If g ( x ) = ∫ 0 x cos 4 t d t , then g ( x + π ) equals

g ( x ) = ∫ 0 x cos 4 t d t ⇒ g ′ ( x ) = cos 4 x ⇒ g ( x ) = 4 s i n 4 x + k ⇒ g ( x ) = 4 s i n 4 x [ ∵ g ( 0 ) = 0 ] g ( x + π ) = g ( x ) + g ( π ) = g ( x ) − g ( π ) ( ∵ g ( π ) = 0 )

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MathematicsDefinite IntegrationDerivatives (Newton- Leibnitz)Hard2 minPYQ_2012

MathematicsHardsingle choice

If $g (x) = \int_{0}^{x} cos 4 t d t$ , then $g (x + π)$ equals

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Solution:

$g (x) = \int_{0}^{x} cos 4 t d t$ $\Rightarrow g^{'} (x) = cos 4 x \Rightarrow g (x) = \frac{s i n 4 x}{4} + k \Rightarrow g (x) = \frac{s i n 4 x}{4} [∵ g (0) = 0]$ $g (x + π) = g (x) + g (π) = g (x) - g (π) (∵ g (π) = 0)$

Stream:JEESubject:MathematicsTopic:Definite IntegrationSubtopic:Derivatives (Newton- Leibnitz)

⏱ 2mℹ️ Source: PYQ_2012

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