
<\/strong>If f(0) = 3, f'(2) = 2 and f'(3) = 12, then our answer is correct.
f(x) = 5x2<\/sup> \u2013 18x + 3 and f'(x) = 10x \u2013 18
f(0) = 5(0)2<\/sup> \u2013 18(0) + 3 = 3
f'(2) = 10(2) \u2013 18 = 2
f'(3) = 10(3) \u2013 18 = 12
Thus, our answer is correct.<\/p>\n\n\n\n

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\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
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Solution:<\/strong><\/p>\n\n\n\n

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\"Maharashtra<\/figure>\n\n\n\n

Download PDF<\/strong><\/h2>\n\n\n\n

Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 2): Chapter 9- Differentiation<\/p>\n\n\n\n

Download PDF<\/strong>: Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 2): Chapter 9- Differentiation PDF<\/a><\/p>\n\n\n\n

Chapterwise Maharashtra Board Solutions Class 11 Arts & Science Maths (Part 2) :<\/strong><\/h2>\n\n\n\n