Question

If $5f\left(x\right)+3f\left(\frac{1}{x}\right)=x+2$ and $y=xf\left(x\right)$ then ${\left[\frac{dy}{dx}\right]}_{x-1}=$

• A. 0
• B. $\frac{7}{8}$
• C. $\frac{1}{8}$
• D. none of these

Answer 1

Answer: (B)

$\frac{7}{8}$

$5f\left(x\right)+3f\left(\frac{1}{x}\right)=x+2\to \left(1\right)\therefore 5f\left(\frac{1}{x}\right)+3f\left(x\right)=\left(\frac{1}{x}\right)+2\to \left(2\right)byreplacingxwith\left(\frac{1}{x}\right)now,multiplyequation\left(1\right)with5andequation\left(2\right)with3,weget25f\left(x\right)+15f\left(\frac{1}{x}\right)=5x+10\to \left(3\right)and9f\left(x\right)+15f\left(\frac{1}{x}\right)=\left(\frac{3}{x}\right)+6\to \left(4\right)subtract\left(4\right)and\left(3\right),weget16f\left(x\right)=5x+10-\left(\frac{3}{x}\right)-616f\left(x\right)=5x-\left(\frac{3}{x}\right)+4\therefore f\left(x\right)=\left(\frac{1}{16}\right)[5x-(\frac{3}{x})+4]theny=xf\left(x\right)=\left(\frac{1}{16}\right)[5x^2-3+4x]\therefore \left(\frac{dy}{dx}\right)=\left(\frac{1}{16}\right)[10x-0+4]\therefore \left(\frac{dy}{dx}\right)atx=1=\left(\frac{1}{16}\right)[10\times 1+4]=\left(\frac{1}{16}\right)\times 14=\left(\frac{7}{8}\right)$