SATHEE: Differential Equations 1 Question 1

Differential Equations 1 Question 1

1. Let $f$ be a differentiable function such that $f (1) = 2$ and $f^{'} (x) = f (x)$ for all $x \in R$ . If $h (x) = f (f (x))$ , then $h^{'} (1)$ is equal to

(2019 Main, 12 Jan II)

(a) $4 e^{2}$

(b) $4 e$

(d) $2 e^{2}$

Show Answer

Answer:

Correct Answer: 1. (b)

Solution:

Given that, $f^{'} (x) = f (x)$

$\begin{aligned} \Rightarrow & \frac{f^{'} (x)}{f (x)} & = 1 \\ \Rightarrow & \int \frac{f^{'} (x)}{f (x)} d x & = \int 1 \cdot d x \end{aligned}$

[by integrating both sides w.r.t. $x$ ]

$\Rightarrow Put f (x) = t \Rightarrow f^{'} (x) d x = d t$

$∴ \int \frac{d t}{t} = \int 1 d x$

$\Rightarrow \ln | t | = x + C ∵ \int \frac{d x}{x} = \ln | x | + C$

$\Rightarrow \ln | f (x) | = x + C$

$[∵ t = \tilde{f (x)}]$

$\begin{aligned} ∵ f (1) = 2 \\ So, \ln (2) = 1 + C \\ \Rightarrow C = \ln 2 - \ln e [∵ \ln e = 1] \\ \Rightarrow C = \ln \frac{2}{e} [∵ \ln A - \ln B = \ln \frac{A}{B}] \end{aligned}$

From Eq. (i), we get

$\ln | f (x) | = x + \ln \frac{2}{e}$

$\Rightarrow \ln | f (x) | - \ln \frac{2}{e} = x$

$\Rightarrow \ln | \frac{e f (x)}{2} | = x [∵ \ln A - \ln B = \ln \frac{A}{B}]$

$\Rightarrow | \frac{e}{2} f (x) | = e^{x} [∵ \ln a = b \Rightarrow a = e^{b}, a > 0]$

$\begin{aligned} \Rightarrow | f (x) | = 2 e^{x - 1} ∵ | \frac{e}{2} f (x) | = \frac{e}{2} | f (x) | \\ f (x) = 2 e^{x - 1} or - 2 e^{x - 1} \\ Now, h (x) = f (f (x)) \\ \Rightarrow h^{'} (x) = f^{'} (f (x)) \cdot f^{'} (x) \\ [on differentiating both sides w.r.t. ’ x ‘] \\ \Rightarrow h^{'} (1) = f^{'} (f (1)) \cdot f^{'} (1) \\ \begin{matrix} = f^{'} (2) \cdot f^{'} (1) [∵ f (1) = 2 (given)] \\ = 2 e^{2 - 1} \cdot 2 e^{1 - 1} [∵ f^{'} (x) = 2 e^{x - 1} or - 2 e^{x - 1}] \\ = 4 e \end{matrix} \end{aligned}$