First slide

Monotonicity

Question

$If g (x) = 2 f (2 x^{3} - 3 x^{2}) + f (6 x^{2} - 4 x^{3} - 3), \forall x \in R and f^{''} (x) > 0, \forall x \in R, then g (x) is incresing in$

Difficult

A

$(- \infty, \frac{- 1}{2}) \cup (0, 1)$
B

$(\frac{- 1}{2}, 0) \cup (1, \infty)$
C

$(0, \infty)$
D

$(- \infty, 1)$

Solution

$\begin{array}{l} For g (x) to be increasing function g^{'} (x) > 0, \forall x \\ Since f^{''} (x) > 0 \end{array}$
$\begin{array}{l} \Rightarrow f^{'} (x) is increasing function \\ Now, g (x) = 2 f (2 x^{3} - 3 x^{2}) + f (6 x^{2} - 4 x^{3} - 3) \\ g^{'} (x) = 2 (6 x^{2} - 6 x) f^{'} (2 x^{3} - 3 x^{2}) + (12 x - 12 x^{2}) f^{'} (6 x^{2} - 4 x^{3} - 3) \\ g^{'} (x) = 12 x (x - 1) [f^{'} (2 x^{3} - 3 x^{2}) - f^{'} (6 x^{2} - 4 x^{3} - 3)] > 0 \end{array}$
$\begin{array}{l} c a s e 1 : i f f^{'} (2 x^{3} - 3 x^{2}) > f^{'} (6 x^{2} - 4 x^{3} - 3) \\ \Rightarrow 2 x^{3} - 3 x^{2} > 6 x^{2} - 4 x^{3} - 3 \\ (∵ f^{'} (x) is increases) \\ \Rightarrow 2 x^{3} - 3 x^{2} + 1 > 0 \\ \Rightarrow (x - 1)^{2} (2 x + 1) > 0 \Rightarrow x > \frac{- 1}{2} \end{array}$
$\begin{array}{l} ∴ g^{'} (x) > 0 \Rightarrow x (x - 1) > 0 a n d^{} (2 x + 1) > 0 \\ ∴ g (x) is increasing on (- \frac{1}{2}, 0) \cup (1, \infty) \end{array}$
$c a s e 2 :$ $i f f^{'} (2 x^{3} - 3 x^{2}) < f^{'} (6 x^{2} - 4 x^{3} - 3) a n d x (x - 1) < 0$
$t h e n t h e r e i s n o v a l u e o f x s a t i s f y i n g g^{'} (x) > 0$

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