JEE MATHEMATICS MCQ: INCREASING & DECREASING

1 Easy

1 of 1

Analyzing +4 | -1

The minimum value of the twice differentiable function

f(x)=\int\limits_0^x e^{x-t} f^{\prime}(t) \mathrm{dt}-\left(x^2-x+1\right) e^x, x \in \mathbb{R}

, is

View answer and explanation

Correct option: A

-\frac{2}{\sqrt{e}}

Option A (-

\frac{{2}}{{\sqrt{{e}}}}

) is correct because it represents the minimum value of the function f(x) found by analyzing its derivatives.

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f\left( x \right) = \frac{{K\sin x + 2\cos x}}{{\sin x + \cos x}}

x

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a

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\sin x-\cos x-a x+b

f(x)=x e^{x(1-x)}, x \in R

f

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fog

f\left( x \right) = {x^3} + b{x^2} + cx + d

0 < {b^2} < c

\left( { - \infty ,\;\infty } \right)