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Question Number 33330 by artibunja last updated on 14/Apr/18
Answered by MJS last updated on 15/Apr/18
∣3tanπx−33tanπx∣⩾2∣32tanπx−33tanπx∣⩾2∣32tanπx−3∣⩾2×3tanπx3tanπx=t1.(t2−3)⩾0t2−3⩾2tt2−2t−3⩾0t⩽−1∨t⩾33tanπx⩾0forx∈R⇒3tanπx⩾3⇒⇒tanπx⩾12.(t2−3)<0t2−3<2tt2−2t−3<0−1<t<33tanπx⩾0forx∈R⇒0⩽3tanπx<3⇒⇒−∞⩽tanπx<1⇒tanπx<1∣3tanπx−31−tanπx∣⩾2istruefor(tanπx⩾1)∨(tanπx<1)⇒⇒trueforallx∈R
Answered by artibunja last updated on 17/Apr/18
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