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nth Term of A.P

Let f'x=ex2...

Question

Let $f^{'} (x) = e^{x^{2}}$ and $f (0) = 10$ . If $A < f (1) < B$ can be concluded from the mean value theorm, then the largest value of $(A - B)$ equals

e

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1 - e

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e - 1

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1 + e

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Solution

The correct option is B $1 - e$
By above theorem
$^{'} c^{'}$ must lies in between $(0, 1)$
$f^{'} (c) = \frac{f (1) - f (0)}{1 - 0}$
$e^{0} 1 = f (1) - 10$
$f (1) = 11 = A$
$e = f (1) - 10$
$f (1) = 10 + e = B$
$A - B = 11 - 10 + e = 1 - e$

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