Verified Solution • 2025–26 Session moderate

QJEEM22Ma962a: JEE Main 2022 Mathematics - Class 12 JEEM22Ma962a Solution: Let f : R R be a continuous function satisfying f(x) + f(x + k) …

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 I1=∫04nkf(x)dx{I_1} = \int\limits_0^{4nk} {f(x)dx} and I2=∫−k3kf(x)dx{I_2} = \int\limits_{ - k}^{3k} {f(x)dx} , then :

Multiple Choice Options

  • A I1+2I2=4nk{I_1} + 2{I_2} = 4nk
  • B I1+2I2=2nk{I_1} + 2{I_2} = 2nk
  • C I1+nI2=4n2k{I_1} + n{I_2} = 4{n^2}k
  • D I1+nI2=6n2k{I_1} + n{I_2} = 6{n^2}k

Step-by-Step Verified Solution

f:R→Rf: R \rightarrow R and f(x)+f(x+k)=n∀x∈Rf(x)+f(x+k)=n \quad \forall x \in R

amp;x→x+kamp;f(x+k)+f(x+2k)=namp;∴f(x+2k)=f(x)\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)f(x) is 2k2 k

amp; Now, I1=∫04nkf(x)dx=2n∫02kf(x)dxamp;=2n[∫0kf(x)dx+∫k2kf(x)dx]amp;x=t+k⇒dx=dt (in second integral) amp;=2n[∫0kf(x)dx+∫0kf(t+k)dt]amp;=2n2k\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, I2=∫−k3kf(x)dx=2∫02kf(x)dxI_2=\int_{-k}^{3 k} f(x) d x=2 \int_{0}^{2 k} f(x) d x

I2=2(nk)I_{2}=2(n k)

∴l1+nl2=4n2k\therefore \quad l_{1}+n l_{2}=4 n^{2} k

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