Find the minimum value of the functionfx - 216 -1- x - -1 xA- 0B- - 2C- - 2D- -

Fx=π216 nbsp;cot 1 nbsp;( x) π cot 1 x nbsp; nbsp; nbsp; nbsp; nbsp; nbsp;= nbsp;cot 1 nbsp; x+ nbsp;π216 nbsp;cot 1 nbsp;( x) nbsp; ...

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Byju's Answer

Standard XII

Mathematics

Properties with Negative X

Find the mini...

Question

Find the minimum value of the function
$f (x) = \frac{π^{2}}{16 {cot}^{- 1} (- x)} - {cot}^{- 1} x .$

\frac{- π}{2}

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\frac{π}{2}

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- π

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Solution

The correct option is A $\frac{- π}{2}$
$f (x) = \frac{π^{2}}{16 {cot}^{- 1} (- x)} - (π - {cot}^{- 1} (- x)) = {cot}^{- 1} (- x) + \frac{π^{2}}{16 {cot}^{- 1} (- x)} - π = {(\sqrt{{cot}^{- 1} (- x)} - \frac{π}{4 \sqrt{{cot}^{- 1} (- x)}})}^{2} + \frac{π}{2} - π \geq \frac{- π}{2} ∴ f_{min} = \frac{- π}{2}$

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Find the minimum value of the function f(x) = π216 cot−1(−x) − cot−1 x.

The correct option is A −π2f(x)=π216 cot−1 (−x)−(π−cot−1(−x)) = cot−1 (−x)+ π216 cot−1 (−x) − π = (√cot−1 (−x) − π4√cot−1 (−x))2+ π2−π ≥ −π2∴ fmin = −π2

Find the minimum value of the function
$f (x) = \frac{π^{2}}{16 {cot}^{- 1} (- x)} - {cot}^{- 1} x .$