Let f:0,1→R be a double differentiable function with f0=f1=0 and f"x+2f'x+fx≥0 ∀ x∈0,1 then

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Let $f : [0, 1] \to R$ be a double differentiable function with $f (0) = f (1) = 0$ and $f " (x) + 2 f' (x) + f (x) \geq 0$ $\forall x \in (0, 1)$ then

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if $e^{x} f (x)$ assumes its minimum value in [0, 1] at $x = \frac{1}{\sqrt{2}}$ then $f (x) + f' (x) < 0$ $\forall x \in (0, \frac{1}{\sqrt{2}})$

$f (x) > 0 \forall x \in [0, 1]$

If $g (x) = e^{x} f (x)$ then number of real solution of the equation $g (g (g (x))) = 0$ is 2

$f (x) \leq 0 \forall x \in [0, 1]$

answer is A, B, C.

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Detailed Solution

$Given {(f (x) e^{x})}^{''} {=e}^{x} (f^{"} (x) + 2 f' (x) + f (x)) \geq 0, So, f (x) e^{x} is concave$ $a n d f (0) = 0 a n d f (1) = 0$

$∴ f (x) \leq 0 \forall x \in [0, 1]$

$o p t i o n (2) : i f f (x) e^{x} h a s m i n i m u m a t x = \frac{1}{\sqrt{2}} t h e n {(f (x) e^{x})}^{'} = e^{x} (f (x) + f^{'} (x)) = 0 a t x = \frac{1}{\sqrt{2}}$