Question

Let $y=f\left(x\right)$ be a differentiable curve satisfying ${\int }_2^{x}f\left(t\right)dt+2=\frac{x^2}{2}+{\int }_x^2{t}^{2}f\left(t\right)dt$,then ${\int }_{-\pi /4}^{\pi /4}\frac{f\left(x\right)+x^9-x^3+x+1}{{\cos }^{2}x}dx$ equals-

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

Answer: (C)

$2$

${\int }_2^{x}f\left(t\right)dt+2=\frac{x^2}{2}+{\int }_x^2{t}^{2}f\left(t\right)dt$

By differentiate on both sides, we get

$⟹f\left(x\right)+0=x-x^{2}f\left(x\right)$

$\therefore f\left(x\right)=\frac{x}{x^2+1}$

$I={\int }_{\frac{-\pi }{4}}^{\frac{\pi }{4}}\frac{\left(\frac{x}{x^2+1}\right)+x^9-x^3+x+1}{{\cos }^{2}x}dx$

as $\frac{x}{x^2+1},x^9,x^3,x$ are odd functions

$I=0+0+0+0+{\int }_{\frac{-\pi }{4}}^{\frac{\pi }{4}}\frac{1}{{\cos }^{2}x}dx$

$I={\int }_{\frac{-\pi }{4}}^{\frac{\pi }{4}}{\sec }^{2}xdx$

$I=2{\int }_0^{\frac{\pi }{4}}{\sec }^{2}xdx$

$I=2{\left[\tan x\right]}_0^{\frac{\pi }{4}}$

$I=2[1-0]$

$I=2$ (Ans)

Option C is correct.