First slide

Methods of integration

Question

$Let f (x) be a polynomial satisfying f (0) = 2, f^{'} (0) = 3, f^{''} (x) = f (x) then f^{2} (4) =$

Difficult

A

$\frac{25 {(e^{8} - 1)}^{2}}{4 e^{8}}$
B

$\frac{e^{8} - 1}{4 e^{8}}$
C

$\frac{{(e^{8} - 1)}^{2}}{4 e^{8}}$
D

$\frac{5 (e^{8} - 1)}{4 e^{8}}$

Solution

$Given f^{''} (x) = f (x)$
$\begin{array}{l} 2 f^{'} (x) f^{''} (x) = 2 f (x) f^{'} (x) \\ \frac{d}{d x} {\{f^{'} (x)\}}^{2} = \frac{d}{d x} {f (x)}^{2} \\ {\{f^{'} (x)\}}^{2} = {f (x)}^{2} + C \\ Now f (0) = 2 and f (0) = 3 \dots (1) \\ Equation from (1) \\ {\{f^{'} (0)\}}^{2} = {f (0)}^{2} + C \end{array}$
$\begin{array}{l} 9 = 4 + C \\ \Rightarrow C = 5 \\ ∴ f^{'} (x) = \sqrt{{f (x)}^{2} + 5} \\ Now \int \frac{1}{\sqrt{(5)^{2} + {f (x)}^{2}}} d {f (x)} = \int 1 d x \\ \log |f (x) + \sqrt{5 + {f (x)}^{2}}| = x + C_{1} \\ f (0) = 2 \Rightarrow \log 5 = C_{1} \end{array}$
$\begin{array}{l} ∴ \log |f (x) + \sqrt{5 + {f (x)}^{2}}| = x + \log 5 \\ \Rightarrow \log \{\frac{f (x) + \sqrt{5 + f^{2} (x)}}{5}\} = x \\ \Rightarrow f (x) + \sqrt{5 + f^{2} (x)} = 5 e^{x} \\ \Rightarrow \sqrt{5 + f^{2} (x)} + f (x) = 5 e^{x} \\ \Rightarrow \sqrt{5 + f^{2} (x)} + f (x) = 5 e^{x} and \Rightarrow \sqrt{5 + f^{2} (x)} - f (x) = 5 e^{- x} \\ \Rightarrow f (x) = \frac{5}{2} (e^{x} - e^{- x}) \\ \Rightarrow f (4) = \frac{5}{2} (e^{4} - e^{- 4}) \\ = \frac{5}{2} (\frac{e^{8} - 1}{e^{4}}) \end{array}$

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