If a function f (x) defined by f(x) = \ cases a e ^ x + b e ^ - x ,...

Continuous at $x = 1, 3$

$f\left({1}^{-}\right) = f\left(1\right) ⇒ ae + b{e}^{- 1} = c . . . ( 1 )$

$f\left(3\right) = f\left({3}^{+}\right) ⇒ 9c = 9a + 6c ⇒ c = 3a . . . \left(2\right)$

From \left(1\right)$and \left(2\right)$

$b = ae ( 3 - e ) . . . \left(3\right)$

{f}^{'}\left(x\right) = \left\{\begin{cases}a{e}^{x} - b{e}^{- x}, & - 1 < x < 1 \\ 2cx, & 1 < x < 3 \\ 2ax + 2c, & 3 < x < 4\end{cases}\right)</p><p>{f}^{'}\left(0\right) = a - b, {f}^{'}\left(2\right) = 4c$</p><p>Given${f}^{'} ( 0 ) + {f}^{'} ( 2 ) = e$</p><p>$a - b + 4c = e . . . \left(4\right)$</p><p>By using eq. \left(1\right), \left(2\right), \left(3\right) \text{ and } \left(4\right)$

$a = \dfrac{e}{{e}^{2} - 3e + 13}$