Verified Solution • 2025–26 Session moderate

QJEEM21Ma600f: JEE Main 2021 Mathematics - Class 11 JEEM21Ma600f Solution: If the coefficients of x 7 in and x 7 in , b 0, are equal, then …

If the coefficients of x7 in (x2+1bx)11{\left( {{x^2} + {1 \over {bx}}} \right)^{11}} and x −- 7 in (x−1bx2)11{\left( {{x} - {1 \over {bx^2}}} \right)^{11}} , b ≠\ne 0, are equal, then the value of b is equal to :

Multiple Choice Options

  • A 2
  • B −- 1
  • C 1
  • D −- 2

Step-by-Step Verified Solution

Coefficient of x7 in (x2+1bx)11{\left( {{x^2} + {1 \over {bx}}} \right)^{11}} :

General Term = 11Cr(x2)11−r.(1bx)r{}^{11}{C_r}{({x^2})^{11 - r}}.{\left( {{1 \over {bx}}} \right)^r}

= 11Crx22−3r.1br{}^{11}{C_r}{x^{22 - 3r}}.{1 \over {{b^r}}}

22−3r=722 - 3r = 7

r=5r = 5

∴\therefore Required Term = 11C5.1b5.x7{}^{11}{C_5}.{1 \over {{b^5}}}.{x^7}

Coefficient of x −- 7 in (x−1bx2)11{\left( {x - {1 \over {b{x^2}}}} \right)^{11}} :

General Term = 11Cr(x)11−r.(−1bx2)r{}^{11}{C_r}{(x)^{11 - r}}.{\left( { - {1 \over {b{x^2}}}} \right)^r}

= 11Crx11−3r.(−1)rbr{}^{11}{C_r}{x^{11 - 3r}}.{{{{( - 1)}^r}} \over {{b^r}}}

11−3r=−711 - 3r = - 7 ∴\therefore r=6r = 6

∴\therefore Required Term = 11C6.1b6x−7{}^{11}{C_6}.{1 \over {{b^6}}}{x^{ - 7}}

According to the question,

11C5.1b5=11C6.1b6{}^{11}{C_5}.{1 \over {{b^5}}} = {}^{11}{C_6}.{1 \over {{b^6}}}

Since, b ≠\ne 0 ∴\therefore b = 1

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