The maximum value of -17-5x2-3-2x2-2 i-

The correct option is D 2π/3sec^ 1 ( 5/2+2/2x^2+2 ); for ^ 1 to be defined x^2+2≥ 2 ⇒1/x^2+2≤1/2 ⇒ ( 5/2+1/x^2+2 )≤ 5/2+1/2 ⇒ sec^ 1 ( 5/2+1/x^2+2 ) ≤ ...

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Byju's Answer

Standard XII

Mathematics

Domain

The maximum v...

Question

The maximum value of ${sec}^{- 1} (\frac{7 - 5 (x^{2} + 3)}{2 (x^{2} + 2)})$ is

\frac{5 π}{6}

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\frac{5 π}{12}

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\frac{7 π}{12}

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\frac{2 π}{3}

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Solution

The correct option is D $\frac{2 π}{3}$
$s e c^{- 1} (\frac{- 5}{2} + \frac{2}{2 x^{2} + 2}); for {sec}^{- 1} to be defined x^{2} + 2 \geq 2$

$\Rightarrow \frac{1}{x^{2} + 2} \leq \frac{1}{2}$

$\Rightarrow (\frac{- 5}{2} + \frac{1}{x^{2} + 2}) \leq \frac{- 5}{2} + \frac{1}{2}$

$\Rightarrow s e c^{- 1} (\frac{- 5}{2} + \frac{1}{x^{2} + 2}) \leq s e c^{- 1} (- 2) = π - s e c^{- 1} (2)$

$= \frac{2 π}{3}$

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The maximum value of sec−1(7−5(x2+3)2(x2+2)) is

The correct option is D 2π3sec−1(−52+22x2+2); for sec−1 to be defined x2+2≥2 ⇒1x2+2≤12 ⇒(−52+1x2+2)≤−52+12 ⇒sec−1(−52+1x2+2)≤sec−1(−2)=π−sec−1(2) =2π3

The maximum value of ${sec}^{- 1} (\frac{7 - 5 (x^{2} + 3)}{2 (x^{2} + 2)})$ is