Question

If $9+f^{\prime \prime }\left(x\right)+f^{\prime }\left(x\right)=x^2+f^2\left(x\right)$ be the differential equation of a curve and let P be the point of minima of this curve then the number of tangents which can be drawn from P to the circle $x^2+y^2=8$ is?

• A. $2$
• B. $1$
• C. $0$
• D. Either $1$ or $2$

Answer 1

Answer: (A)

$2$

Given,

If $f^{\prime \prime }\left(x\right)+f^{\prime }\left(x\right)=x^2+f^2\left(x\right)$

P is the point of the minima in the above curve.

$\text{Therefore}{f}^{\prime }\left(x\right)=0$

$\text{and}{f}^{\prime \prime }\left(x\right)>0\text{[for minima}$

$\text{and concave upward ]}$

$f^{\prime \prime }\left(x\right)=x^2+f^2\left(x\right)-9>0$

$f^{\prime }\left(x\right)>9-x^2$

$f^2\left(x\right)>9-x^2$

$\text{Let}{f}^2\left(x\right)=y$

$\text{Then,}{x}^2+y^2>9$

$\text{Therefore, This is the equation of a circle whose}$

$\text{radius is greater than}{x}^2+y^2=8$

$\text{circle.}$

$\text{Therefore, any point outside this circle}$

$x^2+y^2=8\text{will draw}$ $ 2 tangents$

Therefore option A