(A) We know that, −1≤cosx≤1 ⇒0≤cos2x≤1 ⇒1≤1+cos2x≤2 ∴[1+cos2x]={1,2} So, range of sec−1[1+cos2x]={sec−11,sec−12} (A)→(V) (B) We have, f(x+
1
x
)=x2+
1
x2
f(x+
1
x
)=(x+
1
x
)2−2 Put y=x+
1
x
∴f(y)=y2−2 Now, AM≥GM Case I When x>0,
x+
1
x
2
≥√x⋅
1
x
⇒x+
1
x
≥2 Case II When
x<0,
−|x|+
1
−|x|
2
≥√(−|x|)×
1
−|x|
⇒−(|x|+
1
|x|
)≥2⇒|x|+
1
|x|
≤−2
∴x+
1
x
∈R R−(−2,2) So, domain of f(x) is R−(−2,2) (B) → (IV) (C) We have f(x+y)=f(x)+f(y) Put x=y=0, we get f(0)=f(0)+f(0) ⇒f(0)=0 Now, put y=−x, we get f(x−x)=f(x)+f(−x) ⇒f(0)=f(x)+f(−x) ⇒f(x)+f(−x)=0[∵f(0)=0] ⇒f(−x)=−f(x) So, f(x) is a odd function. (C)→I D. We have sin−1x−cos−1x+sin−1(1−x)=0 ⇒sin−1(1−x)=cos−1x−sin−1x