Question

Let $f$ be a differentiable function satisfying $f^{\prime }(x)=2f(x)+10$ and $f(0)=0$ then the number of roots of the equation $f(x)+5{\sec }^{2}x=0$ in $(0,2\pi )$ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (A)

Given, $f^{\prime }(x)=2(f(x)+5)$ or $\frac{dy}{y+5}=2dx$. On integrating, we get $\ln |(y+5)|=2x+c$, if $x=0;y=0\Rightarrow c=\ln 5$
$\therefore \ln \left(\frac{y+5}{5}\right)=2x\Rightarrow y+5=5e^{2x}\Rightarrow y=5e^{2x}-5\Rightarrow f(x)=5\left(e^{2x}-1\right)$
Now, $f(x)+5{\sec }^{2}x=5\left(e^{2x}+{\tan }^{2}x\right)=0$
So, no solution exist $(0,2\pi )$