Question

If $8f\left(x\right)+6f(1/x)=x+5$and $y=x^{2}f\left(x\right),$ then $\frac{dy}{dx}$ at $x=–1$ is equal to

• A. $0$
• B. $\frac{1}{14}$
• C. $-\frac{1}{14}$
• D. $1$

Answer 1

The correct option is C

$-\frac{1}{14}$

Explanation for the correct option:

Step 1. Finding the value of $\frac{dy}{dx}$ at $x=–1$

Given, $8f\left(x\right)+6f\left(\frac{1}{x}\right)=x+5$ and $y=x^{2}f\left(x\right)$

$8f\left(x\right)+6f\left(\frac{1}{x}\right)=x+5$ …..(1)

Replace $x$ by $\frac{1}{x}$ in equation (1)

$8f\left(\frac{1}{x}\right)+6f\left(x\right)=\frac{1}{x}+5$ …..(2)

Step 2. Multiplying equation (1) by $8$ and equation (2) by $6$

$64f\left(x\right)+48f\left(\frac{1}{x}\right)=8x+40$ …..(3)

$48f\left(\frac{1}{x}\right)+36f\left(x\right)=\frac{6}{x}+30$ …..(4)

Step 3. Subtracting equation (4) from equation (3),

$28f\left(x\right)=8x-\frac{6}{x}+10$ …..(5)

Given that $y=x^{2}f\left(x\right)$

$\Rightarrow$ $f\left(x\right)=\frac{y}{x^2}$

Step 4. Substituting the value of $f\left(x\right)$ in equation (5),

$28\frac{y}{x^2}=8x-\frac{6}{x}+10$

$\Rightarrow$ $28y=8x^3-6x+10x^2$

Step 5. Differentiating it with respect to $x$

$28\frac{dy}{dx}=24x^2-6+20x$

$\Rightarrow$ $\frac{dy}{dx}=\frac{(24x^2-6+20x)}{28}$

Put $x=-1$

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{(24-6-20)}{28}\\ & =& -\frac{2}{28}\\ & =& -\frac{1}{14}\end{array}$

Hence, Option ‘C’ is Correct.