A) \[\left( -\frac{\pi }{2},\,\,0 \right)\]
B) \[(0,\,\,\pi )\]
C) \[\left( \pi ,\,\,\frac{3\pi }{2} \right)\]
D) \[\left( 0,\,\,\frac{\pi }{2} \right)\]
Correct Answer: B
Solution :
Let\[f(x)=\sin x+x\cos x\] Consider\[g(x)=\int\limits_{0}^{x}{(\sin t+t\cos t)dt=t\sin ]_{0}^{x}=x\sin x}\]\[g(x)=x\sin x\]which is differentiable Now,\[g(0)=0\]and\[g(\pi )=0\], using Theorem \[\exists \] atleast one \[c\in (0,\,\,\pi )\] such that\[g'(c)=0\] \[i.e.\,\,c\cos c+\sin c=0\]for atleast one\[c\in (0,\,\,\pi )\]
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