Let be a twice differentiable function such that for all where a is a real number. Let Consider the following two statements: (I) g is increasing in (II) g is decreasing in

Question

Let $f : R \to R$ be a twice differentiable function such that $f'' (x) > 0$ for all $x \in R and$ $f' (a - 1) = 0$ where a is a real number. Let $g (x) = f (\tan^{2} x - 2 \tan x + a), 0 < x < \frac{π}{2} .$ Consider the following two statements:

(I) g is increasing in $(0, \frac{π}{4})$

(II) g is decreasing in $(\frac{π}{4}, \frac{π}{2})$

Then, [2026]

(I) g is increasing in $(0, \frac{π}{4})$

(II) g is decreasing in $(\frac{π}{4}, \frac{π}{2})$

Then, [2026]

Smart Achievers · Accepted Answer

(3)

$g (x) = f ({(\tan x - 1)}^{2} + a - 1)$

$g' (x) = f' ({(\tan x - 1)}^{2} + a - 1) \cdot 2 (\tan x - 1) \sec^{2} x$

$∵$ $f' (a - 1) = 0 and f'' (x) > 0$

$∴$ $f' ({(\tan x - 1)}^{2} + a - 1) > 0$

$g' (x) > 0 if (\tan x - 1) > 0$

$g is increasing in x \in (\frac{π}{4}, \frac{π}{2})$

$g' (x) < 0 if \tan x - 1 < 0$

$g is decreasing in x \in (0, \frac{π}{4})$

Solution

1 Only (I) is True

2 Both (I) and (II) are True

3 Neither (I) nor (II) is True

4 Only (II) is True

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