Let f- 0- 2- - R be continuous and satisfy f- x - 1 1 - cos x for all x - 0- 2- If f0 - 3 then f - 2 has the value equal to

The correct option is C 4 f'(x)=11+cos x=12sec^2x2 f(x)=12∫sec^2x2 dx=tanx2+c f(0)=3 c=3 f(π2)=tanπ4+3=4 ...

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Byju's Answer

Standard XII

Mathematics

Definition of Functions

Let f:[ 0,π...

Question

Let $f : [0, \frac{π}{2}] \to R$ be continuous and satisfy $f^{'} (x) = \frac{1}{1 + cos x}$ for all $x \in [0, \frac{π}{2}]$ . If f(0) = 3 then $f (\frac{π}{2})$ has the value equal to

\frac{13}{4}

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Solution

The correct option is C 4
$f^{'} (x) = \frac{1}{1 + cos x} = \frac{1}{2} {sec}^{2} \frac{x}{2}$
$⟹ f (x) = \frac{1}{2} \int {sec}^{2} \frac{x}{2} d x = tan \frac{x}{2} + c$
$f (0) = 3 ⟹ c = 3$
$f (\frac{π}{2}) = tan \frac{π}{4} + 3 = 4$

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Let f:[0,π2]→R be continuous and satisfy f′(x)=11+cosx for all x∈[0,π2]. If f(0) = 3 then f(π2) has the value equal to

The correct option is C 4f′(x)=11+cosx=12sec2x2⟹f(x)=12∫sec2x2dx=tanx2+cf(0)=3⟹c=3f(π2)=tanπ4+3=4

Let $f : [0, \frac{π}{2}] \to R$ be continuous and satisfy $f^{'} (x) = \frac{1}{1 + cos x}$ for all $x \in [0, \frac{π}{2}]$ . If f(0) = 3 then $f (\frac{π}{2})$ has the value equal to

The correct option is C 4
$f^{'} (x) = \frac{1}{1 + cos x} = \frac{1}{2} {sec}^{2} \frac{x}{2}$
$⟹ f (x) = \frac{1}{2} \int {sec}^{2} \frac{x}{2} d x = tan \frac{x}{2} + c$
$f (0) = 3 ⟹ c = 3$
$f (\frac{π}{2}) = tan \frac{π}{4} + 3 = 4$