Question

If for non-zero x, af (x) + bf $\left(\frac{1}{x}\right)=\frac{1}{x}-25$ where a $\ne$ b then ${\int }_1^2$ f(x) dx is

• A. $\frac{1}{a^2-b^2}\left[a\right(ln2-25)+\frac{47b}{2}]$
• B. $\frac{1}{a^2-b^2}\left[a\right(2ln2-25)+\frac{47b}{2}]$
• C. $\frac{1}{a^2-b^2}\left[a\right(2ln2-25)+\frac{47b}{2}]$
• D. $\frac{1}{a^2-b^2}\left[a\right(ln2-25)-\frac{47b}{2}]$

Answer 1

The correct option is A $\frac{1}{a^2-b^2}\left[a\right(ln2-25)+\frac{47b}{2}]$

Given that

a f(x) +bf $\left(\frac{1}{x}\right)=\frac{1}{x}-25$ ...(1)

Replacing x with $\frac{1}{x}$

a.f $\left(\frac{1}{x}\right)+b.f\left(x\right)=x-25$ ...(2)

Now [(1) $\times$ a - (2) $\times$ b] $\Rightarrow$

$(a^2-b^2)f\left(x\right)=a(\frac{1}{x}-25)-b(x-25)$

$\Rightarrow f\left(x\right)=\frac{1}{a^2-b^2}[a(\frac{1}{x}-25)-b(x-25\left)\right]$

Integrating

${\int }_1^{2}f\left(x\right)dx=\frac{1}{a^2-b^2}{\left\{a\right(logx-25x)-b(\frac{x^2}{2}-25x)\}}_1^2$

$=\frac{1}{a^2-b^2}[\left\{a\right(log2-50)-b(\frac{50}{2}-50)\}-\left\{a\right(0-25)-b(\frac{1}{2}-25)\}]$

= $\frac{1}{a^2-b^2}\left[a\right(log2-25)+\frac{47}{2}b]$