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प्रश्न
If the function f(x) = 2 tan x + (2a + 1) loge | sec x | + (a − 2) x is increasing on R, then
विकल्प
a ∈ (1/2, ∞)
a ∈ (−1/2, 1/2)
a = 1/2
a ∈ R
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उत्तर
\[f(x) = 2 \tan x + \left( 2a + 1 \right) \log_e \left| \sec x \right| + \left( a - 2 \right) x\]
\[\text { When }\sec x > 0 \Rightarrow \left| \sec x \right| = \sec x\]
\[\frac{d}{dx}\left\{ f\left( x \right) \right\} = 2 \sec^2 x + \left( 2a + 1 \right)\frac{1}{\sec x} \times \sec x \tan x + \left( a - 2 \right) \]
\[ = 2 \sec^2 x + \left( 2a + 1 \right)\tan x + \left( a - 2 \right) \]
\[\text { For f(x) to be increasing}, \]
\[2se c^2 x + \left( 2a + 1 \right)\tan x + \left( a - 2 \right) \geqslant 0\]
\[ \Rightarrow 2 + 2 \tan^2 x + \left( 2a + 1 \right)\tan x + a - 2 \geqslant 0\]
\[ \Rightarrow 2 \tan^2 x + \left( 2a + 1 \right)\tan x + a \geqslant 0\]
\[\text { Its discriminant } \leqslant 0 \left[ \because a x^2 + bx + c \geqslant 0 \Rightarrow b^2 - 4ac \leqslant 0 \right]\]
\[ \Rightarrow \left( 2a + 1 \right)^2 - 4 . 2 . a \leqslant 0\]
\[ \Rightarrow 4 a^2 - 4a + 1 \leqslant 0\]
\[ \Rightarrow \left( 2a - 1 \right)^2 \leqslant 0\]
\[ \left( 2a - 1 \right)^2 < 0 \text { cannot be possible } . \]
\[ \therefore \left( 2a - 1 \right)^2 = 0\]
\[ \Rightarrow a = \frac{1}{2}\]
\[\text { When } \sec x < 0 \Rightarrow \left| \sec x \right| = - \sec x\]
\[\frac{d}{dx}\left\{ f\left( x \right) \right\} = 2 \sec^2 x + \left( 2a + 1 \right)\frac{1}{- \sec x} \times \sec x \tan x + \left( a - 2 \right)\]
\[ = 2 \sec^2 x - \left( 2a + 1 \right)\tan x + \left( a - 2 \right) \]
\[\text { For f(x) to be increasing,} \]
\[2se c^2 x - \left( 2a + 1 \right)\tan x + \left( a - 2 \right) \geqslant 0\]
\[ \Rightarrow 2 + 2 \tan^2 x - \left( 2a + 1 \right)\tan x + a - 2 \geqslant 0\]
\[ \Rightarrow 2 \tan^2 x - \left( 2a + 1 \right)\tan x + a \geqslant 0 \]
\[\text { Its discriminant } \leqslant 0 \left[ \because a x^2 + bx + c \geqslant 0 \Rightarrow b^2 - 4ac \leqslant 0 \right]\]
\[ \Rightarrow \left\{ - \left( 2a + 1 \right) \right\}^2 - 4 . 2 . a \leqslant 0\]
\[ \Rightarrow 4 a^2 - 4a + 1 \leqslant 0\]
\[ \Rightarrow \left( 2a - 1 \right)^2 \leqslant 0\]
\[ \left( 2a - 1 \right)^2 < 0 \text { cannot be possible } . \]
\[ \therefore \left( 2a - 1 \right)^2 = 0\]
\[ \Rightarrow a = \frac{1}{2}\]
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