f(x)=sin(πcosx),g(x)=cos(2πsinx),x>0
X={x:f(x)=0},Y={x:f′(x)=0}
Z={x:g(x)=0},W={x:g′(x)=0}
Match X,Y,Z,W with the correct statement (p)–(u).
Choose the only CORRECT combination.
f(x)=0⇒sin(πcosx)=0⇒πcosx=nπ
⇒cosx=n,n∈Z
cosx∈{−1,0,1}⇒x=2nπ,n=1,2,3,…
X={2π,π,23π,2π,25π,…}
common difference=2π ⇒ X is an A.P. (q)
2π,23π,4π,7π ∈X ⇒(p)
X⟶(p), (q)
f′(x)=cos(πcosx)⋅(−πsinx)
f′(x)=−πsinxcos(πcosx)
−πsinxcos(πcosx)=0
sinx=0 or cos(πcosx)=0
x=nπ or cosx=22k+1=±21
Y={3π,32π,π,34π,35π,2π,…}
={3nπ}, n=1,2,3,… ⇒ A.P. (q)
3π,32π,π ∈Y ⇒(t)
Y⟶(q), (t)
g(x)=0⇒cos(2πsinx)=0
2πsinx=2(2k+1)π⇒sinx=±41, ±43
gaps are not constant⇒Z is NOT an A.P. (r)
g′(x)=−sin(2πsinx)⋅(2πcosx)
g′(x)=−2πcosxsin(2πsinx)
−2πcosxsin(2πsinx)=0
cosx=0 or sin(2πsinx)=0
sinx=2n⇒sinx∈{0,±21,±1}
W={x:sinx∈{0,±21,±1}}
residues (units of 6π)={0,1,3,5} ⇒NOT an A.P. (r)
6π,67π,613π ∈W ⇒(s)
SetXYZWColumn II(p), (q)(q), (t)(r)(r), (s)
The only correct combination is option (2)
boxed(I) X → (p), (III) Z → (r)