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### f'(0) = ? for an Even Function f that is Differentiable at 0

Since f is differentiable at 0 we have,

$\mathop {\lim }\limits_{h \to 0} \frac{{f(0 + h) - f(0)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f(0 - h) - f(0)}}{{ - h}}$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{f(h) - f(0)}}{h} = - \mathop {\lim }\limits_{h \to 0} \frac{{f( - h) - f(0)}}{h}$

Since f is an even function, f(-h) = f(h)

$\therefore \mathop {\lim }\limits_{h \to 0} \frac{{f(h) - f(0)}}{h} = - \mathop {\lim }\limits_{h \to 0} \frac{{f(h) - f(0)}}{h}$

$\Rightarrow 2\mathop {\lim }\limits_{h \to 0} \frac{{f(h) - f(0)}}{h} = 0$

$\Rightarrow 2f'(0) = 0$

$\therefore f'(0) = 0$

### Sum of the coefficients in the expansion of $(x+y)^n$ ....

If the sum of the coefficients in the expansion of $(x+y)^n$ is 4096, then the greatest coefficient in the expansion is _ _ _ _ . Solution $C_0 + C_1 + C_2 + C_3 + ......................... + C_n =4096$ $\therefore 2^n = 4096 =2^{12}$ $\Rightarrow n = 12$ Greatest coefficient = ${}^{12}{C_6} = 924$