Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
AP EAMCET202317 May 2023Morning ShiftMathematicsBinomial TheoremActual

If f ( n )= n !(31- n ) ! , where n 0,1,2, , 31 , then the minimum value of f(n) is

Options

  1. A(15!) (15!)
  2. B(15 !)(14 !)
  3. C(14!) (16!)
  4. D(15 !)(16 !)

Correct answer

D. (15 !)(16 !)

Step-by-step solution

aligned & f(n)=n !(31-n) ! & So, f(31-n)=(31-n !)(n !)=f(n) f(0)=f(31) & f(1)=f(30) & Also f(0)>f(1)>f(2)> >f(15) < f(16) < f(17) & < f(30) < f(31) & f(15) is minimum. Then, the minimum value is: & f(15 !)=15 !(31-15) !=(15 !)(16 !) aligned

Practice Binomial Theorem on Quantrex Academy →

More from Binomial Theorem

If n 13 , n 14 and n 15 are in arithmetic progression, then the positive integer value of ' n ' can be 2026If the coefficients of x^2 and x^3 in the expansion of (3 + kx)^9 are equal, then the value of ' k ' is 2026The remainder when 7¹⁰³ is divided by 25 is 2026_ r=1 ¹⁵ r^2 ( ¹⁵ C_r 15 r-1 )= 20251 81^ n - ^ 2 n C ₁ 10 81^ n + ^ 2 n C ₂ 10^2 81^ n - + 10^ 2 n 81^ n = 2025If x is positive real number and the first negative term in the expansion of (1+ x )^ 27 / 5 is t _ k then k = 2025In the binomial expansion of (p-q)¹⁴ , if the sum of 7^ th term and 8^ th term is zero, then p+q p-q = 2025The numerically greatest term in the expansion of (x+3 y)¹³ , when x= 1 2 and y= 1 3 is 2025 Full Binomial Theorem list All AP EAMCET PYQs