Question

Let $f$ be a twice differentiable function on $R$ such that $t^{2}f\left(x\right)-2t{f}^{\prime }\left(x\right)+f^{\prime \prime }\left(x\right)=0$ has two equal values of $t$ for all $x$ and $f\left(0\right)=1,f^{\prime }\left(0\right)=2.$ Then the value of $3\underset{x\to 0}{lim}(\frac{f\left(x\right)-1}{x}-\frac{t}{2})$ is

Answer 1

$t^{2}f\left(x\right)-2t{f}^{\prime }\left(x\right)+f^{\prime \prime }\left(x\right)=0\dots \left(1\right)$
Equation has two equal values of $t$
$\Rightarrow D=0$
$\Rightarrow \left(2f^{\prime }\right(x){)}^2=4f(x\left)f^{\prime \prime }\right(x)\Rightarrow \frac{f^{\prime }\left(x\right)}{f\left(x\right)}=\frac{f^{\prime \prime }\left(x\right)}{f^{\prime }\left(x\right)}$

On integrating both sides,
$\ln f\left(x\right)=\ln f^{\prime }\left(x\right)+C$
Put $x=0,$
$\ln f\left(0\right)=\ln f^{\prime }\left(0\right)+C$
$\Rightarrow 0=\ln 2+C$
$\Rightarrow C=-\ln 2$
$\therefore \ln f\left(x\right)=\ln f^{\prime }\left(x\right)-\ln 2$

$f\left(x\right)=\frac{f^{\prime }\left(x\right)}{2}$
$\Rightarrow \frac{f^{\prime }\left(x\right)}{f\left(x\right)}=2$

On integrating both sides,
$\ln f\left(x\right)=2x+k$
Put $x=0$
$\ln f\left(0\right)=0+k$
$\Rightarrow k=0$
$\therefore \ln f\left(x\right)=2x$
$\Rightarrow f\left(x\right)=e^{2x}$
On differentiating with respect to $x,$
$f^{\prime }\left(x\right)=2e^{2x}$
Again differentiating with respect to $x,$
$f^{\prime \prime }\left(x\right)=4e^{2x}$

Put the value of $f\left(x\right),f^{\prime }\left(x\right),f^{\prime \prime }\left(x\right)$ in equation $\left(1\right),$ we get
$t^2\left(e^{2x}\right)-2t\left(2e^{2x}\right)+4e^{2x}=0\Rightarrow e^{2x}(t^2-4t+4)=0\Rightarrow t=2$

Now, $L=\underset{x\to 0}{lim}\frac{f\left(x\right)-1}{x}-\frac{t}{2}$
$=\underset{x\to 0}{lim}\frac{e^{2x}-1}{x}-1$
$=\underset{x\to 0}{lim}\frac{2(e^{2x}-1)}{2x}-1$
$=2-1$
$=1$