If g(x)=2f2x3−3x2+f6x2−4x3−3,∀x∈R and f′′(x)>0,∀x∈R, then g(x) is incresing in

For g(x) to be increasing function g′(x)>0,∀x Since f′′(x)>0⇒f′(x) is increasing function Now, g(x)=2f2x3−3x2+f6x2−4x3−3g′(x)=26x2−6xf′2x3−3x2+12x−12x2f′6x2−4x3−3g′(x)=12x(x−1)f′2x3−3x2−f′6x2−4x3−3>0 case 1:if f′2x3−3x2>f′6x2−4x3−3⇒2x3−3x2>6x2−4x3−3∵f′(x) is increases ⇒2x3−3x2+1>0⇒(x−1)2(2x+1)>0 ⇒x>-12∴g′(x)>0 ⇒x(x−1)>0 and (2x+1)>0∴g(x) is increasing on −12,0∪(1,∞) case 2:if f′2x3−3x2 0

If g(x)=2f2x3−3x2+f6x2−4x3−3,∀x∈R and f′′(x)>0,∀x∈R, then g(x) is incresing in

For g(x) to be increasing function g′(x)>0,∀x Since f′′(x)>0⇒f′(x) is increasing function Now, g(x)=2f2x3−3x2+f6x2−4x3−3g′(x)=26x2−6xf′2x3−3x2+12x−12x2f′6x2−4x3−3g′(x)=12x(x−1)f′2x3−3x2−f′6x2−4x3−3>0 case 1:if f′2x3−3x2>f′6x2−4x3−3⇒2x3−3x2>6x2−4x3−3∵f′(x) is increases ⇒2x3−3x2+1>0⇒(x−1)2(2x+1)>0 ⇒x>-12∴g′(x)>0 ⇒x(x−1)>0 and (2x+1)>0∴g(x) is increasing on −12,0∪(1,∞) case 2:if f′2x3−3x2 0

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

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

$(0, \infty)$

$(- \infty, 1)$

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

answer is B.

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Detailed Solution

$\begin{array}{l} For g (x) to be increasing function g^{'} (x) > 0, \forall x \\ Since f^{''} (x) > 0 \end{array}$

$\begin{array}{l} \Rightarrow f^{'} (x) is increasing function \\ Now, g (x) = 2 f (2 x^{3} - 3 x^{2}) + f (6 x^{2} - 4 x^{3} - 3) \\ g^{'} (x) = 2 (6 x^{2} - 6 x) f^{'} (2 x^{3} - 3 x^{2}) + (12 x - 12 x^{2}) f^{'} (6 x^{2} - 4 x^{3} - 3) \\ g^{'} (x) = 12 x (x - 1) [f^{'} (2 x^{3} - 3 x^{2}) - f^{'} (6 x^{2} - 4 x^{3} - 3)] > 0 \end{array}$

$\begin{array}{l} c a s e 1 : i f f^{'} (2 x^{3} - 3 x^{2}) > f^{'} (6 x^{2} - 4 x^{3} - 3) \\ \Rightarrow 2 x^{3} - 3 x^{2} > 6 x^{2} - 4 x^{3} - 3 \\ (∵ f^{'} (x) is increases) \\ \Rightarrow 2 x^{3} - 3 x^{2} + 1 > 0 \\ \Rightarrow (x - 1)^{2} (2 x + 1) > 0 \Rightarrow x > \frac{- 1}{2} \end{array}$

$\begin{array}{l} ∴ g^{'} (x) > 0 \Rightarrow x (x - 1) > 0 a n d^{} (2 x + 1) > 0 \\ ∴ g (x) is increasing on (- \frac{1}{2}, 0) \cup (1, \infty) \end{array}$