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$f(x) + f(x + 1) = 2$

$\int\limits_0^8 {f(x)dx + 2\int\limits_{ - 1}^3 {f(x)dx} } = ?$

Let $f:R \to R$ be a continuous function such that $f(x) + f(x + 1) = 2$, for all $x\in R$. If ${I_1} = \int\limits_0^8 {f(x)dx} $ and ${I_2} = \int\limits_{ - 1}^3 {f(x)dx} $, then the value of $I_1 +2I_2 $ is equal to ......

Solution

$x \to t + 1$ in $I_2$ we have ${I_2} = \int\limits_{ - 2}^2 {f(t + 1)dt} $

${I_1} + 2{I_2} = \int\limits_0^8 {f(x)dx}  + 2\int\limits_{ - 2}^2 {f(t + 1)dt} $

$= \int\limits_0^8 {f(x)dx}  + 2\int\limits_{ - 2}^2 {f(x + 1)dx} $

$ = \int\limits_0^8 {f(x)dx}  + 2\int\limits_{ - 2}^2 {[2 - f(x)]dx} $

$ = \int\limits_0^8 {f(x)dx}  - 2\int\limits_{ - 2}^2 {f(x)dx}  + 4\int\limits_{ - 2}^2 {dx} $

$ = \int\limits_0^8 {f(x)dx}  - 2\int\limits_{ - 2}^2 {f(x)dx}  + 16$

$x \to x + 1$ in $f(x) + f(x + 1) = 2$ yields $f(x + 1) + f(x + 2) = 2$

$\therefore f(x) + f(x + 1) = f(x + 1) + f(x + 2)$

$\Rightarrow f(x+2)=f(x)$ which means the function is periodic with period 2. 

So, $I_1 +2I_2$ reduces to $4\int\limits_0^2 {f(x)dx}  - 4\int\limits_0^2 {f(x)dx}  + 16 = 16$

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$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$ $\Rightarrow n=7$