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# Solution for If the Function F(X) = 2 Tan X + (2a + 1) Loge | Sec X | + (A − 2) X Is Increasing on R, Then (A) A ∈ (1/2, ∞) (B) A ∈ (−1/2, 1/2) (C) A = 1/2 (D) A ∈ R - CBSE (Commerce) Class 12 - Mathematics

ConceptIncreasing and Decreasing Functions

#### Question

If the function f(x) = 2 tan x + (2a + 1) loge | sec x | + (a − 2) x is increasing on R, then
(a) a ∈ (1/2, ∞)
(b) a ∈ (−1/2, 1/2)
(c) a = 1/2
(d) a ∈ R

#### Solution

$f(x) = 2 \tan x + \left( 2a + 1 \right) \log_e \left| \sec x \right| + \left( a - 2 \right) x$

$\text { When }\sec x > 0 \Rightarrow \left| \sec x \right| = \sec x$

$\frac{d}{dx}\left\{ f\left( x \right) \right\} = 2 \sec^2 x + \left( 2a + 1 \right)\frac{1}{\sec x} \times \sec x \tan x + \left( a - 2 \right)$

$= 2 \sec^2 x + \left( 2a + 1 \right)\tan x + \left( a - 2 \right)$

$\text { For f(x) to be increasing},$

$2se c^2 x + \left( 2a + 1 \right)\tan x + \left( a - 2 \right) \geqslant 0$

$\Rightarrow 2 + 2 \tan^2 x + \left( 2a + 1 \right)\tan x + a - 2 \geqslant 0$

$\Rightarrow 2 \tan^2 x + \left( 2a + 1 \right)\tan x + a \geqslant 0$

$\text { Its discriminant } \leqslant 0 \left[ \because a x^2 + bx + c \geqslant 0 \Rightarrow b^2 - 4ac \leqslant 0 \right]$

$\Rightarrow \left( 2a + 1 \right)^2 - 4 . 2 . a \leqslant 0$

$\Rightarrow 4 a^2 - 4a + 1 \leqslant 0$

$\Rightarrow \left( 2a - 1 \right)^2 \leqslant 0$

$\left( 2a - 1 \right)^2 < 0 \text { cannot be possible } .$

$\therefore \left( 2a - 1 \right)^2 = 0$

$\Rightarrow a = \frac{1}{2}$

$\text { When } \sec x < 0 \Rightarrow \left| \sec x \right| = - \sec x$

$\frac{d}{dx}\left\{ f\left( x \right) \right\} = 2 \sec^2 x + \left( 2a + 1 \right)\frac{1}{- \sec x} \times \sec x \tan x + \left( a - 2 \right)$

$= 2 \sec^2 x - \left( 2a + 1 \right)\tan x + \left( a - 2 \right)$

$\text { For f(x) to be increasing,}$

$2se c^2 x - \left( 2a + 1 \right)\tan x + \left( a - 2 \right) \geqslant 0$

$\Rightarrow 2 + 2 \tan^2 x - \left( 2a + 1 \right)\tan x + a - 2 \geqslant 0$

$\Rightarrow 2 \tan^2 x - \left( 2a + 1 \right)\tan x + a \geqslant 0$

$\text { Its discriminant } \leqslant 0 \left[ \because a x^2 + bx + c \geqslant 0 \Rightarrow b^2 - 4ac \leqslant 0 \right]$

$\Rightarrow \left\{ - \left( 2a + 1 \right) \right\}^2 - 4 . 2 . a \leqslant 0$

$\Rightarrow 4 a^2 - 4a + 1 \leqslant 0$

$\Rightarrow \left( 2a - 1 \right)^2 \leqslant 0$

$\left( 2a - 1 \right)^2 < 0 \text { cannot be possible } .$

$\therefore \left( 2a - 1 \right)^2 = 0$

$\Rightarrow a = \frac{1}{2}$

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Solution If the Function F(X) = 2 Tan X + (2a + 1) Loge | Sec X | + (A − 2) X Is Increasing on R, Then (A) A ∈ (1/2, ∞) (B) A ∈ (−1/2, 1/2) (C) A = 1/2 (D) A ∈ R Concept: Increasing and Decreasing Functions.
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