Question

If $\varphi \left(x\right)=\varphi \prime \left(x\right)$ and $\varphi \left(1\right)=2$, then $\varphi \left(3\right)$ equals

• A. $e^2$
• B. $2e^2$
• C. $3e^2$
• D. $2e^3$

Answer 1

The correct option is B $2e^2$

$\varphi \left(x\right)=\varphi \left(x\right)$ and $\varphi \left(1\right)=2$

Let $y=\varphi \left(x\right)$

$y=\frac{dy}{dx}$

$\int dx=\int \frac{dy}{y}$

$x+c=lny\Rightarrow y=e^{x+c}$

Given

$\varphi \left(1\right)=2\Rightarrow 2=e^{1+c}$

$\Rightarrow l{n}^2=1+c\Rightarrow c=l{n}^2-1$

$c=ln2-lne$ $[\therefore l{n}_{e}e=1]$

$c=ln\left(\frac{2}{e}\right)$

$\therefore y=e^{x+ln\left(\frac{2}{e}\right)}$

$\varphi \left(3\right)=?y=e^{3+\ln \left(\frac{2}{e}\right)}=e^3.e^{ln\left(\frac{2}{e}\right)}$

$=e^3\times \frac{2}{e}$ $[\because a^{log{a}^b}=b]$

$=\varphi \left(3\right)=2e^2$

options $B$