Let gx-14f2x2-1-12f 1-x2 for all x- where f-x-0 - x- If gx is necessarily strictly increasing in the interval a-0-b- then the value of a-b is

nbsp;f”(x) gt;0⇒ f'(x) is strictly increasing function. Also, nbsp;g(x)=14f(2x^2 1)+12f (1 x^2) g(x) is necessarily strictly increasing if g'(x) gt;0 ⇒ ...

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Byju's Answer

Standard XII

Mathematics

Condition for Strictly Increasing

Let gx=14f2x2...

Question

Let $g (x) = \frac{1}{4} f (2 x^{2} - 1) + \frac{1}{2} f (1 - x^{2})$ for all $x \in R,$ where $f^{''} (x) > 0 \forall x \in R .$ If $g (x)$ is necessarily strictly increasing in the interval $(a, 0) \cup (b, \infty),$ then the value of $(a + b)$ is

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Solution

$f^{''} (x) > 0 \Rightarrow f^{'} (x)$ is strictly increasing function.
Also, $g (x) = \frac{1}{4} f (2 x^{2} - 1) + \frac{1}{2} f (1 - x^{2})$
$g (x)$ is necessarily strictly increasing if
$g^{'} (x) > 0$
$\Rightarrow \frac{1}{4} f^{'} (2 x^{2} - 1) \cdot (4 x) + \frac{1}{2} f^{'} (1 - x^{2}) \cdot (- 2 x) > 0$
$\Rightarrow x {f^{'} (2 x^{2} - 1) - f^{'} (1 - x^{2})} > 0$

Case $1 :$
If $x > 0,$ then $f^{'} (2 x^{2} - 1) - f^{'} (1 - x^{2}) > 0$
$\Rightarrow 2 x^{2} - 1 > 1 - x^{2}$
$\Rightarrow x \in (- \infty, - \sqrt{\frac{2}{3}}) \cup (\sqrt{\frac{2}{3}}, \infty)$
$∴ x \in (\sqrt{\frac{2}{3}}, \infty) \dots (1)$

Case $2 :$
If $x < 0,$ then $f^{'} (2 x^{2} - 1) - f^{'} (1 - x^{2}) < 0$
$\Rightarrow 2 x^{2} - 1 < 1 - x^{2}$
$\Rightarrow x \in (- \sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}})$
$∴ x \in (- \sqrt{\frac{2}{3}}, 0) \dots (2)$
From $(1)$ and $(2),$ we have
$x \in (- \sqrt{\frac{2}{3}}, 0) \cup (\sqrt{\frac{2}{3}}, \infty)$
Hence, $a + b = 0$

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Let g(x)=14f(2x2−1)+12f(1−x2) for all x∈R, where f′′(x)>0 ∀ x∈R. If g(x) is necessarily strictly increasing in the interval (a,0)∪(b,∞), then the value of (a+b) is

Let $g (x) = \frac{1}{4} f (2 x^{2} - 1) + \frac{1}{2} f (1 - x^{2})$ for all $x \in R,$ where $f^{''} (x) > 0 \forall x \in R .$ If $g (x)$ is necessarily strictly increasing in the interval $(a, 0) \cup (b, \infty),$ then the value of $(a + b)$ is