If f x -cos x - 0 - x -2- then the real number - c - of the mean value theorem is A- -6B- cos -12-C- D- sin -12-

The correct option is D sin^ 1(2/π )We know that f'(c)=f(b) f(a)/b a ⇒ f'(c)=0 1/π/2= 2/π ....(i) But f'(x)= sin x ⇒ f'(c) = sin c ....(ii) From (i) and (ii), ...

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Byju's Answer

Standard XII

Mathematics

Functions without Antiderivatives as Known Combination of Basic Functions

If f x =cos x...

Question

If $f (x) = c o s x, 0 \leq x \leq \frac{π}{2}$ , then the real number ‘c’ of the mean value theorem is

\frac{π}{6}

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\frac{π}{4}

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s i n^{- 1} (\frac{2}{π})

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c o s^{- 1} (\frac{2}{π})

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Solution

The correct option is C $s i n^{- 1} (\frac{2}{π})$
We know that $f^{'} (c) = \frac{f (b) - f (a)}{b - a}$
$\Rightarrow f^{'} (c) = \frac{0 - 1}{\frac{π}{2}} = - \frac{2}{π} . . . . (i)$
But $f^{'} (x) = - s i n x \Rightarrow f^{'} (c) = - s i n c . . . . (i i)$
From (i) and (ii), we get $- s i n c = - \frac{2}{π} \Rightarrow c = s i n^{- 1} (\frac{2}{π})$ .

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If f(x)=cosx,0≤x≤π2, then the real number ‘c’ of the mean value theorem is

The correct option is C sin−1(2π)We know that f′(c)=f(b)−f(a)b−a ⇒f′(c)=0−1π2=−2π....(i) But f′(x)=−sinx⇒f′(c)=−sinc....(ii) From (i) and (ii), we get −sinc=−2π⇒c=sin−1(2π).

If $f (x) = c o s x, 0 \leq x \leq \frac{π}{2}$ , then the real number ‘c’ of the mean value theorem is