Question

If $f(x) = a{e^{2x}} + b{e^x} + cx$ , satisfies the conditions $f(0) = - 1,f'(\log 2) = 31$ and $\int\limits_0^{\log 4} {(f(x) - cx)dx = \dfrac{{39}}{2}}$ , then

Answer 1

Hint : Substitute given values of function and then solve the equations.
Initially, substitute the values of the functions corresponding to the variable values given below. After that you will get equations with respect to the variable $x$ . Then after getting the equations, solve the equations further with labelling them as the corresponding equation number and substitute in the value of the limit function given. By doing this process you will get the values of the constants $a,b,c$ mentioned in the question.

Complete step-by-step answer :
$f(x) = a{e^{2x}} + b{e^x} + cx \\\ \Rightarrow f(0) = a{e^{2\left( 0 \right)}} + b{e^{\left( 0 \right)}} + c\left( 0 \right) \\\ \Rightarrow f(0) = a\left( 1 \right) + b\left( 1 \right) + c\left( 0 \right) \\\ \Rightarrow f(0) = a + b \\\ \therefore a + b = - 1 - - - (1)\left( {\because f(0) = - 1} \right) \;$
Now differentiating both sides of the given equation such that we get
$f'(x) = 2a{e^{2x}} + b{e^x} + c$
Now, using $f'(\log 2) = 31$ , we substitute in f(x) and get an equation
$f'(log2) = 2a{e^2}^{log2} + b{e^l}^{og2} + c$
With the properties of log which is $lo{g_e}^e = 1$ and $a{e^{2{{\log }_e}2}} = a{(2 \times 2)^{{{\log }_e}^e}}$ and remaining is accordingly sent to power, which gives us the equation

$$f'(\log 2) = 2a{(2 \times 2)^{{{\log }_e}^e}} + b{(2)^{{{\log }_e}^e}} + c \\\ \Rightarrow f'(\log 2) = 2a(4) + 2b + c = 31 \\\ \Rightarrow 8a + 2b + c = 31\left( {f'(\log 2) = 31} \right) - - - - (2) \;$$

$\int\limits_0^{\log 4} {(f(x) - cx)dx = \dfrac{{39}}{2}} \\\ \Rightarrow \int\limits_0^{\log 4} {(a{e^{2x}} + b{e^x})dx = \dfrac{{39}}{2}} \\\ \Rightarrow \left[ {a\dfrac{{{e^{2x}}}}{2} + b{e^x}} \right] _0^{\log \,4} = \dfrac{{39}}{2} \\\ \Rightarrow a\dfrac{{{e^{2\log 4}}}}{2} + b{e^{\log 4}} - \dfrac{a}{2} - b = \dfrac{{39}}{2} \\\ \Rightarrow \dfrac{{a16}}{2} + b4 - \dfrac{a}{2} - b = \dfrac{{39}}{2} \\\ \Rightarrow 8a + 4b - \dfrac{a}{2} - b = \dfrac{{39}}{2} \\\ \Rightarrow \dfrac{{15a}}{2} + 3b = \dfrac{{39}}{2} \\\ \Rightarrow 15a + 6b = 39 - - - (3) \;$
Solving (1), and (3) simultaneously,
Multiply equation (1) with 6, we get
$6a + 6b = - 6$
We solve them now

$${6a + 6b + 6 = 15a + 6b - 39} \\\ { \Rightarrow - 9a = - 45} \\\ { \Rightarrow a = 5}$$

then substitute in equation (1), we get
$5 + b = - 1 \\\ \Rightarrow b = - 6 \\\$
we have $a = 5,b = - 6$
Using the values of a and b in (2),
$\Rightarrow 8(5) + 2( - 6) + c = 31 \\\ \Rightarrow 40 - 12 + c = 31 \\\ \Rightarrow 28 + c = 31 \\\ \Rightarrow c = 31 - 28 \\\ \Rightarrow c = 3 \;$
Hence we get $c = 3$
So, the correct answer is “ a = 5, b = - 6 and c = 3”.

Note : In this problem we have to be careful while simplifying the equations and also in the integral solving method make sure you write the correct coefficients so that you won’t be getting the solutions or the values. And have a clear idea of the number of variables used and the number of values. When we come across problems like these, we have to first go with the brief reading of the problem and then the understanding of the question then thinking of the correct solution for the given question. The formulas play an important role in getting the equation solved. Without them solving the problem won’t be easy.