Binomial Theorem · JEE Main previous year questions

2005 2 questions

1. Q82005 · 28 Apr 2005 · Shift 1

   If $x$ is so small that $x^3$ and higher powers of $x$ may be neglected, then $\frac{(1+x{)}^{\frac{3}{2}}-{\left(1+\frac{1}{2}x\right)}^3}{(1-x{)}^{\frac{1}{2}}}$ may be approximated as

   1. A.$1-\frac{3}{8}{x}^2$
   2. B.$3x+\frac{3}{8}{x}^2$
   3. C.$-\frac{3}{8}{x}^2$
   4. D.$\frac{x}{2}-\frac{3}{8}{x}^2$
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   Answer: C

2. Q92005 · 28 Apr 2005 · Shift 1

   If the coefficients of $r^{th}$, $(r+1{)}^{th}$, and $(r+2{)}^{th}$ terms in the binomial expansion of $(1+y{)}^m$ are in A.P., then m and r satisfy the equation

   1. A.$m^2-m(4r-1)+4r^2-2=0$
   2. B.$m^2-m(4r+1)+4r^2+2=0$
   3. C.$m^2-m(4r+1)+4r^2-2=0$
   4. D.$m^2-m(4r-1)+4r^2+2=0$
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   Answer: C