JEE Mathematics 2024 Question with Solution

Given:

• $f(x) = ae^{2x} + be^x + cx$
• $f(0) = -1$
• $f'(\log_e 2) = 21$
• $\int_{0}^{\log_e 4} (f(x) - cx) \, dx = \frac{39}{2}$

Find: $|a+b+c|$

From $f(0) = -1$,

$$a \cdot e^0 + b \cdot e^0 + c \cdot 0 = -1$$

so,

$$a + b = -1$$

Now differentiate:

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

Using $f'(\log_e 2) = 21$,

$$2a e^{2\log_e 2} + b e^{\log_e 2} + c = 21$$ $$8a + 2b + c = 21$$

Also,

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

Therefore,

$$\int_{0}^{\log_e 4} (ae^{2x} + be^x) \, dx = \frac{39}{2}$$

Integrating,

$$\left[\frac{a}{2}e^{2x} + be^x\right]_{0}^{\log_e 4} = \frac{39}{2}$$ $$\frac{a}{2} e^{2\log_e 4} + b e^{\log_e 4} - \left(\frac{a}{2} + b\right) = \frac{39}{2}$$ $$8a + 4b - \left(\frac{a}{2} + b\right) = \frac{39}{2}$$ $$\frac{15a + 6b}{2} = \frac{39}{2}$$ $$15a + 6b = 39$$

Using $a+b=-1$, we get $b=-1-a$. Substitute into the last equation:

$$15a + 6(-1-a) = 39$$ $$9a = 45$$ $$a = 5$$

Hence,

$$b = -1-5 = -6$$

Now use $8a+2b+c=21$:

$$8(5) + 2(-6) + c = 21$$ $$40 - 12 + c = 21$$ $$c = -7$$

Thus,

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

So,

$$|a+b+c| = 8$$

Therefore, the correct option is D.

The solution gives the equations directly from the three conditions:

1. From $f(0)=-1$, we get $a+b=-1$.
2. From $f'(\log_e 2)=21$, we get $8a+2b+c=21$.
3. From the integral condition, after simplifying $f(x)-cx$, we get $15a+6b=39$.

Now solve:

$$b=-1-a$$

Substitute into

$$15a+6b=39$$

so,

$$15a+6(-1-a)=39$$ $$9a=45$$ $$a=5$$

Then,

$$b=-6$$

and from

$$8a+2b+c=21$$

we obtain

$$c=-7$$

Hence,

$$a+b+c=-8$$

and therefore,

$$|a+b+c|=8$$

So the correct option is D.