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### f is differentiable function such that$f(x + y) - f(x) = f(y) + 2xy$$x,y \in R$$f(x) = ?$

We have, $f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$

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

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

Putting y=0 in the original equation yields f(0) = 0. So, the limit is of $\frac{0}{0}$ type.

$= 2x + \mathop {\lim }\limits_{h \to 0} f'(h)$ [Using L.H. Rule]

$= 2x + f'(0)$

Now, $f(x) = \int {2x + f'(0)dx}$

$f(x) = {x^2} + f'(0)x + C$

$\because f(0) = 0,C = 0$

$\therefore f(x) = {x^2} + f'(0)x = {x^2} + Bx$ where B is some constant.

### $f(x)=x^6+2x^4+x^3+2x+3 $$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$$n=?$

Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R$. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $\Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$