The correct option is C 12√2
I=3π/4∫−π/4eπ/4 dx(ex+eπ/4)(sinx+cosx)=λπ/2∫−π/2secx dx
⇒I=3π/4∫−π/4eπ/4 dxex(1+eπ/4−x)(sinx+cosx)
⇒I=3π/4∫−π/4eπ/4−x dx(1+eπ/4−x)(1√2sinx+1√2cosx)√2
⇒I=1√23π/4∫−π/4eπ/4−x dx(1+eπ/4−x)[cos(π4−x)]
Let π4−x=t
⇒dx=−dt
I=−1√2−π/2∫π/2et(1+et)costdt
⇒I=1√2π/2∫−π/2et(1+et)costdt ⋯(1)
Applying king's property,
I=1√2π/2∫−π/2e−t(1+e−t)cos(−t)dt
⇒I=1√2π/2∫−π/21(1+et)costdt ⋯(2)
On adding equations (1) and (2),
2I=1√2π/2∫−π/2sect dt
⇒I=12√2π/2∫−π/2secx dx=λπ/2∫−π/2secx dx
On comparing, λ=12√2