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JEE Mains · Maths · STD 12 - 6. Application of derivatives

The minimum value of the twice differentiable function \(f(x)=\int_{0}^{x} e^{x-t} f^{\prime}(t) d t-\left(x^{2}-x+1\right) e^{x}, x \in R\), is.

  1. A \(-\frac{2}{\sqrt{ e }}\)
  2. B \(-2 \sqrt{ e }\)
  3. C \(-\sqrt{ e }\)
  4. D \(\frac{2}{\sqrt{ e }}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(-\frac{2}{\sqrt{ e }}\)

Step-by-step Solution

Detailed explanation

\(f(x)=e^{x} \cdot \int_{0}^{x} \frac{f^{\prime}(t)}{e^{t}} d t\) \(f^{\prime}(x)=e^{x} \cdot \int_{0}^{x} \frac{f^{\prime}(t)}{e^{t}} d t+e^{x} \cdot \frac{f^{\prime}(x)}{e^{x}}\) \(-\left[(2 x-1) \cdot e^{x}+\left(x^{2}-x+1\right) \cdot e^{x}\right]\)…
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