JEE MAIN - Mathematics (2012 - No. 9)
If $$g\left( x \right) = \int\limits_0^x {\cos 4t\,dt,} $$ then $$g\left( {x + \pi } \right)$$ equals
$${{g\left( x \right)} \over {8\left( \pi \right)}}$$
$$g\left( x \right) + g\left( \pi \right)$$
$$g\left( x \right) - g\left( \pi \right)$$
$$g\left( x \right) . g\left( \pi \right)$$
Explanation
$$\left( {b,c} \right)g\left( {x + \pi } \right) = \int\limits_0^{x + \pi } {\cos \,4t\,dt} $$
$$ = \int\limits_0^\pi {\cos 4tdt + \int\limits_\pi ^{\pi + x} {\cos 4t\,dt} } $$
$$ = g\left( \pi \right) + \int\limits_0^x {\cos \,4t\,dt} $$
Putting $$t = \pi + y$$ in second integral, we get
$$\int\limits_x^{\pi + x} {\cos \,4t\,dt = \int\limits_0^\pi {\cos \,4t\,dt} } $$
$$ = g\left( \pi \right) + g\left( x \right) = g\left( x \right) - g\left( \pi \right)$$
$$\therefore$$ $$\left. {g\left( \pi \right) = 0} \right)$$
$$ = \int\limits_0^\pi {\cos 4tdt + \int\limits_\pi ^{\pi + x} {\cos 4t\,dt} } $$
$$ = g\left( \pi \right) + \int\limits_0^x {\cos \,4t\,dt} $$
Putting $$t = \pi + y$$ in second integral, we get
$$\int\limits_x^{\pi + x} {\cos \,4t\,dt = \int\limits_0^\pi {\cos \,4t\,dt} } $$
$$ = g\left( \pi \right) + g\left( x \right) = g\left( x \right) - g\left( \pi \right)$$
$$\therefore$$ $$\left. {g\left( \pi \right) = 0} \right)$$
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