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1/(6!)+1/(7!)=x/(8!),f i n dx...

`1/(6!)+1/(7!)=x/(8!),f i n dx`

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The correct Answer is:
n = 64
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1/(6i)+1/(7i)=x/(8i)

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Knowledge Check

  • If 1/(6!)+1/(7!)=x/(8!) , then the value of x is

    A
    63
    B
    64
    C
    66
    D
    65
  • f'(dx)/((x+6)^(8//7)(x-8)^(6//7)) is equal to

    A
    `((x+6)/(x-8))^(1/7)+e`
    B
    `((x-8)/(x+6))^(1/7)+e`
    C
    `1/2((x-8)/(x+6))^(1/7)+e`
    D
    None of these
  • Let f(x) = {{:((x)/(sin x) ",",x in (0","1)),( 1",", x=0):} Consider the integral I_n = sqrt(n) int_(0)^(1//n) f (x) e^(-nx) dx then lim_( n to oo) I_n

    A
    does not exist
    B
    exists and is 0
    C
    exists and is 1
    D
    exists and is `1 – e^(–1)`
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