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QJEEM21Ma42d1: JEE Main 2021 Mathematics - Class 12 JEEM21Ma42d1 Solution: If is given by , then the value of is :

If f:R→Rf:R \to R is given by f(x)=x+1f(x) = x + 1 , then the value of lim⁡n→∞1n[f(0)+f(5n)+f(10n)+......+f(5(n−1)n)]\mathop {\lim }\limits_{n \to \infty } {1 \over n}\left[ {f(0) + f\left( {{5 \over n}} \right) + f\left( {{{10} \over n}} \right) + ...... + f\left( {{{5(n - 1)} \over n}} \right)} \right] is :

Multiple Choice Options

  • A 32{3 \over 2}
  • B 52{5 \over 2}
  • C 12{1 \over 2}
  • D 72{7 \over 2}

Step-by-Step Verified Solution

f(0)+f(5n)+f(10n)+......+f(5(n−1)n)f(0) + f\left( {{5 \over n}} \right) + f\left( {{{10} \over n}} \right) + ...... + f\left( {{{5(n - 1)} \over n}} \right)

⇒1+1+5n+1+10n+....+1+5(n−1)n\Rightarrow 1 + 1 + {5 \over n} + 1 + {{10} \over n} + .... + 1 + {{5(n - 1)} \over n}

⇒n+5n(n−1)n2=2n+5n−52=7n−52\Rightarrow n + {5 \over n}{{(n - 1)n} \over 2} = {{2n + 5n - 5} \over 2} = {{7n - 5} \over 2}

lim⁡n→∞1n(7n−52)=72\mathop {\lim }\limits_{n \to \infty } {1 \over n}\left( {{{7n - 5} \over 2}} \right) = {7 \over 2}

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