JEE Mains · Maths · STD 12 - 7.2 definite integral
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 \(\int_0^{\log _e 4}(f(x)-c x) d x=\frac{39}{2}\), then the value of \(|a+b+c|\) equals :
- A \(16\)
- B \(10\)
- C \(12\)
- D \(8\)
Answer & Solution
Correct Answer
(D) \(8\)
Step-by-step Solution
Detailed explanation
\(f(x)=a e^{2 x}+b e^x+c x \quad f(0)=-1\) \(a+b=-1\) \(\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{ae}^{2 \mathrm{x}}+\mathrm{be}^{\mathrm{x}}+\mathrm{c} \quad \mathrm{f}^{\prime}(\ln 2)=21\) \(8 a+2 b+c=21\) \(\int_0^{\ln 4}\left(a e^{2 x}+b e^x\right) d x=\frac{39}{2}\)…
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