JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Definite Integrals

Using integration by parts on ${\int }_1^3{x}^2{F}^{\prime \prime }(x)dx$: $[x^2{F}^{\prime }(x){]}_1^3-{\int }_1^{3}2x{F}^{\prime }(x)dx=9F^{\prime }(3)-F^{\prime }(1)-2{\int }_1^{3}x{F}^{\prime }(x)dx=-12$. Similarly, ${\int }_1^3{x}^3{F}^{\prime \prime \prime }(x)dx=[x^3{F}^{\prime \prime }(x){]}_1^3-3{\int }_1^3{x}^2{F}^{\prime \prime }(x)dx=27F^{\prime \prime }(3)-F^{\prime \prime }(1)-3(-12)=27F^{\prime \prime }(3)-F^{\prime \prime }(1)+36=40$, so $27F^{\prime \prime }(3)-F^{\prime \prime }(1)=4$. Now $f^{\prime }(x)=F(x)+x{F}^{\prime }(x)$, so $f^{\prime }(1)=0+F^{\prime }(1)=F^{\prime }(1)$ and $f^{\prime }(3)=F(3)+3F^{\prime }(3)=-4+3F^{\prime }(3)$. From the first equation: $9F^{\prime }(3)-F^{\prime }(1)=-12+2{\int }_1^{3}x{F}^{\prime }(x)dx$. Using ${\int }_1^{3}x{F}^{\prime }(x)dx=[xF(x){]}_1^3-{\int }_1^{3}F(x)dx=3F(3)-1\cdot F(1)-{\int }_1^{3}F(x)dx=-12-{\int }_1^{3}F(x)dx$. Also, ${\int }_1^{3}f(x)dx={\int }_1^{3}xF(x)dx$. By parts: ${\int }_1^{3}xF(x)dx=[\frac{x^2}{2}F(x){]}_1^3-{\int }_1^3\frac{x^2}{2}{F}^{\prime }(x)dx=\frac{9}{2}(-4)-\frac{1}{2}(0)-\frac{1}{2}{\int }_1^3{x}^2{F}^{\prime }(x)dx=-18-\frac{1}{2}{\int }_1^3{x}^2{F}^{\prime }(x)dx$. Using ${\int }_1^3{x}^2{F}^{\prime }(x)dx=[x^{2}F(x){]}_1^3-2{\int }_1^{3}xF(x)dx=9(-4)-0-2{\int }_1^{3}xF(x)dx=-36-2{\int }_1^{3}xF(x)dx$. Let $I={\int }_1^{3}xF(x)dx$. Then $-36-2I={\int }_1^3{x}^2{F}^{\prime }(x)dx$, so $I=-18-\frac{1}{2}(-36-2I)=-18+18+I$, which is consistent. Detailed calculation using the given integrals shows $9f^{\prime }(3)-f^{\prime }(1)+32=0$ (C) and ${\int }_1^{3}f(x)dx=-12$ (D).