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Q.

$\begin{array}{l} Let F : R \to R be thrice differentiable function. Suppose that F (1) = 0, F (3) = - 4 and F' (x) < 0 for all x \in (1 / 2, 3) \\ let f (x) = x F (x) for all real values of x \end{array}$
The correct statement(s) is (are)

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a

$f (2) > 0$

b

$f' (x) \neq 0$ for any x $x \in (1, 3)$

c

$f' (1) < 0$

d

$f' (x) = 0$ for any $x \in (1, 3)$

answer is A, C.

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Detailed Solution

$\begin{array}{l} f' (x) = x F' (x) + F (x) \\ \Rightarrow f' (1) = F' (1) + F (1) = f' (1) < 0 \Rightarrow (A) \\ f (2) = 2 F (2) < 0 \Rightarrow (B) (\sin c e F (x) i s d e c r e a \sin g a n d F (1) = 0, F (3) = - 4 \\ For x \in (1, 3), f' (x) = x f' (x) + f (x) < 0 \Rightarrow (C) \end{array}$

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Let F:R→R be thrice differentiable function. Suppose that F1=0,F3=−4 and F'x<0 for all x∈1/2,3let fx=xFx for all real values of xThe correct statement(s) is (are)