R T i RW 100 <fin o oos | 0 a8 .

1.	R T i RW 100 <fin o oos \| 0 a8 .
Answer» if f(x+a)=f(x), then a is the period of the function f(x). Here, f(x)=\|sin x\|+\|cos x\| Here the range of \|sin x\| is [0,1]. Similarly, range of \|cos x\| is also [0,1] So, when we consider a=π/2. f(a+x)=\|sin(x+π/2)\|+\|cos(x+π/2)\|. sin(x+π/2) is equal to cosx. cos(x+π/2) is equal to -sinx. However, we have the modulus function to convert the -sinx to sinx. While, cosx will remain the same. Here, the range of both functions will be the same as it was, for f(x). So, f(x)=f(x+{π/2}). So, it also applies for integral multiples of π/2. So, period of the function is π/2. If x = –π we know sin –π = 0 but cos –π = – 1 so y = 0 + 1 = 1

Answer»

if f(x+a)=f(x), then a is the period of the function f(x).

Here,

f(x)=|sin x|+|cos x|

Here the range of |sin x| is [0,1].

Similarly, range of |cos x| is also [0,1]

So, when we consider a=π/2.

f(a+x)=|sin(x+π/2)|+|cos(x+π/2)|.

sin(x+π/2) is equal to cosx.

cos(x+π/2) is equal to -sinx.

However, we have the modulus function to convert the -sinx to sinx.

While, cosx will remain the same.

Here, the range of both functions will be the same as it was, for f(x).

So, f(x)=f(x+{π/2}).

So, it also applies for integral multiples of π/2.

So, period of the function is π/2.

If x = –π we know sin –π = 0 but cos –π = – 1 so y = 0 + 1 = 1

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