1. Given function

$f(x)=ae^{2x}+be^x+cx$

We are given:

• $f(0)=-1$
• $f'(\ln 2)=21$
• $\int_0^{\ln 4}(f(x)-cx)\,dx=\frac{39}{2}$

We need to find $|a+b+c|$.

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2. Use $f(0)=-1$

Since

$f(0)=ae^0+be^0+c\cdot 0=a+b$

so,

$a+b=-1 \quad \text{...(1)}$

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3. Differentiate $f(x)$

$f'(x)=2ae^{2x}+be^x+c$

Now at $x=\ln 2$:

$e^{\ln 2}=2, \qquad e^{2\ln 2}=4$

Hence,

$f'(\ln 2)=2a(4)+b(2)+c=8a+2b+c$

Given this equals 21, so

$8a+2b+c=21 \quad \text{...(2)}$

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4. Use the integral condition

We have

$f(x)-cx=ae^{2x}+be^x$

Therefore,

$\int_0^{\ln 4}(f(x)-cx)\,dx=\int_0^{\ln 4}(ae^{2x}+be^x)\,dx$

Integrate:

$\int ae^{2x}dx=\frac{a}{2}e^{2x}, \qquad \int be^x dx=be^x$

So,

$\int_0^{\ln 4}(ae^{2x}+be^x)dx=\left[\frac{a}{2}e^{2x}+be^x\right]_0^{\ln 4}$

Now,

$e^{\ln 4}=4, \qquad e^{2\ln 4}=16$

Thus,

$\left(\frac{a}{2}\cdot 16+b\cdot 4\right)-\left(\frac{a}{2}\cdot 1+b\cdot 1\right)=\frac{39}{2}$

$8a+4b-\frac{a}{2}-b=\frac{39}{2}$

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

Multiply by 2:

$15a+6b=39$

Divide by 3:

$5a+2b=13 \quad \text{...(3)}$

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5. Solve for $a,b,c$

From (1):

$a+b=-1$

So,

$b=-1-a$

Substitute into (3):

$5a+2(-1-a)=13$

$5a-2-2a=13$

$3a=15$

$a=5$

Then,

$b=-1-5=-6$

Now use (2):

$8a+2b+c=21$

$8(5)+2(-6)+c=21$

$40-12+c=21$

$28+c=21$

$c=-7$

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6. Compute $|a+b+c|$

$a+b+c=5-6-7=-8$

Hence,

$|a+b+c|=8$

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7. Check with options

The correct option is:

C: 8

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8. Compare with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.