Question

A line through origin is tangent to the curve $y=x^3+x+16$. If the slope of line is $5k+3,$ then $k$ equals

Answer 1

${\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=3x_1^2+1$

Also slope of line = $\frac{y_1-0}{x_1-0}=\frac{y_1}{x_1}$

$\Rightarrow 3x_1^2+1=\frac{y_1}{x_1}$

$\Rightarrow 3x_1^3+x_1=y_1$

$\Rightarrow 3x_1^3+x_1=x_1^3+x_1+16$

$\Rightarrow x_1=2\Rightarrow y_1=26$

$\Rightarrow$ Slope of the line is $3\times 4+1=13$

$\Rightarrow 5k+3=13$

$\Rightarrow k=2$