JEE Main 2022 — Mathematics Calculus

Given,

$f(x)+f(x+k)=n$f(x)+f(x+k)=n ....(i)

Replacing $x\rightarrow x+k$x→x+k in above equation we get,

$f(x+k)+f(x+2k)=n$f(x+k)+f(x+2k)=n .....(ii)

Subtraction equation (i) from equation (ii) we get,

$$\Rightarrow f(x)=f(x+2k)$$⇒f(x)=f(x+2k)

$f(x)$f(x) is periodic with period $2k$2k

Now, $\displaystyle {I}_{1}={\int }_{0}^{4nk}f(x)dx=2n{\int }_{0}^{2k}f(x)dx$I1​=∫04nk​f(x)dx=2n∫02k​f(x)dx

$${I}_{2}={\int }_{-k}^{3k}f(x)dx=2{\int }_{0}^{2k}f(x)dx$$I2​=∫−k3k​f(x)dx=2∫02k​f(x)dx

Now integrating both side of $f(x)+f(x+k)=n$f(x)+f(x+k)=n with limit $\displaystyle {\int }_{0}^{k}$∫0k​ we get,

$$\Rightarrow {\int }_{0}^{k}f(x)dx+{\int }_{0}^{k}f(x+k)dx=nk$$⇒∫0k​f(x)dx+∫0k​f(x+k)dx=nk

$$\Rightarrow {\int }_{0}^{k}f(x)dx+{\int }_{k}^{2k}f(x)dx=nk$$⇒∫0k​f(x)dx+∫k2k​f(x)dx=nk

$$\Rightarrow {\int }_{0}^{2k}f(x)dx=nk$$⇒∫02k​f(x)dx=nk

$$\Rightarrow {I}_{1}=2{n}^{2}k,{I}_{2}=2nk$$⇒I1​=2n2k,I2​=2nk

$$\Rightarrow {I}_{1}+n{I}_{2}=4{n}^{2}k$$⇒I1​+nI2​=4n2k