Let f: R → R be defined f(x) = a e ^ 2x + b e ^ x + cx . If f ( 0 )...

Given: $f\left(x\right) = a{e}^{2x} + b{e}^{x} + cx, f\left(0\right) = - 1, {f}^{'}\left(\log 2\right) = 21, {∫}_{0}^{\log 4}\left(f\left(x\right) - cx\right) = \dfrac{39}{2}$

$⇒ f\left(0\right) = a + b = - 1 . . . \left(i\right)$

$⇒ {f}^{'}\left(x\right) = 2a{e}^{2x} + b{e}^{x} + c$

$⇒ {f}^{'}\left(\log 2\right) = 2a\left(4\right) + b\left(2\right) + c = 21$

$⇒ 8a + 2b + c = 21 . . . \left(ii\right)$

Also, ${∫}_{0}^{\log 4}\left(a{e}^{2x} + b{e}^{x} + cx - cx\right) = \dfrac{39}{2}$

$⇒ {∫}_{0}^{\log 4}\left(a{e}^{2x} + b{e}^{x}\right) = \dfrac{39}{2}$

$⇒ {\left[\dfrac{a{e}^{2x}}{2} + b{e}^{x}\right]}_{0}^{\log 4} = \dfrac{39}{2}$

$⇒ \dfrac{a{e}^{2\log 4}}{2} + b{e}^{\log 4} - \dfrac{a}{2} - b = \dfrac{39}{2}$

$⇒ \dfrac{15a}{2} + 3b = \dfrac{39}{2}$

$⇒ 15a + 6b = 39$

Using equation \left(i\right)$,</p><p>$⇒ 15a + 6\left(- 1 - a\right) = 39$</p><p>$⇒ 15a - 6 - 6a = 39$</p><p>$⇒ 9a = 45$</p><p>$⇒ a = 5$</p><p>$⇒ b = - 1 - 5 = - 6$</p><p>$⇒ c = 21 - 8a - 2b$</p><p>$⇒ c = 21 - 40 + 12 = - 7$</p><p>$⇒ c = - 7$</p><p>$⇒ \left|a + b + c\right| = 8