Show that the function x - log cos2 x has maximum at x - 0 and minimum at x - - - 3 - Given x-2-2-

f(x)=secx+log cosx #178; f (x) = secx + 2 log cosx Therefore, f #8242; (x) = secx tanx #8211; 2 tanx = tanx (secx #8211;2) f #8 ...

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Standard XII

Mathematics

Sufficient Condition for an Extrema

Show that the...

Question

Show that the function $sec x + l o g {cos}^{2} x$ has maximum at $x = 0$ and minimum at $x = π / 3 . G i v e n x ϵ [\frac{- π}{2}, \frac{π}{2}]$

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Solution

$f (x) = s e c x + l o g c o s x ²$
$f (x) = s e c x + 2 l o g c o s x$
Therefore, $f' (x) = s e c x t a n x - 2 t a n x = t a n x (s e c x - 2)$
$f' (x) = 0 \Rightarrow t a n x = 0 o r s e c x = 2 o r c o s x = 1 / 2$
$f^{'} (x) > 0 f o r x ϵ (\frac{- π}{2}, 0) \cup (\frac{π}{3}, \frac{π}{2})$
$f^{'} (x) < 0 f o r x ϵ (0, \frac{π}{3})$
Hence, F(x) has maxima at $x = 0$ and has minima at $x = \frac{π}{3}$

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Show that the function secx+logcos2x has maximum at x=0 and minimum at x=π/3.Given xϵ[−π2,π2]

f(x)=secx+logcosx²f(x)=secx+2logcosxTherefore, f'(x)=secx tanx–2tanx=tanx(secx–2)f'(x)=0⇒tanx=0 or secx=2 or cosx=1/2 f′(x)>0 for xϵ(−π2,0)∪(π3,π2)f′(x)<0 for xϵ(0,π3)Hence, F(x) has maxima at x=0 and has minima at x=π3

Show that the function $sec x + l o g {cos}^{2} x$ has maximum at $x = 0$ and minimum at $x = π / 3 . G i v e n x ϵ [\frac{- π}{2}, \frac{π}{2}]$