Question

Let $I_1={\int }_{se{c}^{2}z}^{2-ta{n}^{2}z}xf\left(x\right(3-x\left)\right)dxandlet{I}_2={\int }_{se{c}^{2}z}^{2-ta{n}^{2}z}f\left(x\right(3-x\left)\right)dx$
where 'f' is a continuous function and 'z' is any real number, then $\frac{I_1}{I_2}=$

• A. $\frac{3}{2}$
• B. $\frac{1}{2}$
• C. $1$
• D. $\frac{2}{3}$

Answer 1

The correct option is A $\frac{3}{2}$

$I_1={\int }_{se{c}^{2}z}^{2-ta{n}^{2}z}xf\left(x\right(3-x\left)\right)dx{I}_2={\int }_{se{c}^{2}z}^{2-ta{n}^{2}z}f\left(x\right(3-x\left)\right)dx{I}_1={\int }_{se{c}^{2}z}^{2-ta{n}^{2}z}(3-x)f\left(\right(3-x\left)x\right)dx2I_1=3{\int }_{se{c}^{2}z}^{2-ta{n}^{2}z}\left(\right(x\left)\right(3-x\left)\right)dx2I_1=3I_2$