Jee Main 2024 30 01 2024 Shift 2 - Question17

Answer: (25)

Solution:

$\because f(x)=a e^{2 x}+b e^{x}+c x$

$$ \Rightarrow f(x)=2 a e^{2 x}+b e^{x}+c $$

$\because f^{\prime}(\ln 2)=4$

$\Rightarrow 4=2 a(4)+b(2)+c$

$\Rightarrow 8 a+2 b+c=4$

$\because \quad \int _0^{\ln 4}\left(a e^{2 x}+b e^{x}\right) d x=\frac{39}{2}$

$\Rightarrow \frac{a}{2}\left[e^{2 x}\right] _0^{\ln 4}+b\left(e^{x}\right) _0^{\ln 4}=\frac{39}{2}$

$\Rightarrow \frac{a}{2}[16-1]+b(4-1)=\frac{39}{2}$ $\Rightarrow \frac{15 a}{2}+3 b=\frac{39}{2}$

$\Rightarrow \frac{5 a}{2}+b=\frac{13}{2}$

$\Rightarrow 5 a+2 b=13$

Also $f(0)=-1$

$\Rightarrow-1=a+b$

From (ii) & (iii)

$5 a+5 b=-5$

$5 a+2 b=13$

$3 b=-18$

$\Rightarrow b=-6$

$\Rightarrow \quad a=5$

$\therefore c=-24$

$|a+b+c|=25$