The range of f(x) = cos [x], for -π/2< x <π/2 isA. {-1,1,0} B. {cos

1.	The range of f(x) = cos [x], for -π/2< x <π/2 isA. {-1,1,0} B. {cos 1, cos 2,1} C. {cos 1, -cos 1,1} D. [-1,1]
Answer» Option : (B) Given, f(x) = cos [x], for -π/2< x <π/2 For -π/2< x <-1, [x]= -2 f(x)= cos[x]= cos(-2) = cos2 Because, cos(-x) = cos(x) For-1 ≤x<0 [x] = -1 f(x) = cos[x] =cos(-1) = cos1 For 0 ≤x< 1, [x] = 0 f(x) = cos 0 = 1 For 1≤ x <π/2, [x] = 1 f(x) = cos 1 Therefore, R(f) = {1, cos 1,cos 2} Option B is correct.

The range of f(x) = cos [x], for -π/2< x <π/2 isA. {-1,1,0} B. {cos 1, cos 2,1} C. {cos 1, -cos 1,1} D. [-1,1]

Answer»

Option : (B)

Given,

f(x) = cos [x], for -π/2< x <π/2

For -π/2< x <-1,

[x]= -2

f(x)= cos[x]= cos(-2)

= cos2

Because,

cos(-x) = cos(x)

For-1 ≤x<0

[x] = -1

f(x) = cos[x]

=cos(-1)

= cos1

For 0 ≤x< 1,

[x] = 0

f(x) = cos 0

= 1

For 1≤ x <π/2,

[x] = 1

f(x) = cos 1

Therefore,

R(f) = {1, cos 1,cos 2}

Option B is correct.