Let f(x)=x+aπ2−4sin⁡x+bπ2−4cos⁡x,x∈ℝ be a function which satisfies f(x)=x+∫0π/2 sin⁡(x+y)f(y)dy. Then a+bis equal to

f(x)=x+∫0π/2 (sin⁡xcos⁡y+cos⁡xsin⁡y)f(y)dy=x+sin⁡x⋅∫0π/2 cos⁡yf(y)dy+cos⁡x∫0π/2 sin⁡yf(y)dyf(x)=x+Asin⁡x+Bcos⁡xA=∫0π/2 cos⁡yf(y)dy=∫0π/2 cos⁡y(y+Asin⁡y+Bcos⁡y)dy∫0π/2 ycos⁡ydy+A∫0π/2 sin⁡ycos⁡ydy+B∫0π/2 cos2⁡ydyA=∫0π/2 y⋅cos⁡ydy+A2∫0π/2 sin⁡2y+B∫0π/2 cos2⁡ydyA=π2−1+A2+Bπ4A2=π2−1+B4πA=π−2+B2πB=∫0π/2 sin⁡yf(y)dy=∫0π/2 sin⁡y(y+Asin⁡y+Bcos⁡y)=∫0π/2 ysin⁡y+Asin2⁡y+B2sin⁡2ydy=1+A⋅12⋅π2+B2(1)B2=1+A⋅π4B=2+π2AA=π−2+π22+π2A=π−2+π+π24A1−π24A=2π−2⇒A=2(π−1)4−π2/4A=8(π−1)4−π2B=2+π28(π−1)4−π2=24−π2+4π2−4π4−π2=8+2π2−4π4−π24π−8−2π2π2−4a=−8(π−1)=−8π+8,b=4π−8−2π2a+b=−4π−2π2=−2π(π+2)

Let f(x)=x+aπ2−4sin⁡x+bπ2−4cos⁡x,x∈ℝ be a function which satisfies f(x)=x+∫0π/2 sin⁡(x+y)f(y)dy. Then a+bis equal to

f(x)=x+∫0π/2 (sin⁡xcos⁡y+cos⁡xsin⁡y)f(y)dy=x+sin⁡x⋅∫0π/2 cos⁡yf(y)dy+cos⁡x∫0π/2 sin⁡yf(y)dyf(x)=x+Asin⁡x+Bcos⁡xA=∫0π/2 cos⁡yf(y)dy=∫0π/2 cos⁡y(y+Asin⁡y+Bcos⁡y)dy∫0π/2 ycos⁡ydy+A∫0π/2 sin⁡ycos⁡ydy+B∫0π/2 cos2⁡ydyA=∫0π/2 y⋅cos⁡ydy+A2∫0π/2 sin⁡2y+B∫0π/2 cos2⁡ydyA=π2−1+A2+Bπ4A2=π2−1+B4πA=π−2+B2πB=∫0π/2 sin⁡yf(y)dy=∫0π/2 sin⁡y(y+Asin⁡y+Bcos⁡y)=∫0π/2 ysin⁡y+Asin2⁡y+B2sin⁡2ydy=1+A⋅12⋅π2+B2(1)B2=1+A⋅π4B=2+π2AA=π−2+π22+π2A=π−2+π+π24A1−π24A=2π−2⇒A=2(π−1)4−π2/4A=8(π−1)4−π2B=2+π28(π−1)4−π2=24−π2+4π2−4π4−π2=8+2π2−4π4−π24π−8−2π2π2−4a=−8(π−1)=−8π+8,b=4π−8−2π2a+b=−4π−2π2=−2π(π+2)

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Let $f (x) = x + \frac{a}{π^{2} - 4} \sin x + \frac{b}{π^{2} - 4} \cos x, x \in ℝ$ be a function which satisfies $f (x) = x + \int_{0}^{π / 2} \sin (x + y) f (y) dy$ . Then $(a + b)$ is equal to

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$- 2 π (π + 2)$

$- π (π - 2)$

$- 2 π (π - 2)$

$- π (π + 2)$

answer is D.

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Detailed Solution

$\begin{array}{l} f (x) = x + \int_{0}^{π / 2} (\sin xcos y + \cos xsin y) f (y) dy \\ = x + \sin x \cdot \int_{0}^{π / 2} \cos yf (y) dy + \cos x \int_{0}^{π / 2} \sin yf (y) dy \\ f (x) = x + Asin x + Bcos x \\ A = \int_{0}^{π / 2} \cos yf (y) dy \\ = \int_{0}^{π / 2} \cos y (y + Asin y + Bcos y) dy \\ \int_{0}^{π / 2} ycos ydy + A \int_{0}^{π / 2} \sin ycos ydy + B \int_{0}^{π / 2} \cos^{2} ydy \\ A = \int_{0}^{π / 2} y \cdot \cos ydy + \frac{A}{2} \int_{0}^{π / 2} \sin 2 y + B \int_{0}^{π / 2} \cos^{2} ydy \\ A = \frac{π}{2} - 1 + \frac{A}{2} + B \frac{π}{4} \\ \frac{A}{2} = \frac{π}{2} - 1 + \frac{B}{4} π \\ A = π - 2 + \frac{B}{2} π \end{array}$
$\begin{array}{l} B = \int_{0}^{π / 2} \sin yf (y) dy \\ = \int_{0}^{π / 2} \sin y (y + Asin y + Bcos y) \\ = \int_{0}^{π / 2} (ysin y + {Asin}^{2} y + \frac{B}{2} \sin 2 y) dy \\ = 1 + A \cdot \frac{1}{2} \cdot \frac{π}{2} + \frac{B}{2} (1) \\ \frac{B}{2} = 1 + A \cdot \frac{π}{4} \\ B = 2 + \frac{π}{2} A \\ A = π - 2 + \frac{π}{2} (2 + \frac{π}{2} A) \end{array}$
$\begin{array}{l} = π - 2 + π + \frac{π^{2}}{4} A \\ (1 - \frac{π^{2}}{4}) A = 2 π - 2 \Rightarrow A = \frac{2 (π - 1)}{(4 - π^{2}) / 4} \\ A = \frac{8 (π - 1)}{4 - π^{2}} \\ B = 2 + \frac{π}{2} (\frac{8 (π - 1)}{4 - π^{2}}) \\ = \frac{2 (4 - π^{2}) + 4 π^{2} - 4 π}{4 - π^{2}} \\ = \frac{8 + 2 π^{2} - 4 π}{4 - π^{2}} \\ \frac{4 π - 8 - 2 π^{2}}{π^{2} - 4} \\ a = - 8 (π - 1) = - 8 π + 8, b = 4 π - 8 - 2 π^{2} \\ a + b = - 4 π - 2 π^{2} \\ = - 2 π (π + 2) \end{array}$