Question

If the line $3x-4y=0$ is tangent in the first quadrant to the curve $y=x^3+k,$ then value of $k$ is

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{-1}{8}$
• D. $\frac{-1}{2}$

Answer 1

The correct option is B $\frac{1}{4}$

$y=x^3+k$
$\frac{dy}{dx}=3x^2=\frac{3}{4}\Rightarrow x=\pm \frac{1}{2}$
When $x=\frac{1}{2},y=\frac{3}{8},\text{and}(\frac{1}{2},\frac{3}{8})$ lies on the curve $y=x^3+k$
$\text{So,}\frac{3}{8}=\frac{1}{8}+k\Rightarrow k=\frac{1}{4}$
When $x=\frac{-1}{2},y=\frac{-3}{8},$ and $(\frac{-1}{2},\frac{-3}{8})$ lies on the curve
So, $\frac{-3}{8}=\frac{-1}{8}+k\Rightarrow k=\frac{-1}{4}$
But we need the point lying in the first quadrant. Hence, $k=\frac{1}{4}$