Mathematics · 2024
JEE Main · 30 January 2024, Shift 2 · Q10
Let f: R → R be defined as f(x)=a e^2 x+b e^x+c x. If f(0)=-1, f^′( log_e 2)=21 and ∫_0^log_e 4(f(x)-c x) d x=39/2, then the value of |a+b+c| equals
Let $\displaystyle f: \mathbb{R} \rightarrow \mathbb{R}$ be defined as $\displaystyle f(x)=a e^{2 x}+b e^x+c x$. If $\displaystyle f(0)=-1, f^{\prime}\left(\log _e 2\right)=21$ and $\displaystyle \int_0^{\log _e 4}(f(x)-c x) d x=\frac{39}{2}$, then the value of $\displaystyle |a+b+c|$ equals
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle 8$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.