A) \[4\pi \] sq. units
B) \[8\pi \] sq. units
C) 4 sq. units
D) 8 sq. units
Correct Answer: D
Solution :
\[\because \,f:\,\left[ 0,\,2\pi \right]\to \left[ 0,\,2\pi \right]\] is given by \[f(x)=x+\sin x\] is a bijective function. The graph of \[{{f}^{-1}}(x)\] is the mirror image of the graph of \[f(x)\] in the line \[y=x\]. \[\therefore \] Required Area = 2 (Area of one loop) \[=2\times 2\left[ \int\limits_{0}^{\pi }{(x+\sin x)}dx-\int\limits_{0}^{\pi }{xdx} \right]\] \[=4\left[ \int\limits_{0}^{\pi }{\sin xdx} \right]\] = 8 sq. units.You need to login to perform this action.
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