f(x)=23Tan−1⁡2x+13−loge⁡x2+x+1+b2−5b+3x+c,f1(x)<0∀x∈R Then range of ‘b’ is

f1(x)=2311+2x+13223−2x+1x2+x+1+b2−5b+3<0=−2xx2+x+1+b2−5b+3<0b2−5b+3<2xx2+x+1b2−5b+3<−2b2−5b+5<0

f(x)=23Tan−1⁡2x+13−loge⁡x2+x+1+b2−5b+3x+c,f1(x)<0∀x∈R Then range of ‘b’ is

f1(x)=2311+2x+13223−2x+1x2+x+1+b2−5b+3<0=−2xx2+x+1+b2−5b+3<0b2−5b+3<2xx2+x+1b2−5b+3<−2b2−5b+5<0

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$f (x) = \frac{2}{\sqrt{3}} {Tan}^{- 1} (\frac{2 x + 1}{\sqrt{3}}) - \log_{e} (x^{2} + x + 1) + (b^{2} - 5 b + 3) x + c, f^{1} (x) < 0 \forall x \in R$ Then range of ‘b’ is

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(0,1)

$(\frac{5 + \sqrt{5}}{2}, \infty)$

$(0, \frac{5 - \sqrt{5}}{2})$

$(\frac{5 - \sqrt{5}}{2}, \frac{5 + \sqrt{5}}{2})$

answer is C.

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Detailed Solution

$\begin{array}{l} f^{1} (x) = \frac{2}{\sqrt{3}} \frac{1}{1 + {(\frac{2 x + 1}{\sqrt{3}})}^{2}} \frac{2}{\sqrt{3}} - \frac{2 x + 1}{x^{2} + x + 1} + b^{2} - 5 b + 3 < 0 \\ = \frac{- 2 x}{x^{2} + x + 1} + b^{2} - 5 b + 3 < 0 \\ b^{2} - 5 b + 3 < \frac{2 x}{x^{2} + x + 1} \\ b^{2} - 5 b + 3 < - 2 \\ b^{2} - 5 b + 5 < 0 \end{array}$