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← JEE Main 2022 · 29 Jun · Shift 2
MathematicsInteger+4 / -1Question 27 of 90

JEE Main 2022 · 29 Jun · Shift 2, Mathematics question 27

Let the coefficients of x−1x^{-1} and x−3x^{-3} in the expansion of (2x15−1x15)15\left(2x^{\frac15}-\dfrac1{x^{\frac15}}\right)^{15}, x>0x>0, be mm and nn respectively. If rr is a positive integer such that mn2=15Cr⋅2rmn^2={}^{15}C_r\cdot2^r, then the value of rr is equal to ______.

Answer and worked solution

Official answer: 5

General term. 15Cr215−r(−1)rx(15−2r)/5{}^{15}C_r2^{15-r}(-1)^rx^{(15-2r)/5}. x−1x^{-1}: r=10r=10; x−3x^{-3}: r=15r=15.

Values. m=15C1025m={}^{15}C_{10}2^5, n=−1n=-1, so mn2=15C5 25mn^2={}^{15}C_5\,2^5 and r=5r=5.

Answer: 5