Let f(x)=1-x2sin2x+x2 for all x∈R, and let g(x)=integral from 1 to x2(t-1)[t+1]-lntf(t)dt for all x∈(1,∞]. Which of the following is true
g is increasing on (1,∞)
g is decreasing on (1,∞)
g is increasing on (1,2) and decreasing on(2,∞)
g is decreasing on (1,2) and increasing on (2,∞)
Explanation for the correct option:
Given that:
f(x)=(1-x)2sin2x+x2g(x)=∫1x2(t-1)(t+1)-lntf(t)dt
Now, g(x)=2(t-1)t+1-lnt.f(x)
Let φ=2(t-1)t+1-lnt
φ1=2(t+1)·(1)-(t-1)(t)2(t+1)2-lnt
=48(t+1)2-1t
φ1x=4(x+1)-1x=4x-x2-1-2xx(x+1)2=-x2+2x-1x(x+1)2
φ1(x)=-(x-1)2x(x+1)2
φ1(x)<0forallx∈(1,∞]
Therefore, g is decreasing on (1,∞]
Hence, the correct answer is option (B).