{"id":556444,"date":"2021-11-17T09:56:44","date_gmt":"2021-11-17T09:56:44","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=556444"},"modified":"2021-11-25T09:13:10","modified_gmt":"2021-11-25T09:13:10","slug":"maharashtra-board-solutions-class-12-arts-science-maths-part-2-chapter-2-applications-of-derivatives","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-12-arts-science-maths-part-2-chapter-2-applications-of-derivatives\/","title":{"rendered":"Maharashtra Board Solutions Class 12-Arts & Science Maths (Part 2): Chapter 2- Applications of Derivatives"},"content":{"rendered":"\n

Class 12: Maths Chapter 2 solutions. Complete Class 12 Maths Chapter 2 Notes.<\/p>\n\n\n\n

Maharashtra Board Solutions Class 12-Arts & Science Maths (Part 2): Chapter 2- Applications of Derivatives<\/h2>\n\n\n\n

Maharashtra Board 12th Maths Chapter 2, Class 12 Maths Chapter 2 solutions<\/p>\n\n\n\n

Ex 2.1<\/h4>\n\n\n\n

Question 1.
Find the equations of tangents and normals to the curve at the point on it.
(i) y = x2<\/sup> + 2ex<\/sup> + 2 at (0, 4)
Solution:<\/strong><\/p>\n\n\n\n

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Ex 2.2<\/h4>\n\n\n\n

Question 1.
Find the approximate value of given functions, at required points.
(i) \u221a8.95
Solution:<\/strong><\/p>\n\n\n\n

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\u221a8.95 = 2.9917<\/p>\n\n\n\n

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Ex 2.3<\/h4>\n\n\n\n

Question 1.
Check the validity of the Rolle\u2019s theorem for the following functions.
(i) f(x) = x2<\/sup> \u2013 4x + 3, x \u2208 [1, 3]
Solution:<\/strong>
The function f given as f(x) = x2<\/sup> \u2013 4x + 3 is polynomial function.
Hence, it is continuous on [1, 3] and differentiable on (1, 3).
Now, f(1) = 12<\/sup> \u2013 4(1) + 3 = 1 \u2013 4 + 3 = 0
and f(3) = 32<\/sup> \u2013 4(3) + 3 = 9 \u2013 12 + 3 = 0
\u2234 f(1) = f(3)
Thus, the function f satisfies all the conditions of Rolle\u2019s theorem.<\/p>\n\n\n\n

(ii) f(x) = e-x<\/sup> sin x, x \u2208 [0, \u03c0].
Solution:
The functions e-x<\/sup> and sin x are continuous and differentiable on their domains.
\u2234 f(x) = e-x<\/sup> sin x is continuous on [0, \u03c0] and differentiable on (0, \u03c0).
Now, f(0) = e0<\/sup> sin 0 = 1 \u00d7 0 = 0
and f(\u03c0) = e-\u03c0<\/sup> sin \u03c0 = e-\u03c0<\/sup> \u00d7 0 = 0
\u2234 f(0) = f(\u03c0)
Thus, the function f satisfies all the conditions of the Rolle\u2019s theorem.<\/p>\n\n\n\n

(iii) f(x) = 2x2<\/sup> \u2013 5x + 3, x \u2208 [1, 3].
Solution:<\/strong>
The function f given as f(x) = 2x2<\/sup> \u2013 5x + 3 is a polynomial function.
Hence, it is continuous on [1, 3] and differentiable on (1, 3).
Now, f(1) = 2(1)2<\/sup> \u2013 5(1) + 3 = 2 \u2013 5 + 3 = 0
and f(3) = 2(3)2<\/sup> \u2013 5(3) + 3 = 18 \u2013 15 + 3 = 6
\u2234 f(1) \u2260 f(3)
Hence, the conditions of Rolle\u2019s theorem are not satisfied.<\/p>\n\n\n\n

(iv) f(x) = sin x \u2013 cos x + 3, x \u2208 [0, 2\u03c0].
Solution:<\/strong>
The functions sin x, cos x and 3 are continuous and differentiable on their domains.
\u2234 f(x) = sin x \u2013 cos x + 3 is continuous on [0, 2\u03c0] and differentiable on (0, 2\u03c0).
Now, f(0) = sin 0 \u2013 cos 0 + 3 = 0 \u2013 1 + 3 = 2
and f(2\u03c0) = sin 2\u03c0 \u2013 cos 2\u03c0 + 3 = 0 \u2013 1 + 3 = 2
\u2234 f(0) = f(2\u03c0)
Thus, the function f satisfies all the conditions of the Rolle\u2019s theorem.<\/p>\n\n\n\n

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Ex 2.4<\/h4>\n\n\n\n
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Download PDF<\/strong><\/h2>\n\n\n\n

Maharashtra Board Solutions Class 12-Arts & Science Maths (Part 2): Chapter 2- Applications of Derivatives<\/p>\n\n\n\n

Download PDF<\/strong>: Maharashtra Board Solutions Class 12-Arts & Science Maths (Part 2): Chapter 2- Applications of Derivatives PDF<\/a><\/p>\n\n\n\n

Chapterwise Maharashtra Board Solutions Class 12 Arts & Science Maths (Part 2) :<\/strong><\/h2>\n\n\n\n