JEE MAIN - Mathematics (2022 - 28th June Evening Shift - No. 8)
Let f : R $$\to$$ R be a continuous function satisfying f(x) + f(x + k) = n, for all x $$\in$$ R where k > 0 and n is a positive integer. If $${I_1} = \int\limits_0^{4nk} {f(x)dx} $$ and $${I_2} = \int\limits_{ - k}^{3k} {f(x)dx} $$, then :
$${I_1} + 2{I_2} = 4nk$$
$${I_1} + 2{I_2} = 2nk$$
$${I_1} + n{I_2} = 4{n^2}k$$
$${I_1} + n{I_2} = 6{n^2}k$$
Explanation
$f: R \rightarrow R$ and $f(x)+f(x+k)=n \quad \forall x \in R$
$$ \begin{aligned} &x \rightarrow x+k \\\\ &f(x+k)+f(x+2 k)=n \\\\ &\therefore \quad f(x+2 k)=f(x) \end{aligned} $$
So, period of $f(x)$ is $2 k$
$$ \begin{aligned} &\text { Now, } I_{1}=\int_{0}^{4 n k} f(x) d x = 2 n \int_{0}^{2 k} f(x) d x \\\\ &=2 n\left[\int_{0}^{k} f(x) d x+\int_{k}^{2 k} f(x) d x\right] \\\\ &x=t+k \Rightarrow d x=d t \text { (in second integral) } \\\\ &=2 n\left[\int_{0}^{k} f(x) d x+\int_{0}^{k} f(t+k) d t\right] \\\\ &=2 n^{2} k \end{aligned} $$
Now, $I_2=\int_{-k}^{3 k} f(x) d x=2 \int_{0}^{2 k} f(x) d x$
$$ I_{2}=2(n k) $$
$\therefore \quad l_{1}+n l_{2}=4 n^{2} k$
$$ \begin{aligned} &x \rightarrow x+k \\\\ &f(x+k)+f(x+2 k)=n \\\\ &\therefore \quad f(x+2 k)=f(x) \end{aligned} $$
So, period of $f(x)$ is $2 k$
$$ \begin{aligned} &\text { Now, } I_{1}=\int_{0}^{4 n k} f(x) d x = 2 n \int_{0}^{2 k} f(x) d x \\\\ &=2 n\left[\int_{0}^{k} f(x) d x+\int_{k}^{2 k} f(x) d x\right] \\\\ &x=t+k \Rightarrow d x=d t \text { (in second integral) } \\\\ &=2 n\left[\int_{0}^{k} f(x) d x+\int_{0}^{k} f(t+k) d t\right] \\\\ &=2 n^{2} k \end{aligned} $$
Now, $I_2=\int_{-k}^{3 k} f(x) d x=2 \int_{0}^{2 k} f(x) d x$
$$ I_{2}=2(n k) $$
$\therefore \quad l_{1}+n l_{2}=4 n^{2} k$
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