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Question

Let f be a function defined as f(x)=11+(tanx)m, where m is a constant. Then 
  1. π/40xf(x)cos2xdx=π4ln2, when m=1
  2. π/40xf(x)cos2xdx=π8ln2, when m=1
  3. π/20f(x)dx=π4, when m=2  
  4. π/20f(x)dx=π2, when m=2   


Solution

The correct options are
B π/40xf(x)cos2xdx=π8ln2, when m=1
C π/20f(x)dx=π4, when m=2  
For m=1
I=π/40xf(x)cos2xdx 
  =π/40x(sinx+cosx)cosxdx

sinx+cosx=2cos(π4x) 

I=π/40x2cos(π4x)cosxdx   ...(1)

Using a0f(x)dx=a0f(ax)dx

I=π/40π4x2cos(π4x)cosxdx   ...(2)

Adding eqn (1) and (2), we get

2I=π/40π42cos(π4x)cosxdx 

I=π8π/4012cos(π4x)cosxdx 

I=π8π/40sin(π4x+x)cos(π4x)cosxdx

I=π8π/40sin(π4x)cosx+cos(π4x)sinxcos(π4x)cosxdx

I=π8π/40[tan(π4x)+tanx]dx

π/40tan(π4x)dx=π/40tanx dx

I=π4π/40tanx dx

I=π4[ln|secx|]π/40

I=π4ln2

I=π8ln2


Now, when m=2
I=π/2011+(tanx)2dx   ...(3)

Using a0f(x)dx=a0f(ax)dx

I=π/2011+(cotx)2dx

I=π/20(tanx)21+(tanx)2dx   ...(4)

Adding eqn (3) and (4), we get

2I=π/201 dx

I=π4 

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