Application of Derivatives — Class 12 Maths Solution

We have,
$y = {x^3} - 11x + 5$ …(i)
and $y = x - 11$ …(ii)

Slope of (ii) is $1$ …(iii)

From (i), $\cfrac{{dy}}{{dx}} = 3{x^2} - 11$

Slope of tangent is $\cfrac{{dy}}{{dx}} = 1$ …[from(iii)]

$\Rightarrow 3{x^2} - 11 = 1 \Rightarrow x = \pm \,2$

When $x = 2$, then from (i), $y = {2^3} - 11 \times 2 + 5 = - 9$

When $x = - 2$, then from (i), $y = {( - 2)^3} - 11( - 2) + 5 = 19$

So, we find that at $\left( {2, - 9} \right)$and at $\left( { - 2,19} \right)$ the slope of tangent is $1$.

But only $\left( {2, - 9} \right)$ satisfies given equation of tangent.

Therefore the point at which the line (ii) is tangent is $\left( {2, - 9} \right)$.