Question

Let $I_1={\int }_{{\sec }^{2}z}^{2-{\tan }^{2}z}xf\left(x(3-x)\right)dx$ and $I_2={\int }_{{\sec }^{2}z}^{2-{\tan }^{2}z}f\left(x(3-x)\right)dx$, where $f$ is a continuous function and $z$ is any real number, $\frac{I_1}{I_2}=$

• A. $\frac{3}{2}$
• B. $\frac{1}{2}$
  
• C. $1$
• D. none of these

Answer 1

Answer: (A)

$\frac{3}{2}$

We have, $I_1={\int }_{{\sec }^{2}z}^{2-{\tan }^{2}z}xf\left(x(3-x)\right)dx$

$={\int }_{{\sec }^{2}z}^{2-{\tan }^{2}z}(3-x)f\left((3-x)(3-(3-x))\right)dx$ $[\because {\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx]$

$={\int }_{{\sec }^{2}z}^{2-{\tan }^{2}z}(3-x)f\left(x(3-x)\right)dx$

$=3{\int }_{{\sec }^{2}z}^{2-{\tan }^{2}z}f\left(x(3-x)\right)dx-{\int }_{{\sec }^{2}z}^{2-{\tan }^{2}z}xf\left(x(3-x)\right)dx$

$=3I_2-I_1$

$\therefore 2I_1=3I_2\Rightarrow \frac{I_1}{I_2}=\frac{3}{2}$