Question

The number of values of c such that the straight line $3x+4y=c$ touches the curve $\frac{x^4}{2}=x+y$ is

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is B $1$

$3x+4y=c$
Or
$3+4y^{\prime }=0$
Or
$y^{\prime }=\frac{-3}{4}$.
This is the slope of the tangent.
Now
Differentiating the equation of the curve with respect to $x$, we get
$2x^3=1+y^{\prime }$
Now
$y^{\prime }=\frac{-3}{4}$
Hence
$2x^3=\frac{1}{4}$
Or
$x^3=\frac{1}{8}$
Hence
$x=0.5$
Substituting in the equation of the curve we get
$\frac{1}{32}=\frac{1}{2}+y$
Or
$y=\frac{-15}{32}$.
Hence substituting the above points on the equation of the line, we get
$\frac{3}{2}+\frac{-15}{8}=c$
Or
$c=\frac{-3}{8}$
Hence there is only $1$ value of $c$.