Let $f: R \rightarrow R$ be defined $f(x)=a e^{2 x}+b e^{x}+c x$. If $f(0)=-1, f^{\prime}\left(\log _{e} 2\right)=21$ and $\int_{0}^{\log _{c} 4}(f(x)-c x) d x=\frac{39}{2}$, then the value of $|a+b+c|$ equals :

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Definite Integrals

Let

f : R \to R

be defined

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

. If

f (0) = - 1, f^{'} (lo g_{e} 2) = 21

and

\int_{0}^{l o g_{c} 4} (f (x) - c x) d x = \frac{39}{2}

, then the value of

∣ a + b + c ∣

equals :

(A)16
(B)10
(C)12
(D)8✓ Correct

Answer: (D)

Solution

From f(0)=-1 we get a+b=-1; f'(ln2)=21 gives 8a+2b+c=21; and the integral =39/2 gives 7.5a+3b=19.5. Solving, a=5, b=-6, c=-7, so |a+b+c|=8. Option D (value 8) is correct.

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Single correct option

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