Question

The point at which the tangent to the curve $y=x^3+5$ is perpendicular to the line $x+3y=2$ are

• A. $(6,1),(-1,4)$
• B. $(6,1)(4,-1)$
• C. $(1,6),(1,4)$
• D. $(1,6),(-1,4)$

Answer 1

Answer: (D)

$(1,6),(-1,4)$

Let point be $(x_1,y_1)$
$y=x^3+5$
$\Rightarrow \frac{dy}{dx}=3x^2$
Now, the slope of tangent at this point is $={\left(\frac{dy}{dx}\right)}_{x_1{y}_1}=3x_1^2$
It is $\perp$ to $x+3y-2=0$
So $3x_1^2\times -\frac{1}{3}=-1\Rightarrow x_1=\pm 1$
Thus,
at $x_1=1$, $\Rightarrow$ $y_1=6$
and at $x_1=-1$ $\Rightarrow$ $y_1=4$
Thus, the points are $(1,6)$ and $(-1,4)$
Hence, option 'D' is correct.