Question

Let $f$ be a differentiable function satisfying $f ^{\prime}( x )=2 f ( x )+10$ 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 $\frac{ dy }{ y +5}=2 dx$. On integrating, we get $\ln |( y +5)|=2 x + c$, if $x =0 ; y =0 \Rightarrow c =\ln 5$
$\therefore \ln \left(\frac{ y +5}{5}\right)=2 x \Rightarrow y +5=5 e ^{2 x } \Rightarrow y =5 e ^{2 x }-5 \Rightarrow f ( x )=5\left( e ^{2 x }-1\right)$
Now, $f(x)+5 \sec ^2 x=5\left(e^{2 x}+\tan ^2 x\right)=0$
So, no solution exist $(0,2 \pi)$