AP EAMCET202317 May 2023Evening ShiftMathematicsDefinite IntegrationActual
If m Z ⁺, n =2 ~m and ₀^ 2 ^ m x ^ n x d x = K ( m ) ₀^ 2 ^m x d x , then 2^ m-1 (m-1) ! (2 m-1) ! K(m)=
Options
- A1 m+2 1 m+4 1 m+r 1 3 m
- B1 2 m+2 1 2 m+4 1 3 m
- C2 1 m+2 1 m+4 1 m+r 1 3 m
- D2 1 2 m+2 1 2 m+4 1 3 m
Correct answer
A. 1 m+2 1 m+4 1 m+r 1 3 m
Step-by-step solution
From eq ^ n (i) k ( m )= 2 m-1 3 m 2 m-3 3 m-2 2 m-5 3 m-4 3 m+4 1 m+2 Now, 2^ m-1 ( ~m -1) ! (2 ~m -1) ! k( ~m ) aligned & = 2^ m-1 (m-1) ! (2 m-1) ! (2 m-1)(2 m-3)(2 m-5) 3.1 (3 m)(3 m-2)(3 m-4) (m+4)(m+2) & = 1 m+2 1 m+4 1 m+r 1 3 m aligned