CBSE Class 12-science Answered
Verify Rolle’s theorem for each of the following functions on indicated intervals : f(x) = sin x + cos x – on ![open square brackets 0 comma straight pi over 2 close square brackets](http://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=42b52afeb63c4492309d1c8dd9815261.png)
![open square brackets 0 comma straight pi over 2 close square brackets](http://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=42b52afeb63c4492309d1c8dd9815261.png)
Asked by Topperlearning User | 04 Aug, 2014, 16:16: PM
Since sin x and cos x are everywhere continuous and differentiable. Therefore, f (x) = sin x + cos x – 1 is continuous on and differentiable on
Also, f (0) = sin 0 + cos 0 – 1 = 0 and ![f open parentheses pi over 2 close parentheses equals sin space pi over 2 plus cos space pi over 2 minus 1 equals 1 minus 1 equals 0](https://images.topperlearning.com/topper/tinymce/cache/7c25a4a26c1c430bc496b51bbb6f2e5c.png)
![f open parentheses pi over 2 close parentheses equals sin space pi over 2 plus cos space pi over 2 minus 1 equals 1 minus 1 equals 0](https://images.topperlearning.com/topper/tinymce/cache/7c25a4a26c1c430bc496b51bbb6f2e5c.png)
\ f (0) = ![f open parentheses pi over 2 close parentheses](https://images.topperlearning.com/topper/tinymce/cache/b17fbd637b890f7b48d7f95151669341.png)
![f open parentheses pi over 2 close parentheses](https://images.topperlearning.com/topper/tinymce/cache/b17fbd637b890f7b48d7f95151669341.png)
Thus, f (x) satisfied conditions of Rolle’s theorem on
Therefore, there exists c ε
such that f ¢ (c) = 0
![open square brackets 0 comma straight pi over 2 close square brackets](https://images.topperlearning.com/topper/tinymce/cache/42b52afeb63c4492309d1c8dd9815261.png)
![open parentheses 0 comma pi over 2 close parentheses](https://images.topperlearning.com/topper/tinymce/cache/cdedba1abbc078931a4addaa9be48994.png)
Þ cos x – sin x = 0 Þ sin x = cos x Þ tan x = 1 Þ x = ![straight pi over 4](https://images.topperlearning.com/topper/tinymce/cache/324e7746562ff8467266ed6352434e91.png)
![straight pi over 4](https://images.topperlearning.com/topper/tinymce/cache/324e7746562ff8467266ed6352434e91.png)
Thus, c =
such that f ¢ (c) = 0
![pi over 4 element of open parentheses 0 comma pi over 2 close parentheses](https://images.topperlearning.com/topper/tinymce/cache/dad17eeb7068f38ee108f5b61736ea55.png)
Hence, Rolle’s theorem is verified
Answered by | 04 Aug, 2014, 18:16: PM
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