Q78 / 125 · BITSAT 2018 · Mathematics

Let Px=∫dxex+8e-x+4e-3x, Qx=∫dxe3x+8ex+4e-x and Rx=Px-2Qx. If Rx=12AB+2e-xC+K, then the value of A,B,C is

  • A
    tan-1,2,ex
  • Correct answer
    B
    tan-1,ex,2
  • C
    tan-1,12,1ex
  • D
    tan-1,1ex,12

Solution

We have, Rx=∫dxex+8e-x+4e-3x-2∫dxe3x+8ex+4e-x

⇒Rx=∫exe2x-2e4x+8e2x+4dx

we get Rt=∫t2-2dtt4+8t2+4=∫1-2t-2dtt+2t-12+4

=12tan-1t+2t-12+K

⇒Rx=12tan-1ex+2e-x2+K

Hence, A,B,C=tan-1,ex,2

∴ Option (b) is correct.

Hence, the correct answer is option: 'B'.

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