Question

The slope of the tangent to the curve $y=x^3+x+54$ which also passes through the origin is

Answer 1

Answer: 28

Let point $P\left(x_1,y_1\right)$ be the point on this curve such that the tangent of $P$ passes through the origin.
Equation of tangent at $P$ is $y-y_1=\left(3x_1^2+1\right)\left(x-x_1\right)$ It passes through (0,0) so $y_1=\left(3x_1^2+1\right)x_1$ $\Rightarrow x_1^3+x_1+54=3x_1^3+x_1$
$\Rightarrow x_1^3=27\Rightarrow x_1=3$
So, ${\frac{dy}{dx}|}_{\left(x_1,y_1\right)}=3x_1^2+1=3(9)+1=28$