Question:
The point on the curve $y=6 x-x^{2}$ at which the tangent to the curve is inclined at $\frac{\pi}{4}$ to the line $x+y=0$ is
A. $(-3,-27)$
B. $(3,9)$
C. $\left(\frac{7}{2}, \frac{35}{4}\right)$
D. $(0,0)$
Solution:
The curve $y=6 x-x^{2}$ has a point at which the tangent to the curve is inclined at to $\frac{\pi}{4}$ the line $x+y=0$.
Differentiating w.r.t. $x$,
$\frac{\mathrm{dy}}{\mathrm{dx}}=6-2 \mathrm{x}=\mathrm{m}_{1}$ and $\frac{\mathrm{dy}}{\mathrm{dx}}=-1=\mathrm{m}_{2}$
$\tan \theta=\left|\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{1+\mathrm{m}_{1} \mathrm{~m}_{2}}\right|$
$\Rightarrow \tan \frac{\pi}{4}=\left|\frac{6-2 \mathrm{x}+1}{1+2 \mathrm{x}-6}\right|$
On solving we get $x=3$
Thus $y=9$
Hence, option B is correct.