### $f(x) + f(x + 1) = 2$$\int\limits_0^8 {f(x)dx + 2\int\limits_{ - 1}^3 {f(x)dx} } = ? Let f:R \to R be a continuous function such that f(x) + f(x + 1) = 2, for all x\in R. If {I_1} = \int\limits_0^8 {f(x)dx} and {I_2} = \int\limits_{ - 1}^3 {f(x)dx} , then the value of I_1 +2I_2 is equal to ...... Solution x \to t + 1 in I_2 we have {I_2} = \int\limits_{ - 2}^2 {f(t + 1)dt} {I_1} + 2{I_2} = \int\limits_0^8 {f(x)dx} + 2\int\limits_{ - 2}^2 {f(t + 1)dt} = \int\limits_0^8 {f(x)dx} + 2\int\limits_{ - 2}^2 {f(x + 1)dx} = \int\limits_0^8 {f(x)dx} + 2\int\limits_{ - 2}^2 {[2 - f(x)]dx} = \int\limits_0^8 {f(x)dx} - 2\int\limits_{ - 2}^2 {f(x)dx} + 4\int\limits_{ - 2}^2 {dx} = \int\limits_0^8 {f(x)dx} - 2\int\limits_{ - 2}^2 {f(x)dx} + 16 x \to x + 1 in f(x) + f(x + 1) = 2 yields f(x + 1) + f(x + 2) = 2 \therefore f(x) + f(x + 1) = f(x + 1) + f(x + 2) \Rightarrow f(x+2)=f(x) which means the function is periodic with period 2. So, I_1 +2I_2 reduces to 4\int\limits_0^2 {f(x)dx} - 4\int\limits_0^2 {f(x)dx} + 16 = 16 ### Popular posts from this blog ### f(x)=x^6+2x^4+x^3+2x+3$$\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, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $\Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$