CBSE Class 12-science Answered
Verify Rolle's theorem for the function f (x) =
on [a, b], where 0 < a < b.
![log space open curly brackets fraction numerator straight x squared plus ab over denominator straight a left parenthesis straight a plus straight b right parenthesis end fraction close curly brackets](http://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=f00e1bff71f7bf2b30e98df62cb846f4.png)
Asked by Topperlearning User | 04 Aug, 2014, 15:48: PM
We have,
f (x) =
= log (x2 + ab) - log x - log (a + b)
![log space open curly brackets fraction numerator straight x squared plus ab over denominator straight x left parenthesis straight a plus straight b right parenthesis end fraction close curly brackets](https://images.topperlearning.com/topper/tinymce/cache/b2f14289eeb7eb3fd3288e5896044b1b.png)
Since logarithimic function is differentiable and so continuous on its domain. Therefore, f (x) is continuous on [a, b] and differentiable on (a, b).
Also, f (a) =
= log 1 = 0, and f (b) =
= log 1 = 0
![log open curly brackets fraction numerator a squared plus a b over denominator a left parenthesis a plus b right parenthesis end fraction close curly brackets](https://images.topperlearning.com/topper/tinymce/cache/8683228708dd36c20eb2de84d1498ffc.png)
![log open curly brackets fraction numerator b squared plus a b over denominator b left parenthesis a plus b right parenthesis end fraction close curly brackets](https://images.topperlearning.com/topper/tinymce/cache/a2196778f4e70c07796a7228c0fe8502.png)
\ f (a) = f (b)
Thus, all the three conditions of Rolle's theorem are satisfied.
Now, we have to show that there exists c ε (a, b) such that f '(c) = 0
We have,
f (x) = log (x2 + ab) - log x - log (a + b)
f '(x) = ![fraction numerator 2 straight x over denominator straight x squared plus ab end fraction minus 1 over straight x equals fraction numerator straight x squared minus ab over denominator straight x left parenthesis straight x squared plus ab right parenthesis end fraction](https://images.topperlearning.com/topper/tinymce/cache/01b5aa6e61a5bf7bb10f12e37a5dc860.png)
![fraction numerator 2 straight x over denominator straight x squared plus ab end fraction minus 1 over straight x equals fraction numerator straight x squared minus ab over denominator straight x left parenthesis straight x squared plus ab right parenthesis end fraction](https://images.topperlearning.com/topper/tinymce/cache/01b5aa6e61a5bf7bb10f12e37a5dc860.png)
\ f '(x) = 0 &
= 0 =>x2 = ab => x = ![square root of ab](https://images.topperlearning.com/topper/tinymce/cache/e0786b38761798f385e20f8547fff748.png)
![fraction numerator straight x squared minus ab over denominator straight x left parenthesis straight x squared plus ab right parenthesis end fraction](https://images.topperlearning.com/topper/tinymce/cache/f62edc3885f1b19e4e4e33ffe326ba73.png)
![square root of ab](https://images.topperlearning.com/topper/tinymce/cache/e0786b38761798f385e20f8547fff748.png)
Since a<
<b Therefore, c =
ε (a, b) is such that f '(c) = 0
![square root of ab](https://images.topperlearning.com/topper/tinymce/cache/e0786b38761798f385e20f8547fff748.png)
![square root of ab](https://images.topperlearning.com/topper/tinymce/cache/e0786b38761798f385e20f8547fff748.png)
Hence, Rolle's theorem is verified.
Answered by | 04 Aug, 2014, 17:48: PM
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