Question

$3a{\int }_0^1{\left[\frac{(ax-1)}{(a-1)}\right]}^{2}dx=?$

• A. $(a–1)+{(a–1)}^{-2}$
• B. $a+a^{-2}$
• C. $a–a^{-2}$
• D. $a^2+\frac{1}{a^2}$

Answer 1

The correct option is A

$(a–1)+{(a–1)}^{-2}$

Explanation for the correct option:

Step 1. Integrate the given equation:

${\int }_0^{1}3a{\left(\frac{ax-1}{a-1}\right)}^{2}dx=\frac{3a}{{(a-1)}^2}{\int }_0^1{\left(ax-1\right)}^{2}dx$

$\Rightarrow \frac{3a}{{(a-1)}^2}{\int }_0^1\left(a^2{x}^2+1-2ax\right)dx$

$\Rightarrow \frac{3a}{{(a-1)}^2}{\left[a^2\frac{x^3}{3}+x-2a\frac{x^2}{2}\right]}_0^1$

$\Rightarrow \frac{3a}{{(a-1)}^2}\left(\frac{a^2}{3}+1-a\right)$

$\Rightarrow \frac{a^3+3a-3a^2}{{\left(a-1\right)}^2}$

Step 2. Add and Subtract 1 on numerator:

$\Rightarrow \frac{a^3+3a-3a^2-1+1}{{\left(a-1\right)}^2}$

$\Rightarrow \frac{{\left(a-1\right)}^3+1}{{\left(a-1\right)}^2}$

$\Rightarrow \left(a-1\right)\mathbf{+}{\left(a-1\right)}^{\mathbf{-}\mathbf{2}}$

Hence, Option ‘A’ is Correct.