JEE Main 2024 — Mathematics Calculus

Given: $\displaystyle f(x)=a{e}^{2x}+b{e}^{x}+cx,f(0)=-1,{f}^{'}(\mathrm{log}2)=21,{\int }_{0}^{\mathrm{log}4}(f(x)-cx)=\frac{39}{2}$f(x)=ae2x+bex+cx,f(0)=−1,f′(log2)=21,∫0log4​(f(x)−cx)=239​

$$\Rightarrow f(0)=a+b=-1...(i)$$⇒f(0)=a+b=−1...(i)

$$\Rightarrow {f}^{'}(x)=2a{e}^{2x}+b{e}^{x}+c$$⇒f′(x)=2ae2x+bex+c

$$\Rightarrow {f}^{'}(\mathrm{log}2)=2a(4)+b(2)+c=21$$⇒f′(log2)=2a(4)+b(2)+c=21

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

Also, $\displaystyle {\int }_{0}^{\mathrm{log}4}(a{e}^{2x}+b{e}^{x}+cx-cx)=\frac{39}{2}$∫0log4​(ae2x+bex+cx−cx)=239​

$$\Rightarrow {\int }_{0}^{\mathrm{log}4}(a{e}^{2x}+b{e}^{x})=\frac{39}{2}$$⇒∫0log4​(ae2x+bex)=239​

$$\Rightarrow {[\frac{a{e}^{2x}}{2}+b{e}^{x}]}_{0}^{\mathrm{log}4}=\frac{39}{2}$$⇒[2ae2x​+bex]0log4​=239​

$$\Rightarrow \frac{a{e}^{2\mathrm{log}4}}{2}+b{e}^{\mathrm{log}4}-\frac{a}{2}-b=\frac{39}{2}$$⇒2ae2log4​+belog4−2a​−b=239​

$$\Rightarrow \frac{15a}{2}+3b=\frac{39}{2}$$⇒215a​+3b=239​

$$\Rightarrow 15a+6b=39$$⇒15a+6b=39

Using equation $(i)$(i),

$$\Rightarrow 15a+6(-1-a)=39$$⇒15a+6(−1−a)=39

$$\Rightarrow 15a-6-6a=39$$⇒15a−6−6a=39

$$\Rightarrow 9a=45$$⇒9a=45

$$\Rightarrow a=5$$⇒a=5

$$\Rightarrow b=-1-5=-6$$⇒b=−1−5=−6

$$\Rightarrow c=21-8a-2b$$⇒c=21−8a−2b

$$\Rightarrow c=21-40+12=-7$$⇒c=21−40+12=−7

$$\Rightarrow c=-7$$⇒c=−7

$$\Rightarrow |a+b+c|=8$$⇒∣a+b+c∣=8