- f-x- sin x- and - g-x- x- then - f- -f - 3-2 -f-0-g- -g-0-g - -3 - 12- 13- 1- 2-

The correct option is A 12 f ( π #160; )=0;f ( 3π2 )= 1;f ( 0 )=0 g ( π #160; )= 1; g ( 0 )=1; g ( π3 )=2 ∴f ( π #160; ) f ( 3π2 )+ ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Compound Angles

f(x)= sin x ...

Question

$f (x) = sin x$ and $g (x) = sec x$ then $\frac{f (π) - f (\frac{3 π}{2}) + f (0)}{g (π) + g (0) + g (\frac{π}{3})}$

\frac{1}{2}

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\frac{1}{3}

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Solution

The correct option is A $\frac{1}{2}$
$f (π) = 0; f (\frac{3 π}{2}) = - 1; f (0) = 0$
$g (π) = - 1; g (0) = 1; g (\frac{π}{3}) = 2$
$∴ \frac{f (π) - f (\frac{3 π}{2}) + f (0)}{g (π) + g (0) + g (\frac{π}{3})}$ $= \frac{1}{2}$

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Q. If

F (x) = \int \frac{x + sin x}{1 + cos x} d x

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Q. If

f : R \to R, f (x) = ⎧ ⎨ ⎩ \begin{matrix} 1 x > 0 0 x = 0 - 1 x < 0 \end{matrix}

and

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, then

(f \circ g) (π)

is:

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g (x) = \int x^{x} {log}_{e} (e x) d x

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$f (x) = sin x$ and $g (x) = sec x$ then $\frac{f (π) - f (\frac{3 π}{2}) + f (0)}{g (π) + g (0) + g (\frac{π}{3})}$

The correct option is A $\frac{1}{2}$
$f (π) = 0; f (\frac{3 π}{2}) = - 1; f (0) = 0$
$g (π) = - 1; g (0) = 1; g (\frac{π}{3}) = 2$
$∴ \frac{f (π) - f (\frac{3 π}{2}) + f (0)}{g (π) + g (0) + g (\frac{π}{3})}$ $= \frac{1}{2}$

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f(x)=sinx and g(x)=secx then f(π)−f(3π2)+f(0)g(π)+g(0)+g(π3)

The correct option is A 12f(π)=0;f(3π2)=−1;f(0)=0g(π)=−1;g(0)=1;g(π3)=2∴f(π)−f(3π2)+f(0)g(π)+g(0)+g(π3) =12

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$f (x) = sin x$ and $g (x) = sec x$ then $\frac{f (π) - f (\frac{3 π}{2}) + f (0)}{g (π) + g (0) + g (\frac{π}{3})}$

The correct option is A $\frac{1}{2}$
$f (π) = 0; f (\frac{3 π}{2}) = - 1; f (0) = 0$
$g (π) = - 1; g (0) = 1; g (\frac{π}{3}) = 2$
$∴ \frac{f (π) - f (\frac{3 π}{2}) + f (0)}{g (π) + g (0) + g (\frac{π}{3})}$ $= \frac{1}{2}$