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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

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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

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

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

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

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

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

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

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

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

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

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

\[ \Rightarrow 4 a^2 - 4a + 1 \leqslant 0\]

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

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

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

\[ \Rightarrow a = \frac{1}{2}\]

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

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

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

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

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

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

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

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

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

\[ \Rightarrow 4 a^2 - 4a + 1 \leqslant 0\]

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

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

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

\[ \Rightarrow a = \frac{1}{2}\]

  Is there an error in this question or solution?
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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