Solution:
Solution
f(x)=−x3?/ 3 + x2sin1.5 a−xsina⋅sin2a−5arcsin(a2−8a+17)
(x)=−x3?/ 3 + x2sin6 −xsin4sin8−5sin−1((a−4)2+1)
f′(x)=−x2+2xsin6−sin4.sin8 .......( a=4 considering the domain of inverse sine function)
f′(sin8)=−sin28+2sin6sin8−sin4sin8
=sin8(−sin8+2sin6−sin4)
=−sin8(sin8+sin4−2sin6)
=−sin8(2sin6cos2−2sin6)
=2sin8sin6(1−cos2)
Now,
sin8>0 (Since,2π<8<3π )
sin6<0 (Since, π<6<2π)
(1−cos2)>0
Hence, f′(sin8)<0
