JEE Main 2023 — Mathematics Calculus

Given,

$$f(x)=x+{\int }_{0}^{\pi /2}\mathrm{sin}(x+y)f(y)dy$$f(x)=x+∫0π/2​sin(x+y)f(y)dy

$$\Rightarrow f(x)=x+{\int }_{0}^{\pi /2}(\mathrm{sin}x\mathrm{cos}y+\mathrm{cos}x\mathrm{sin}y)f(y)dy$$⇒f(x)=x+∫0π/2​(sinxcosy+cosxsiny)f(y)dy

$$\Rightarrow f(x)=x+{\int }_{0}^{\pi /2}((\mathrm{cos}yf(y)dy)\mathrm{sin}x+(\mathrm{sin}yf(y)dy)\mathrm{cos}x)$$⇒f(x)=x+∫0π/2​((cosyf(y)dy)sinx+(sinyf(y)dy)cosx)

And,$f(x)=x+\frac{a}{{\pi }^{2}-4}\mathrm{sin}x+\frac{b}{{\pi }^{2}-4}\mathrm{cos}x,x\in \mathbb{R}$f(x)=x+π2−4a​sinx+π2−4b​cosx,x∈R

On comparing both the equations of $f(x)$f(x) we get,

$$\Rightarrow \frac{a}{{\pi }^{2}-4}={\int }_{0}^{\pi /2}\mathrm{cos}yf(y)dy\ldots (1)$$⇒π2−4a​=∫0π/2​cosyf(y)dy…(1)

$$\Rightarrow \frac{b}{{\pi }^{2}-4}={\int }_{0}^{\pi /2}\mathrm{sin}yf(y)dy\ldots (2)$$⇒π2−4b​=∫0π/2​sinyf(y)dy…(2)

Adding equation $(1)$(1) and equation $(2)$(2) we get,

$$\frac{a+b}{{\pi }^{2}-4}={\int }_{0}^{\pi /2}(\mathrm{sin}y+\mathrm{cos}y)f(y)dy\ldots (3)$$π2−4a+b​=∫0π/2​(siny+cosy)f(y)dy…(3)

$$\because {\int }_{0}^{a}f(x)dx={\int }_{0}^{a}f(a-x)dx$$∵∫0a​f(x)dx=∫0a​f(a−x)dx

$$\therefore \frac{a+b}{{\pi }^{2}-4}={\int }_{0}^{\pi /2}(\mathrm{sin}y+\mathrm{cos}y)f(\frac{\pi }{2}-y)dy\ldots (4)$$∴π2−4a+b​=∫0π/2​(siny+cosy)f(2π​−y)dy…(4)

Add equation $(3)$(3) and equation $(4)$(4) we get,

$$\frac{2(a+b)}{{\pi }^{2}-4}={\int }_{0}^{\pi /2}(\mathrm{sin}y+\mathrm{cos}y)(\frac{\pi }{2}+\frac{(a+b)}{{\pi }^{2}-4}(\mathrm{sin}y+\mathrm{cos}y))dy$$π2−42(a+b)​=∫0π/2​(siny+cosy)(2π​+π2−4(a+b)​(siny+cosy))dy

$$\Rightarrow \frac{2(a+b)}{{\pi }^{2}-4}=\pi +\frac{a+b}{{\pi }^{2}-4}(\frac{\pi }{2}+1)$$⇒π2−42(a+b)​=π+π2−4a+b​(2π​+1)

$$\Rightarrow (a+b)=-2\pi (\pi +2)$$⇒(a+b)=−2π(π+2)