JEE MAIN - Mathematics (2021 - 20th July Evening Shift - No. 6)
If $$f:R \to R$$ is given by $$f(x) = x + 1$$, then the value of $$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\left[ {f(0) + f\left( {{5 \over n}} \right) + f\left( {{{10} \over n}} \right) + ...... + f\left( {{{5(n - 1)} \over n}} \right)} \right]$$ is :
$${3 \over 2}$$
$${5 \over 2}$$
$${1 \over 2}$$
$${7 \over 2}$$
Explanation
$$f(0) + f\left( {{5 \over n}} \right) + f\left( {{{10} \over n}} \right) + ...... + f\left( {{{5(n - 1)} \over n}} \right)$$
$$ \Rightarrow 1 + 1 + {5 \over n} + 1 + {{10} \over n} + .... + 1 + {{5(n - 1)} \over n}$$
$$ \Rightarrow n + {5 \over n}{{(n - 1)n} \over 2} = {{2n + 5n - 5} \over 2} = {{7n - 5} \over 2}$$
$$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\left( {{{7n - 5} \over 2}} \right) = {7 \over 2}$$
$$ \Rightarrow 1 + 1 + {5 \over n} + 1 + {{10} \over n} + .... + 1 + {{5(n - 1)} \over n}$$
$$ \Rightarrow n + {5 \over n}{{(n - 1)n} \over 2} = {{2n + 5n - 5} \over 2} = {{7n - 5} \over 2}$$
$$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\left( {{{7n - 5} \over 2}} \right) = {7 \over 2}$$
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