Differential Equations 1 Question 1
1. Let $f$ be a differentiable function such that $f(1)=2$ and $f^{\prime}(x)=f(x)$ for all $x \in R$. If $h(x)=f(f(x))$, then $h^{\prime}(1)$ is equal to
(2019 Main, 12 Jan II)
(a) $4 e^{2}$
(b) $4 e$
(c) $2 e$
(d) $2 e^{2}$
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Answer:
Correct Answer: 1. (b)
Solution:
- Given that, $f^{\prime}(x)=f(x)$
$$ \begin{array}{rlrl} \Rightarrow & \frac{f^{\prime}(x)}{f(x)} & =1 \\ \Rightarrow & & \int \frac{f^{\prime}(x)}{f(x)} d x & =\int 1 \cdot d x \end{array} $$
[by integrating both sides w.r.t. $x$ ]
$\Rightarrow \quad \operatorname{Put} f(x)=t \Rightarrow f^{\prime}(x) d x=d t$
$$ \therefore \quad \int \frac{d t}{t}=\int 1 d x $$
$\Rightarrow \quad \ln |t|=x+C \quad \because \int \frac{d x}{x}=\ln |x|+C$
$\Rightarrow \ln |f(x)|=x+C$
$$ [\because t=\widetilde{f(x)}] $$
$$ \begin{aligned} & \because \quad f(1)=2 \\ & \text { So, } \quad \ln (2)=1+C \\ & \Rightarrow \quad C=\ln 2-\ln e \quad[\because \ln e=1] \\ & \Rightarrow \quad C=\ln \frac{2}{e} \quad\left[\because \ln A-\ln B=\ln \frac{A}{B}\right] \end{aligned} $$
From Eq. (i), we get
$$ \ln |f(x)|=x+\ln \frac{2}{e} $$
$\Rightarrow \ln |f(x)|-\ln \frac{2}{e}=x$
$\Rightarrow \quad \ln \left|\frac{e f(x)}{2}\right|=x \quad\left[\because \ln A-\ln B=\ln \frac{A}{B}\right]$
$\Rightarrow\left|\frac{e}{2} f(x)\right|=e^{x} \quad\left[\because \ln a=b \Rightarrow a=e^{b}, a>0\right]$
$$ \begin{aligned} & \Rightarrow \quad|f(x)|=2 e^{x-1} \quad \because\left|\frac{e}{2} f(x)\right|=\frac{e}{2}|f(x)| \\ & f(x)=2 e^{x-1} \text { or }-2 e^{x-1} \\ & \text { Now, } \quad h(x)=f(f(x)) \\ & \Rightarrow \quad h^{\prime}(x)=f^{\prime}(f(x)) \cdot f^{\prime}(x) \\ & \text { [on differentiating both sides w.r.t. ’ } x \text { ‘] } \\ & \Rightarrow \quad h^{\prime}(1)=f^{\prime}(f(1)) \cdot f^{\prime}(1) \\ & \begin{array}{l} =f^{\prime}(2) \cdot f^{\prime}(1) \quad[\because f(1)=2 \text { (given) }] \\ =2 e^{2-1} \cdot 2 e^{1-1} \quad\left[\because f^{\prime}(x)=2 e^{x-1} \text { or }-2 e^{x-1}\right] \\ =4 e \end{array} \end{aligned} $$
