Question

Question 44 and 45 plz answer

Answer 1

44) f( x ) = x/( e^x -1) + x/2 +1

put x = - x

f( -x) = (-x)/( e^-x -1) + (-x)/2 + 1

= -x/( 1/e^x -1) -x/2 +1

=-x.e^x /( 1-e^x) -x/2 +1

= x.e^x/( e^x -1 ) -x/2 +1

now, subtract f(x ) and f( -x)

f(x ) - f(-x) = x/( e^x -1) +x/2 +1 -x.e^x/( e^x-1) + x/ 2 -1

= x( 1- e^x)/(e^x -1 ) + x

= -x + x = 0

hence , f( x ) -f( -x) = 0
f(x ) = f( -x)

so, f(x) is an even function .


45) f( x) = { 2x( sinx + tanx )}/{ 2[ ( x +21π/π) ] -41 }

we know , by property of fractional part

[x ] + [ -x ] = { 0 when, x€ integer

= { 1 when, x doesn't belongs to integer

now,
f(x) = {2x (sinx +tanx)}/{ 2[x/π] +42 -41 }
={x ( sinx + tanx)}/{ [x/π] +1/2}

f(-x ) = {(-x)( -sinx - tanx)}/{ -[x/π] +1/2 }
when, x/π € I
= -x(sinx + tanx)/{-[x/π]+ 1/2} when, x € I × π

but when, x€ i× π
then, f( -x) = 0


again , when, x doesn't belong to i× π
f( -x) =x(sinx + tanx)}/{ -1-[x/π] +1/2}

f( -x ) = -x(sinx + tanx)/{ [x/π] +1/2 }


hence,
f(-x) = { 0 when, x€ i × π

={ -f(x) when , x doesn't belong to i× π

we know , zeros is even and odd both function ,

but second one f(-x) = -f(x)
this is an odd function property so,
f(x) is an odd function .